text:
Taylor's series

A well-behaved function can be expanded into a power series. This means that for all non-negative integers $$k$$ there are real numbers $$a_k$$ such that

f(x)=\sum_{k=0}^\infty a_kx^k=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+\cdots$$

Let us calculate the first four derivatives using $$(x^n)'=n\,x^{n-1}$$:

f'(x)=a_1+2\,a_2x+3\,a_3x^2+4\,a_4x^3+5\,a_5x^4+\cdots$$

f''(x)=2\,a_2+2\cdot3\,a_3x+3\cdot4\,a_4x^2+4\cdot5\,a_5x^3+\cdots $$

f'''(x)=2\cdot3\,a_3+2\cdot3\cdot4\,a_4x+3\cdot4\cdot5\,a_5x^2+\cdots $$

f(x)=2\cdot3\cdot4\,a_4+2\cdot3\cdot4\cdot5\,a_5x+\cdots$$

Setting $$x$$ equal to zero, we obtain

f(0)=a_0,\quad f'(0)=a_1,\quad f''(0)=2\,a_2,\quad f(0)=2\times3\,a_3,\quad f'(0)=2\times3\times4\,a_4. $$

Let us write $$f^{(n)}(x)$$ for the $$n$$-th derivative of $$f(x).$$ We also write $$f^{(0)}(x)=f(x)$$ — think of $$f(x)$$ as the "zeroth derivative" of $$f(x).$$ We thus arrive at the general result $$f^{(k)}(0)=k!\,a_k,$$ where the factorial $$k!$$ is defined as equal to 1 for $$k=0$$ and $$k=1$$ and as the product of all natural numbers $$n\leq k$$ for $$k>1.$$ Expressing the coefficients $$a_k$$ in terms of the derivatives of $$f(x)$$ at $$x=0,$$ we obtain

This is the Taylor series for $$f(x).$$

A remarkable result: if you know the value of a well-behaved function $$f(x)$$ and the values of all of its derivatives at the single point $$x=0$$ then you know $$f(x)$$ at all points $$x.$$ Besides, there is nothing special about $$x=0,$$ so $$f(x)$$ is also determined by its value and the values of its derivatives at any other point $$x_0$$:

cos(x)
$$f(x) = \cos(x);\ f(0) = \cos(0) = 1$$

$$f'(x) = -\sin(x);\ f'(0) = -\sin(0) = 0$$

$$f(x) = -\cos(x);\ f(0) = -\cos(0) = -1$$

$$f(x) = \sin(x);\ f(0) = \sin(0) = 0$$

$$f'(x) = \cos(x);\ f'(0) = \cos(0) = 1$$

$$a_0 = f(0) = 1$$

$$a_1 = f'(0) = 0 $$

$$a_2 = \frac{f''(0)}{2} = \frac{-1}{2!} $$

$$a_3 = \frac{f'''(0)}{2*3} = \frac{0}{3!} = 0 $$

$$a_4 = \frac{f(0)}{2*3*4} = \frac{1}{4!} $$

$$\cos(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + \cdots $$ $$= 1 + (0)x + \frac{-1}{2}x^2 + 0x + \frac{1}{4!}x^4 + \cdots$$ $$= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \cdots$$

Some basic checking:

$$\cos(0) = 1 - \frac{0^2}{2!} + \frac{0^4}{4!} - \frac{0^6}{6!} + \frac{0^8}{8!} - \cdots = 1$$

$$\cos(-x) = 1 - \frac{(-x)^2}{2!} + \frac{(-x)^4}{4!} - \frac{(-x)^6}{6!} + \frac{(-x)^8}{8!} - \cdots$$ $$= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \cdots = \cos(x)$$

$$\frac{d(\cos(x))}{dx} = 0 - \frac{2x}{2!} + \frac{4x^3}{4!} - \frac{6x^5}{6!} + \frac{8x^7}{8!} \cdots$$ $$= -x + \frac{x^3}{3!} - \frac{x^5}{5!} + \frac{x^7}{7!} \cdots$$ $$= -\sin(x)$$

$$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$$

$$\frac{d(\sin(x))}{dx} = 1 - \frac{3x^2}{3!} + \frac{5x^4}{5!} - \frac{7x^6}{7!} + \cdots$$ $$ = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$$ $$ = \cos(x)$$

arctan(x)
$$y = \arctan(x)$$

$$x = \tan(y)$$

$$\frac{dx}{dy} = \sec^2(y)$$

$$y' = \frac{dy}{dx} = \cos^2(y) = \frac{1}{1 + x^2}$$. See.

Second derivative y″
$$(1 + x^2)y' = 1$$

$$(1 + x^2)y'' + y'(2x) = 0$$

$$y'' = \frac{-y'(2x)}{1 + x^2}$$ $$= \frac{-2x}{1 + x^2}.y'$$ $$= \frac{-2x}{1 + x^2}.\frac{1}{1 + x^2}$$ $$= \frac{-2x}{(1+x^2)^2}$$

Third derivative y¹¹¹
$$(1+2x^2+x^4)y'' = -2x$$

$$(1+2x^2+x^4)y' + y(4x+4x^3) = -2$$

$$y' = \frac{-2-y(4x+4x^3)}{1+2x^2+x^4}$$ $$= \frac{-2-y''4x(1+x^2)}{1+2x^2+x^4}$$ $$= \frac{-2-    \frac{-2x}{(1+x^2)^2}. 4x(1+x^2)}{1+2x^2+x^4}$$ $$= \frac{-2-    \frac{-2x}{1+x^2}. 4x}{1+2x^2+x^4}$$ $$= \frac{-2+    \frac{8x^2}{1+x^2}  }      {1+2x^2+x^4}$$ $$= \frac{           \frac{-2(1+x^2) + 8x^2}{1+x^2}       }{1+2x^2+x^4}$$ $$= \frac{           \frac{-2-2x^2 + 8x^2}{1+x^2}       }{1+2x^2+x^4}$$ $$= \frac{           \frac{-2+6x^2 }{1+x^2}       }{1+2x^2+x^4}$$ $$= \frac{-2+6x^2     }{(1+2x^2+x^4)(1+x^2)}$$ $$= \frac{-2+6x^2     }{(1+x^2)^3}$$

(continued)
If you continue to calculate derivatives, you will produce the following sequence:

$$f(0) = 0$$

$$f'(0) = 1$$

$$f''(0) = 0$$

$$f'''(0) = -2 = -(2!)$$

$$f(0) = 0$$

$$f^v(0) = 24 = 4!$$

$$f^{vi}(0) = 0$$

$$f^{vii}(0) = -720 = -(6!)$$

$$f^{viii}(0) = 0$$

$$f^{viiii}(0) = 40320 = 8!$$

$$f^{x}(0) = 0$$

$$f^{xi}(0) = -3628800 = -(10!)$$

$$f^{xii}(0) = 0$$

$$f^{xiii}(0) = 479001600 = 12!$$

$$\arctan(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + \cdots $$

$$= 0 + (1)x + (0)x^2 + \frac{-(2!)}{3!}x^3 + (0)x^4 + \frac{4!}{5!}x^5 + (0)x^6 + \frac{-(6!)}{7!}x^7 + (0)x^8 + \frac{8!}{9!}x^9 + (0)x^{10} + \frac{-(10!)}{11!}x^{11} + \cdots$$

$$= x - \frac{1}{3}x^3 + \frac{1}{5}x^5 - \frac{1}{7}x^7  +     \frac{1}{9}x^9 - \frac{1}{11}x^{11}  +     \cdots$$

Some basic checking:

$$\arctan(0) = 0$$

$$\arctan(-x) = -\arctan(x).$$

$$\frac{d(\arctan(x))}{dx}$$ $$= 1 - \frac{3x^2}{3} + \frac{5x^4}{5} - \frac{7x^6}{7} +  \frac{9x^8}{9} - \frac{11x^{10}}{11} +     \cdots$$ $$= 1 - x^2 + x^4 - x^6 + x^8 - x^{10} + \cdots$$

Also, $$\frac{d(\arctan(x))}{dx} = \frac{1}{1 + x^2}$$

Show that $$\frac{1}{1 + x^2}$$ $$= 1 - x^2 + x^4 - x^6 + x^8 - x^{10} + \cdots$$

or that $$1 = (1 + x^2)(1 - x^2 + x^4 - x^6 + x^8 - x^{10} + \cdots)$$

$$(1 + x^2)(1 - x^2 + x^4 - x^6 + x^8 - x^{10} + \cdots)$$

$$= 1(1 - x^2 + x^4 - x^6 + x^8 - x^{10} + \cdots) + x^2(1 - x^2 + x^4 - x^6 + x^8 - x^{10} +  \cdots)$$

$$= 1 - x^2 + x^4 - x^6 + x^8 - x^{10} + \cdots + x^2 - x^4 + x^6 - x^8 + x^{10} - x^{12} +  \cdots$$

$$= 1\ \cdots\ \pm\ x^{2n}$$

If abs$$(x) < 1,\ \lim_{n\rightarrow \infty}x^{2n} = 0.$$

In the diagram to the right, $$t(x)$$ is the Taylor series representing $$\arctan (x)$$ for $$x$$ close to $$0.$$

In the box above the proof that $$t(x)$$ is an accurate representation of $$\arctan(x)$$ is valid for abs$$(x) < 1.$$

When abs$$(x) > 1,$$ the diagram vividly illustrates that the series rapidly diverges.

To be accurate, the line $$y = \frac{\pi}{4}$$ should be $$y = \frac{\pi}{4}$$ rad or $$y = \frac{\pi}{4}^c$$ meaning $$y = \frac{\pi}{4}$$ radians. In theoretical work a value such as $$\frac{\pi}{4}$$ is understood to be $$\frac{\pi}{4}$$ radians or $$45^{\circ}$$ meaning $$45$$ degrees.

In practice
The expansion of $$\arctan(x)$$ above is theoretically valid for $$|x| < 1.$$ However, if $$|x|$$ is close to $$1,$$ the calculation of $$\arctan(x)$$ will take forever.

This section uses $$x_0$$ so that $$|x - x_0|$$ is small enough to make time of calculation acceptable.

Let $$\theta = 36\tfrac{3}{4}^{\circ}.$$ To calculate $$\tan(\theta):$$

$$\tan\frac{11\pi}{60}=\tan 33^\circ=\tfrac{1}{4}\left[2-\left(2-\sqrt3\right)\left(3+\sqrt5\right)\right]\left[2+\sqrt{2\left(5-\sqrt5\right)}\,\right]\,$$

$$\tan(147^\circ) = -\tan(33^\circ)$$

Using the half-angle formula $$\tan(\tfrac{\theta}{2}) = \csc(\theta) - \cot(\theta),$$

calculate $$\tan(73\tfrac{1}{2}) = \tan(\tfrac{147}{2})$$ and $$\tan(36\tfrac{3}{4}) = \tan(\tfrac{      73\tfrac{1}{2}     }{2})$$

$$\theta = 36\tfrac{3}{4}^\circ.\ x_0 = \tan(\theta) = 0.7467354177837216717375001402.$$

$$f(x_0) = \frac{36\tfrac{3}{4}\pi}{180} = 0.6414085001079161195194563572.$$

This value $$(36\tfrac{3}{4}^\circ)$$ was chosen for $$\theta$$ because $$x_0 = \tan(\theta)$$ is close to $$0.75.$$ For $$1 \ge x \ge 0.5,\ |x - x_0| \le 0.25$$ approx.

$$0.25^{50} = 7.888609052210118e-31.$$ If $$|x - x_0| == 0.25,$$ the code below is accurate to $$30$$ places of decimals.

