text:
User:Proof2012

user:adamsaleel

user:proof2013

Page 14

$$arctan{\left(\frac{1}{\phi^{2n}}\right)}-arctan{\left(\frac{1}{\phi^{2n+2}}\right)}= arctan{\left(\frac{1}{\phi^{2n+1}+\phi^{-2n-1}}\right)} $$

$$\phi^mF_n-\phi^nF_m=F_{m-n}$$

-

$$arctan\left(\frac{F_n}{\phi^n}\right)-arctan\left(\frac{F_m}{\phi^m}\right)= (-1)^{n-1}arctan\left(\frac{F_{m-n}}{F_{n}F_{m}+\phi^{m+n}}\right)$$

n <= m

---

$$\pi=3arctan\left(\frac{1}{\phi}\right)+6arctan\left(\frac{1}{\phi^3}\right) +arctan\left(\frac{1}{\phi^5}\right)$$

$$\frac{\pi}{2}=arctan\left(\frac{1}{\phi}\right)+4arctan\left(\frac{1}{\phi^3}\right) +arctan\left(\frac{1}{\phi^5}\right)$$

$$\frac{\pi}{4}=3arctan\left(\frac{1}{\phi}\right)-3arctan\left(\frac{1}{\phi^3}\right) -2arctan\left(\frac{1}{\phi^5}\right)$$

$$\frac{\pi}{4}=2arctan\left(\frac{1}{\phi}\right)-arctan\left(\frac{1}{\phi^3}\right) -arctan\left(\frac{1}{\phi^5}\right)$$

I realised the above formulae is the variation of these three below

$$\frac{\pi}{4}=arctan\left(\frac{1}{\phi}\right)+arctan\left(\frac{1}{\phi^3}\right)$$

$$\frac{\pi}{4}=3arctan\left(\frac{1}{\phi^3}\right)+arctan\left(\frac{1}{\phi^5}\right)$$

$$\frac{\pi}{2}=3arctan\left(\frac{1}{\phi}\right)-arctan\left(\frac{1}{\phi^5}\right)$$

---

Page 13

$$F_1=1$$, $$F_2=1$$

$$F_{n+2}=F_n+F_{n+1}$$

$$\phi^2F_{n+k}-F_{n+k-2}=\phi^{n+k}$$

$$arctan\left(\frac{1}{\phi^{n}F_{n}}\right)-arctan\left(\frac{1}{\phi^{n+2}F_{n+2}}\right)= arctan\left(\frac{1}{\phi^{-2n-2}+F_{n}F_{n+2}}\right)$$

$$arctan\left(\frac{1}{\phi^{2n-1}}\right)+arctan\left(\frac{1}{\phi^{2n+1}}\right)=arctan\left(\frac{1}{F_{2n+1}}\right)+ arctan\left(\frac{1}{F_{2n+2}}\right) $$

$$2arctan\left(\frac{1}{\phi^{2n-1}}\right)+arctan\left(\frac{1}{\phi^{2n+1}}\right)-arctan\left(\frac{1}{\phi^{2n+3}}\right)= arctan\left(\frac{1}{F_{2n}}\right)+arctan\left(\frac{1}{F_{2n+1}}\right) $$

$$arctan\left(\frac{1}{\phi^{2n-1}}\right)-arctan\left(\frac{1}{\phi^{2n+1}}\right)=arctan\left(\frac{1}{L_{2n+1}}\right) $$

$$arctan\left(\frac{1}{\phi^{2n-1}}\right)-arctan\left(\frac{1}{\phi^{2n+3}}\right)=arctan\left(\frac{1}{F_{2n+1}}\right) $$

$$arctan\left(\frac{1}{\phi^{2n-3}}\right)=arctan\left(\frac{1}{F_{2n-1}}\right)+arctan\left(\frac{1}{F_{2n+3}}\right) +arctan\left(\frac{1}{F_{2n+7}}\right)+\cdots$$

$$\frac{\pi}{2}=arctan\left(\frac{1}{F_{2n-1}}\right)+arctan\left(\frac{1}{F_{-2n+5}}\right) +arctan\left(\frac{1}{F_{2n+3}}\right)+arctan\left(\frac{1}{F_{-2n+9}}\right)+ arctan\left(\frac{1}{F_{2n+7}}\right)\cdots+$$

$$arctan\left(\frac{1}{\phi}\right)+arctan\left(\frac{1}{\phi^3}\right)+arctan\left(\frac{1}{\phi^5}\right)=arctan\left(\frac{1}{1}\right)+ arctan\left(\frac{1}{2}\right)-arctan\left(\frac{1}{3}\right)-arctan\left(\frac{1}{5}\right)+ arctan\left(\frac{1}{8}\right)+arctan\left(\frac{1}{13}\right)-\cdots$$

$$arctan\left(\frac{1}{\phi}\right)=arctan\left(\frac{1}{2}\right)+ arctan\left(\frac{1}{3}\right)-arctan\left(\frac{1}{5}\right)-arctan\left(\frac{1}{8}\right)+ arctan\left(\frac{1}{13}\right)+arctan\left(\frac{1}{21}\right)-\cdots$$

In general

$$arctan\left(\frac{1}{\phi^{2n-1}}\right)=arctan\left(\frac{1}{F_{2n+1}}\right)+ arctan\left(\frac{1}{F_{2n+2}}\right)-arctan\left(\frac{1}{F_{2n+3}}\right)-arctan\left(\frac{1}{F_{2n+4}}\right)+ arctan\left(\frac{1}{F_{2n+5}}\right)+arctan\left(\frac{1}{F_{2n+6}}\right)-\cdots$$

$$arctan\left(\frac{1}{\phi^{2n-1}}\right)=arctan\left(\frac{1}{F_{2n}}\right)-arctan\left(\frac{1}{F_{2n+2}}\right) +arctan\left(\frac{1}{F_{2n+4}}\right)- arctan\left(\frac{1}{F_{2n+6}}\right) +\cdots$$

---

$$arctan\left(\frac{1}{2}\right)=arctan\left(\frac{1}{1}\right)-arctan\left(\frac{1}{2}\right)+ arctan\left(\frac{1}{5}\right)-arctan\left(\frac{1}{13}\right)+\cdots$$

$$arctan\left(\frac{1}{3}\right)=arctan\left(\frac{1}{2}\right)-arctan\left(\frac{1}{5}\right)+ arctan\left(\frac{1}{13}\right)-arctan\left(\frac{1}{34}\right)+\cdots$$

n > = 2

$$arctan\left(\frac{1}{\phi^{2n}}\right)=arctan\left(\frac{1}{F_{2n+1}}\right)-arctan\left(\frac{1}{F_{2n+3}}\right) +arctan\left(\frac{1}{F_{2n+5}}\right)-\cdots$$

Page 12

From Ramanujan 6 - 10 - 8 Identity

Let $$ad=bc$$

I have noticed the following relationship between Ramanujan and Hirschhorn identities

$$(a+b+c)^4+(b+c+d)^4-(c+d+a)^4-(d+a+b)^4+(a-d)^4-(b-c)^4=0$$

$$(b+c+d)^4+(a-d)^4-(c+d+a)^4-(b-c)^4=(d+a+b)^4-(a+b+c)^4$$

$$2(d+a+b)^4-2(a+b+c)^4=K\left[(b+c+d)^2+(a-d)^2-(a+b+c)^2-(c+d+a)^2-(b-c)^2+(d+a+b)^2\right]$$

$$2(b+c+d)^4+2(a-d)^4-2(c+d+a)^4-2(b-c)^4=K\left[(b+c+d)^2+(a-d)^2-(a+b+c)^2-(c+d+a)^2-(b-c)^2+(d+a+b)^2\right]$$

and

$$(a+b+c)^2+(b+c+d)^2-(c+d+a)^2-(d+a+b)^2+(a-d)^2-(b-c)^2=0$$

$$(b+c+d)^2+(a-d)^2-(c+d+a)^2-(b-c)^2=(d+a+b)^2-(a+b+c)^2$$

$$2(d+a+b)^2-2(a+b+c)^2=K\left[(b+c+d)^4+(a-d)^4-(a+b+c)^4-(c+d+a)^4-(b-c)^4+(d+a+b)^4\right]$$

$$2(b+c+d)^2+2(a-d)^2-2(c+d+a)^2-2(b-c)^2=K\left[(b+c+d)^4+(a-d)^4-(a+b+c)^4-(c+d+a)^4-(b-c)^4+(d+a+b)^4\right]$$

From Hirschhorn 3 - 7 - 10 Identity

Let $$ad=bc$$

where K is an integer

$$\frac{(b+c+d)^4+(a-d)^4-(a+b+c)^4-(c+d+a)^4-(b-c)^4+(d+a+b)^4}{(b+c+d)^2+(a-d)^2-(a+b+c)^2-(c+d+a)^2-(b-c)^2+(d+a+b)^2}=K$$

--

Page 11

Ramanujan's equation

$$\frac{x}{2x-1}=1-\frac{x-1}{x+1}+\frac{(x-1)(x-2)}{(x+1)(x+2)}-\frac{(x-1)(x-2)(x-3)}{(x+1)(x+2)(x+3)}+\cdots$$

-

Section a

Let

$$R(x)=1-\frac{x-1}{x+1}+\frac{(x-1)(x-2)}{2(x+1)(x+2)}-\frac{(x-1)(x-2)(x-3)}{3(x+1)(x+2)(x+3)}+\cdots$$,

$$H_x^{-}=1-\frac{1}{2}+\frac{1}{3}-\cdots-\frac{1}{x}$$

where x be any integers greater or equal to 1

Therefore we have

$$R(x)=H_{2x-1}^{-}$$

and

$$\lim_{x\to \infty}R(x)=\ln2$$

---

Section b

Let

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

Where $$C_x$$ is Catalan numbers

$$C_x=\frac{1}{x+1}$$

and $$x$$ is any integers greater than 2

Therefore we have

$$R_s(x)=(-1)^x\frac{C_{x-2}}$$

--

Section c

Let

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

Where $$C_x$$ is Catalan numbers

$$C_x=\frac{1}{x+1}$$

and $$x$$ is any integers greater than or equal to 2

Therefore we have

$$R_t(x)=-\frac{C_{x-2}}$$

--

Section d

Let

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

$$x$$ is any integers greater than or equal to 2

Therefore we have

$$R_u(x)=\frac$$

Page 10

Ramanujan's equation

$$\frac{1}{2}=\sqrt{1-\sqrt{1-\frac{1}{2}\sqrt{1-\frac{1}{4}\sqrt{1-\frac{1}{8}\sqrt{1-\cdots}}}}}$$

I noticed the following formula from the above,

$$\frac{x}{2}+\sqrt{y}=\sqrt{y+x\sqrt{y+\frac{x}{2}\sqrt{y+\frac{x}{4}\sqrt{y+\frac{x}{8}\sqrt{y+\cdots}}}}}$$

Page 9

$$\sum_{i=1}^{n}\cos{\left[\frac{(3i-1)\pi}{2(2n+1)}\right]}\cos{\left[\frac{(i-1)\pi}{2(2n+1)}\right]}=\sum_{i=1}^{n}\sin{\left[\frac{3i\pi}{2(2n+1)}\right]}\sin{\left[\frac{i\pi}{2(2n+1)}\right]}= \frac{1}{4\sin{\left[\frac{\pi}{2(2n+1)}\right]}}$$

$$\sum_{i=1}^{n}\cos{\left[\frac{3i\pi}{2(2n+1)}\right]}\cos{\left[\frac{i\pi}{2(2n+1)}\right]}= \cos{\left[\frac{2\pi}{2(2n+1)}\right]}$$

Page 8

Adamchik and Wagon

$$\pi$$ : A 2000-Year Search Changes Direction

Alternating

$$\tan^{-1}(2)=\sum_{k=0}^{\infty}\frac{(-1)^k}{4^k}\left[\frac{1}{4k+1}+\frac{1}{2(4k+3)}\right]$$

Non - alternating

$$\tan^{-1}(2)=\sum_{k=0}^{\infty}\frac{1}{4^{2k}}\left[\frac{1}{8k+1}+\frac{1}{2(8k+3)}-\frac{1}{4(8k+5)}-\frac{1}{8(8k+7)}\right]$$

I noticed a pattern of the above formulae, the next one will be Alternating and Non - alternating and so on, ...

$$\tan^{-1}(2)=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{4^{3k}}\left[\frac{1}{16k+1}+\frac{1}{2(16k+3)}-\frac{1}{4(16k+5)}-\frac{1}{8(16k+7)}+ \frac{1}{16(16k+9)}+\frac{1}{32(16k+11)}\right]$$

$$\tan^{-1}(2)=\sum_{k=0}^{\infty}\frac{1}{4^{4k}}\left[\frac{1}{32k+1}+\frac{1}{2(32k+3)}-\frac{1}{4(32k+5)}-\frac{1}{8(32k+7)}+ \frac{1}{16(32k+9)}+\frac{1}{32(32k+11)}-\frac{1}{64(32k+13)}-\frac{1}{128(32k+15)}\right]$$

---

General formulae,

-

$$2^{2k+1}\sum_{i=1}^{n}\sin^{2k+1}{\left(\frac{2i\pi}{2n+1}\right)}=\sum_{j=0}^{k}(-1)^j\tan{\left[\frac{(3-j)\pi}{2n+1}\right]}$$

$$2^{2k}\sum_{i=1}^{n}\sin^{2k}{\left(\frac{2i\pi}{2n+1}\right)}=n+\sum_{j=0}^{k-1}(-1)^{j+k-1}$$

$$2^{2k}\sum_{i=1}^{n}\cos^{2k}{\left(\frac{2i\pi}{2n+1}\right)}=n-\sum_{j=0}^{k-1}$$

$$\sum_{i=1}^{n}\cos^{2k+1}{\left(\frac{2i\pi}{2n+1}\right)}=-\frac{1}{2}$$

--

examples of odd powers

$$\sum_{i=1}^{n}\sin^7{\left(\frac{2i\pi}{2n+1}\right)}=\frac{35}{128}\tan{\left(\frac{3\pi}{2n+1}\right)}-\frac{21}{128}\tan{\left(\frac{2\pi}{2n+1}\right)}+ \frac{7}{128}\tan{\left(\frac{\pi}{2n+1}\right)}-\frac{1}{128}\tan{\left(\frac{0\pi}{2n+1}\right)}$$

$$\sum_{i=1}^{n}\sin^5{\left(\frac{2i\pi}{2n+1}\right)}=\frac{10}{32}\tan{\left(\frac{3\pi}{2n+1}\right)}-\frac{5}{32}\tan{\left(\frac{2\pi}{2n+1}\right)}+ \frac{1}{32}\tan{\left(\frac{\pi}{2n+1}\right)}$$

$$\sum_{i=1}^{n}\sin^3{\left(\frac{2i\pi}{2n+1}\right)}=\frac{3}{8}\tan{\left(\frac{3\pi}{2n+1}\right)}-\frac{1}{8}\tan{\left(\frac{2\pi}{2n+1}\right)}$$

$$\sum_{i=1}^{n}\sin{\left(\frac{2i\pi}{2n+1}\right)}=\frac{1}{2}\tan{\left(\frac{3\pi}{2n+1}\right)}$$