This section uses the whole sequence of derivatives:

$$y000 = \arctan(x)$$ where $$y000 = y^{(0)} = y$$

$$y001 = \frac{1}{U}$$ where $$U = 1 + x^2,\ y001 = y^{(1)} = y'$$

$$y002 = -\frac{(2)*(x)*(y001)}{U}$$ where $$y002 = y^{(2)} = y''$$

$$y003 = -\frac{  (4)*(x)*(y002)+(2)*(y001)   }{U}$$ where $$y003 = y^{(3)} = y'''$$ and so on.

$$y004 = -\frac{  (6)*(x)*(y003)+(6)*(y002)  }{U}$$

$$y005 = -\frac{ (8)*(x)*(y004)+(12)*(y003) }{U}$$

$$y006 = -\frac{ (10)*(x)*(y005)+(20)*(y004) }{U}$$

$$y007 = -\frac{ (12)*(x)*(y006)+(30)*(y005) }{U}$$

$$y008 = -\frac{ (14)*(x)*(y007)+(42)*(y006) }{U}$$

$$y009 = -\frac{ (16)*(x)*(y008)+(56)*(y007) }{U}$$

$$y010 = -\frac{ (18)*(x)*(y009)+(72)*(y008) }{U}$$

$$\cdots$$

$$y050 = -\frac{ (98)*(x)*(y049)+(2352)*(y048) }{U}$$

Using $$f(x) = f(x_0)(x - x_0)^0 + f'(x_0)(x - x_0)^1 + \frac{f''(x_0)}{2}(x - x_0)^2 + \frac{f'''(x_0)}{2 \cdot 3}(x - x_0)^3 + \cdots$$

then $$f(x) = y000 + y001(x - x_0) + \frac{y002}{2}(x - x_0)^2 + \cdots$$ and:

y = 0.6414085001079161195194563572 +(0.6420076723519613087221948458)(x-(0.7467354177837216717375001402)) +(-0.3077848130939266477182675970)(x-0.7467354177837216717375001402)^2 +(0.05934881813852229894809158807)(x-0.7467354177837216717375001402)^3 +(0.05612149216561873709345633871)(x-0.7467354177837216717375001402)^4 +(-0.0659097533448882821311588572)(x-0.7467354177837216717375001402)^5 +(0.02864269115336634046783964776)(x-0.7467354177837216717375001402)^6 +(0.006684824489389227404750195292)(x-0.7467354177837216717375001402)^7 +(-0.01939996954693863883077829889)(x-0.7467354177837216717375001402)^8 +(0.01319629210955273736079467214)(x-0.7467354177837216717375001402)^9 +(-0.001423635337528918097834676738)(x-0.7467354177837216717375001402)^10 +(-0.005690817314508170664127596721)(x-0.7467354177837216717375001402)^11 +(0.005763416294060825609852171147)(x-0.7467354177837216717375001402)^12 +(-0.002009530403041012685757678258)(x-0.7467354177837216717375001402)^13 +(-0.001382413103546475118963526286)(x-0.7467354177837216717375001402)^14 +(0.002355235379425975362106687309)(x-0.7467354177837216717375001402)^15 +(-0.001340525935442931206538139095)(x-0.7467354177837216717375001402)^16 +(-0.0001244720120416846251034920203)(x-0.7467354177837216717375001402)^17 +(0.0008777184853106580549638629701)(x-0.7467354177837216717375001402)^18 +(-0.0007257802485492202930793702930)(x-0.7467354177837216717375001402)^19 +(0.0001539460026510816727324277808)(x-0.7467354177837216717375001402)^20 +(0.0002810020934892180446689969911)(x-0.7467354177837216717375001402)^21 +(-0.0003470330774958963760466009045)(x-0.7467354177837216717375001402)^22 +(0.0001535570475871531716152621841)(x-0.7467354177837216717375001402)^23 +(0.00006313260945054237374661397478)(x-0.7467354177837216717375001402)^24 +(-0.0001488094986598041280906554962)(x-0.7467354177837216717375001402)^25 +(0.00009977993191704606200503722720)(x-0.7467354177837216717375001402)^26 +(-0.000003667561779685224841381874106)(x-0.7467354177837216717375001402)^27 +(-0.00005609286432922550484209657985)(x-0.7467354177837216717375001402)^28 +(0.00005412057738460028511566507574)(x-0.7467354177837216717375001402)^29 +(-0.00001655090242419904039018979491)(x-0.7467354177837216717375001402)^30 +(-0.00001714674178231985986067601799)(x-0.7467354177837216717375001402)^31 +(0.00002588855802866968641107644970)(x-0.7467354177837216717375001402)^32 +(-0.00001372909690493026279553133838)(x-0.7467354177837216717375001402)^33 +(-0.000002866406864208033772447118585)(x-0.7467354177837216717375001402)^34 +(0.00001098036048658105543109288040)(x-0.7467354177837216717375001402)^35 +(-0.000008497717911244361532280438636)(x-0.7467354177837216717375001402)^36 +(0.000001259146436001274560243296183)(x-0.7467354177837216717375001402)^37 +(0.000003992939704019955177003526706)(x-0.7467354177837216717375001402)^38 +(-0.000004497268683100848779169934291)(x-0.7467354177837216717375001402)^39 +(0.000001768945188244137235636524921)(x-0.7467354177837216717375001402)^40 +(0.000001091706749083768850937760502)(x-0.7467354177837216717375001402)^41 +(-0.000002103423912310375893571195410)(x-0.7467354177837216717375001402)^42 +(0.000001301617082996039555612971998)(x-0.7467354177837216717375001402)^43 +(6.937967909808721515382339295E-8)(x-0.7467354177837216717375001402)^44 +(-8.635525611989332402947366709E-7)(x-0.7467354177837216717375001402)^45 +(7.673857631132879175874596987E-7)(x-0.7467354177837216717375001402)^46 +(-1.893140553441536683377149770E-7)(x-0.7467354177837216717375001402)^47 +(-2.944033068289732156296704644E-7)(x-0.7467354177837216717375001402)^48 +(3.930991061512879804635643270E-7)(x-0.7467354177837216717375001402)^49 +(-1.879241426421737899718180888E-7)(x-0.7467354177837216717375001402)^50

A faster version
The calculation of $$\arctan (x)$$ above is suitable as input to application

The following python code has precision set to $$33.$$ If it is desired to calculate $$\arctan (x)$$ for one value of $$x,$$ the following python code is much faster than the code supplied to  above.

x close to x0
When $$x$$ is close to $$x_0,$$ result is achieved with only 12 passes through loop.

Testing with known values
Check results using known combinations of $$x$$ and $$\arctan(x):$$

$$\tan\frac{3\pi}{20}=\tan 27^\circ = \sqrt5 - 1 -\sqrt{5 - 2\sqrt5}.$$

For $$\tan 27^\circ$$ and other exact values of $$\tan (x),$$ see Exact Values for Common Angles.

Verifying the recommended limits
For $$1.05 \ge x \ge 0.45$$ the above calculation of $$\arctan (x)$$ is accurate to more than 30 places of decimals.

Input outside Recommended Limits
If input is outside recommended limits, this does not necessarily mean that result is invalid.

If $$\tan 51^\circ \ge x \ge \tan 9^\circ,$$ result is accurate to precision of python floats, 15 places of decimals.

arcsin(x)
$$y = \arcsin(x)$$

$$x = \sin(y)$$

$$\frac{dx}{dy} = \cos(y) = \sqrt{1 - x^2}$$

$$\frac{dy}{dx} = \frac{1}{\cos(y)} = \frac{1}{\sqrt{1 - x^2}}$$

Simple differential equations eliminate the square root and make calculations so much easier.

Let $$a = \frac{dy}{dx}$$

Then $$a^2U = 1$$ where $$U=1-x^2$$ and $$U^' = -2x.$$

Differentiating both sides:

$$a^2(-2x) + U2aa^' = 0$$

$$U2aa^' = 2xa^2$$

$$a^' = \frac{2xaa}{2aU} = \frac{ax}{U}.$$

Let $$b = a^'$$

Then $$bU=ax$$

Differentiating both sides:

$$b(-2x) + Ub^' = a + xa^' = a + xb$$

Let $$c = b^'$$

Then $$c = \frac{a+3bx}{U}$$

When $$x = 0,\ U =1.$$ Calculation of more derivatives yields:

$$y = f(x) = \arcsin (x)$$

$$a = f^'(x) = \frac{1}{\sqrt{1-x^2}}$$

$$b = f^{''}(x) = ax$$

$$c = f^{'''}(x) = a + 3bx$$

$$d = f^{}(x) = 4b + 5cx$$

$$g = f^{v}(x) = 9c + 7dx$$

$$h = f^{vi}(x) = 16d + 9gx$$

$$j = f^{vii}(x) = 25g + 11hx$$

$$k = f^{viii}(x) = 36h + 13jx$$

$$l = f^{viiii}(x) = 49j + 15kx$$

and so on.

$$\cdots = k = h = d = b = 0$$

$$a = 1;\ c = 1^2a;\ g = 3^2c;\ j = 5^2g;\ l = 7^2j;\ \cdots.$$

$$\arcsin(x) = x + \frac{1}{3!}x^3 + \frac{3^2}{5!}x^5 + \frac{3^2\cdot5^2}{7!}x^7 + \frac{3^2\cdot5^2\cdot7^2}{9!}x^9+ \frac{3^2\cdot5^2\cdot7^2\cdot9^2}{11!}x^{11} + \cdots $$

$$\arcsin(x) = x + \frac{1}{3!}x^3 + \frac{3}{(2!)(2^2)5}x^5 + \frac{3\cdot5}{(3!)(2^3)7}x^7 + \frac{3\cdot5\cdot7}{(4!)(2^4)9}x^9 + \frac{3\cdot5\cdot7\cdot9}{(5!)(2^5)11}x^{11} + \cdots $$

As programming algorithm:
$$\arcsin(x) = x + \frac{top=1}{(bottom=2)3}(X = x^3 )$$ $$+ \frac{top=top(5-2)}{(bottom=bottom(5-1))5}(X=X(x^2))$$ $$+ \frac{top=top(7-2)}{(bottom=bottom(7-1))7}(X=X(x^2)) $$ $$+ \cdots $$ $$+ \frac{top=top(n-2)}{(bottom=bottom(n-1))n}(X=X(x^2)) $$

As implemented in Python:
0.707106781186547524400844362 True 171

In practice
If $$|x|$$ is close to $$1,$$ the calculation of $$\arcsin (x)$$ will take forever.

If you limit $$x$$ to $$\sin 45^\circ >= x >= -\sin 45^\circ,$$ then $$x^2 <= 0.5$$ and each term is guaranteed to be less than half the preceding term.

If $$x > \sin 45^\circ,$$ let $$t = \frac{\pi}{2} - y.$$

Then $$\sin t = \cos y = \sqrt{1 - x^2};\ t = \arcsin(\cos y);\ y = \frac{\pi}{2} -t.$$

Integral of expression

 * $$\frac{dy}{dx} = a^{-x^2}$$

According to the reference "this expression cannot be integrated..." However, if we convert the expression to a Taylor series, the integral of the series is quite easily calculated.

Let $$y = f(x) = a^{-x^2} = \frac{1}{a^{x^2}}$$

When $$x = 0,\ y = 1$$ and the following sequence can be produced.

$$y^{''} = -2Ay$$ where $$A = \ln(a).$$

$$y^{(4)} = -6Ay^{''}$$

$$y^{(6)} = -10Ay^{(4)}$$

$$y^{(8)} = -14Ay^{(6)}$$

$$y^{(10)} = -18Ay^{(8)}$$

$$y^{(12)} = -22Ay^{(10)}$$ and so on.