--

$$\sum_{i=1}^{n}(-1)^{i+1}\sin{\left(\frac{2i\pi}{2n+1}\right)}=\frac{1}{2}\tan{\left(\frac{\pi}{2n+1}\right)}$$

$$\sum_{i=1}^{n}\sin{\left(\frac{5i\pi}{2n+1}\right)}=\frac{1}{2}\tan{\left(\frac{\pi}{2n+1}\right)}$$

$$\sum_{i=1}^{n}\sin{\left(\frac{3i\pi}{2n+1}\right)}=\frac{1}{2}\tan{\left(\frac{2\pi}{2n+1}\right)}$$

$$\sum_{i=1}^{n}\sin^4{\left(\frac{i\pi}{2n+1}\right)}=\frac{6n+3}{16}$$

$$\sum_{i=1}^{n}\cos^4{\left(\frac{i\pi}{2n+1}\right)}=\frac{6n-5}{16}$$

In general, solve for X and Y

$$X=4^k\sum_{i=1}^{n}\sin^{2k}{\left(\frac{i\pi}{2n+1}\right)}$$

$$Y=4^k\sum_{i=1}^{n}\cos^{2k}{\left(\frac{i\pi}{2n+1}\right)}$$

$$\tan^2\left(\frac{\pi}{5}\right)+\tan^2\left(\frac{2\pi}{5}\right)+\tan^2\left(\frac{3\pi}{5}\right)=2\sqrt{5}+15$$

$$\sum_{i=1}^{n}\sin{\left[\frac{(4i-1)\pi}{2(2n+1)}\right]}=\frac{1}{2\sin\left[\frac{\pi}{2(2n+1)}\right]}$$

$$\sum_{i=1}^{n}\sin{\left[\frac{(4i-1)\pi}{2(2n+1)}+A\right]}=\frac{\cos{A}}{2\sin\left[\frac{\pi}{2(2n+1)}\right]}$$

$$\sum_{i=1}^{n}\cos\left(\frac{(2i-1)\pi}{2n+1}\right)=\frac{1}{2}$$

$$\sum_{i=1}^{n}\cos\left(\frac{2i\pi}{2n+1}\right)=-\frac{1}{2}$$

$$\sum_{i=1}^{n}\sin^2\left({\frac{i\pi}{2n+1}}\right)=\frac{2n+1}{4}$$

$$\sum_{i=1}^{n}\cos^2\left({\frac{i\pi}{2n+1}}\right)=\frac{2n-1}{4}$$

$$\sum_{i=1}^{n}\tan^2\left({\frac{i\pi}{2n+1}}\right)=T_{2n}$$

Where $$T_n=1,3,6,10,15,\cdots$$

are triangle numbers

Page 7

Where,

$$A=49+20\sqrt{6}$$

$$B=49-20\sqrt{6}$$

$$x_0=10$$

$$x_1=2\sqrt{3}$$

$$x_{n+1}=\sqrt{2+x_n}=A^{\frac{1}{2^{n+2}}}+B^{\frac{1}{2^{n+2}}}$$

$$\sqrt{\left(9+\sqrt[3]{9}\right)^2-36\sqrt[3]{9}}=9-\sqrt[3]{9}$$

$$\sqrt[4]{49+20\sqrt{6}}=\sqrt{3}+\sqrt{2}$$

$$\sqrt[4]{49-20\sqrt{6}}=\sqrt{3}-\sqrt{2}$$

Page 6

$$\frac{sin{A_1}}{cos{A_1}}=tan{A_1}$$

General form

$$\frac =tan\left(\frac{\sum_{j=1}^{n}A_j}{n}\right)$$

Page 5

$$\frac{tan\left(\frac{81}{2}\right)+tan\left(\frac{9}{2}\right)}{tan\left(\frac{81}{2}\right)-tan\left(\frac{9}{2}\right)}  =\sqrt{\frac{5+\sqrt{5}}{5}}$$

$$\frac{tan\left(\frac{75}{2}\right)+tan\left(\frac{15}{2}\right)}{tan\left(\frac{75}{2}\right)-tan\left(\frac{15}{2}\right)}  =\sqrt{2}$$

$$\frac{tan\left(\frac{63}{2}\right)+tan\left(\frac{27}{2}\right)}{tan\left(\frac{63}{2}\right)-tan\left(\frac{27}{2}\right)}  =\sqrt{3+\sqrt{5}}$$

Page 4

(1)

L(1) = 1 and L(2) = 3

$$L_{n+2}=L_n+L_{n+1}$$

Let,

$$L_{n}^k=a_1L_{n}^{k-2}+a_2L_{n}^{k-4}+\cdots+L_{kn}$$

Her are a few

$$Ln_{n}^2=2(-1)^n+L_{2n}$$

$$L_{n}^3=3L_{n}(-1)^n+L_{3n}$$

$$L_{n}^4=4L_{n}^2(-1)^n-2+L_{4n}$$

$$L_{n}^5=5L_{n}^3(-1)^n-5L_{n}+L_{5n}$$

$$L_{n}^6=6L_{n}^4(-1)^n-9L_{n}^2(-1)^{n-1}+2(-1)^n+L_{6n}$$

(2)

$$L_{2^{n+1}}=L_{2^n}F_{2^n}$$

(3)

$$\frac{tan\left[2arctan\left(\frac{1}{\phi^{2^n}}\right)\right]}{tan\left[2arctan\left(\frac{1}{\phi^{2^{n+1}}}\right)\right]}=L_{2^n}$$

(4)

$$\sum_{k=1}^{n}{n\choose k}\frac{L_{k+1}}{k+1}=\frac{L_{n+1}^2}{n+1}$$

page 3

$$\phi=\frac{1+\sqrt{5}}{2}$$

Where $$n = 2,4,6,8,...$$

$$\frac{{n\choose 0}\phi^n+{n\choose 2}\phi^{n-2}+\cdots} {{n\choose 1}\phi^{n-1}+{n\choose 3}\phi^{n-3}+\cdots}\times{\sqrt{5}}=\frac{a}{b}$$

$$\lim_{n \to \infty}\frac{a}{b}=\sqrt{5}$$

Pgae 2

(1)

Let $$N_0=1,3,5,7,...$$

$$X=\left({N_0\choose 0}\phi^{N_0}+{N_0\choose 2}\phi^{N_0-2}+{N_0\choose 4}\phi^{N_0-4}+\cdots\right) \left({N_0\choose 1}\phi^{N_0-1}+{N_0\choose 3}\phi^{N_0-3}+{N_0\choose 5}\phi^{N_0-5}+\cdots\right)$$

$$2arctan\left(\frac{1}{\phi^{3N_0}}\right)=arctan\left(\frac{\phi^{N_0}}{2X}\right)$$

(2)

$$arctan\left(\frac{1}{\phi}\right)=2arctan\left(\frac{1}{\phi^3}\right)+ arctan\left(\frac{1}{\phi^5}\right)$$

(3)

$$4arctan\left(\frac{1}{\phi}\right)+arctan\left(\frac{1}{7}\right)=\frac{3\pi}{4}$$

(4)

$$2arctan\left(\frac{1}{\phi}\right)+arctan\left(\frac{1}{2}\right)=\frac{\pi}{2}$$

(5)

$$3arctan\left(\frac{1}{\phi}\right)+arctan\left(\frac{1}{\phi^5}\right)=\frac{\pi}{2}$$

(6)

$$2arctan\left(\frac{1}{\sqrt{\phi^3}}\right)+arctan\left(\frac{1}{\sqrt{\phi}}\right)=\frac{\pi}{2}$$

(7)

$$2arctan\left(\frac{1}{\sqrt{\phi}}\right)+arctan\left(\frac{1}{\sqrt{\phi^6-1}}\right)=\frac{\pi}{2}$$

(8)

$$4arctan\left(\frac{1}{\sqrt{\phi^3}}\right)-arctan\left(\frac{1}{\sqrt{\phi^6-1}}\right)=\frac{\pi}{2}$$

Page 1

(1)

Where a < b, Let

$$x=\frac{{n\choose 0}a^nb^0+{n\choose 2}a^{n-2}b^2+{n\choose 4}a^{n-4}b^4+\cdots}{{n\choose 1}a^nb+{n\choose 3}a^{n-3}b^3+{n\choose 5}a^{n-5}b^5+\cdots}$$

$$Artanh\left(x^{(-1)^{n+1}}\right)=nArtanh\left(\frac{a}{b}\right)$$

(2)

Let

$$y=\frac{{n\choose 1}a^{n-1}b-{n\choose 3}a^{n-3}b^3+{n\choose 5}a^{n-5}b^5-\cdots}{{n\choose 0}a^{n}b^0-{n\choose 2}a^{n-2}b^2+{n\choose 4}a^{n-4}b^4-\cdots}$$

$$\arctan(y)=n\arctan\left(\frac{b}{a}\right)$$

--

$$\sum_{k=0}^{n}{n\choose k}{n+1\choose k+1}={2n+1\choose n+1}$$

$$\sum_{k=0}^{2n}(-1)^k{2n\choose k}{2n+1\choose k+1}=(-1)^n{2n\choose n}$$

$$\sum_{k=0}^{2n}{n\choose k}{2n\choose k}={6n\choose 2n}$$

$$\sum_{k=0}^{2n}\frac=\frac{4n+1}{2n+1}$$

---

Defn:

$$H_{n,x}=\sum_{k=1}^{n}\frac{1}{k^x}$$

$$H_{n,x}'=\sum_{k=1}^{n}\frac{(-1)^{k-1}}{k^x}$$

$$\sum_{k=0}^{n}\left(\frac\right)^x \frac{1}{(k+1)^y}=\frac{H_{n+1,y-x}}{(n+1)^x}$$

$$\sum_{k=0}^{n}(-1)^k\left(\frac\right)^x \frac{1}{(k+1)^y}=\frac{H_{n+1,y-x}'}{(n+1)^x}$$

Invserse sum of triangular numbers

$$\sum_{k=0}^{n}\left(\frac\right)^2\frac{1}{k+1}=1-\frac{1}{4}\sum_{k=1}^{n}\frac{1}{T_k}$$

Where $$T_k$$ = 1, 3, 6, 10, 15,...

$$\lim_{n\to \infty}Y_{n+1}=\tan^{-1}\left(\frac{1}{\sin{Y_n}}\right)$$

$$\sqrt{\phi}=\frac{1}{\sin{Y_{n+1}}}$$

Yn is any values except zero

$$\sum_{j=1}^{n}(-1)^{j+1}{n\choose j}\frac{1}{j}=H_n$$

$$\sum_{j=1}^{n}(-1)^{j+1}{n+1\choose j+1}H_j=H_n$$

--

$$\sum_{j=0}^{n}(-1)^j{n\choose j}\frac{1}{k+j}=\frac{1}{(n+1){n+k \choose k-1}}$$

$$\sum_{j=0}^{n}(-1)^j{n\choose j}\frac{H_{j+1}'}{j+1}=\frac{2^{n+1}-1}{(n+1)^2}$$

---

$$\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{1}{k+1}=\frac{1}{1+n}$$

$$\sum_{k=0}^{n}(-1)^k{n\choose k}\frac{H_{k+1}}{k+1}=\frac{1}{(1+n)^2}$$

$$\sum_{k=0}^{n}(-1)^k{n\choose k} \frac{1}{k+1}\sum_{j=0}^{k}\frac{H_{j+1}}{j+1}=\frac{1}{(1+n)^3}$$

$$\frac{1}{2(q^2-p^2)}\left[H_{-\frac{1}{q}}+H_{\frac{1}{q}}-\left(H_{-\frac{1}{p}}+H_{\frac{1}{p}}\right)\right]=\sum_{n=1}^{\infty}\frac{n}{(p^2n^2-1)(q^2n^2-1)}$$

--

$$\lim_{n,m \to \infty}\left(\frac{\pi}{2}\right)^2\left[(1+n)^2+(1+m)^2-(n-m)(n-m+1)\right]=(m+1)^2\prod_{k=1}^{n}\left(\frac{2k}{2k-1}\right)^2\prod_{k=1}^{m}\left(\frac{2k}{2k+1}\right)^2+ (n+1)^2\prod_{k=1}^{n}\left(\frac{2k}{2k+1}\right)^2\prod_{k=1}^{m}\left(\frac{2k}{2k-1}\right)^2 $$

let n = m

and we have Wallis formula

$$\frac{\pi}{2}=\prod_{k=1}^{\infty}\frac{2k}{2k-1}\cdot\frac{2k}{2k+1}$$

-

$$\frac{1}{n(m+1)_n}+\sum_{k=1}^{m}(-1)^{k-1}{m\choose k}\frac{(k-1)!(n-1)!}{(n+1)_k(m+1)_n}=\sum_{k=1}^{n}(-1)^{k-1}{n-1\choose k-1}\frac{H_{m+k}}{m+k}$$

$$\sum_{k=0}^{n}(-1)^k{n\choose k}\sum_{j=0}^{k}\frac{H_{j+1}}{j+1}=\frac{1}{(n+1)^2}-\frac{1}{n^2}$$

---

$$\sum_{k=1}^{x}\frac{1}{k+y-1}=\sum_{k=y}^{x+y-1}\frac{1}{k}$$

-

Digamma and Harmonic functions

$$n\psi(nx)=\psi(x)+\psi\left(x+\frac{1}{n}\right)+\psi\left(x+\frac{2}{n}\right)+\cdots+\psi\left(x+\frac{n-1}{n}\right)+n\ln{n}$$

$$\psi(x+n)=\psi(x+m)+\frac{1}{x+m}+\frac{1}{x+m+1}+\cdots+\frac{1}{x+n-1}$$

$$H_{1-x}+H_{x-1}=\frac{1}{x}-\frac{1}{1-x}+2\gamma+\psi(1+x)+\psi(1-x)$$

$$H_{1-x}-\psi(1-x)+\frac{1}{1-x}=H_{x}-\psi(x)+\frac{1}{x}$$

(Guillera and Sondow)

$$\psi(z)=\sum_{n=0}^{\infty}\frac{1}{n+1}\sum_{k=0}^{n}(-1)^k{n\choose k}\ln(z+k)$$

we just write is slightly different here

$$\psi(z)=\sum_{n=0}^{\infty}\frac{1}{n+1}\sum_{k=0}^{y}(-1)^k{y\choose k}\sum_{j=0}^{x}(-1)^j{x\choose j}\ln(z+k+j)$$

where n = x + y

$$\psi(kn)-2\ln{k}=\frac{1}{k}\sum_{i=1}^{k}\left[H_{n-\frac{i}{k}}+\psi\left(\frac{i}{k}\right)\right]$$

$$\psi(kn)+2\gamma=\frac{1}{k}\sum_{i=1}^{k}\left[H_{n-\frac{i}{k}}-\psi\left(\frac{i}{k}\right)\right]$$

---

$$\sum_{k=1}^{m}(-1)^{k-1}{m\choose k}\frac{(k-1)!}{(n+1)_k}=H_{m+n}-H_n$$

$$\sum_{k=1}^{m}(-1)^{k-1}{m\choose k}\frac{(k-1)!}{(n)_k}=\psi(m+n)-\psi(n)$$

$$\psi(m+n)=\psi(n)+\sum_{k=1}^{m}\frac{1}{k+n-1}$$

$$\sum_{k=1}^{m}\psi(k+n)=m\psi(n)+\sum_{k=1}^{m}(-1)^{k-1}{m+1\choose k+1}\frac{(k-1)!}{(n)_k}$$

$$\psi(n+x)=-\gamma+H_{n+x-1}$$

'''Reverse order and summations '''

Reverse order

Def:

$$(n)_{k}=n(n+1)(n+2)\cdots(n+k-1)$$

$$\sum_{k=0}^{m}(-1)^k{m\choose k}\frac{n}{k+n}\left(H_{n-1}-H_{k+n-1}\right)=\frac{H_m}$$

$$\sum_{k=0}^{m}(-1)^k{m\choose k}\left(\frac{nH_{n-1}}{k+n}+\frac{n}{(k+n)^2} \right)=\frac{H_{n+m}}$$

$$\sum_{k=0}^{m}(-1)^k{m\choose k}\frac{nH_{k+n}}{k+n}=\frac{H_{n+m}-H_m}$$

$$\sum_{k=0}^{m}(-1)^k{m\choose k}\frac{(n-1)!}{(k+1)_{(n)}}\left(H_{k+n}-H_{n-1}\right)=\frac{1}{(m+n)^2}$$

$$\sum_{k=1}^{m}(-1)^{k-1}{m-1\choose k-1}\frac{(k-1)!}{(n+1)_{(k)}}=\frac{1}{m+n}$$

$$\sum_{k=1}^{m}(-1)^{k-1}{m-1\choose k-1}\frac{(k-1)!}{(n+1)_{(k)}}\left(H_{k+n}-H_{k-1}\right)=\frac{H_{m+n}}{m+n}$$

Summations

Def:

$$\sum_{k=1}^{m}\frac{1}{k+n}=H_m(n)$$

$$\sum_{k=1}^{m}(-1)^{k}{m\choose k}\frac{(k-1)!}{(n+1)_{(k)}}=H_{m}(n)$$

$$\sum_{k=1}^{m}(-1)^{k}{m\choose k}\frac{(k-1)!}{(n+1)_{(k)}}\left(H_{k+n}-H_{k-1}\right)=\sum_{k=1}^{m}\frac{H_{k+n}}{k+n}$$

Others

$$\sum_{k=0}^{m}(-1)^k{m\choose k}\frac{H_{k+m}}{k+m}=\frac{H_{2m}'}{m{2m\choose m}}$$

$$\sum_{k=1}^{m}(-1)^(k-1){m\choose k}\frac{(k-1)!}{(\frac{1}{2})_k}=H_{m-\frac{1}{2}}+\ln4$$

$$\sum_{k=1}^{m}(-1)^{k-1}{m\choose k}\frac{(k-1)!}{(n+1)_k}=H_{m+n}-\psi(n+1)-\gamma$$

--

$$\sum_{j=0}^{n}\left(nH_j-jH_j-jH_{n-j}\right){n\choose j}^k=0$$

$$\sum_{j=0}^{n}jH_{n-j}{n\choose j}=\frac{1}{2}\left(1+2^n\left(-1+nH_{n}-n\sum_{j=0}^{n}\frac{1}{j2^j}\right)\right)$$

$$\sum_{j=0}^{n}jH_{n-j}{n\choose j}^2=\frac{1}{4}{2n\choose n}\left(-1+4nH_n-2nH_{2n}\right)$$

-

Def:

$$H_{n}'=\sum_{k=1}^{n}(-1)^{k-1}\frac{1}{k}$$

$$H_{n,m}=\sum_{k=1}^{n}\frac{1}{n^m}$$

$$\sum_{k=1}^{n}(-1)^{k-1}{2n\choose k+n}\frac{1}{k}={2n\choose n}H_{2n}'$$

$$\sum_{k=1}^{n}(-1)^{k-1}{2n+1\choose k+n+1}\frac{1}{k}=\sum_{j=1}^{n}{2j\choose j}H_{2j}'$$

$$\sum_{j=1}^{n}\frac{H_{j+n-1}-H_{n-1}}{j\Beta(j,n)}={2n\choose n}H_{2n}'$$

$$\sum_{j=1}^{n}(-1)^{j-1}{n-1\choose j-1}\frac{H_{j+n-1}-H_{n-1}}{j\Beta(j,n)}=(-1)^{n-1}H_n$$

$$\sum_{j=1}^{n}(-1)^{j-1}{n\choose j}\frac{H_{j+n-1}-H_{n-1}}{j\Beta(j,n)}=\frac{(-1)^{n-1}}{n}$$

$$\sum_{j=1}^{n}(-1)^{j-1}{n-1\choose j}\frac{H_{j+n-1}-H_{n-1}}{j\Beta(j,n)}=(-1)^nH_{n-1}$$

$$\sum_{j=0}^{n}(-1)^{j-n}{n\choose j}H_{k,-j}=H_{k-1,-n}$$

$$\sum_{j=0}^{n}{n\choose j}H_{k,-j}=H_{k+1,-n}-1$$

$$\sum_{j=1}^{n}(-1)^{j-1}{n-1\choose j-1}H_{k,-j}=(-1)^{n-1}\left(H_{k-1,1-n}+H_{k-1,-n}\right)$$

$$\frac{2^{n+1}}{n+1}\sum_{j=0}^{n}{n\choose j}^{-1}-\sum_{j=0}^{n}{n+1\choose j+1}\frac{1}{j+1}=H_{n+1}$$

$$\sum_{j=0}^{n}H_j{n\choose j}^2-\sum_{j=1}^{n}(-1)^{j-1}{2n \choose j+n}\frac{1}{j}={2n\choose n}\left(3H_{n}-2H_{2n}\right)$$

$$\sum_{j=0}^{n}H_j{n\choose j}^2+\sum_{j=1}^{n}(-1)^{j-1}{2n \choose j+n}\frac{1}{j}={2n\choose n}H_{n}$$

$$\sum_{j=0}^{n}H_j{n\choose j}^2+2\sum_{j=1}^{n}(-1)^{j-1}{2n \choose j+n}\frac{1}{j}={2n\choose n}H_{2n}$$

$$(n+1)+\sum_{j=0}^{n}{n+1\choose j+1}H_{k,-j}=\sum_{j=0}^{n}H_{k+1,-j}$$

$$\sum_{j=0}^{n}{n\choose j}H_{k,-j-x}=\sum_{i=0}^{x}(-1)^i{x\choose i}H_{k+1,-n-x+i}$$

$$\sum_{k=1}^{n}(-1)^{k-1}{n-1\choose k-1}\frac{1}{2k+1}=\frac{(2n-2)!!}{(2n+1)!!}$$

$$\sum_{k=1}^{n}(-1)^{k-1}{n\choose k}\frac{1}{2k+1}=\sum_{j=1}^{n}\frac{(2j-2)!!}{(2j+1)!!}$$

$$\sum_{k=1}^{n}(-1)^{k-1}{n+1\choose k+1}\frac{1}{2k+1}=\sum_{j=1}^{n}\sum_{i=1}^{j}\frac{(2i-2)!!}{(2i+1)!!}$$

$$\sum_{j=1}^{2n}(-1)^j\sum_{k=1}^{j}(-1)^{k-1}{j\choose k}\frac{1}{2k+1}= \sum_{i=1}^{n}\frac{(4i-2)!!}{(4i+1)!!}$$

Where

$$\Beta(n,m)=\frac{\Gamma(n)\Gamma(m)}{\Gamma(n+m)}$$

$$\sum_{i=0}^{n}(-1)^i{n\choose i}H_{n+m-i}=(-1)^{n-1}\frac{m}{n+m}\Beta(n,m)$$

$$\sum_{k=0}^{n}{n\choose k}\sum_{i=0}^{k}(-1)^i{k\choose i}H_{n+k-i}=H_{2n}$$

$$\sum_{k=1}^{n}{n\choose k}\sum_{i=0}^{k}(-1)^i{k\choose i}H_{n+k-i}=H_{2n}'$$

$$\sum_{n=1}^{\infty}\sum_{i=0}^{n}(-1)^{i}{n\choose i}H_{n+1-i}=\ln4-1$$

$$\frac{1}{m(m+n)}=\sum_{k=1}^{n}(-1)^{k-1}{n-1\choose k-1}\frac{\Beta(k,m)}{k+m}$$

$$\frac{\Beta(n,m)}{n+m}=\sum_{k=1}^{n}(-1)^{k-1}{n-1\choose k-1}\frac{1}{m(m+k)}$$

$$\sum_{j=1}^{n}\frac{1}{m(m+j)}=\sum_{k=1}^{n}(-1)^{k-1}{n\choose k}\frac{\Beta(k,m)}{k+m}$$

$$\sum_{j=1}^{n}\frac{\Beta(j,m)}{j+m}=\sum_{k=1}^{n}(-1)^{k-1}{n\choose k}\frac{1}{m(m+k)}$$

$${n+m\choose m}\left(H_{n+m}-H_m\right)=\sum_{k=1}^{n}(-1)^{k-1}{n+m\choose k+m}\frac{1}{k}$$

-

Zeta function

Where n = x + y - 1 and y = > 1

$$\zeta(s)=\frac{1}{1-2^{1-s}}\sum_{n=0}^{\infty}\frac{1}{2^{n+1}}\sum_{i=0}^{x}(-1)^i {x\choose i}\sum_{j=1}^{y}(-1)^{j-1}{y-1\choose j-1}(i+j)^{-s}$$

$$\zeta(2)=\sum_{n=0}^{\infty}\frac{H_{n+1}}{2^n(n+1)}$$

$$3\zeta(3)=\sum_{n=0}^{\infty}\frac{H_{n+1}^2+H_{n+1,2}}{2^n(n+1)}$$

$$\frac{3\zeta(3)}{2}=\sum_{n=1}^{\infty}\frac{H_n^2+H_{n,2}}{2^nn}$$

$$\frac{7\zeta(3)}{8}=\sum_{n=1}^{\infty}\frac{H_n^2}{2^nn}$$

$$\frac{7\zeta(4)}{4}=\sum_{n=1}^{\infty}\frac{H_{n,2}}{n^2}$$

-

Where F is function

$$\sum_{k=0}^{n_1+n_2+n_3+\cdots+n_p+m-1}{n_1+n_2+n_3+\cdots+n_p+m-1\choose k}(F_{k+1})^{-s}= \sum_{i_1=0}^{n_1}{n\choose i_1}\sum_{i_2=0}^{n_2}{n\choose i_2}\sum_{i_3=0}^{n_3}{n\choose i_3}\cdots\sum_{i_p=0}^{n_p}{n\choose i_p}\sum_{j=1}^{m}{m-1\choose j-1}(F_{i_1+i_2+i_3+\cdots+i_p+j})^{-s} $$

easy version

$$\sum_{k=0}^{n_1+n_2+n_3+\cdots+n_p+m-1}{n_1+n_2+n_3+\cdots+n_p+m-1\choose k}(k+1)^{-s}= \sum_{i_1=0}^{n_1}{n\choose i_1}\sum_{i_2=0}^{n_2}{n\choose i_2}\sum_{i_3=0}^{n_3}{n\choose i_3}\cdots\sum_{i_p=0}^{n_p}{n\choose i_p}\sum_{j=1}^{m}{m-1\choose j-1}(i_1+i_2+i_3+\cdots+i_p+j)^{-s} $$

$$\sum_{j=0}^{\sum_{p=1}^{k}n_p}{\sum_{p=1}^{k}n_p\choose j}=\sum_{i_1=0}^{n_1}{n_1\choose i_1}\sum_{i_2=0}^{n_2}{n_2\choose i_2}\sum_{i_3=0}^{n_3}{n_3\choose i_3}\cdots\sum_{i_k=0}^{n_k}{n_k\choose i_k}$$

$$\sum_{i=0}^{kn}{kn\choose i}=\left[\sum_{j=0}^{n}{n\choose j}\right]^k$$

$$\sum_{i=0}^{n}{n\choose i}^k+2\sum_{j=0}^{n-1}{n\choose j}\sum_{k=1}^{n-j}{n\choose j+k}=\sum_{l=0}^{nk}{nk\choose l}$$

$$\sum_{i=0}^{2n}{2n\choose i}={2n\choose n}+2\sum_{j=0}^{n-1}{n\choose j}\sum_{k=1}^{n-j}{n\choose k+j}$$

$${j\choose m}=\sum_{k=0}^{n}(-1)^k{n\choose k}{j+n-k \choose m+n}$$

$$\sum_{k=0}^{n}(-1)^k{n\choose k}{2m-k\choose m}(k+1)^{-1}=\frac{1}{n}\sum_{i=0}^{n}{2m-i\choose m}$$

--

Harmonics

Let:

$$H_{n,m}=\sum_{k=1}^{n}\frac{1}{k^m}$$

and

$$H_{n,m}'=\sum_{k=1}^{n}\frac{(-1)^{k-1}}{k^m}$$

(1)

$$\sum_{k=1}^{n}\left[\frac\right]^m=n^mH_{n,m}$$

(2)

$$\sum_{k=1}^{n}(-1)^{k-1}\left[\frac\right]^m=n^mH_{n,m}'$$

(3)

$$\sum_{k=1}^{n}\left[\frac\right]^m =\frac{1}{n^m}\sum_{j=1}^{n}j^m$$

(4)

$$\sum_{k=1}^{n}(-1)^{k-1}\left[\frac\right]^m =\frac{1}{n^m}\sum_{j=1}^{n}(-1)^{k-1}j^m$$

(6)

$$\sum_{k=1}^{n}\frack^{m-1}=\frac{1}{n} \sum_{j=1}^{n}j^m$$

(7)

$$\sum_{k=1}^{n}(-1)^{k+1}\frack^{m-1}=\frac{1}{n} \sum_{j=1}^{n}(-1)^{k+1}j^m$$

(8)

$$\sum_{k=1}^{n}\frac{1}{k^{m-1}}\cdot\frac= nH_{n,m}$$

(10)

$$\sum_{k=1}^{n}\frac{(-1)^{k+1}}{k^{m-1}}\cdot\frac= nH_{n,m}'$$

(11)

Where, Bj {Bernoulli numbers}

j = 0, 1, 2, 3, 4, 5, ...