Taylor series of $$f(x)$$ for $$x$$ close to $$0 = t(x) = 1 + c02\cdot x^2 + c04\cdot x^4 + c06\cdot x^6 + \cdots$$

where $$c02 = \frac{y^{''}}{2!};\ c04 = \frac{y^{(4)}}{4!};\ c06 = \frac{y^{(6)}}{6!} \cdots$$

For $$a = 2$$ python code produces the following:

c02 = -0.6931471805599453094172321215 c04 = 0.2402265069591007123335512632 c06 = -0.05550410866482157995314226378 c08 = 0.009618129107628477161979071575 c10 = -0.001333355814642844342341222199 c12 = 0.0001540353039338160995443709734 c14 = -0.00001525273380405984028002543902 c16 = 0.000001321548679014430948840375823 c18 = -1.017808600923969972749000760E-7 c20 = 7.054911620801123329875392184E-9 c22 = -4.445538271870811497596408561E-10 c24 = 2.567843599348820514199480240E-11 c26 = -1.369148885390412888089195400E-12 c28 = 6.778726354822545633449104318E-14 c30 = -3.132436707088428621634944443E-15 c32 = 1.357024794875514719311296624E-16 c34 = -5.533046532458242043485546100E-18 c36 = 2.130675335489117996020398479E-19 c38 = -7.773008428857356419088997166E-21 c40 = 2.693919438465583416972861154E-22 c42 = -8.891822206800239171648619811E-24

For $$a$$ close to $$1$$ or $$x$$ close to $$0,$$ the Taylor series is a quite accurate representation of the original expression. When abs$$(x) \le 1.6,$$ the abs(maximum difference) between expression and Taylor series is $$< 1e-15.$$

For greater accuracy, greater precision may be specified in python or more terms after $$c42$$ may be added.

The integral $$= I(x) = x + C02\cdot x^3 + C04\cdot x^5 + C06\cdot x^7 + \cdots$$

where $$C02 = \frac{c02}{3};\ C04 = \frac{c04}{5};\ C06 = \frac{c06}{7} \cdots$$

In figure to right, $$t(x) = 1.5 + \cdots,$$ separating $$t(x)$$ from $$f(x)$$ to illustrate shapes of curves.

The correct value of $$t(x) = 1 + \cdots$$.

When $$x = 1,\ f(x) = 0.5$$ and $$t(x) = 0.4999999999999999999999997281$$.

To 24 places of decimals $$t(x) = 0.5000$$_$$0000$$_$$0000$$_$$0000$$_$$0000$$_$$0000$$.

If it were important to calculate the area under $$f(x)$$ from $$x = 0$$ to $$x = 1,\ I(x)$$ returns $$0.8100254543909558266292852604,$$ accurate to about 26 places of decimals.

sin(x) using (x - x0)
$$f(x) = \sin(x)$$

$$f'(x) = \cos(x)$$

$$f''(x) = -\sin(x)$$

$$f'''(x) = -\cos(x)$$

$$f^{(4)} = \sin(x)$$

Let $$x_0 = \frac{\pi}{4} = 45^{\circ}$$

$$f(x_0) = f(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$$

$$f'(x_0) = f'(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$$

$$f(x_0) = f(\frac{\pi}{4}) = -\sin(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}$$

$$f(x_0) = f(\frac{\pi}{4}) = -\cos(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}$$

$$f^{(4)}(x_0) = f^{(4)}(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$$

$$t(x) = \frac{\sqrt{2}}{2}\cdot(x - x_0)^0$$ $$+ \frac{\sqrt{2}}{2}\cdot(x - x_0)^1$$ $$- \frac{\sqrt{2}}{2}\cdot\frac{1}{2!}\cdot(x - x_0)^2$$ $$- \frac{\sqrt{2}}{2}\cdot\frac{1}{3!}\cdot(x - x_0)^3$$ $$+ \frac{\sqrt{2}}{2}\cdot\frac{1}{4!}\cdot(x - x_0)^4 + \cdots$$

Let $$X = (x - x_0) = (x - \frac{\pi}{4})$$

Then $$t(x)= \frac{\sqrt{2}}{2}($$ $$1 + X$$ $$- \frac{X^2}{2!} - \frac{X^3}{3!}$$ $$+ \frac{X^4}{4!} + \frac{X^5}{5!}$$ $$- \frac{X^6}{6!} - \frac{X^7}{7!}$$ $$\ \cdots$$ $$)$$

where $$t(x)$$ is the Taylor series representing $$f(x) = \sin(x)$$ for values of $$x$$ close to $$45^{\circ}$$ or $$\frac{\pi}{4}.$$

If $$\frac{\pi}{2}(90^{\circ}) \ge x \ge 0$$, then $$t(x)$$ containing powers of $$X$$ through $$15$$ is sufficient to keep the error to $$< 1e-15.$$

Almost a sine curve
Graph to right was produced by Grapher on a Mac.

A python script produced the following data: ( (2^(0.5))/2 )( 1 +(x-.785398163397448)

-((x-.785398163397448)^2)/2 -((x-.785398163397448)^3)/(2(3))

+((x-.785398163397448)^4)/(24) +((x-.785398163397448)^5)/(120)

-((x-.785398163397448)^6)/(720) -((x-.785398163397448)^7)/(5040)

+((x-.785398163397448)^8)/(40320) +((x-.785398163397448)^9)/(362880)

-((x-.785398163397448)^10)/(3628800) -((x-.785398163397448)^11)/(39916800)

+((x-.785398163397448)^12)/(479001600) +((x-.785398163397448)^13)/(6227020800)

-((x-.785398163397448)^14)/(87178291200) -((x-.785398163397448)^15)/(1307674368000)

+((x-.785398163397448)^16)/(20922789888000) +((x-.785398163397448)^17)/(355687428096000)

-((x-.785398163397448)^18)/(6402373705728000) -((x-.785398163397448)^19)/(121645100408832000)

+((x-.785398163397448)^20)/(2432902008176640000) +((x-.785398163397448)^21)/(51090942171709440000)

-((x-.785398163397448)^22)/(1124000727777607680000) -((x-.785398163397448)^23)/(25852016738884976640000)

+((x-.785398163397448)^24)/(620448401733239439360000) +((x-.785398163397448)^25)/(15511210043330985984000000)

-((x-.785398163397448)^26)/(403291461126605635584000000) -((x-.785398163397448)^27)/(10888869450418352160768000000)

+((x-.785398163397448)^28)/(304888344611713860501504000000) +((x-.785398163397448)^29)/(8841761993739701954543616000000) ) I highlighted the data, copied it with command C and pasted it into the input area of Grapher. Well done! Grapher.

Integral of 1/x
The Taylor series for $$f(x) = \frac{1}{x}$$ for $$x$$ close to $$2$$ is:

$$t(x) = \frac{1}{2} - \frac{1}{2^2}\cdot (x - 2) + \frac{1}{2^3}\cdot (x - 2)^2 - \frac{1}{2^4}\cdot (x - 2)^3 \cdots$$

The integral of this series is:

$$I(x) = \frac{1}{2}\cdot (x - 2) - ((\frac{1}{2^2})/2)\cdot (x - 2)^2 + ((\frac{1}{2^3})/3)\cdot (x - 2)^3 - ((\frac{1}{2^4})/4)\cdot (x - 2)^4 \cdots + C$$

The integral of $$\frac{1}{x} = \ln(x).$$

Therefore $$\ln(x) = I(x)$$ but what is the value of $$C?$$

Without $$C,$$ when $$x = 2, I(x) = 0.\ I(x)$$ should be $$\ln(2).$$

Therefore, for $$x$$ close to $$2:$$

$$\ln(x) = I(x)$$ where $$C = \ln(2).$$

But what is the value of $$\ln(2)?$$

Without $$C,$$ when $$x = 1,\ I(x) = -0.693147180559945.\ I(x)$$ should be $$0.$$

Therefore $$C = 0.693147180559945$$ or $$\ln(2).$$

For $$x$$ close to $$2:$$

$$\ln(x) = \ln(2) + \frac{1}{2}\cdot (x - 2) - ((\frac{1}{2^2})/2)\cdot (x - 2)^2 + ((\frac{1}{2^3})/3)\cdot (x - 2)^3 - ((\frac{1}{2^4})/4)\cdot (x - 2)^4 \cdots$$

where $$\ln(2) = 0.693147180559945\dots$$

y = 0.693147180559945 + ((1/(2^1))/1)(x - 2)^1 - ((1/(2^2))/2)(x - 2)^2 + ((1/(2^3))/3)(x - 2)^3 - ((1/(2^4))/4)(x - 2)^4 + ((1/(2^5))/5)(x - 2)^5 - ((1/(2^6))/6)(x - 2)^6 + ((1/(2^7))/7)(x - 2)^7 ........................... ........................... - ((1/(2^42))/42)(x - 2)^42 + ((1/(2^43))/43)(x - 2)^43 - ((1/(2^44))/44)(x - 2)^44 + ((1/(2^45))/45)(x - 2)^45 - ((1/(2^46))/46)(x - 2)^46 + ((1/(2^47))/47)(x - 2)^47 - ((1/(2^48))/48)(x - 2)^48 + ((1/(2^49))/49)(x - 2)^49

Calculating ln(x)
This section presents a system for calculating $$\ln(x)$$ for $$10 \ge x \ge 1,$$ knowing only that $$\ln(1) = 0.$$

where  is a list containing values of $$x_0$$ in which each value after the first is $$\le 10$$% more than the preceding value.

The performance of the above code is better than logarithmic to base $$10$$. This means, for example, if $$x$$ contains 60 significant decimal digits, the above code produces a result with fewer than 30 passes through the loop because each iteration of the lop performs two operations.

is designed so that  $$= \frac{x\_minus\_x_0}{x_0}$$ is always $$\le \frac{1}{10}.$$ When $$x$$ is very close to $$x_0,$$ time to calculate $$\ln(x)$$ is greatly reduced.

y = ln(7.9) + ((1/((7.9)^(1)))/(1))((x - 7.9)^(1)) - ((1/((7.9)^(2)))/(2))((x - 7.9)^(2)) + ((1/((7.9)^(3)))/(3))((x - 7.9)^(3)) - ((1/((7.9)^(4)))/(4))((x - 7.9)^(4)) + ((1/((7.9)^(5)))/(5))((x - 7.9)^(5)) - ((1/((7.9)^(6)))/(6))((x - 7.9)^(6)) + ((1/((7.9)^(7)))/(7))((x - 7.9)^(7)) - ((1/((7.9)^(8)))/(8))((x - 7.9)^(8)) + ((1/((7.9)^(9)))/(9))((x - 7.9)^(9)) ..................... ..................... - ((1/((7.9)^(22)))/(22))((x - 7.9)^(22)) + ((1/((7.9)^(23)))/(23))((x - 7.9)^(23)) - ((1/((7.9)^(24)))/(24))((x - 7.9)^(24)) + ((1/((7.9)^(25)))/(25))((x - 7.9)^(25)) - ((1/((7.9)^(26)))/(26))((x - 7.9)^(26)) + ((1/((7.9)^(27)))/(27))((x - 7.9)^(27)) - ((1/((7.9)^(28)))/(28))((x - 7.9)^(28)) + ((1/((7.9)^(29)))/(29))((x - 7.9)^(29))

The next piece of code progressively calculates $$\ln(1.1),\ \ln(1.2),\ \ln(1.3)\ \dots$$ and puts the calculated values in dictionary dict2.

dict2 = { Decimal(' 1.0'): +Decimal('0'), Decimal(' 1.1'): +Decimal('0.095310179804324860043952123280765092220605365308644199'), Decimal(' 1.2'): +Decimal('0.18232155679395462621171802515451463319738933791448698'), Decimal(' 1.3'): +Decimal('0.26236426446749105203549598688095439720416645613143414'), Decimal(' 1.4'): +Decimal('0.33647223662121293050459341021699209011148337531334347'), Decimal(' 1.5'): +Decimal('0.40546510810816438197801311546434913657199042346249420'), Decimal(' 1.6'): +Decimal('0.47000362924573555365093703114834206470089904881224805'), Decimal(' 1.7'): +Decimal('0.53062825106217039623154316318876232798710152395697182'), Decimal(' 1.8'): +Decimal('0.58778666490211900818973114061886376976937976137698120'), Decimal(' 1.9'): +Decimal('0.64185388617239477599103597720348932963627777267035586'), Decimal(' 2.0'): +Decimal('0.69314718055994530941723212145817656807550013436025527'), Decimal(' 2.2'): +Decimal('0.78845736036427016946118424473894166029610549966889947'), Decimal(' 2.4'): +Decimal('0.87546873735389993562895014661269120127288947227474225'), Decimal(' 2.6'): +Decimal('0.95551144502743636145272810833913096527966659049168941'), Decimal(' 2.8'): +Decimal('1.0296194171811582399218255316751686581869835096735987'), Decimal(' 3.0'): +Decimal('1.0986122886681096913952452369225257046474905578227494'), Decimal(' 3.3'): +Decimal('1.1939224684724345514391973602032907968680959231313936'), Decimal(' 3.6'): +Decimal('1.2809338454620643176069632620770403378448798957372364'), Decimal(' 3.9'): +Decimal('1.3609765531356007434307412238034801018516570139541836'), Decimal(' 4.2'): +Decimal('1.4350845252893226218998386471395177947589739331360929'), Decimal(' 4.6'): +Decimal('1.5260563034950493162059934985840084789167789605719180'), Decimal(' 5.0'): +Decimal('1.6094379124341003746007593332261876395256013542685177'), Decimal(' 5.5'): +Decimal('1.7047480922384252346447114565069527317462067195771619'), Decimal(' 6.0'): +Decimal('1.7917594692280550008124773583807022727229906921830047'), Decimal(' 6.6'): +Decimal('1.8870696490323798608564294816614673649435960574916489'), Decimal(' 7.2'): +Decimal('1.9740810260220096270241953835352169059203800300974917'), Decimal(' 7.9'): +Decimal('2.0668627594729758101549540867970467145724397357938367'), Decimal(' 8.6'): +Decimal('2.1517622032594620488720831801196593960335348306130377'), Decimal(' 9.3'): +Decimal('2.2300144001592102533064181067805187074963279996745685'), Decimal('10.0'): +Decimal('2.3025850929940456840179914546843642076011014886287730'), }