Bj = 1, -1/2, 1/6, 0, -1/30, 0, 1/42, ...

$$\sum_{k=1}^{n}\left[\frac\right]^m= \frac{1}{m+1}\sum_{j=0}^{m}(-1)^j{m+1\choose j}B_jn^{1-j}$$

(12)

z < 1

$$\frac{z-1}{z+1}=\frac{\sum_{n=1}^{\infty}(-z)^nH_n'} {\sum_{n=1}^{\infty}z^nH_n}$$

Sun's curious identity

$$(x+m+1)\sum_{i=0}^{m}(-1)^i{x+y+i\choose m-i}{y+2i\choose i}-\sum_{i=0}^{m}{x+i\choose m-i}(-4)^i=(x-m) {x\choose m}$$

Let y = 0 and we rewrite sun's identity as

$$(x+m+1)\sum_{i=0}^{m}(-1)^i{x+i\choose m-i}{2i\choose i}-\sum_{i=0}^{m}(-1)^i{x+i\choose m-i}(4)^i=(x-m) {x\choose m}$$

and we have arrived at another two variation of its,

(1)

$$(x+m+1)\sum_{i=0}^{m}(-1)^i{x+i\choose m-i}{2i+2\choose i+1}-\sum_{i=0}^{m}(-1)^i{x+i\choose m-i}(4)^{i+1}=2(x-m-1) {x-1\choose m}$$

(2)

$$(x+m+1)\sum_{i=0}^{m}(-1)^i{x+i\choose m-i}{2i+4\choose i+2}-\sum_{i=0}^{m}(-1)^i{x+i\choose m-i}(4)^{i+2}=2(3x-m-5) {x-2\choose m}$$

Where Ci is Catalan numbers

$$C_i=\frac{1}{i+1}{2i\choose i}$$

Ci = {1, 1, 2, 5, 14, 42, 132, ...}

We have another two sum

(3)

$$(x+m+1)\sum_{i=0}^{m}(-1)^iC_i{x+i\choose m-i}=2(x-m){x\choose m}-(x-1-m){x-1\choose m}$$

(4)

$$(x+m+1)\sum_{i=0}^{m}(-1)^iC_{i+1}{x+i\choose m-i}=4(x-1-m){x-1\choose m}-(3x-m-5){x-2\choose m}$$

Others

(5)

$$\sum_{i=1}^{m+1}(-1)^{i+1}{x+m+1-i\choose 2m}{m\choose i-1}={x\choose m}$$

(6)

$$\sum_{i=1}^{m}(-1)^{i+1}{x-i\choose m}{m-1\choose i-1}=(x-m)$$

(7)

$$(x-m){x\choose m}=(m+1){x\choose m+1}$$

-

(8)

$$(x+m+1)\sum_{i=1}^{m}(-1)^i{x+y+i\choose m-i}{2i+y\choose i-1}- \sum_{i=1}^{m}{x+i\choose m-i}(-4)^i={x\choose m}-(x-m){x\choose m}+(x-1-m){x-1\choose m}$$

(9)

$$2(x+m+1)\sum_{i=1}^{m}(-1)^i{x+i\choose m-i}{2i+1\choose i-1}- \sum_{i=1}^{m}(-1)^i{x+i\choose m-i}4^{i+1}=4{x\choose m}-6(x-1-m){x-1\choose m}+2(3x-m-5){x-2\choose m}$$

--

Proposed Formulas

Bernoulli numbers

B(2n) = {1/6, -1/30, 1/42, -1/30, 5/66,...}

B(2) = 1/6, B(4) = -1/30 ...

$$\sum_{k=0}^{n-1}{n+1\choose k+1}B_{k+1}=-1$$

$$\sum_{k=0}^{n-1}{n+2\choose k+2}B_{k+1}=\frac{n}{2}$$

$$\sum_{k=0}^{n-1}{n+4\choose k+4}B_{k+1}=-\frac{n(n+1)}{12}$$

$$\sum_{k=1}^{n}{2n+2\choose 2k}B_{2k}=n$$

$$\sum_{k=1}^{n}{2n+1\choose 2k}B_{2k}=\frac{2n-1}{2}$$

$$\sum_{k=1}^{n}{2n\choose 2k}B_{2k}=B_{2n}+n-1$$

$$\sum_{k=1}^{n}{2n\choose 2k-2}B_{2k}=B_{2n}$$

$$\sum_{k=1}^{n}\left[{2n+3\choose 2k}-{2n+1\choose 2k-2}\right]B_{2k}=\frac{2n+1}{2}$$

$$\sum_{k=1}^{n}{2n+x-2\choose 2k-2}B_{2k}=\sum_{k=1}^{n}\left[{2n+x\choose2k} -2{2n+x-1\choose2k} +{2n+x-2\choose 2k}\right]B_{2k}$$

-

(1)

$$\gamma=\lim_{n \to\infty}\left(\sum_{k=1}^{n-1}\frac{1}{k}-\sum_{k=n}^{(n-1)^2}\frac{1}{k}\right)$$

(2)

Telescoping sum

$$\sum_{k=1}^{m-1}(-1)^{k+1}\frac{ak-(a-1)}= \sum_{n=1}^{x}\sum_{k=1}^{m}(-1)^{k+1}\frac{an+ak-(2a-1)}+\sum_{k=1}^{m-1}(-1)^{k+1}\frac{ax+ak-(a-1)}$$

General formula

(3)

$$\frac{f(m)}{(2m-1)!!}-2^{m-3}\pi=\sum_{n=1}^{\infty}\sum_{k=0}^{m}(-1)^{k}\frac{4n+2k-1}$$

Let

m = 1, f(m) = 1

$$1-\frac{\pi}{4}=\sum_{n=1}^{\infty}\left(\frac{1}{4n-1}-\frac{1}{4n+1}\right)$$

m = 2, f(m) = 5

$$\frac{5}{1\times{3}}-\frac{\pi}{2}=\sum_{n=1}^{\infty}\left(\frac{1}{4n-1}-\frac{2}{4n-1}+\frac{1}{4n+1}\right)$$

m = 3, f(m) = 48

$$\frac{48}{1\times{3}\times{5}}-\pi=\sum_{n=1}^{\infty}\left(\frac{1}{4n-1}-\frac{3}{4n+1}+\frac{3}{4n+3}-\frac{1}{4n+5}\right)$$

m = 4, f(m) = 664

$$\frac{664}{1\times{3}\times{5}\times{7}}-2\pi= \sum_{n=1}^{\infty}\left(\frac{1}{4n-1}-\frac{4}{4n+1}+\frac{6}{4n+3}-\frac{4}{4n+5}+\frac{1}{4n+7}\right)$$

m = 5, f(m) = 11904

$$\frac{11904}{1\times{3}\times{5}\times{7}\times{9}}-4\pi= \sum_{n=1}^{\infty}\left(\frac{1}{4n-1}-\frac{5}{4n+1}+\frac{10}{4n+3}-\frac{10}{4n+5}+\frac{5}{4n+7}+\frac{1}{4n+9}\right)$$

m = 6, f(m) = 261504

$$\frac{261504}{(11)!!}-8\pi=\sum_{n=1}^{\infty}\sum_{k=0}^{6}(-1)^{k+1}\frac{4n+2k-1}$$

(4)

$$\frac{g(m)}{(2m)!!}-2^{m-2}\ln{2}=\sum_{n=1}^{\infty}\sum_{k=0}^{m}(-1)^{k}\frac{4n+2k}$$

Let

m = 1, g(m) = 1

$$\frac{1}{2}-\frac{\ln{2}}{2}=\sum_{n=1}^{\infty}\left(\frac{1}{4n}-\frac{1}{4n+2}\right)$$

m = 2, g(m) = 6

$$\frac{6}{2\times{4}}-\ln{2}=\sum_{n=1}^{\infty}\left(\frac{1}{4n}-\frac{2}{4n+2}+\frac{1}{4n+4}\right)$$

(5)

$$\sum_{k=1}^{\infty}\sum_{m=1}^{2k-1}(-1)^{m+1}\frac{m+1}=\ln{2}$$

(6)

$$\lim_{m\to\infty}\left(\prod_{n=1}^{m}\left(1+\frac{1}{2n-1}\right)^2-\prod_{n=1}^{m-k}\left(1+\frac{1}{2n-1}\right)^2\right)=k\pi$$

(7)

$$\lim_{n\to\infty}\left[\left(n\sum_{k=1}^{n}(-1)^{k+1}\frac{2k-1}\right)^2- \left(m\sum_{k=1}^{m}(-1)^{k+1}\frac{2k-1}\right)^2\right]=(n-m)\cdot\frac{\pi}{4}$$

(8)

$$\prod_{k=1}^{n-1}\left(\frac{1}{a}-\frac{1}{a+kd}\right)=\frac{1}{d-a}\cdot\frac{1}{a^2} \left(\prod_{k=1}^{n}\frac{kd}{kd-(d-a)}-\prod_{k=1}^{n-1}\frac{kd}{kd-(d-a)}\right)= \frac{1}{a^2}\sum_{k=1}^{n}(-1)^{k+1}\frac{a+d(k-1)}$$

(9)

$$\frac{1}{2}\ln{2}=1+\frac{1}{3}-\frac{2}{2}+\frac{1}{5}+\frac{1}{7}-\frac{2}{6}+\frac{1}{9}+\frac{1}{11}-\frac{2}{10}+\cdots$$

(10)

x < 1

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

(11)

$$\frac{1}{2(1+x)}\frac{\ln{(1-x)}}{1+x}=x-x^2+ \left(1+\frac{1}{3}\right)x^3-\left(1+\frac{1}{3}+\frac{1}{5}\right)x^4-\cdots$$

(12)

$$-\frac{\ln{(1-x^2)}}{1+x}=x^2-x^3+\left(1+\frac{1}{2}\right)x^4- \left(1+\frac{1}{2}\right)x^5+\left(1+\frac{1}{2}+\frac{1}{3}\right)x^6- \left(1+\frac{1}{2}+\frac{1}{3}\right)x^7+\cdots$$

-

(13)

$$\sum_{k=1}^{n+1}\frac{2k-1}=\prod_{k=2}^{2n-1}\left(\frac{k^2}{k^2-1}\right)^{2n+1-k}$$

(14)

$$\frac{\pi}{2}=\lim_{n\to\infty}\frac{\sum_{k=1}^{n+1}\frac{2k-1}}{\prod_{k=2}^{n-1}\left(\frac{2k}{2k-1}\cdot\frac{2k+1}{2k+2}\right)^{2(n+1-k)}}$$

(15) where

$$H_n=1+\frac{1}{2}+\cdots+\frac{1}{n}$$

$$\sum_{k=1}^{n}(-1)^{k+1}{n-1 \choose k-1}\frac{H_{x+k}}{x+k}=\sum_{k=1}^{x+1}(-1)^{k+1}\frac{(k+n-1)^2}$$

(16) Three nice sums

$$\sum_{k=1}^{n}(-1)^{k+1}\frac{k}=\frac{1}{n}$$

$$\sum_{k=1}^{n}(-1)^{k+1}\frac{k^2}=\frac{H_n}{n}$$

$$\sum_{k=1}^{n}(-1)^{k+1}\frac{k^3}=\frac{H_n^2}{2n}+\frac{1}{2n}\sum_{k=1}^{n}\frac{1}{k^2}$$

(17)

$$\sum_{k=1}^{n}(-1)^{k+1}\frac{k^2}=\frac{1+\sqrt{1+4n^2H_nH_{n-1}}}{2n^2}$$

(18)

$$\frac{H_{n-x}}{n}=\sum_{k=1}^{n}(-1)^{k+1}\frac{k^2}- \sum_{k=1}^{x}(-1)^{k+1}\frac{(n-x+k)^2}$$

Section Ten

An inspiring from Sondow's exotic symmetric formula for pi

Sondow's symmetric formula for pi

$$\pi=\frac{\prod_{n=1}^{\infty}\left(1+\frac{1}{4n^2-1}\right)} {\sum_{n=1}^{\infty}\frac{1}{4n^2-1}}$$

Proof,

$$\frac{4}{\pi}=\frac{2^2}{2^2-1}\cdot\frac{4^2-1}{4^2}\cdot\frac{6^2}{6^2-1}\frac{8^2-1}{8^2}\cdots= \prod_{n=1}^{\infty}\left({1+\frac{1}{4n^2-1}}\right)^{(-1)^{n+1}}$$

From Paul Loya (chapter 5) page 247 wallis - products

(1)

$$\sqrt{\pi}=\lim_{n\to\infty}\frac{1}{\sqrt{n}}\cdot\frac{2}{1}\cdot\frac{4}{3}\cdot\frac{6}{5}\cdots\frac{2n}{2n-1}$$

Let

$$P=\frac{2^2}{2^2-1}\cdot\frac{4^2-1}{4^2}\cdot\frac{6^2}{6^2-1}\frac{8^2-1}{8^2}\cdots$$

Expand

$$P=\frac{2\cdot2}{1\cdot3}\cdot\frac{3\cdot5}{4\cdot4}\cdot\frac{6\cdot6}{5\cdot7}\cdot\frac{7\cdot9}{8\cdot8}\cdots$$

Simplify

$$P=\lim_{n\to\infty}(4n+1)\times{\frac{2\cdot2}{4\cdot4}\cdot\frac{6\cdot6}{8\cdot8}\cdots\left(\frac{4n-2}{4n}\right)^2}$$

So we have

$$P=\lim_{n \to\infty}\left(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdots\frac{2n-1}{2n}\right)^2\cdot{4n+1}$$

(2)

$$\sqrt{P}=\lim_{n \to\infty}\sqrt{4n+1}\cdot\left(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdots\frac{2n-1}{2n}\right)$$

(1) * (2)

$$\sqrt{\pi}\times{\sqrt{P}}=\lim_{n\to\infty}\frac{\sqrt{4n+1}}{\sqrt{n}}=\lim_{n\to\infty}\sqrt{\frac{4n+1}{n}}= \lim_{n\to\infty}\sqrt{4+\frac{1}{n}}$$

$$\sqrt{\pi}\times{\sqrt{P}}=2$$

$$\pi\times{P}=4$$

Therefore we have P,

$$P=\frac{4}{\pi}$$

$$\frac{4}{\pi}=\frac{2^2}{2^2-1}\cdot\frac{4^2-1}{4^2}\cdot\frac{6^2}{6^2-1}\frac{8^2-1}{8^2}\cdots= \prod_{n=1}^{\infty}\left({1+\frac{1}{4n^2-1}}\right)^{(-1)^{n+1}}$$

-- Proof,

$$\frac{1}{4}=\frac{1}{1\cdot3}-\frac{2}{3\cdot5}+\frac{3}{5\cdot7}-\cdots=\sum_{n=1}^{\infty}(-1)^{n+1} \frac{n}{4n^2-1}$$

where

$$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\cdots$$

Let

$$A=\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{5\cdot7}-\cdots$$

Split into partial fractions

$$A=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}\right)-\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}\right)+ \frac{1}{2}\left(\frac{1}{5}-\frac{1}{7}\right)-\cdots$$

Simplify to

$$2A=1-\frac{2}{3}+\frac{2}{5}-\frac{2}{7}+\cdots$$

add 1 to both sides

$$2A+1=2-\frac{2}{3}+\frac{2}{5}-\frac{2}{7}+\cdots$$

Divide 2 by both sides

$$A+\frac{1}{2}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots$$

and we have

$$A+\frac{1}{2}=\frac{\pi}{4}$$

therefore

$$A=\frac{\pi}{4}-\frac{1}{2}$$

Let

$$Q=\frac{1}{1\cdot3}-\frac{2}{3\cdot5}+\frac{3}{5\cdot7}-\cdots$$

Multiply 2 by Q

$$2Q=\frac{2}{1\cdot3}-\frac{4}{3\cdot5}+\frac{6}{5\cdot7}-\cdots$$

recall series A

$$A=\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{5\cdot7}-\cdots$$

2Q + A

$$2Q+A=\frac{3}{1\cdot3}-\frac{5}{3\cdot5}+\frac{7}{5\cdot7}-\cdots$$

Simplify and we have

$$2Q+A=\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\cdots$$

$$2Q+A=\frac{\pi}{4}$$

recall the result of A

$$A=\frac{\pi}{4}-\frac{1}{2}$$

$$2Q=\frac{\pi}{4}-\frac{\pi}{4}+\frac{1}{2}$$

therefore Q

$$Q=\frac{1}{4}$$

recall series of Q

$$Q=\frac{1}{1\cdot3}-\frac{2}{3\cdot5}+\frac{3}{5\cdot7}-\cdots$$

and therefore we have

$$\frac{1}{4}=\frac{1}{1\cdot3}-\frac{2}{3\cdot5}+\frac{3}{5\cdot7}-\cdots=\sum_{n=1}^{\infty}(-1)^{n+1} \frac{n}{4n^2-1}$$

-

Proof,

$$-\pi=\frac{\prod_{n=1}^{\infty}\left({1+\frac{1}{4n^2-1}}\right)^{(-1)^{n}}}{\sum_{n=1}^{\infty}(-1)^{n}\frac{n}{4n^2-1}}$$

Recall

$$\frac{4}{\pi}=\frac{2^2}{2^2-1}\cdot\frac{4^2-1}{4^2}\cdot\frac{6^2}{6^2-1}\frac{8^2-1}{8^2}\cdots= \prod_{n=1}^{\infty}\left({1+\frac{1}{4n^2-1}}\right)^{(-1)^{n+1}}$$

Change it to (inverse the product)

$$\frac{\pi}{4}=\frac{2^2}{2^2-1}\cdot\frac{4^2-1}{4^2}\cdot\frac{6^2}{6^2-1}\frac{8^2-1}{8^2}\cdots= \prod_{n=1}^{\infty}\left({1+\frac{1}{4n^2-1}}\right)^{(-1)^{n}}$$

and

$$\frac{1}{4}=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{n}{4n^2-1}$$

Change it to

$$\frac{1}{4}=-\sum_{n=1}^{\infty}(-1)^{n}\frac{n}{4n^2-1}$$

Divide them and we have

$$-\pi=\frac{\prod_{n=1}^{\infty}\left({1+\frac{1}{4n^2-1}}\right)^{(-1)^{n}}}{\sum_{n=1}^{\infty}(-1)^{n}\frac{n}{4n^2-1}}$$

-- ---

Recall

$$1=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{2n+2n}{(2n-1)(2n+1)}$$

and

$$\frac{4}{\pi}= \prod_{n=1}^{\infty}\left(\frac{2n}{2n-1}\cdot\frac{2n}{2n+1}\right)^{(-1)^{n+1}}$$

Divide them and we have

$$\frac{4}{\pi}=\frac{\prod_{n=1}^{\infty}\left(\frac{2n}{2n-1}\cdot\frac{2n}{2n+1}\right)^{(-1)^{n+1}}   }{\sum_{n=1}^{\infty}(-1)^{n+1}\frac{2n+2n}{(2n-1)(2n+1)}}$$

-

By summing both series it is no longer diverging

$$\frac{8}{\pi^2}=\lim_{k\to\infty}\frac{\prod_{n=1}^{\infty}\left(1+\frac{1}{4n^2-1}\right)^{(-1)^{n+1}}}{\sum_{n=1}^{2k-1}(-1)^{n+1}\left(1+\frac{1}{4n^2-1}\right)+ \sum_{n=1}^{2k}(-1)^{n+1}\left(1+\frac{1}{4n^2-1}\right) }$$

Section Nine

$$\lim_{n\to\infty}\left[\frac{\ln{\left(1+a-a^{\frac{a}{n}}+\ln{x^{\frac{1}{n}}}\cdot{\ln{a^{ya}}}\right)}}{\ln{a}}\right]^{n}=\frac{x^2}{e}$$

$$\lim_{n\to\infty}\left[\frac{\ln{\left(1+\frac{x}{n}\right)^n}}{x}\right]^{n}=\frac{1}{e^{\frac{x}{y}}}$$

Section Eight

Seidel's formula

{[pdf] more on the infinite products and partial fractions}

$$\ln{2}=\frac{2}{1+2^{\frac{1}{2}}}\cdot\frac{2}{1+2^{\frac{1}{4}}}\cdot\frac{2}{1+2^{\frac{1}{8}}}\cdots$$

$$\frac{\ln{x}}{x-1}=\prod_{n=1}^{\infty}\frac{2}{1+x^{\frac{1}{2^n}}}$$

The more general form for it is:

$$\frac{2\ln{x}}{y(x^{\frac{2}{y}}-1)}=\prod_{n=0}^{\infty}\frac{2}{1+x^{\frac{1}{2^ny}}}$$

$$\frac{1}{\pi}\cdot\frac{1}{e^{\frac{\pi}{2}}-1}=\frac{\frac{2}{1+e^{\frac{1}{\pi}}}\cdot\frac{2}{1+e^{\frac{1}{2\pi}}}\cdot \frac{2}{1+e^{\frac{1}{4\pi}}}\cdot\frac{2}{1+e^{\frac{1}{8\pi}}}\cdots}{1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots}$$

Section Seven

--

Double infinite products

(1)

$$\prod_{k=1}^{\infty}\left[\prod_{i=1}^{\infty}\frac{(2i)^{2^{k}}-1}{(2i)^{2^{k}}+1}\right]^{\frac{1}{2^{k}}}=\frac{2}{\pi}$$

(2)

$$\prod_{k=1}^{\infty}\left[\prod_{i=1}^{\infty}\frac{(2i)^{2^{k}}-1}{(2i)^{2^{k}}+1}\right]^{\frac{(-1)^{k-1}}{2^{k}}}=\frac{\pi}{4}$$

Section six

$$\gamma=\lim_{n\to\infty}\sum_{k=1}^{n}\frac{2}{2k-1}-\ln{4n}$$

-

Section Five

Ramanujan identity (From : Genius of Ramanujan VS. Modern Mathematical Technology)

$$\sum_{k=1}^{n}\frac{1}{n+k}=\frac{n}{2n+1}+\sum_{k=1}^{n}\frac{1}{(2k)^3-2k}$$

I have noticed another one

$$\sum_{k=1}^{n}\frac{1}{2n+k}+\sum_{k=1}^{n}\frac{1}{n+k}=\frac{3n}{3n+1}+\sum_{k=1}^{n}\frac{2}{(3k)^3-3k}$$

other

$$\gamma=\lim_{x\to \infty}\left(\sum_{i=1}^{x}\frac{1}{i}-\lim_{n\to \infty}\sum_{j=1}^{n}\sum_{k=1}^{x-1}\frac{1}{kn+j}\right)$$

Section four

Not really Wallis's product, but look alike,

(1)

$$\gamma-1+\ln4=\sum_{k=1}^{n\to \infty}\frac{2k+2k}{(2k-1)(2k+1)}-\ln{n}=\frac{2+2}{1\cdot3}+\frac{4+4}{3\cdot5}+ \frac{6+6}{5\cdot7}+\cdots+\frac{2n+2n}{(2n-1)(2n+1)}-\ln{n}$$

(2)

$$12=\frac{3^3+5}{1\cdot1!}-\frac{5^3+11}{2\cdot2!}+\frac{7^3+17}{3\cdot3!}-\frac{9^3+23}{4\cdot4!}+\cdots$$

(3)

$$0=\frac{1^2+1}{0!}-\frac{3^2-1}{1!}+\frac{5^2-3}{2!}-\frac{7^2-5}{3!}+\cdots$$

(4)

$$0=\frac{3^3+1}{1!}-\frac{5^3+3}{2!}+\frac{7^3+5}{3!}-\frac{9^3+7}{4!}+\cdots$$

(5)

$$1=\frac{2^2+1^2+0}{2!}-\frac{3^2+2^2+1}{3!}+\frac{4^2+3^2+2}{4!}-\frac{5^2+4^2+3}{5!}+\cdots$$

(6)

$$81+\frac{1}{e}+\frac{2e}{4+e}=\sum_{k=1}^{\infty}(-1)^k\frac{1-(2k+1)^6}{k\cdot{k!}}$$

(7)

$$0=\sum_{k=1}^{\infty}(-1)^{k-1}\frac{k^2(2k-1)^2}{k!}$$

Ramanujan' series

$$\frac{1}{2}+1\sum_{k=1}^{\infty}\frac{1}{(2k)^3-2k}=\ln2$$ and

$$1+2\sum_{k=1}^{\infty}\frac{1}{(3k)^3-3k}=\ln3$$

Let

$$a+b\sum_{k=1}^{\infty}\frac{1}{(pk)^3-pk}=\ln{x}$$

$$c+d\sum_{k=1}^{\infty}\frac{1}{(qk)^3-qk}=\ln{y}$$

$$e+f\sum_{k=1}^{\infty}\frac{1}{(rk)^3-rk}=\ln{z}$$

Where p < q < r

We have two series

(8)

$$\frac{1}{q^2-p^2}\left(\frac{cq}{d}-\frac{ap}{b}+\ln{\frac{x^{\frac{p}{b}}}{y^{\frac{q}{d}}}}\right)=\sum_{k=1}^{\infty}\frac{k}{[(pk)^2-1][(qk)^2-1]}$$

(9)

$$\frac{1}{(q^2-p^2)(r^2-q^2)}\left(\frac{cq}{d}-\frac{ap}{b}+\ln{\frac{x^{\frac{p}{b}}}{y^{\frac{q}{d}}}}\right)- \frac{1}{(r^2-p^2)(r^2-q^2)}\left(\frac{er}{f}-\frac{ap}{b}+\ln{\frac{x^{\frac{p}{b}}}{z^{\frac{r}{f}}}}\right) =\sum_{k=1}^{\infty}\frac{k^3}{[(pk)^2-1][(qk)^2-1][(rk)^2-1]}$$

Examples for (8) and (9)

(8)

$$\frac{1}{10}+\frac{1}{5}\ln\left(\frac{4}{3\sqrt{3}}\right)= \sum_{k=1}^{\infty}\frac{k}{(4k^2-1)(9k^2-1)}$$

(9)

$$\frac{1}{420}+\frac{1}{7}\ln{2^{\frac{1}{12}}\left(\frac{4}{3\sqrt{3}}\right)^{\frac{1}{5}}}=\sum_{k=1}^{\infty}\frac{k^3}{(4k^2-1)(9k^2-1)(16k^2-1)}$$

(10)

$$\frac{1}{12}\left(1-\ln2\right)= \sum_{k=1}^{\infty}\frac{k}{(4k^2-1)(16k^2-1)}$$

-

Section three

Ramanujan's series

From book : Mathematical Mysteries, by Calvin C. Clawson) on page 223

(1)

$$\sum_{k=1}^{\infty}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{k}\right)\frac{x^k}{k!}= e^x\sum_{k=1}^{\infty}\frac{(-1)^{k-1}x^k}{k!k}$$

-

Let x = 1 and we have

and I am just moving the e on the LHS

$$\frac{1}{e}\sum_{k=1}^{\infty}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{k}\right)\frac{1}{k!}= \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k!k}$$

From this series I have noticed

(2)

$$1-\frac{1}{e}+\frac{1}{e}\sum_{k=1}^{\infty}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{k}\right)\frac{1}{k!}= \sum_{k=1}^{\infty}\frac{(-1)^{k-1}(k+1)}{k!k}$$

(3)

$$2-\frac{2}{e}+\frac{1}{e}\sum_{k=1}^{\infty}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{k}\right)\frac{1}{k!}= \sum_{k=1}^{\infty}\frac{(-1)^{k-1}(2k+1)}{k!k}$$

--

Others

(4)

$$4=\frac{3^2-1}{1!\cdot1}-\frac{5^2-1}{2!\cdot2}+\frac{7^2-1}{3!\cdot3}-\frac{9^2-1}{4!\cdot4}+\cdots$$

(5)

$$8=\frac{3^4-1}{1!\cdot1}-\frac{5^4-1}{2!\cdot2}+\frac{7^4-1}{3!\cdot3}-\frac{9^4-1}{4!\cdot4}+\cdots$$

---

Ramanujan's series

From book : Mathematical Mysteries, by Calvin C. Clawson) on page 221

(6)

$$\frac{1}{2}+\sum_{k=1}^{\infty}\frac{1}{(2k)^3-2k}=\frac{1}{2}+\frac{1}{2^3-2}+\frac{1}{4^3-4}+\frac{1}{6^3-6}+\cdots=\ln2$$

and

(7)

$$1+2\sum_{k=1}^{\infty}\frac{1}{(3k)^3-3k}=1+2\left[\frac{1}{3^3-3}+\frac{1}{6^3-6}+\frac{1}{9^3-9}+\cdots\right]=\ln3$$

From the above series {(6) and (7)} I have noticed

(8)

$$\frac{3}{2}+3\sum_{k=1}^{\infty}\frac{1}{(3k)^3-3k}=\frac{3}{2}+3\left[\frac{1}{3^3-3}+\frac{1}{6^3-6}+\frac{1}{9^3-9}+\cdots\right]=\ln3\sqrt{3}$$

(9)

$$2+4\sum_{k=1}^{\infty}\frac{1}{(4k)^3-4k}=2+4\left[\frac{1}{4^3-4}+\frac{1}{8^3-8}+\frac{1}{12^3-12}+\cdots\right]=\ln8$$

(10)

$$2+8\sum_{k=1}^{\infty}\frac{1}{(5k)^3-5k}=2+8\left[\frac{1}{5^3-5}+\frac{1}{10^3-10}+\frac{1}{15^3-15}+\cdots\right]=\ln8$$