A quick check: ln(2.2) - (ln(1.1) + ln(2.0)) = 0E-50 ln(2.4) - (ln(1.2) + ln(2.0)) = 0E-50 ln(2.6) - (ln(1.3) + ln(2.0)) = 0E-50 ln(2.8) - (ln(1.4) + ln(2.0)) = 0E-50 ln(3.0) - (ln(1.5) + ln(2.0)) = -0E-50 ln(3.3) - (ln(1.1) + ln(3.0)) = 0E-50 ln(3.6) - (ln(1.2) + ln(3.0)) = 0E-50 ln(3.6) - (ln(1.8) + ln(2.0)) = -0E-50 ln(3.9) - (ln(1.3) + ln(3.0)) = 0E-50 ln(4.2) - (ln(1.4) + ln(3.0)) = 0E-50 ln(5.5) - (ln(1.1) + ln(5.0)) = 0E-50 ln(6.0) - (ln(1.2) + ln(5.0)) = 0E-50 ln(6.0) - (ln(2.0) + ln(3.0)) = 0E-50 ln(6.6) - (ln(1.1) + ln(6.0)) = 0E-50 ln(6.6) - (ln(2.2) + ln(3.0)) = 0E-50 ln(6.6) - (ln(3.3) + ln(2.0)) = 0E-50 ln(7.2) - (ln(1.2) + ln(6.0)) = 0E-50 ln(7.2) - (ln(2.4) + ln(3.0)) = 0E-50 ln(10.0) - (ln(5.0) + ln(2.0)) = 0E-50

Put the data from  into 2 tuples

Calculate the decision points.

Display the three tuples.

Previous code was used to produce three tuples. Operational code follows:

Values of $$x_0:$$

Values of $$\ln(x_0):$$

Decision points:

At each decision point $$x$$ is assigned to the next low value or the next high value of $$x_0.$$ For example, if $$x$$ is between $$1.2,\ 1.3,$$ the decision point is $$1.248.$$ This means that the ratio $$\frac{1.248 - 1.2}{1.2} = \frac{1.3 - 1.248}{1.3} = 0.04$$ and the maximum value of abs$$(\frac{x - x_0}{x_0}) = 0.04.$$

During creation of  the maximum value of $$\frac{x - x_0}{x_0} = \frac{1}{10}.$$ During normal operations after creation of  maximum value of abs$$(\frac{x - x_0}{x_0}) = \frac{1.047619\dots - 1}{1}$$ between $$1.0,\ 1.1.$$

Choose a suitable value of  with the value of its natural log.

Ready to calculate, for example, $$\ln(3.456789)$$

ln(3.456789) = 1.24034_01234_96758_02986_53847_82231_30004_00340_53893_89110 # displayed with 50 places of decimals.

Testing ln(x)
Choose random numbers $$a,\ b$$ so that $$10 \ge a \ge b \ge 1.$$

Produce values $$\ln(a),\ \ln(b).$$

Calculate product $$p = a\cdot b.$$

Produce value $$\ln(p).$$

If $$p > 10,\ p = (\frac{p}{10})\cdot 10$$ and $$\ln(p) = \ln(\frac{p}{10}) + \ln(10).$$

Verify that $$\ln(p) = \ln(a) + \ln(b).$$

For example: During testing, successive invocations of the above code produced:

d1         = 3.300463847393627263496303126765085976697315885228780009201595937 ln_d1_     = 1.1940630184110798505583266934968432937656468440595029 d2         = 4.727915623201914684885711302927600487326893972103794963997766615 ln_d2_     = 1.5534844337520634527664958773360448454701186698422347 ln_product_ = 2.7475474521631433033248225708328881392357655139017377

d1         = 6.56429212435850275252301147228535243835226966080458915176241218 ln_d1_     = 1.8816446762531860392218213681767770852191644273705970 d2         = 8.15468991518212749204100104755219361919087392341006662123706307 ln_d2_     = 2.0985932114606734087366302984138612677420896519457258 ln_product_ = 3.9802378877138594479584516665906383529612540793163228

title: Taylor's series


text:
Applying definite integrals

For a discussion on the basic theory of integrals see: Riemann sum

Imagine an object $$\mathcal{O}$$ that is free to move in one dimension — say, along the $$x$$ axis. Like every physical object, it has a more or less fuzzy position (relative to whatever reference object we choose). For the purpose of describing its fuzzy position, quantum mechanics provides us with a probability density $$\rho(x).$$ This depends on actual measurement outcomes, and it allows us to calculate the probability of finding the particle in any given interval of the $$x$$ axis, provided that an appropriate measurement is made.

We call $$\rho(x)$$ a probability density because it represents a probability per unit length. The probability of finding $$\mathcal{O}$$ in the interval between $$x_1$$ and $$x_2$$ is given by the area $$A$$ between the graph of $$\rho(x),$$ the $$x$$ axis, and the vertical lines at $$x_1$$ and $$x_2,$$ respectively. How do we calculate this area? The trick is to cover it with narrow rectangles of width $$\Delta x.$$

The area of the first rectangle from the left is $$\rho(x_1+\Delta x)\,\Delta x,$$ the area of the second is $$\rho(x_1+2\,\Delta x)\,\Delta x,$$ and the area of the last is $$\rho(x_1+12\,\Delta x)\,\Delta x.$$ For the sum of these areas we have the shorthand notation

\sum_{k=1}^{12}\rho(x+k\,\Delta x)\,\Delta x.$$ It is not hard to visualize that if we increase the number $$N$$ of rectangles and at the same time decrease the width $$\Delta x$$ of each rectangle, then the sum of the areas of all rectangles fitting under the graph of $$\rho(x)$$ between $$x_1$$ and $$x_2$$ gives us a better and better approximation to the area $$A$$ and thus to the probability of finding $$\mathcal{O}$$ in the interval between $$x_1$$ and $$x_2.$$ As $$\Delta x$$ tends toward 0 and $$N$$ tends toward infinity ($$\infty$$), the above sum tends toward the integral

\int_{x_1}^{x_2}\rho(x)\,dx.$$

We sometimes call this a definite integral to emphasize that it's just a number. (As you can guess, there are also indefinite integrals, about which more later.) The uppercase delta has turned into a $$d$$ indicating that $$dx$$ is an infinitely small (or infinitesimal) width, and the summation symbol (the uppercase sigma) has turned into an elongated S indicating that we are adding infinitely many infinitesimal areas.

Don't let the term "infinitesimal" scare you. An infinitesimal quantity means nothing by itself. It is the combination of the integration symbol $$\textstyle\int$$ with the infinitesimal quantity $$dx$$ that makes sense as a limit, in which $$N$$ grows above any number however large, $$dx$$ (and hence the area of each rectangle) shrinks below any (positive) number however small, while the sum of the areas tends toward a well-defined, finite number.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: Applying definite integrals


text:
User:Karen Thurmond

--Karen Thurmond 18:59, 8 March 2007 (UTC)

title: User:Karen Thurmond


text:
Calculus II/Moments

The moment of inertia is sometimes called and referred to as "angular mass." It is the "amount of resistance" to changes in rotational motion.

The moment of inertia of a point mass m at a distance r from the axis of rotation is

 * $$I \ \stackrel{\mathrm{def}}{=}\ m r^2\,\!$$

The moment of inertia is additive, therefore, for a collection of $$N$$ point masses $$m_{i}$$ with distances $$r_{i}$$ to the rotation axis, the total moment of inertia is the sum of the point-mass moments of inertia

 * $$I \ \stackrel{\mathrm{def}}{=}\ \sum_{i=1}^{N} {m_{i} r_{i}^2}\,\!$$

The moment of inertia is commonly used to calculate torque when multiplied by the angular acceleration. This is analogous to F = ma.

 * $$Torque = Ia\,$$

title: Calculus II/Moments


text:
User:Kisama

hi

title: User:Kisama


text:
Talk:Peace Studies in Africa

This project is an extension of a qualitative study concerning internet access and use in Africa.

Category
How should this learning project be categorized? I see it is linked and included as a division news item at Topic:Peace Studies. Perhaps this could be a first entry to Category:Peace Studies.

title: Talk:Peace Studies in Africa


text:
User talk:Karen Thurmond

Image copyright problems

 * Your currently affected images are listed below. For each, please provide a license template and source. Thanks. – Mike.lifeguard | @en.wb 23:30, 13 February 2008 (UTC)

 * 2007-03-08 18:33, Karen Thurmond: Krista wedding 025.jpg: untagged! ; 0 (0)
 * 2007-03-08 18:38, Karen Thurmond: Karen and Brothers.jpg: untagged! ; 0 (0)
 * 2007-03-08 18:41, Karen Thurmond: Thurmonds2006ChristmasPicture.JPG: untagged! ; 0 (0)
 * 2007-03-08 18:44, Karen Thurmond: Karen and Family.JPG: untagged! ; 2 (0)
 * 2007-03-08 18:54, Karen Thurmond: Karen and Brothers1.jpg: untagged! ; 1 (0)

title: User talk:Karen Thurmond


text:
David Watts, short papers

Sources of Law, Short Paper #2,

The sources of law are an important consideration. We need to look at where are laws have come from and where they are going. I will consider different sources of law, and then propose what I take to be the most important ones. When we brainstorm sources of law, we end up with a list of things which very much related to each other. We may refer to different sources which all fall under constitutionally based law. Or we may not distinguish between common law and precedence. I propose that we group these sources so that we can handle them more clearly. In the end, I think those sources of law which guide and promote change are the most important ones, namely, judges’ rulings and meta-law, in our case, the constitution. -Law falls into a broad dichotomy. There is positive and normative law. When we say "sources of law," we are talking about sources of positive law, or the law that actually is enforced. The sources themselves fall into the same dichotomy. There are sources which already are written law, with fixed crimes and their punishments, and those which insist they ought to be law, based on some criteria other than written law. -Positive law sources include common law, precedence, and written law in all its different forms. Common law is what the outcomes of cases were prior to being written down, so it is a historical precedence. Precedence usually refers to the history of what has been the case and judgment, but written down. Written is the basis for most court judgments, as most cases fall within parameters and conditions stipulated by those laws. The difference between written law and positive law is that positive law is the activity of law in the present and the phenomenon of what happens and what is decided in court. A judge may or may not use the written law exactly in a case, although, most of the time, positive law reflects written law. -Normative sources include natural law, religious or moral codes, judgments made by judges that don’t align themselves with positive law, history, and meta-law. The last two can be contested as positive law, but I will defend myself soon. Natural law is usually appealed to as a law which exists in the world, and as seen in nature, or a state of nature between men. Kind of vague, I know. Religious and moral codes are rather self-explanatory; people who follow these codes want them to be enforced legally. Judgments made by judges can be seen as a positive law source, because that power is granted to them. As such, their discretion is positive law itself. However, I consider it a normative source because it a choice based on what the law ought to be, in the judge’s mind, rather than what the written law is. Meta-law, like the constitution, is something that acts a guiding principle for the making of law. It can be considered law itself, but as a source of law, it says what law ought to be or how it should be decided and doesn’t explicitly spell it out. -I think that the most important sources of law are meta-law and judgment. This is because one of the more important aspects of the law is its evolution. We all admit that the law isn’t perfect. We can be pessimistic about it and say this is the best we can do, or it’s impossible to achieve a perfect legal system, or we can think about it as a community effort, evolving towards an ideal. To avoid existential resignation, it’s important to always maintain the state of striving-towards. Meta-law defines what we believe that ideal is. It helps guide the evolution of law towards that ideal. Judgments are where positive law is able to change. The written law is altered slightly by a judge making a decision that is different from it. Hopefully, the judge is keeping the ideal in mind, and helping the evolutionary process towards the ideal.