(11)

$$3+16\sum_{k=1}^{\infty}\frac{1}{(6k)^3-6k}=3+16\left[\frac{1}{6^3-6}+\frac{1}{12^3-12}+\frac{1}{18^3-18}+\cdots\right]=\ln22$$

Section one:

From book : Mathematical Mysteries, by Calvin C. Clawson)

Ramanujan's series

(1)

$$\sum_{k=0}^{\infty}(-1)^k\frac{(2k+1)^3+(2k+1)^2}{k!}=0$$

From (1) I have noticed the following series

(2)

$$\sum_{k=0}^{\infty}(-1)^k\frac{(2k+1)^2-(2k+1)}{k!}=-\frac{2}{e}$$

(3)

$$\sum_{k=0}^{\infty}(-1)^k\frac{(2k+1)^3+(2k+1)}{k!}=\frac{2}{e}$$

(2) + (3) = (1)

(4)

$$\sum_{k=0}^{\infty}(-1)^k\frac{(2k+1)^3-(2k+1)}{k!}=\frac{4}{e}$$

(5)

$$\sum_{k=0}^{\infty}(-1)^k\frac{(2k+1)^2+(2k+1)}{k!}=-\frac{4}{e}$$

(4) + (5) = (1)

---

(6)

$$\sum_{k=0}^{\infty}(-1)^k\frac{2k+1}{k!}=-\frac{1}{e}$$

(7)

$$\sum_{k=0}^{\infty}(-1)^k\frac{(2k+1)^2}{k!}=-\frac{3}{e}$$

(8)

$$\sum_{k=0}^{\infty}(-1)^k\frac{(2k+1)^3}{k!}=\frac{3}{e}$$

(9)

$$\sum_{k=0}^{\infty}(-1)^k\frac{k+1}{k!}=0$$

(10)

$$\sum_{k=1}^{\infty}(-1)^{k+1}\frac{k^3+k^2}{(k+1)!}=0$$

(11)

$$\sum_{k=1}^{\infty}(-1)^{k+1}\frac{k^4+k}{(k+1)!}=0$$

-

(12)

$$\frac{-1d+1}{e}=\frac{1}{0!}-\frac{1+d}{1!}+\frac{1+2d}{2!}-\frac{1+3d}{3!}+\cdots$$

(13)

$$\frac{-3d+1}{e}+d=\frac{1}{2!}-\frac{1+d}{3!}+\frac{1+2d}{4!}-\frac{1+3d}{5!}+\cdots$$

(14)

$$\frac{4d-1}{e}-\frac{3d-1}{2}=\frac{1}{3!}-\frac{1+d}{4!}+\frac{1+2d}{5!}-\frac{1+3d}{6!}+\cdots$$

(15)

$$\frac{-5d+1}{e}-\frac{9d-2}{6}=\frac{1}{4!}-\frac{1+d}{5!}+\frac{1+2d}{6!}-\frac{1+3d}{7!}+\cdots$$

---

Section two

From book : Mathematical Mysteries, by Calvin C. Clawson)

Ramanujan's series

(1)

$$\frac{x}{2x-1}=1-\frac{x-1}{x+1}+\frac{(x-1)(x-2)}{(x+1)(x+2)}-\frac{(x-1)(x-2)(x-3)}{(x+1)(x+2)(x+3)}-\cdots$$

I have noticed the general form of it

(2)

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

Let k = 0 we have (1)

-

Let k = x we have

$$\frac{2x}{2x-1}=1+\frac{1}{2x+1}+\frac{1\cdot2}{(2x+1)(2x+2)}+\frac{1\cdot2\cdot3}{(2x+1)(2x+2)(2x+3)}+\cdots$$

Let 2x = y to make it simple

$$\frac{y}{y-1}=1+\frac{1}{y+1}+\frac{1\cdot2}{(y+1)(y+2)}+\frac{1\cdot2\cdot3}{(y+1)(y+2)(y+3)}+\cdots$$ Example

Let x = 1

$$2=1+\frac{1}{3}+\frac{1\cdot2}{3\cdot4}+\frac{1\cdot2\cdot3}{3\cdot4\cdot5}+\cdots$$

rearrange and we have

$$1=\frac{0!}{2!}+\frac{1!}{3!}+\frac{2!}{4!}+\frac{3!}{5!}+\cdots$$

and simplify to

$$1=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\cdots$$

others

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

$$\frac{1}{y}=\frac{1}{1!}+\frac{1-y}{2!}+\frac{(1-y)(2-y)}{3!}+\frac{(1-y)(2-y)(3-y)}{4!}+\cdots$$

-

Pi

$$\sqrt{\pi}=\frac{\Gamma\left(\frac{3}{2}\right)}{2!}+\frac{\Gamma\left(\frac{5}{2}\right)}{3!}+ \frac{\Gamma\left(\frac{7}{2}\right)}{4!}+\frac{\Gamma\left(\frac{9}{2}\right)}{5!}+\cdots$$

Hypergeometric function

$$_2F_1(1,1;c;1)=\frac{c-1}{c-2}$$

$$_2F_1(-a,b;b;1)=0$$

- -

k > 1 or = 1

$$\frac{16^{k+1}-1}{240}=\sum_{n=1}^{\infty}\frac{n^3}{e^{\frac{n\pi}{2^k}}-1}$$

-- ---

$$\frac{\Gamma\left(\frac{3}{4}\right)}{\pi^{\frac{1}{4}}}=\prod_{n=1}^{\infty} \left(\frac{e^{n\pi}-1}{e^{n\pi}+1}\right)^{(-1)^{n+1}}$$

$$\frac{\Gamma\left(\frac{3}{4}\right)}{\pi^{\frac{1}{4}}}\cdot\sqrt{2-\sqrt{2}} =\prod_{n=1}^{\infty} \left(\frac{e^{\frac{n\pi}{2}}-1}{e^{\frac{n\pi}{2}}+1}\right)^{(-1)^{n+1}}$$

-- -

Wallis-type infinite products

(1)

$$\left(\frac{\Gamma\left(\frac{1}{4}\right)}{2}\right)^2\frac{1}{\sqrt{\pi}}=\frac{2}{1}\cdot\frac{4}{5}\cdot\frac{6}{5}\cdot\frac{8}{9}\cdot\frac{10}{9}\cdots$$

(2)

$$\left(\frac{\pi}{\Gamma\left(\frac{1}{4}\right)}\right)^2\frac{2}{\sqrt{\pi}}=\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{6}{7}\cdot\frac{8}{7}\cdot\frac{10}{11}\cdots$$

$$(1)\times(2)=$$ Wallis's product

--

(3)

$$\frac{\Gamma^2\left(\frac{1}{4}\right)}{2^{\frac{5}{2}} \sqrt{\pi}}=\frac{2}{1}\cdot\frac{3}{5}\cdot \frac{4}{3}\cdot\frac{7}{9}\cdot\frac{6}{5}\cdot\frac{11}{13}\cdot\frac{8}{7}\cdot\frac{15}{17}\cdots$$

(4)

$$\frac{\pi^{\frac{3}{2}}\sqrt{2}}{\Gamma^2(\frac{1}{4})}=\frac{2}{1}\cdot\frac{1}{3}\cdot \frac{4}{3}\cdot\frac{5}{7}\cdot\frac{6}{5}\cdot\frac{9}{11}\cdot\frac{8}{7}\cdot\frac{13}{15}\cdots$$

$$2(3)\times(4)=$$ Wallis's product

--- --

(5)

$$\frac{\pi}{2^\frac{5}{2}}=\left(\frac{4}{1}\cdot\frac{1\cdot5}{3\cdot7}\right) \left(\frac{4}{5}\cdot\frac{9\cdot13}{11\cdot15}\right) \left(\frac{8}{5}\cdot\frac{17\cdot21}{19\cdot23}\right) \left(\frac{8}{9}\cdot\frac{25\cdot29}{27\cdot31}\right)\cdots$$

--

(6)

$$\frac{\pi}{2}=\lim_{n\to\infty}2^{\frac{1}{4n}}\prod_{j=1}^{n}\frac{(2j)^2}{(2j)^2-1}\cdot\prod_{k=n+1}^{n-1}\left[\frac{(2k)^2}{(2k)^2-1}\right]^{\frac{2n-k}{n}}$$

-- --

-- -- --

I have noticed some patterns in the following three products from your website.

Pattern of Wallis's product

$$\frac{\pi}{2}=\left(\frac{2}{1}\right)^{\frac{1}{2}}\left(\frac{2^2}{1\cdot3}\right)^{\frac{1}{4}} \left(\frac{2^3\cdot4}{1\cdot3^3}\right)^{\frac{1}{8}}\left(\frac{2^4\cdot4^4}{1\cdot3^6\cdot5}\right)^{\frac{1}{16}}\cdots$$

Even number

$$\frac{\pi\cdot1\cdot3}{2\cdot2\cdot2}=\left(\frac{4}{3}\right)^{\frac{1}{2}}\left(\frac{4^2}{3\cdot5}\right)^{\frac{1}{4}} \left(\frac{4^3\cdot6}{3\cdot5^3}\right)^{\frac{1}{8}}\left(\frac{4^4\cdot6^4}{3\cdot5^6\cdot7}\right)^{\frac{1}{16}}\cdots$$

$$\frac{\pi\cdot1\cdot3\cdot3\cdot5}{2\cdot2\cdot2\cdot4\cdot4}=\left(\frac{6}{5}\right)^{\frac{1}{2}}\left(\frac{6^2}{5\cdot7}\right)^{\frac{1}{4}} \left(\frac{6^3\cdot8}{5\cdot7^3}\right)^{\frac{1}{8}}\left(\frac{6^4\cdot8^4}{5\cdot7^6\cdot9}\right)^{\frac{1}{16}}\cdots$$

$$\frac{\pi\cdot1\cdot3\cdot3\cdot5\cdot5\cdot7}{2\cdot2\cdot2\cdot4\cdot4\cdot6\cdot6}=\left(\frac{8}{7}\right)^{\frac{1}{2}}\left(\frac{8^2}{7\cdot9}\right)^{\frac{1}{4}} \left(\frac{8^3\cdot10}{7\cdot9^3}\right)^{\frac{1}{8}}\left(\frac{8^4\cdot10^4}{7\cdot9^6\cdot11}\right)^{\frac{1}{16}}\cdots$$

and so on ...

--

Odd number

$$\frac{2\cdot2}{\pi\cdot1}=\left(\frac{3}{2}\right)^{\frac{1}{2}}\left(\frac{3^2}{2\cdot4}\right)^{\frac{1}{4}} \left(\frac{3^3\cdot5}{2\cdot4^3}\right)^{\frac{1}{8}}\left(\frac{3^4\cdot5^4}{2\cdot4^6\cdot6}\right)^{\frac{1}{16}}\cdots$$

$$\frac{2\cdot2\cdot2\cdot4}{\pi\cdot1\cdot3\cdot3}=\left(\frac{5}{4}\right)^{\frac{1}{2}}\left(\frac{5^2}{4\cdot6}\right)^{\frac{1}{4}} \left(\frac{5^3\cdot7}{4\cdot6^3}\right)^{\frac{1}{8}}\left(\frac{5^4\cdot7^4}{4\cdot6^6\cdot8}\right)^{\frac{1}{16}}\cdots$$

$$\frac{2\cdot2\cdot2\cdot4\cdot4\cdot6}{\pi\cdot1\cdot3\cdot3\cdot5\cdot5}=\left(\frac{7}{6}\right)^{\frac{1}{2}}\left(\frac{7^2}{6\cdot8}\right)^{\frac{1}{4}} \left(\frac{7^3\cdot9}{6\cdot8^3}\right)^{\frac{1}{8}}\left(\frac{7^4\cdot9^4}{7\cdot8^6\cdot10}\right)^{\frac{1}{16}}\cdots$$

and so on ...

Pattern of Euler constant

$$\frac{1e^{\gamma}}{e^0}=\left(\frac{2}{1}\right)^{\frac{1}{2}}\left(\frac{2^2}{1\cdot3}\right)^{\frac{1}{3}} \left(\frac{2^3\cdot4}{1\cdot3^3}\right)^{\frac{1}{4}}\left(\frac{2^4\cdot4^4}{1\cdot3^6\cdot5}\right)^{\frac{1}{5}}\cdots$$

$$\frac{2e^{\gamma}}{e^1}=\left(\frac{3}{2}\right)^{\frac{1}{2}}\left(\frac{3^2}{2\cdot4}\right)^{\frac{1}{3}} \left(\frac{3^3\cdot5}{2\cdot4^3}\right)^{\frac{1}{4}}\left(\frac{3^4\cdot5^4}{2\cdot4^6\cdot6}\right)^{\frac{1}{5}}\cdots$$

$$\frac{3e^{\gamma}}{e^{1+\frac{1}{2}}}=\left(\frac{4}{3}\right)^{\frac{1}{2}}\left(\frac{4^2}{3\cdot5}\right)^{\frac{1}{3}} \left(\frac{4^3\cdot6}{3\cdot5^3}\right)^{\frac{1}{4}}\left(\frac{4^4\cdot6^4}{3\cdot5^6\cdot7}\right)^{\frac{1}{5}}\cdots$$

$$\frac{4e^{\gamma}}{e^{1+\frac{1}{2}+\frac{1}{3}}}=\left(\frac{5}{4}\right)^{\frac{1}{2}}\left(\frac{5^2}{4\cdot6}\right)^{\frac{1}{3}} \left(\frac{5^3\cdot7}{4\cdot6^3}\right)^{\frac{1}{4}}\left(\frac{5^4\cdot7^4}{4\cdot6^6\cdot8}\right)^{\frac{1}{5}}\cdots$$

The right side will approaches to 1 as n tends to infinity

Take the log on both sides and we have

$$\gamma=\lim_{n\to\infty}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n-1}-\ln{n}\right)$$

---

Pattern of e

$$e^{\frac{1}{1}}=\left(\frac{2}{1}\right)^{\frac{1}{1}}\left(\frac{2^2}{1\cdot3}\right)^{\frac{1}{2}} \left(\frac{2^3\cdot4}{1\cdot3^3}\right)^{\frac{1}{3}}\left(\frac{2^4\cdot4^4}{1\cdot3^6\cdot5}\right)^{\frac{1}{4}}\cdots$$

$$e^{\frac{1}{2}}=\left(\frac{3}{2}\right)^{\frac{1}{1}}\left(\frac{3^2}{2\cdot4}\right)^{\frac{1}{2}} \left(\frac{3^3\cdot5}{2\cdot4^3}\right)^{\frac{1}{3}}\left(\frac{3^4\cdot5^4}{2\cdot4^6\cdot6}\right)^{\frac{1}{4}}\cdots$$

$$e^{\frac{1}{3}}=\left(\frac{4}{3}\right)^{\frac{1}{1}}\left(\frac{4^2}{3\cdot5}\right)^{\frac{1}{2}} \left(\frac{4^3\cdot6}{3\cdot5^3}\right)^{\frac{1}{3}}\left(\frac{4^4\cdot6^4}{3\cdot5^6\cdot7}\right)^{\frac{1}{4}}\cdots$$

$$e^{\frac{1}{4}}=\left(\frac{5}{4}\right)^{\frac{1}{1}}\left(\frac{5^2}{4\cdot6}\right)^{\frac{1}{2}} \left(\frac{5^3\cdot7}{4\cdot6^3}\right)^{\frac{1}{3}}\left(\frac{5^4\cdot7^4}{4\cdot6^6\cdot8}\right)^{\frac{1}{4}}\cdots$$

 and so on ...