-David Watts

title: David Watts, short papers


text:
Category talk:Business learning projects

would anyone be interested in learning social networking, social marketing or negotiation skills?
I'm just testing the waters here. I know anyone can add pretty much anything, though I'm curious if anyone would be interested in these particular avenues of business?

If not, is there anything the consensus feels we should start with? --Bema Self 03:26, 14 February 2012 (UTC)


text:
User talk:SueB

Hi Sue, thanks for joining in on the Action research/AREOL25 page. If you've any hesitations or uncertainties about the use of wikis, I can hopefully be of help - you can reply here or on my talk page. Wikis do take a bit of getting used to, but they can produce some interesting learning and results - if I can be of any help, just let me know.. Cheers, Cormaggio beep 00:29, 9 March 2007 (UTC)

Thanks Cormac
The Wik university thing is very interesting! I am struggling with the editing but will get the hang of it. My interest are in Feminist methods and Action research and I am finding the AEROL25 not very friendly or engaging. Very linear and controlled. I am quite interested to develop resources around my work areas and feel like I have a new toy.--SueB 05:14, 9 March 2007 (UTC)

 * Gosh, that's very interesting! You know, I think that many people would have come to the wiki and felt that they could not participate, and that it was too controlling - I suppose I wouldn't necessarily agree with the latter (!), but I would certainly appreciate where they were coming from, and it's just interesting to me to see the diversity of perceptions of different socio-technical spaces. Just on the lists, what group(s) are you a part of? (Apologies if we've "met" there already - it's just that with a whole host of new names and email addresses, I can't quite place your "wiki name". :-)) Is your frustration to do with the way particular lists are managed - or is it something to do with the limitations of the medium itself? I suppose it also occurs to me to ask whether you think your feminist interests and sensitivities inform this discontent? (I am appalled at some of the chauvinism that I see in some online fora - but I haven't perceived anything but mutual and equal engagement from the emails so far..) But, in any case, please let the wiki be that toy - that's a lovely description - and long may this spirit of playfulness and discovery last! Cormaggio talk 01:35, 10 March 2007 (UTC)

title: User talk:SueB


text:
Computer-aided design/Computer Aided Drafting

Introduction
Computer Aided Drafting (CAD) means using the computer, instead of the classical tools (pencil, ink, rulers, paper) to create drawings. There are several advantages to this, for example the drawing can be subdivided in smaller parts, that can be reused or be worked on by several architects, updating the drawing is much faster than with hand-made drawings (where you often needed to redraw the whole drawing), and several tools can help you to check your drawing for errors. Besides this, computer generally allows you to work in real-world units, and does the scaling automatically, so your drawing fits on the printer sheet.

CAD Applications are nowadays complex and very carefully conceived. Therefore most of them are highly expensive software. Among the most used worldwide, are Autocad, Archicad, Inventor, Microstation, Vectorworks or Allplan. Almost all engineering and architecture offices on the planet use one of them, so knowing at least some of them is very important in the professional world, and usually the most important requirement on architecture job offerings.

Those applications are generally used to design whole architecture or engineering projects, often from scratch, and to produce the printed drawings that will be used to discuss the project with involved people such as project partners, authorities and clients, and the execution documents that will be used by the building team to actually build the project. Nowadays, all of them can be used to make 2-dimensional drawings directly (similarly to drawing on a sheet of paper), or to build a 3-dimensional model of your project, from which the software will extract 2D drawings that will be printed on paper.

Since CAD applications are focused on technical drawings, additional pieces of software are often used together, for rendering realistic images of the project. Among the most used in architecture are 3DS Max, Maya, Rhino, V-Ray, or Mental ray. Usually, file transfers between CAD software and rendering software is straightforward.

Besides this, several other types of software, not specifically made for architecture, are often used in conjunction with the ones shown above, for example for modeling more organic geometries, or using specific engineering techniques. Among the application that can sometimes be found in architecture offices, are Cinema 4D, Blender, Lightwave, Form Z, Real3D, SketchUp, XSI, SolidWorks or CATIA. Some offices even use them instead of traditional CAD applications.

Finally, it must be noted that the term CAD is an old concept, born when the first computer drawing systems appeared, and people switched from the drawing table to the computer, but continued to work the same way. Today, more and more, big CAD software manufacturers are switching from "drawing-oriented" solution like the software listed above, to a more "full-featured architecture solution", that allows you not only to design your project, but also to manage the enormous quantity of information (such as materials, prices, uses), that goes in an architecture project. There is even a new term for that, called BIM (Building Information Modeling). Advanced BIM software already exist, such as Revit, Archicad and Microstation (last two were CAD software that evolved and can today be called BIM), and a transitional CAD/BIM application based on Autocad called Architectural Desktop.

Most of the above mentioned applications are professional, expensive applications. Some of them offer a learning version for free or at low cost to architecture students. Some have a free version, that has a couple of features less than the full version, such as SketchUp or Solid Edge, and there exist also several free or low-cost solutions, usually very similar (but very limited) to most well-known software like Autocad. Searching for "free CAD application" on the internet should give you many options. There is also a list of free CAD applications at the bottom of this page. In all the software listed above, only Blender is open-source, but still not much used in architecture. There are also a couple of ongoing projects of open-source CAD programs such as FreeCAD but that have until today not reached a level suitable for production work.

Software-specific Courses
This is a list of learning material specially oriented to architecture.

Rhinoceros

 * Features and Overview
 * Learn
 * Grasshopper
 * Grasshopper Tutorials

AutoCAD

 * Introduction to AutoCAD
 * AutoCAD 2D

SketchUp

 * Getting started: General introduction to sketchup
 * Drawing a room: Tutorial showing how to draw a simple room

Blender

 * Official blender tutorials
 * Architecture modeling tutorial: This is a general introduction to modeling architecture with Blender
 * Precision modeling tutorial: This tutorial focuses on modeling architecture with precision
 * Urban design tutorial: This tutorial shows how to use Blender for doing urban design
 * Tree making tutorial: a tutorial oriented at modeling vegetation with Blender and several other open-source tools

FreeCAD

 * Many FreeCAD tutorials including Architecture, Part Design, Product Design, Drafting, 3D Printing, 3D Modeling and more.

Free CAD programs

 * Sketchup (free limited browser-based version, windows, mac osx): http://sketchup.google.com
 * Blender (opensource, windows, linux, mac osx)): http://www.blender.org
 * Solid Edge 2D (free limited version, windows): http://www.solidedge.eu.com/isapi/pagegen.dll/pages?page=free_2d
 * HyCAD (freeware, windows): http://www.drawease.com/index-en.htm
 * LiteCAD (freeware, windows): http://www.litecad.com
 * QCAD (opensource on linux, shareware on windows): http://www.qcad.org
 * FreeCAD (full version free on windows, mac, linux): https://freecadweb.org/


text:
Anthropomorphic Robotics Project/Brainstorming

Subpages

 * /Basic functions/
 * /Visual systems/
 * /Physical systems/
 * Logic systems
 * /Human interaction/

Efficiency is the key component. Efficiency of code. Efficiency of mechanical design. Of energy use.

For sight, breaking things down into chunks. Maybe, have vision translate things into 3D, robots then process and orient themselves in 3D space in bits, and this translates into mechanical movement. But would that take up too much computing power? Could algorithms be written efficiently enough?

Air muscles - could they be designed well enough to make them aesthetically pleasing, but also powerful? Noise seems like a problem with air muscles. Also, if they are not designed well, they may be jerky in movement. How far along is the design of chemically driven artificial muscles? The cost and the state of the art seems prohibitive.

Suppose you tell a robot to do a task. It must break that task down into chunks, and create an executable system of steps to complete the task. Language processing is a must.

The robot must look non threatening. The robot must look at least like nothing out of the ordinary. You see a chair, and it looks normal - just like a chair. You see an iPod and it looks nice, it looks expensive, but it is just an iPod. It is functional. A robot must be like this.

What ways could voice synthesis be made to sound okay? Not like a robot, but like a friendly, anthropomorphic-like voice? Why is it so robotic sounding at the moment? Maybe this is a good starting point because decent code may exist, and it may be a necessary first step that requires little funding.

Object specifiers---
 * computer vision
 * http://sourceforge.net/search/?type_of_search=soft&type_of_search=soft&words=computer+vision

Matching materials to objects. Instead of using texture, sheen or other properties of a surface to determine the material, discover what the object is that is being examined and then reference what objects of this or that sort is usually composed of....


text:
User:Jeff D

Welcome to Zetetikon

This wikiversity page is devoted to the aim of publishing new work in philosophy that is written, composed or constructed primarily by undergraduates interested in philosophy. The journal includes traditional philosophical essays as well as poetry, comedy, graphic art, photography, and any other medium that might be used to express philosophical ideas. We publish works in all areas of philosophy. The journal is edited by students studying philosophy at Northern Arizona University.

/Zetetikon: volume 2, Spring 2007

Category:Journals Category:Philosophy

title: User:Jeff D


text: User talk:B52skj

title: User talk:B52skj


text:
User:Dfordivam

Hello this is Divam. i am currently doing my studies in electronics and communication engineering. i also have a background of biology, and have special interest in microbiology, biochemistry and ecology.

title: User:Dfordivam


text:
User:Viswa

Toxins & Poisons: Phenacetin is metabolized to acetaminophen. Acetammophen is metabolized by enzymes of the cytochrome P-450 series to intermediate products: nonreactive glucuronides and sulfates (which are conjugated and eliminated in the urine), and reactive metabolites, which are metabolized with glutathione to nontoxic mercapturic acid (which is eliminated). If the toxic metabolites accumulate as a result of insufficient glucuronide or sulfate metabolism or insufficient glutathione, they are converted to toxic macromolecules that directly cause cellular death. Cats lack glucuronyl transferase and inefficiently form glucuronic acid and sulfate conjugates, leaving more acetaminophen or phenacetin to be metabolized to toxic metabolites. The glutathione stores are rapidly depleted in cats, leaving a large amount of toxic metabolites. Methemoglobinernia occurs.

title: User:Viswa


text:
Film editing/Analyzing film dailies/Morning Surprise

{| cellpadding="20" cellspacing="5" style="width: 99%; background-color: inherit; margin-left: auto; margin-right: auto"
 * style="background-color: #ffffaa; border: 1px solid #777777; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" colspan="2" |

 * style="width: 50%; background-color: Lavender; border: 1px solid #777777; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" rowspan="2" |
 * style="width: 50%; background-color: Lavender; border: 1px solid #777777; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" rowspan="2" |

"Morning Surprise"

 * This scene seems surprisingly short. Yet, many things happens in this scene which is not obvious by just looking at the dailies.

 * The location

 * This story takes place in children's bedroom in the house of their mother and father early one morning. It also takes place in an airplane in the sky.

 * The characters

 * The two young ladies are sleeping. A man in a biplane wakes them.

 * The story

 * Unfortunately, this scene does not include the script. You have to guess what is happening :

 * The two ladies are sleeping and, as the director says, the sound of an airplane can be heard.

 * What the director does not says is next the audience sees a man flying a biplane.