Infinite product

The formula from your Website

(1)

$$\frac{\pi}{2}=\left(\frac{2}{1}\right)^{\frac{1}{2}}\left(\frac{2^2}{1\cdot3}\right)^{\frac{1}{4}} \left(\frac{2^3\cdot4}{1\cdot3^3}\right)^{\frac{1}{8}}\left(\frac{2^4\cdot4^4}{1\cdot3^6\cdot5}\right)^{\frac{1}{16}}\cdots$$

Re-write is as wallis form $$\frac{\pi}{2}=\left[\left(\frac{2}{1}\right)\right]^{\frac{1}{2}} \left[\left(\frac{2}{1}\right)\left(\frac{2}{3}\right)\right]^{\frac{1}{4}} \left[\left(\frac{2}{1}\right)\left(\frac{2}{3}\right)^2\left(\frac{4}{3}\right)\right]^{\frac{1}{8}} \left[\left(\frac{2}{1}\right)\left(\frac{2}{3}\right)^3\left(\frac{4}{3}\right)^3\left(\frac{4}{5}\right)\right]^{\frac{1}{16}}\cdots$$

You and Guillera wrote an interesting paper, out of curiosity I have to play around with it and these are the product I came across:

--- -

(2)

$$\frac{\pi}{2}=\prod_{n=0}^{\infty}\left(\prod_{k=0}^{n}\left(\frac{k+2}{k+1}\right)^{(-1)^k(n-2k)\tbinom nk}\right)^{\frac{1}{2^{n+1}}}$$

Expand the above and we have

$$\frac{\pi}{2}=\left[\left(\frac{2}{1}\right)^0\right]^{\frac{1}{2}} \left[\left(\frac{2}{1}\right)^{1(1)}\left(\frac{2}{3}\right)^{-1(1)}\right]^{\frac{1}{4}} \left[\left(\frac{2}{1}\right)^{2(1)}\left(\frac{2}{3}\right)^{0(2)}\left(\frac{4}{3}\right)^{-2(1)}\right]^{\frac{1}{8}} \left[\left(\frac{2}{1}\right)^{3(1)}\left(\frac{2}{3}\right)^{1(3)}\left(\frac{4}{3}\right)^{-1(3)} \left(\frac{4}{5}\right)^{-3(1)}\right]^{\frac{1}{16}}\left[\left(\frac{2}{1}\right)^{4(1)}\left(\frac{2}{3}\right)^{2(4)}\left(\frac{4}{3}\right)^{0(6)} \left(\frac{4}{5}\right)^{-2(4)}\left(\frac{6}{5}\right)^{-4(1)}\right]^{\frac{1}{32}}\cdots$$

Simplify the exponents and we have

$$\frac{\pi}{2}=\left[\left(\frac{2}{1}\right)^0\right]^{\frac{1}{2}} \left[\left(\frac{2}{1}\right)^1\left(\frac{2}{3}\right)^{-1}\right]^{\frac{1}{4}} \left[\left(\frac{2}{1}\right)^{2}\left(\frac{2}{3}\right)^{0}\left(\frac{4}{3}\right)^{-2}\right]^{\frac{1}{8}} \left[\left(\frac{2}{1}\right)^{3}\left(\frac{2}{3}\right)^{3}\left(\frac{4}{3}\right)^{-3} \left(\frac{4}{5}\right)^{-3}\right]^{\frac{1}{16}} \left[\left(\frac{2}{1}\right)^{4}\left(\frac{2}{3}\right)^{8}\left(\frac{4}{3}\right)^{0} \left(\frac{4}{5}\right)^{-8}\left(\frac{6}{5}\right)^{-4}\right]^{\frac{1}{32}}\cdots$$

Derivation

---

The formula from your Paper Under the section 5. on infinite product.

(3)

$$\frac{A^{12}}{2^{\frac{4}{3}}e}=\left(\frac{2}{1}\right)^{\frac{2}{2}}\left(\frac{2^2}{1\cdot3}\right)^{\frac{3}{4}} \left(\frac{2^3\cdot4}{1\cdot3^3}\right)^{\frac{4}{8}}\left(\frac{2^4\cdot4^4}{1\cdot3^6\cdot5}\right)^{\frac{5}{16}}\cdots$$

Re-write is as wallis form $$\frac{A^{12}}{2^{\frac{4}{3}}e}=\left[\left(\frac{2}{1}\right)\right]^{\frac{2}{2}} \left[\left(\frac{2}{1}\right)\left(\frac{2}{3}\right)\right]^{\frac{3}{4}} \left[\left(\frac{2}{1}\right)\left(\frac{2}{3}\right)^2\left(\frac{4}{3}\right)\right]^{\frac{4}{8}} \left[\left(\frac{2}{1}\right)\left(\frac{2}{3}\right)^3\left(\frac{4}{3}\right)^3\left(\frac{4}{5}\right)\right]^{\frac{5}{16}}\cdots$$

By the method of trial and error we obtained two of the following products

Let

$$x=\frac{A^6}{2^{\frac{1}{6}}\pi^{\frac{1}{2}}e^{\frac{1}{2}}}$$

$$x=\left[\left(\frac{2}{1}\right)^{1(1)}\right]^{\frac{1}{2}} \left[\left(\frac{2}{1}\right)^{1(1)}\left(\frac{2}{3}\right)^{2(1)}\right]^{\frac{1}{4}} \left[\left(\frac{2}{1}\right)^{1(1)}\left(\frac{2}{3}\right)^{2(2)}\left(\frac{4}{3}\right)^{3(1)}\right]^{\frac{1}{8}} \left[\left(\frac{2}{1}\right)^{1(1)} \left(\frac{2}{3}\right)^{2(3)}\left(\frac{4}{3}\right)^{3(3)}\left(\frac{4}{5}\right)^{4(1)}\right]^{\frac{1}{16}}\cdots$$

$$y=\frac{A^6\pi^{\frac{1}{2}}}{2^{\frac{7}{6}}e^{\frac{1}{2}}}$$

$$y=\left[\left(\frac{2}{1}\right)^{1(1)}\right]^{\frac{1}{2}} \left[\left(\frac{2}{1}\right)^{2(1)}\left(\frac{2}{3}\right)^{1(1)}\right]^{\frac{1}{4}} \left[\left(\frac{2}{1}\right)^{3(1)}\left(\frac{2}{3}\right)^{2(2)}\left(\frac{4}{3}\right)^{1(1)}\right]^{\frac{1}{8}} \left[\left(\frac{2}{1}\right)^{4(1)}\left(\frac{2}{3}\right)^{3(3)}\left(\frac{4}{3}\right)^{2(3)}\left(\frac{4}{5}\right)^{1(1)}\right]^{\frac{1}{16}}\cdots$$

Evaluate $$x\times{y}$$ and we have

-

$$\frac{A}{2^{\frac{4}{3}}e}=\left[\left(\frac{2}{1}\right)^{2(1)}\right]^{\frac{1}{2}} \left[\left(\frac{2}{1}\right)^{3(1)}\left(\frac{2}{3}\right)^{3(1)}\right]^{\frac{1}{4}} \left[\left(\frac{2}{1}\right)^{4(1)}\left(\frac{2}{3}\right)^{4(2)}\left(\frac{4}{3}\right)^{4(1)}\right]^{\frac{1}{8}} \left[\left(\frac{2}{1}\right)^{5(1)}\left(\frac{2}{3}\right)^{5(3)}\left(\frac{4}{3}\right)^{5(3)}\left(\frac{4}{5}\right)^{5(1)}\right]^{\frac{1}{16}}\cdots$$

Simplify and therefore we have (3)

$$\frac{A}{2^{\frac{4}{3}}e}=\left[\left(\frac{2}{1}\right)^{(1)}\right]^{\frac{2}{2}} \left[\left(\frac{2}{1}\right)^{(1)}\left(\frac{2}{3}\right)^{(1)}\right]^{\frac{3}{4}} \left[\left(\frac{2}{1}\right)^{(1)}\left(\frac{2}{3}\right)^{(2)}\left(\frac{4}{3}\right)^{(1)}\right]^{\frac{4}{8}} \left[\left(\frac{2}{1}\right)^{(1)}\left(\frac{2}{3}\right)^{(3)}\left(\frac{4}{3}\right)^{(3)}\left(\frac{4}{5}\right)^{(1)}\right]^{\frac{5}{16}}\cdots$$

Evaluation of $$y\div{x}$$ and therefore we have (2)

---

Recall the two products

$$y=\left[\left(\frac{2}{1}\right)^{1(1)}\right]^{\frac{1}{2}} \left[\left(\frac{2}{1}\right)^{2(1)}\left(\frac{2}{3}\right)^{1(1)}\right]^{\frac{1}{4}} \left[\left(\frac{2}{1}\right)^{3(1)}\left(\frac{2}{3}\right)^{2(2)}\left(\frac{4}{3}\right)^{1(1)}\right]^{\frac{1}{8}} \left[\left(\frac{2}{1}\right)^{4(1)}\left(\frac{2}{3}\right)^{3(3)}\left(\frac{4}{3}\right)^{2(3)}\left(\frac{4}{5}\right)^{1(1)}\right]^{\frac{1}{16}}\cdots$$

$$x=\left[\left(\frac{2}{1}\right)^{1(1)}\right]^{\frac{1}{2}} \left[\left(\frac{2}{1}\right)^{1(1)}\left(\frac{2}{3}\right)^{2(1)}\right]^{\frac{1}{4}} \left[\left(\frac{2}{1}\right)^{1(1)}\left(\frac{2}{3}\right)^{2(2)}\left(\frac{4}{3}\right)^{3(1)}\right]^{\frac{1}{8}} \left[\left(\frac{2}{1}\right)^{1(1)} \left(\frac{2}{3}\right)^{2(3)}\left(\frac{4}{3}\right)^{3(3)}\left(\frac{4}{5}\right)^{4(1)}\right]^{\frac{1}{16}}\cdots$$

--

'''Rule of indices (same base subtract the exponents) ''' $$y\div{x}= \left[\left(\frac{2}{1}\right)^{1(1)-1(1)}\right]^{\frac{1}{2}}

\left[\left(\frac{2}{1}\right)^{2(1)-1(1)}\left(\frac{2}{3}\right)^{1(1)-2(1)}\right]^{\frac{1}{4}}

\left[\left(\frac{2}{1}\right)^{3(1)-1(1)}\left(\frac{2}{3}\right)^{2(2)-2(2)}\left(\frac{4}{3}\right)^{1(1)-3(1)}\right]^{\frac{1}{8}} \left[\left(\frac{2}{1}\right)^{4(1)-1(1)}\left(\frac{2}{3}\right)^{3(3)-2(3)}\left(\frac{4}{3}\right)^{2(3)-3(3)}\left(\frac{4}{5}\right)^{1(1)-4(1)}\right]^{\frac{1}{16}}\cdots$$

Simplify and we have (2)

$$\frac{\pi}{2}=\left[\left(\frac{2}{1}\right)^0\right]^{\frac{1}{2}} \left[\left(\frac{2}{1}\right)^{1(1)}\left(\frac{2}{3}\right)^{-1(1)}\right]^{\frac{1}{4}} \left[\left(\frac{2}{1}\right)^{2(1)}\left(\frac{2}{3}\right)^{0(2)}\left(\frac{4}{3}\right)^{-2(1)}\right]^{\frac{1}{8}} \left[\left(\frac{2}{1}\right)^{3(1)}\left(\frac{2}{3}\right)^{1(3)}\left(\frac{4}{3}\right)^{-1(3)} \left(\frac{4}{5}\right)^{-3(1)}\right]^{\frac{1}{16}}\left[\left(\frac{2}{1}\right)^{4(1)}\left(\frac{2}{3}\right)^{2(4)}\left(\frac{4}{3}\right)^{0(6)} \left(\frac{4}{5}\right)^{-2(4)}\left(\frac{6}{5}\right)^{-4(1)}\right]^{\frac{1}{32}}\cdots$$

$$\frac{\pi}{2}=\left[\left(\frac{3}{1}\right)^{1(1)}\right]^{\frac{1}{2}} \left[\left(\frac{3}{1}\right)^{1(1)}\left(\frac{2}{4}\right)^{2(1)}\right]^{\frac{1}{4}} \left[\left(\frac{3}{1}\right)^{1(1)}\left(\frac{2}{4}\right)^{2(2)}\left(\frac{5}{3}\right)^{3(1)}\right]^{\frac{1}{8}} \left[\left(\frac{3}{1}\right)^{1(1)}\left(\frac{2}{4}\right)^{2(3)}\left(\frac{5}{3}\right)^{3(3)}\left(\frac{4}{6}\right)^{4(1)}\right]^{\frac{1}{16}}\cdots$$

---

$$\frac{\pi}{2}=\prod_{n=0}^{\infty}\left(\prod_{k=0}^{n}\left(\frac{k+3}{k+1}\right)^{(-1)^k(3-2n+5k)\tbinom nk}\right)^{\frac{1}{2^{n+1}}}$$ Bold text

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University of Florida/Egm4313/s12.team5.R2

Question
Given the two roots and the initial conditions:

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$$\displaystyle \lambda_1=-2, \lambda_2 = +5$$ (1.0)
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$$\displaystyle y(0)=1, y'(0)=0$$ (1.1)
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Part 1.

Find the non-homogeneous L2-ODE-CC in standard form. Find the solution in terms of the initial conditions and the general excitation $$r(x).\!$$ Now with no excitation and plot the solution:

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$$\displaystyle r(x) = 0$$ (1.2) Part 2.
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Generate 3 non-standard (and non-homogeneous) L2-ODE-CC that admit the two values $$\lambda_1=-2, \lambda_2 = +5\!$$ as the two roots of the corresponding characteristic equation.

Solution
Part 1.