 * Then the audience sees one of the ladies wake up.

 * Then the audience sees the biplane begin to dive.

 * Then the audience sees the lady say, "Oh, he wouldn't."

 * Then the audience sses the man drop a water balloon from the biplane.

 * Then the audience sees the lady get up and leave. Finally we hear the water balloon hit the house and the younger lady goes back to sleep.

 * The challenge

 * The story is very simple except for the missing footage. This is a low-budget movie so all the missing footage (a man in a biplane) is done with simple 3D animation.

 * style="width: 50%; background-color: BlanchedAlmond; border: 1px solid #777777; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" rowspan="1" |

Edit this scene

 * Order the disk

 * Order the disk from the Star Movie Shop. Remember, if you have earned enough points at the Wikiversity Film School, the disk is free.  If you do not have enough points, the Star Movie Shop will accept PayPal. To contact the Star Movie Shop, you simply @undefined [[Image:Crystal Clear app xfmail.png|24px]]

 * The Overview Movie

 * Start by looking at the overview movie for "Out for a Walk". (This QuickTime movie file is in a different folder than the film dailies.  Don't get the two folders confused.)  The overview movie shows you the important parts of the film dailies in chronological order.

 * Transfer the dailies

 * Copy all the film dailies (not the overview movie) to your hard drive. If necessary, convert the dailies to whatever format your editing program prefers. Import them into your editing program.

 * Calculate the missing elements

 * You cannot edit this scene with the missing footage. To start, create still frames (storyboard frames) for the missing shots.  Then edit the scene together for a rough estimate of the timing of the missing animation shots.  Since the timing will depend on the music, you must also create the music or at least a rough of the musical score.

 * Create the missing elements

 * Now that you know the timing of the missing animation, you need to use a program like Poser or DAZ Studio to create the animation.

Edit the scene

 * Finally you edit the whole thing together.

 * Submit your finished scene
 * This scene is worth 40 points for the scene including all the missing elements. @undefined  [[Image:Crystal Clear app xfmail.png|24px]]

 * style="width: 50%; background-color: #ffff77; border: 1px solid Red; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" |
 * style="width: 50%; background-color: #ffff77; border: 1px solid Red; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" |

Analyze the Dailies

 * Multiple shots in one

 * In this little story, the action goes from interior to exterior to interior to exterior. Yet, for all the interior shots, the filmmaker simply shoots one continuous shot which will be cut apart to create all the interior elements.

 * }

{| cellpadding="20" cellspacing="5" style="width: 99%; background-color: inherit; margin-left: auto; margin-right: auto"
 * style="width: 100%; background-color: LemonChiffon; border: 1px solid #777777; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" rowspan="2" |

The next scene

 * Now look a scene which is extremely simple... or is it?

 * "Doctor, Doctor!" on the disk "24 Unedited Scenes" DVD-Data disk [[Image:Crystal Clear action forward.png|24px]]

Contact your instructor

 * Your instructor for this class is Robert Elliott. To email me, you simply  Click Here. [[Image:Crystal Clear app xfmail.png|24px]]

 * }


text:
User:Ali~enwikiversity

Hi,

I'm Alisdair - my research interests lie primarily in the areas of digital literacies, social software and socio-constructivist learning.

I am also interested in issues of accessibility, inclusivity and representation in media.

I'm a Research Assistant for the Academic Analytics Group, within the Teaching, Learning and Assessment Office of the University of Manchester. Our work involves process improvement in e-learning, using benchmarking tools.

title: User:Ali~enwikiversity


text:
User:Thuvhack

I am a Junior Electrical Engineer in trainning for a Consulting Firm in S.A

title: User:Thuvhack


text:
Student Collaboration: Africa Project

Aims and Objectives of the project
This project is a collaboration of students. The project is designed to demonstrate the extent to which online collaboration gives voice to previously unheard voices.

Introduction to Peace Studies and Research in Africa

 * What is the focus of your studies in Africa?

While studying Peace in Africa, we should focus on both the traditional and post-modern realities of African land. The way African people relate and solve their conflictual issues is different from other people world wide. With regard to the genocide that happened in Rwanda in 1994, we may not be right to assume that we may use any methods from Europe or Asia while solving the interpersonal problems and use them to reconcile the Rwandans conflict analysis training course (2008) In this regard, Peace in Africa is multidimensional and requires various interventions: human security by the government and non-state actors, regional security and defense by the police and the military, and solving inter-tribal loggerheads. This needs to take holistic approach by building coalition networks.

I argue that African political and military elites together with the affected populations have the power and potentials to bring about peace in their local areas especially in the Sub-Saharan Africa. We will look at it in more details in the upcoming days.

Posted by Nelson J M, a Human Rights Advocate -- The researcher worked on human rights and peace building in Africa and has authored "The Impact of World Vision on child protection rights in Rwanda 2000-2005"

Learning Materials

 * What learning materials are used to study in Africa? to study peace in Africa?
 * What texts have you read when studying in Africa? when studying peace in Africa?

Assignments

 * What are some of the activities (inside and outside the classroom) that support your learning in Africa? that support learning about peace in Africa?

Plans

 * What do you plan to do when you complete your studies in Africa?

title: Student Collaboration: Africa Project


text:
What is Blocking

Module by: Nos'milo Dlamini

Summary
Providing sufficient equipment to carry all traffic that could be offered in a telecommunications system would be very uneconomical.As a result of this, telecommunications systems are likely to experience problems at times where a call will not be able to get through. This phenomenon is termed blocking. Certain amounts of the measure of blocking are aceptable in such systems since it is impractical for telecommunication providers to provide sufficient equipment to carry all possible traffic.

Definition
Blocking in telecommunication systems is when a circuit group if fully occupied and unable to accept further calls [1]. It also referred to as congestion.

title: What is Blocking


text: File:Voltage.jpg

title: File:Voltage.jpg


text: User talk:Dfordivam

title: User talk:Dfordivam


text:
User talk:Ali~enwikiversity

Hey Ali! Good to see you made it over here! Yes, I can meet for lunch - time is pretty tight these days (with the panel etc), but I'm uni anyway next Wednesday - would that suit? In any case, feel free to ring or text me - not sure if you have my mobile number - its 07946372535. Alternatively I'm "cormaggio" on Skype/Gizmo/IRC etc. Cheers! Cormaggio talk 14:41, 9 March 2007 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Ali. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Ali~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:27, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:02, 19 April 2015 (UTC)

title: User talk:Ali~enwikiversity


text:
User talk:DavidWDon

Hi David - thanks for coming to Wikiversity - as well as being honest about your experience. I'm very much aware that wikis can be off-putting to newcomers - in fact, I imagine that addressing that question will constitute a significant part of my/our research. Actually, what you did was quite normal - I have spent my whole time learning about wikis through copying and modifying other people's work! I look forward to working with you - here and on-list. :-) Cormaggio talk 14:47, 9 March 2007 (UTC)

Draft charter
Hi David, thanks for your draft charter - I've commented on it and moved it to Talk:Action_research/AREOL25. Cheers! Cormaggio talk 23:46, 12 March 2007 (UTC)

 * I was just posting the above note at the moment you left your note on my page! That's a wonderful observation - and I fully appreciate what you're saying. I think for me the most nerve-wracking thing would be that I would somehow "break" a page - but when I realised that wasn't possible, I just went ahead and broke everything I could - though not out of any malice! I think wikis are a great example of learning through mistakes - you can just go back and edit your mistake until it's how you wanted it (which I can see you've been doing). Or, you can look through a page's history (click the "history" button to the right of the "edit" button) to see how somebody else fixed it - have you discovered that yet? Cormaggio talk 23:54, 12 March 2007 (UTC)

title: User talk:DavidWDon


text:
Template:Proofread2

This template is a helper template for Template:Proofread and shouldn't be used on its own.

title: Template:Proofread2


text:
Talk:Astronomy outline

"Wikiversity Astronomy textbook" <-- I think it would be great if Wikiversity participants created material for astronomy-related textbooks. Textbooks should be kept at Wikibooks. --JWSchmidt 15:52, 9 March 2007 (UTC)

title: Talk:Astronomy outline


text:
Dark matter

Welcome to the Wikiversity learning project for Dark matter.

Reading

 * Dark matter - Wikipedia article
 * Observations that support the existence of dark matter - article by Douglas Clowe, et al. (2006)
 * DAMA/NaI - laboratory experiment attempting to detect dark matter
 * Non-baryonic dark matter matter - review by Paolo Gondolo (2004).
 * Investigation of possible dark matter direct detection in electron accelerators - by Junji Hisano, et al (2006).
 * Astronomy/Current Unsolved Mysteries - Wikibooks
 * Observational Evidence for Extra Dimensions from Dark Matter by Bo Qin, Ue-Li Pen and Joseph Silk (2006).

Discussion
According to this essay by Joe Silk, "There is at most an amount of dark matter in the disk equal to the amount of luminous matter." I would like to find some published articles about how this estimate is made. --JWSchmidt 17:10, 9 March 2007 (UTC)

title: Dark matter


text:
Intermediate value theorem

The intermediate value theorem (sometimes abbreviated IVT) is a theorem about continuous functions. Recall that we call a function f continuous at the point c if $$\lim_{x \to c} f(x) = f(c)$$.

Formal statement
If f is a continuous function on a closed interval [a, b], then for all w between f(a) and f(b), there exists a c with the property that $$a \leq c \leq b$$ such that f(c) = w.

Meaning
Since the formal mathematical statement is sometimes hard to understand, we can illustrate the theorem with an example. If the temperature outside is 55 degrees at 7 in the morning, and rises to 85 degrees by noon, we know that at some point during the day, the temperature must have been 67 degrees. This is essentially what the intermediate value theorem is stating. While we might not have any idea when it was 67 degrees, we know it had to be 67 degrees sometime between 7 am and noon, because temperature doesn't "jump" from degree to degree. It fluctuates in a continuous fashion.

How is it used?
The illustration in the above section makes the the IVT seem a bit trivial. In mathematics, we often encounter such examples of seemingly trivial statements that end up being very useful. We'll show how to use the IVT below.

Example
Show that the equation $$\sin x + x = 1$$ has a solution between x = 0 and x = π.

Solution
We will use the IVT to solve this problem. Before we begin, we note that the question is asking us to show that the equation has a solution on the interval [0, π], not actually find the value of x that would make the equation true.

Let $$f(x) = \sin x + x$$. We want to use the IVT with this particular f, so we first have to check that f is continuous. Since $$g(x) = \sin x$$ is continuous everywhere and $$h(x) = x$$ is also continous everywhere, and the sum of two continuous functions is continuous, $$f = g + h$$ is a continous function.

Note further that f(0) = 0 and f(π) = π. So, if we're following along with the formal statement of the IVT, we would say that our a is 0 and our b is π, and thus our f(a) = 0 and our f(b) is π. Notice that $$0 = f(0) \leq 1 \leq \pi = f(\pi)$$, so we can assign the value 1 to the variable w in the statement of the IVT. Thus the conclusion of the theorem is true, and there exists a c, somewhere between 0 and π with the property that f(c) = 1. Since $$f(x) = \sin x + x$$, we now have that there is a c between 0 and π with the property that $$\sin c + c = 1$$. We have finished solving the problem.

Commentary
Notice that we didn't actually find the value of c that would "solve" the equation, but we didn't have to. The IVT doesn't help us find these values, but just tells us that they exist. If we wanted to actually find a solution to the equation in question, then we would have to try another approach. What the IVT tells us about the equation $$\sin x + x = 1$$ is that if we wanted to find a solution, we're not searching for something that doesn't even exist.

The Theory Behind This
The truth of the IVT depends on the numbers being "Dedekind complete", which is a property of the real numbers. The theorem is not true when only rational numbers are used. See Real Numbers and Dedekind Cut for a discussion of how the construction of the real numbers makes this theorem true.

Resources

 * Intermediate value theorem
 * Search for Intermediate value theorem at Wikiversity

title: Intermediate value theorem


text:
Generating functions

Generating functions are often used in combinatorics as a formal way of expressing certain integer-valued sequences. For a comprehensive discussion of the topic, see Wilf's book, "generatingfunctionology." The second edition of this book is available for free online as a PDF download from this link.

Mathematical maturity
The reader probably wants a strong mathematical background in calculus, and should have an understanding of abstract algebra and some analysis (for when convergence issues do arise).

Topics

 * Introduction to generating functions
 * Calculus of generating functions

title: Generating functions


text:
Introduction to generating functions

Overview
The main idea here is that we're going to work in the ring of formal power series. A ring, recall, is just a certain type of algebraic structure. For us, all we need to concern ourselves with right now is that the sum of two power series is a power series and that the product of two power series is a power series. Every power series has an additive inverse (almost trivially), and some have multiplicative inverses (which will be discussed in later sections). The most important thing to remember is that we consider these power series as elements of a certain ring. As such, we are not concerned with the convergence of any power series (yet). There will come a time and place to discuss these ideas (as we will want to sum certain sequences and pick off certain terms, both of which demand us "plugging in values for x"), but for now, we'll just interpret them formally.

As a brief example, note that the expression $$\sum_{n \geq 0} n! x^n$$ does not have a positive radius of convergence (converges only when x = 0), but it is still a perfectly good element of our ring of formal power series.

Ordinary generating functions
Given a sequence $$\left\{a_n\right\}$$, the ordinary generating function of  $$\left\{a_n\right\}$$ f(x) is the formal power series
 * $$f(x) = \sum_{n\geq 0} a_n x^n.$$

If f(x) is the ordinary generating function for the sequence $$\left\{a_n\right\}$$, we often write
 * $$f(x) \longleftrightarrow\!\!\!\!\!\!\!\!\!\!^{\text{ops}}\ \left\{a_n\right\}$$

We don't often concern ourselves with the convergence of these series; we interpret them strictly in a formal sense.

For an example, recall from calculus that the function $$f(x) = \dfrac{1}{1-x}$$ has power series representation
 * $$\dfrac{1}{1-x} = 1 + x + x^2 + \cdots = \sum_{n \geq 0} x^n.$$

The sequence of coefficients of the powers of x in the above power series is the sequence that is constantly 1. Hence, we have that
 * $$\dfrac{1}{1 - x} \longleftrightarrow\!\!\!\!\!\!\!\!\!\!^{\text{ops}}\ \{1\}_{n \geq 0}.$$

Exponential generating functions
Another type of generating function, the exponential generating function is similar to the ordinary generating function, but with a twist. If f(x) is the exponential generating function for the sequence $$\left\{a_n\right\}$$, we write
 * $$f(x) \longleftrightarrow\!\!\!\!\!\!\!\!\!\!^{\text{egf}}\ \left\{a_n\right\},$$

and this means that the function f(x) has power series representation
 * $$f(x) = \sum_{n\geq 0} a_n \dfrac{x^n}{n!}.$$

We saw above that the function $$\dfrac{1}{1-x}$$ is the ordinary generating function for the sequence that is constantly 1; but for what sequence is it the exponential generating function?

Note that for all n, we have
 * $$\dfrac{1}{1-x} = \sum_{n \geq 0} x^n = \sum_{n \geq 0} n! \dfrac{x^n}{n!}.$$

If we set $$\left\{a_n\right\} = \left\{n!\right\}$$, we have that
 * $$\dfrac{1}{1-x} = \sum_{n \geq 0} n! \dfrac{x^n}{n!}$$,

so we can write
 * $$\dfrac{1}{1-x} \longleftrightarrow\!\!\!\!\!\!\!\!\!\!^{\text{egf}}\ \left\{n!\right\}_{n \geq 0}.$$

So what is the exponential generating function of the sequence that is constantly 1? Recall that ex has the power series representation
 * $$e^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!} + \cdots = \sum_{n \geq 0} \dfrac{x^n}{n!}$$,

so we have
 * $$e^x \longleftrightarrow\!\!\!\!\!\!\!\!\!\!^{\text{egf}}\ \{1\}_{n \geq 0}.$$

This should give you a feel as to why the name for such a generating function is "exponential" generating function.

Continuation
This lesson continues on the page titled calculus of generating functions.

title: Introduction to generating functions


text:
Calculus of generating functions

The main idea behind the calculus of generating functions is to figure out how to compute the generating function of a sequence that is the product (or sum, etc.) of two other sequences whose generating functions are known. There are different rules for dealing with different types of generating functions; for instance, the rules that hold for manipulating ordinary generating functions may have analogues with their exponential generating function counterparts, but the rules are often not the same. Hence, we must be careful when manipulating formal power series, and note which type of generating function we're working with.

Ordinary generating functions
Similar to differentiation in the subject of calculus, we have a lot of rules we can use to manipulate ordinary power series. We'll start with the shift rule.

Shift rule
To start, let's consider an easy example. Suppose we have that
 * $$f(x) \longleftrightarrow\!\!\!\!\!\!\!\!\!\!^{\text{ops}}\ \left\{a_n\right\}$$,

and we want to find the generating function for the sequence $$\left\{a_{n+1}\right\}_{n\geq 0}$$. Formally, we want the function g(x) that has power series representation
 * $$g(x) = \sum_{n\geq 0} a_{n+1} x^n$$.

Since we know that
 * $$f(x) = \sum_{n \geq 0} a_n x^n$$,

We have that
 * $$f(x) - a_0 = \sum_{n \geq 1} a_n x^n$$,

and reindexing gives
 * $$f(x) - a_0 = \sum_{n \geq 0} a_{n+1} x^{n+1}$$.

Finally, dividing both sides by x, we achieve
 * $$\dfrac{f(x) - a_0}{x} = \sum_{n \geq 0} a_{n+1} x^n$$,

which was precisely the function we wanted. Setting $$g(x) = \dfrac{f(x) - a_0}{x}$$, we now have
 * $$g(x)\longleftrightarrow\!\!\!\!\!\!\!\!\!\!^{\text{ops}}\ \left\{a_{n+1}\right\}_{n \geq 0}$$, as desired.

From this example, we can derive a rule. If $$f(x) \longleftrightarrow\!\!\!\!\!\!\!\!\!\!^{\text{ops}}\ \left\{a_n\right\}$$, then the ordinary generating function for the sequence $$\left\{a_{n+k}\right\}_{n \geq 0}$$ is the function
 * Shift rule
 * $$g(x) = \dfrac{f(x) - a_0 - a_1x - a_2x^2 - \cdots - a_{k-1}x^{k-1}}{x^k}.$$

Polynomial multiplier rule
Again, let
 * $$f(x) \longleftrightarrow\!\!\!\!\!\!\!\!\!\!^{\text{ops}}\ \left\{a_n\right\}$$.

Suppose we want the generating function for $$\left\{na_n\right\}_{n \geq 0}$$. How can we get this to come about?

Here, we have to be a bit more clever than we were with the shift rule above. Notice that if $$f(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots$$ then
 * $$f'(x) = 0 + a_1 + 2a_2x + 3a_3x^2 + \cdots,$$

which is almost what we want. If we multiply both sides by x, we get
 * $$xf'(x) = 0 + a_1x + 2a_2x^2 + 3a_3x^3 + \cdots,$$

which is exactly what we want. So, if
 * $$f(x) \longleftrightarrow\!\!\!\!\!\!\!\!\!\!^{\text{ops}}\ \left\{a_n\right\},$$

then
 * $$xf'(x) \longleftrightarrow\!\!\!\!\!\!\!\!\!\!^{\text{ops}}\ \left\{na_n\right\}.$$

Immediately, we see the use for some operator notation; notation that will tell us to perform the operation "differentiate f once with respect to x, then multiply the result by x".

We elect to use notation similar to Euler's differential operator (for ease); that is, D. To express this more precisely, we'll stipulate that
 * $$D f = f'(x)$$

and that
 * $$(xD)f = xf'(x).$$

Applying a similar idea to the one above, if we want to find the OPS for $$\left\{n^2a_n\right\}_{n \geq 0}$$ given the generating function for $$\left\{a_n\right\}_{n \geq 0}$$, we would apply the "xD" operator twice. Notation for this would be
 * $$(xD)^2f,$$

and in words means "differentiate f, multiply by x, differentiate again, and then multiply by x". This leads to the intermediate rule that $$(xD)^kf \longleftrightarrow\!\!\!\!\!\!\!\!\!\!^{\text{ops}}\ \left\{n^ka_n\right\}_{n \geq 0}$$ whenever f is the generating function for the sequence $$\left\{a_n\right\}_{n \geq 0}$$. Now that we have a rule for multiplying a sequence by nk, by default we have a rule for multiplying a sequence by any polynomial in n. If
 * Polynomial multiplier rule
 * $$f(x) \longleftrightarrow\!\!\!\!\!\!\!\!\!\!^{\text{ops}}\ \left\{a_n\right\}_{n \geq 0}$$

and p is any polynomial, then
 * $$p(xD) f(x) \longleftrightarrow\!\!\!\!\!\!\!\!\!\!^{\text{ops}}\ \left\{p(n) a_n\right\}_{n \geq 0}.$$

title: Calculus of generating functions


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User:Drysdale 1

Wizov, Nanette - Spanish I

title: User:Drysdale 1


text:
Talk:Introduction to generating functions

Formal sense
What does this mean when we say "formal sense"? Does it mean I can put those coefficients to be anything without worrying whether or not the function actually converges? --HappyCamper 19:47, 9 March 2007 (UTC)

 * "Formal" here means "in virtue of the form." So, it's the "form" of the power series that we're using. It's a nice, concise way to express a sequence.
 * In practice, we don't care at all whether these series converge. There will be times where we have to consider radii of convergence, but in the introduction, I wanted to just talk about basically what these things are. –King Bee (τ • γ) 19:51, 9 March 2007 (UTC)
 * I guess to really answer your question, the answer is a plain yes. =) –King Bee (τ • γ) 19:58, 9 March 2007 (UTC)

 * Okay I see, thank you. I started reading that book you recommended. Very nice :-) --HappyCamper 02:06, 10 March 2007 (UTC)

First tutorial question?
Alright, so I have finally constructed a series which is of interest to me...Let's say we have a sequence that satisfies

 * $$c_{k+2} = - {{c_k} \over 2} \sqrt{\frac{k+1}{k+2}}$$

and say, $$c_0$$ and $$c_1$$ are some fixed constants a and b. Let's also take $$k \ge 0$$. What's the generating function for this sequence of numbers? I was thinking we can walk through something like this, and put it on a page somewhere. --HappyCamper 04:54, 17 March 2007 (UTC)

So just what IS a power series?
If I am looking at an expression, how can I determine whether it is, or is not, a power series? Is a power series always defined by a summation? Is it required that a power series be a sum over an infinity? Must the only variables involved be the summation one and x? Can the series be a summation over more than 1 variable or with an expression involving more than 1 variable? (29 September 2015 by 76.89.22.244)

 * You might try asking this question at Help Desk. -- Dave Braunschweig (discuss • contribs) 13:32, 3 October 2015 (UTC)


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Template:French exercises

EXERCISE • Exercise Appendix • •

SOLUTION • •

 * Add the following to both the lesson where the exercise goes and the appropriate section on this page. Replace [...] with the specified lesson info.

 * Example:

title: Template:French exercises


text: User talk:Bruhsmith

title: User talk:Bruhsmith


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Sustainability/Organic agriculture

This is an area where research and learning related to organic agriculture can take place.

Discussion questions

 * How can organic agriculture positively affect society?
 * How can organic agriculture and global warming interact?
 * What research can take place online related to organic agriculture?

Readings

 * Sustainability
 * Organic cotton
 * Biological pest control
 * Biodynamic agriculture

title: Sustainability/Organic agriculture


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Sustainability

Please contribute an essay or article on . Enter title that is not already on the list below and start your draft. If you wish to change your title, or have any other questions, visit us at Colloquium.
 * Call for action
 * List of contributions

Sustainability is an attempt to provide the best outcomes for the human and natural environments both now and into the indefinite future. It relates to the continuity of economic, social, institutional and environmental aspects of human society, as well as the non-human environment. It is intended to be a means of configuring civilization and human activity so that society, its members and its economies are able to meet their needs and express their greatest potential in the present, while preserving biodiversity and natural ecosystems, and planning and acting for the ability to maintain these ideals in a very long term. Sustainability affects every level of organization, from the local neighborhood to the entire planet.

Intended Learning Outcomes

 * Understand the concept
 * Be aware of the importance in multiple spheres ...
 * Determine possible directions for further study, for example:
 * Triple bottom line reporting

/Handbook/
A Living Handbook for Tertiary Sustainability Educators.

Subpages
Some of these subpages are just stubs. Please improve Wikiversity by improving them. Go to the Colloquium if you have any questions.

title: Sustainability


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Category:Sustainability

Sustainability is an attempt to provide the best outcomes for the human and natural environments both now and into the indefinite future. It relates to the continuity of economic, social, institutional and environmental aspects of human society, as well as the non-human environment. It is intended to be a means of configuring civilization and human activity so that society, its members and its economies are able to meet their needs and express their greatest potential in the present, while preserving biodiversity and natural ecosystems, and planning and acting for the ability to maintain these ideals in a very long term. Sustainability affects every level of organization, from the local neighborhood to the entire planet.

title: Category:Sustainability


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User:JWSchmidt/Blog/10 March 2007

Verified credentials  Jimmy Wales again raised the issue of credentials verification for Wikipedians who wish to claim that they have credentials. The first thing to say about such a system is that it would be voluntary. Nobody is talking about making this a BIG DEAL for the Wikimedia Foundation. In fact, it is clear that the Wikimedia Foundation is not going to get involved with any credential verification process for typical Wikipedians (see). The Foundation does have a process for verification of the real world identity of those few members of the community who are given access to information covered by the Foundation's privacy policy. However, any credential verification process for typical Wikipedians will have to be run by the community.

The second thing to be said is that there are really two limited reasons for such a credential verification process. The first reason is to try to prevent future problems in which a Wikipedian falsely claims to have a set of credentials. It has been suggested that Wikipedia have a template saying "this claim of credentials has not been verified" that could be placed on any user page where a Wikipedian claims to have a set of credentials. The second reason to verify credentials is the original reason given by Jimmy two years ago; as a way to let the world know that there are experts who edit Wikipedia.

But how can a community of volunteer wiki editors verify the identities of their fellow wiki editors? I am most concerned with verification of the identities of wiki editors who have a professional website, particularly in the .edu domain. I wonder if it would be possible to make modifications to the "Email this user" system in order to create a semi-automated system for verifying that wiki users are the owners of .edu email address. I'm not sure how hard it is to fake access to a .edu email address (see), but in any case I'm not sure that verification of a .edu email address is itself very useful. What is really needed is a webpage at a .edu website that lists the credentials of a person. If that person wants to be identified as editing Wikipedia as user FooBar, they that fact could be included on their professional credentials webpage. Again, I'm not sure if it is possible to fake the control of a URL in the .edu domain.

Relevance to Wikiversity  If Wikiversity does develop a system that allows for some editing outside of NPOV, then there will be a need for participants who can help guide the community in making sure that good scholarship is used. Sometimes it takes an expert to spot spoofs and bogus research methodologies. Can most people have confidence in Wikiversity without some formal evidence that there are experts helping with the project?

title: User:JWSchmidt/Blog/10 March 2007


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Film editing/Analyzing film dailies/Trivia Quiz

{| cellpadding="15" cellspacing="5" style="width: 99%; background-color: inherit; margin-left: auto; margin-right: auto"
 * style="background-color: #ffffaa; border: 1px solid #777777; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" colspan="2" | [[Image:Crystal Clear action apply.png|32px]] Please take this pop quiz! [[Image:Crystal Clear app clock.svg|32px]]

 * style="width: 50%; background-color: #ffffe0; border: 1px solid #777777; vertical-align: center; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" |
 * style="width: 50%; background-color: #ffffe0; border: 1px solid #777777; vertical-align: center; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" |

Trivia Question #1
What is the name of the space station in "Out for a Walk".

(Hint: It is located in deep space but it is not number nine!)

@undefined. (1/2 point)

 * style="width: 50%; background-color: #ffe0ff; border: 1px solid #777777; vertical-align: center; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" |

Trivia Question #2
What is the name of the family in the scenes of the "Family Reunion".

(Hint: They have a mountain named after them.)

@undefined. (1/2 point)
 * }

{| cellpadding="15" cellspacing="5" style="width: 99%; background-color: inherit; margin-left: auto; margin-right: auto"
 * style="width: 50%; background-color: #e0ffff; border: 1px solid #777777; vertical-align: center; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" |

Trivia Question #3
The doctor in "Doctor, Doctor!" was a child actor. What famous role did she play?

(Hint: The TV series featured a very small house at the edge of the Indian Territories on very flat ground with lots of grass on it.)

@undefined. (1/2 point)

 * style="width: 50%; background-color: #e0ffe0; border: 1px solid #777777; vertical-align: center; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" |

Trivia Question #4
The male actor in "Wedding Bells" once dreamed of who?

(Hint: This is the name of a popular TV show about a beautiful girl who lived in a bottle and he was an astronaut.)

@undefined. (1/2 point)
 * }

{| cellpadding="20" cellspacing="5" style="width: 99%; background-color: inherit; margin-left: auto; margin-right: auto"
 * style="width: 100%; background-color: LemonChiffon; border: 1px solid #777777; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" rowspan="2" |

The next page

 * Now you will being to look at 24 unedited scenes filmed in Hollywood. The first one is:
 * "Out for a Walk" on the "24 Unedited scenes" Editing Workshop CD [[Image:Crystal Clear action forward.png|24px]]

 * }


text:
Talk:Fire and emergency management/Who is who in the department

Hello Mfinney: the link to your User page is no good;  it appears the page got left behind on wikibooks.


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Film editing/Analyzing film dailies/Doctor Doctor

{| cellpadding="20" cellspacing="5" style="width: 99%; background-color: inherit; margin-left: auto; margin-right: auto"
 * style="background-color: #ffffaa; border: 1px solid #777777; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" colspan="2" |

 * style="width: 50%; background-color: Lavender; border: 1px solid #777777; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" rowspan="2" |
 * style="width: 50%; background-color: Lavender; border: 1px solid #777777; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" rowspan="2" |

"Doctor, Doctor!"

 * This is a very simple scene to edit. But this scene will not be believable without a huge number of sound effects which are mostly background sound of a hospital.

 * Also, music is needed to make the scene more easy for the audience to accept.

 * The location

 * This story takes place in a large hospital.

 * The characters

 * The lady who is reading the magazine is a patient. The mute person is the doctor.

 * The story

 * This scene is from an educational afternoon TV special. This special TV movie explains about hormones of the human body.

 * The lady did not feel normal so she went to the hospital. They took many tests.  Now the lady waits for the results of the tests to find out what is wrong with her.

 * To keep herself happy, the lady reads magazines which show pictures of very beautiful young women. The lady wonders what she would look like if she were young again.

 * When the doctor enters the room, the lady jokes with the doctor that she would like to have the body of the young and beautiful women in the magazine. The doctor (and the audience) does not find this funny but this emphasizes the lady's worries about getting old.

 * The doctor cannot speak so the doctor gives the lady a note pad with the name of the problem that the lady has. The lady reads the note which only says, "Menopause".  This means that the lady's horemones have changes now that she is getting older.  That was why the lady felt sick.

 * The doctor tries to comfort the lady with a back rub. The lady remains embarased and unhappy.

 * The challenge

 * The story is very simple but the emotions are not. The subject of the story is not appealing to the audience.  You must use music (and editing) to make the scene seem warm and friendly.  That is not going to be easy.

 * style="width: 50%; background-color: BlanchedAlmond; border: 1px solid #777777; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" rowspan="1" |

Edit this scene

 * Order the disk

 * Order the disk from the Star Movie Shop. Remember, if you have earned enough points at the Wikiversity Film School, the disk is free.  If you do not have enough points, the Star Movie Shop will accept PayPal. To contact the Star Movie Shop, you simply @undefined [[Image:Crystal Clear app xfmail.png|24px]]

 * The Overview Movie

 * For most scenes, you look at the overview movies. This scene is so short, there is no point in having a overview movie.  Instead, you look at the two good shots and the two rejected shots.

 * Transfer the dailies

 * Copy all the film dailies to your hard drive. If necessary, convert the dailies to whatever format your editing program prefers. Import them into your editing program.

 * Write the music

 * You have to decide what emotions you want to create and you must create musical sound effects which create these moods.

Edit the scene

 * Finally you edit the picture to match the timing of your music.

 * Submit your finished scene
 * This scene is worth 10 points for the scene and a possible 36 points for the music. @undefined [[Image:Crystal Clear app xfmail.png|24px]]

 * style="width: 50%; background-color: #ffff77; border: 1px solid Red; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" |
 * style="width: 50%; background-color: #ffff77; border: 1px solid Red; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" |

Analyze the Dailies

 * Who is the director?

 * When you watch the dailies, you are suprised that the doctor does not say anything except to yell, "Cut!" This is because the director is performing the roll of the doctor and is keeping the cost of this union production down by being an extra (extras are not allowed to say anything) rather than being a full union actor.

 * }

{| cellpadding="20" cellspacing="5" style="width: 99%; background-color: inherit; margin-left: auto; margin-right: auto"
 * style="width: 100%; background-color: LemonChiffon; border: 1px solid #777777; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" rowspan="2" |

The next scene

 * Now look a scene which is extremely simple... or is it?

 * "The Girl and the Golden Jewel" on the disk "24 Unedited Scenes" DVD-Data disk [[Image:Crystal Clear action forward.png|24px]]

Contact your instructor

 * Your instructor for this class is Robert Elliott. To email me, you simply  Click Here. [[Image:Crystal Clear app xfmail.png|24px]]

 * }


text:
Introductory Classical Mechanics/Units of Measurement

In general a Unit of Measurement is a term which describes a physical phenomena. The system of measurements used almost universally in physics is the SI, or metric system. The fundamental  units of any system quantify distance, mass, time, temperature, electric current, brightness of light, and quantity of matter. The units and abbreviations are the meter(m), kilogram (kg), second (s), kelvin (K), ampere (A), candela (cd), and the mole (mol).

The meter
The meter is the fundamental unit of distance. It is defined as the distance a beam of light travels in a perfect vacuum in 3.33564095 billionths of a second.

The kilogram
The kilogram is the fundamental unit of mass. A kilogram is the exact mass of a sample of platinum iridium kept in the International Bureau of Weights and Measures.

The second
The second is the fundamental unit of time. It is the time taken for an atom of cesium to oscillate through 9.192631770 billion cycles.

The kelvin
The kelvin is the fundamental unit of temperature, equivalent to the temperature increment of .003661 part of the thermodynamic temperature of the triple point of water.

The ampere
The ampere, or amp, is the fundamental unit of current. The definition is that an ampere is the amount of constant charge carrier flow through two straight, parallel, infinitely thin wires placed 1 meter apart in a vacuum that results in a force between the conductors of 10 millionths of a Newton per linear meter.

The candela
The candela is the unit of luminous intensity. It equals 1/683 of a watt of radiant energy at a frequency of 54 terahertz in a solid angle of one steradian.

The mole
The mole is the quantity of particles that has a mass in grams equal to the molecular weight of said particle. It is equal to Avogadro's number, or 6.02×10^23 particles

Derived units
Derived units are units that are products of fundamental units

The newton
The newton is the unit of force, equivalent to one kg*m/s^2.

The joule
The joule is the unit of energy or the result of a force acting through a distance. One joule is equal to one newton meter. $$Nm=kgm^2/s^2$$.


text: User:Okyea

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User:Chuboltite

Who am I?
My actual user page can be found at Wikipedia.

Status at Wikiversity
Chuboltite 10:03, 10 March 2007 (UTC) Currently studying Wikiversity's C++ course.

Chuboltite 15:42, 1 April 2007 (UTC)Signed up for the Ruby course.

title: User:Chuboltite


text: Template:EuroHistDisc

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