Characteristic Equation:
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$$\displaystyle (\lambda-\lambda_1)(\lambda-\lambda_2) = 0$$ (1.3)
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$$\displaystyle (\lambda-(-2))(\lambda-5) = 0$$ (1.4)
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$$\displaystyle \lambda^2-3\lambda-10=0$$ (1.5) Non-Homogeneous L2-ODE-CC:
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$$ \displaystyle y''-3y'-10y=r(x)\rightarrow standard form$$ (1.6)
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Homogenous Solution:
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$$\displaystyle y_h(x)=c_1e^{-2x} + c_2e^{5x}$$ (1.7) Overall Solution:
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$$\displaystyle y(x)=c_1e^{-2x} + c_2e^{5x}+y_p(x)$$ (1.8)
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$$\displaystyle y'(x)=-2c_1e^{-2x} + 5c_2e^{5x}+y'_p(x)$$ (1.9) Satisfy Initial Conditions:
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$$\displaystyle y(0)=1=c_1 + c_2 +y_p(0)$$ (1.10)
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$$\displaystyle y'(0)=0=-2c_1 + 5c_2 +y'_p(0)$$ (1.11) No excitation:
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$$\displaystyle r(x) = 0 \rightarrow y_p(x)=0 \rightarrow y'_p(x)=0$$ (1.12) From (1.10):
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$$\displaystyle c_1+c_2 = 1 \rightarrow c_1=1-c_2$$ (1.13) From (1.11):
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$$\displaystyle -2c_1+5c_2 = 0$$ (1.14)
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$$\displaystyle 5c_2=2c_1$$ (1.15)
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$$\displaystyle \frac{5}{2}c_2=c_1$$ (1.16) Plug (1.13) into (1.16):
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$$\displaystyle \frac{5}{2}c_2=1-c_2$$ (1.17) Solve for $$c_2\!$$:
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$$\displaystyle c_2=\frac{2}{7}$$ (1.18) Plug (1.18) into (1.13) and solve for $$c_1\!$$:
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$$\displaystyle c_1=\frac{5}{7}$$ (1.19) Therefore, the final solution in terms of the initial conditions and the general excitation $$r(x)\!$$ is:
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$$\displaystyle y(x)=\frac{5}{7}e^{-2x}+\frac{2}{7}e^{5x}$$ (1.20)
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Plot of Solution:

Figure 1

Part 2.

Three non-standard and non-homogeneous solutions using the same roots given above:

1.
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$$\displaystyle 2(\lambda-(-2))(\lambda-5) = 0$$ (1.21)
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$$\displaystyle 2\lambda^2-6\lambda-20=0$$ (1.22)
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2.
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$$\displaystyle 3(\lambda-(-2))(\lambda-5) = 0$$ (1.23)
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$$\displaystyle 3\lambda^2-9\lambda-30=0$$ (1.24)
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3.
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$$\displaystyle 4(\lambda-(-2))(\lambda-5) = 0$$ (1.25)
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$$\displaystyle 4\lambda^2-12\lambda-40=0$$ (1.26)
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 * }

Author
This problem was solved and uploaded by [Derik Bell]

This problem was proofread by: David Herrick

Question
From Section 5 in the notes, pg. 5-6

Solve the L2-ODE-CC from pg. 5-5, equation (4)

$$ \displaystyle \ y'' - 10y' +25y = r(x)\ $$

For initial conditions:

$$ \displaystyle \ y(0) = 1, y'(0) = 0\ $$

Where there is no excitation, i.e. $$ \displaystyle \ r(x) = 0\ $$

Solution
$$ \displaystyle \ \lambda^2 - 10\lambda + 25 = 0\ $$

Factoring:

$$ \displaystyle \ (\lambda - 5)^2 = 0\ $$

Thus, there is a double root, where $$ \displaystyle \ \lambda = 5\ $$

Because there is no excitation, the homogenous solution will be the same as the final solution. Thus, the solution will be:

$$ \displaystyle \ y_h(x) = y(x) = C_1e^{5x} + C_2xe^{5x}\ $$

Plugging in the initial condition,

$$ \displaystyle \ y(0) = 1 = C_1e^{0} + C_2 (0) e^{0}\ $$

$$ \displaystyle \ C_1 = 1\ $$

In order to solve for the second constant, we need to take a derivative.

$$ \displaystyle \ y'_h(x) = y'(x) = 5C_1e^{5x} + C_2e^{5x} + 5C_2xe^{5x}\ $$

Plugging in C1 and the other initial condition:

$$ \displaystyle \ y'_h(0) = y'(0) = 0 = 5(1)e^{0} + C_2e^{0} + 5C_2(0)e^{0} \ $$

$$ \displaystyle \ 5 + C_2 = 0 \Rightarrow C_2 = -5\ $$

So, the final solution is:

$$ \displaystyle \ y(x) = e^{5x} - 5xe^{5x}\ $$

Matlab Code:

x = 0:0.001:1;

f = exp(5*x) - 5*x.*exp(5*x);

plot(x,f)

xlabel ('x')

ylabel ('y(x)')

Author
This problem was solved and uploaded by [David Herrick] This problem was checked by [William Knapper]

Question
Complete problems 3 and 4 from p.59 of K 2011.

Solution
The two problems have the same instructions: "Find a general solution. Check your answer by substitution."

Problem 3
$$ \displaystyle 3. \; {y}''+6{y}' +8.96y = 0 $$

We start by using the characteristic equation of this ODE in order to find the roots. We use:

$$ \displaystyle \lambda ^{2} + a\lambda + b = 0$$

where a = 6 and b = 8.96. Taking the discriminant, we find that

$$ \displaystyle 6^2 - 4(8.96) = 0.16 > 0 $$

A positive discriminant implies that there exists two real roots, with a general solution of the form:

$$ \displaystyle y = c_{1}e^{\lambda _{1}x}+c_{2}e^{\lambda _{2}x} $$

To find the unknowns, we find the roots of the above equation:

$$ \displaystyle \lambda ^{2} + 6\lambda + 8.96 = 0$$

$$ \displaystyle \lambda _{1} = -3.2, \lambda _{2} = -2.8 $$

Therefore, the correct general solution is:

$$ \displaystyle y=c_{1}e^{-3.2x} + c_{2}e^{-2.8x} $$

To check that this solution is correct, we take the first and second derivatives:

$$ \displaystyle {y}' = -3.2c_{1}e^{-3.2x} - 2.8c_{2}e^{-2.8x} $$

$$ \displaystyle {y}'' = 10.24c_{1}e^{-3.2x} + 7.84c_{2}e^{-2.8x} $$

Now, we plug these values into the original equation:

$$ \displaystyle 10.24c_{1}e^{-3.2x} + 7.84c_{2}e^{-2.8x} - 19.2c_{1}e^{-3.2x} - 16.8c_{2}e^{-2.8x} + 8.96c_{1}e^{-3.2x} + 8.96c_{2}e^{-2.8x} = 0 $$

Upon further inspection, all the terms on the left side of the equation cancel out and equal zero.

Problem 4
$$ \displaystyle 4. \; {y}'' + 4{y}' + (\pi ^{2}+4)y = 0$$

We start by using the characteristic equation of this ODE in order to find the roots. We use:

$$ \displaystyle \lambda ^{2} + a\lambda + b = 0$$

where a = 4 and b = (π^2 + 4). Taking the discriminant, we find that

$$ \displaystyle 4^{2} - 4(\pi ^{2} + 4) = -4\pi ^{2} < 0 $$

A negative discriminant implies that there exists complex conjugate roots to this equation. The general solution for this type of equation is

$$ \displaystyle y = e^{-2x}(Acos(\omega x)+Bsin(\omega x)) $$

Where

$$ \displaystyle \omega = \sqrt{b-\frac{1}{4}a^{2}} = \sqrt{(\pi ^{2}+4)-(\frac{1}{4})4^{2}} = \pi  $$

This yields

$$ \displaystyle y = e^{-2x}(Acos(\pi x)+Bsin(\pi x)) $$

To verify that this is the correct solution, we first find the first and second derivatives of the solution:

$$ \displaystyle y = e^{-2x}(Acos(\pi x)+Bsin(\pi x)) $$ $$ \displaystyle {y}' = e^{-2x}(-A\pi sin(\pi x)+B\pi cos(\pi x))-2e^{-2x}(Acos(\pi x)+Bsin(\pi x)) $$ $$ \displaystyle {y}''=e^{-2x}[-A\pi ^{2}cos(\pi x)-B\pi ^{2}sin(\pi x)+2A\pi sin(\pi x)-2B\pi sin(\pi x)]-2e^{-2x}[-A\pi sin(\pi x)+B\pi cos(\pi x)-2Acos(\pi x)-2Bsin(\pi x)] $$

We then plug these values into the original problem to get this somewhat lengthy equation:

$$ \displaystyle e^{-2x}A\pi ^{2}cos(\pi x)-e^{-2x}B\pi ^{2}sin(\pi x) +2e^{-2x}A\pi sin(\pi x)-2e^{-2x}B\pi sin(\pi x)+2e^{-2x}A\pi sin(\pi x)-2e^{-2x}B\pi cos(\pi x)+$$ $$ \displaystyle 4e^{-2x}Acos(\pi x)+4e^{-2x}Bsin(\pi x)-4e^{-2x}A\pi sin(\pi x)+4e^{-2x}B\pi cos(\pi x)-$$ $$ \displaystyle 8e^{-2x}Acos(\pi x)-8e^{-2x}Bsin(\pi x)-e^{-2x}A\pi ^{2}cos(\pi x)+e^{-2x}B\pi ^{2}sin(\pi x)+4e^{-2x}Acos(\pi x)+4e^{-2x}Bsin(\pi x) = 0 $$

Upon further inspection, all the terms on the left side of the equation cancel out to 0.

Author
This problem was solved and uploaded by: (Will Knapper)

This problem was proofread by: David Herrick

Question
K 2011 p.59 pbs.5,6. Find a general solution. Check your answer by substitution.

Problem 5

 * $$ y'' + 2\pi y' + \pi^2y = 0 \ $$

Solution
Determine the characteristic equation and the nature of the determinate:
 * $$ y = e^{-\lambda x} \ $$
 * $$ y' = -\lambda e^{-\lambda x} \ $$
 * $$ y'' = \lambda^2 e^{-\lambda x} \ $$

Characteristic Equation:
 * $$ \lambda^2 - 2\pi \lambda + \pi^2\lambda = 0 \ $$
 * $$ a^2 - 4b = (2\pi)^2 - 4(\pi^2) = 4\pi^2 - 4\pi^2 = 0 \ $$

0 indicates a real double root, therefore:
 * $$ y_g = e^{-\frac{a}{2}x} (C_1 + C_2 x) \ $$

$$ a = 2\pi \ $$ from the original equation. Therefore: $$ y_g = e^{-\pi x} (C_1 + C_2 x) \ $$
 * $$ y'_g = e^{-\pi x} (C_2 - \pi C_1 - \pi C_2 x) \ $$
 * $$ y''_g = e^{-\pi x} (C_1 \pi^2 + C_2 \pi^2 x - 2C_2 \pi) \ $$

Substituting into the original equation gives:
 * $$ e^{-\pi x} [(C_1 \pi^2 + C_2 \pi^2 x - 2C_2 \pi) + 2\pi (C_2 - \pi C_1 - \pi C_2 x) + \pi^2 (C_1 + C_2 x)] = 0 \ $$
 * $$ = e^{-\pi x} [C_1\pi^2 + C_2\pi^2 x - 2C_2\pi + 2C_2\pi -2C_1\pi^2 - 2C_2\pi^2 x +C_1\pi^2 + C_2\pi^2 x] \ $$

Canceling gives 0 = 0, confirming the general solution found above.

Author
This problem was solved and uploaded by: (John North)

This problem was proofread by: David Herrick

Problem 5

 * $$ 10y'' - 32y' +25.6y = 0 \ $$

Solution
Reduce original equation so that the first coefficient is 1:
 * $$ y'' - 3.2y' + 2.56y = 0 \ $$

Determine the characteristic equation and the nature of the determinate:
 * $$ y = e^{-\lambda x} \ $$
 * $$ y' = -\lambda e^{-\lambda x} \ $$
 * $$ y'' = \lambda^2 e^{-\lambda x} \ $$

Characteristic Equation:
 * $$ \lambda^2 - 3.2\lambda + 2.56 = 0 \ $$
 * $$ a^2 - 4b = 3.2^2 - 4(2.56) = 10.24 - 10.24 = 0 \ $$

0 indicates a real double root, therefore:
 * $$ y_g = e^{-\frac{a}{2}x} (C_1 + C_2 x) \ $$

$$ a = \frac{-(-3.2)}{2} = 1.6 $$ from the original equation. Therefore: $$ y_g = e^{1.6x} (C_1 + C_2 x) \ $$
 * $$ y'_g = e^{1.6x} (1.6C_1 + 1.6C_2 x + C_2) \ $$
 * $$ y''_g = e^{1.6x} (2.56C_1 + 2.56C_2 x + 3.2C_2) \ $$

Substituting into the original equation gives:
 * $$ e^{1.6x} [(2.56C_1 + 2.56C_2 x + 3.2C_2) -3.2(1.6C_1 + 1.6C_2 x + C_2) + 2.56(C_1 + C_2 x)] = 0 \ $$
 * $$ = e^{1.6x} [(2.56C_1 + 2.56C_2 x + 3.2C_2 - 5.12C_1 - 5.12C_2 x - 3.2C_2 + 2.56C_1 + 2.56C_2 x] = 0 \ $$

Canceling gives 0 = 0, confirming that general solution found above.

Author
This problem was solved and uploaded by: (John North)

This problem was proofread by: David Herrick

Question
Problems 16 and 17 from pg. 59 of the textbook. The two questions have the same instructions: Find an ODE of the form $$ \displaystyle {y}'' + a{y}' + by = 0 $$ for the given basis.

Problem 16
$$ \displaystyle 16. \; e^{2.6x}, e^{-4.3x} $$ This basis implies that there are two distinct, real roots: 2.6 and -4.3. We put these roots into the characteristic equation to get: $$ \displaystyle (\lambda - 2.6)(\lambda + 4.3)=0 \rightarrow \lambda ^{2} + 1.7\lambda -11.18=0 $$ Then, you convert the characteristic equation to a regular ODE: $$ \displaystyle {y}'' + 1.7{y}' -11.18y = 0 $$

Problem 17
$$ \displaystyle 17. \; e^{-\sqrt{5}x} \;, xe^{-\sqrt{5}x} $$ A basis of this form implies that the solution is of the form: $$ \displaystyle y = (c_{1} + c_{2}x)e^{-ax/2} $$ The real double root can be found by this equation: $$ \displaystyle \lambda = -a/2 = -\sqrt{5} $$ That means that the characteristic equation can be found by solving: $$ \displaystyle (\lambda + \sqrt{5} )^{2} = \lambda ^{2} + 2 \sqrt{5}\lambda + 5 $$ Converting the characteristic equation to the ODE, we get: $$ \displaystyle {y}'' + 2 \sqrt{5} {y}' + 5y = 0 $$

Author
This problem was solved and uploaded by: (Will Knapper) This problem was proofread by (Josh House)

Question
Realize spring-dashpot-mass system in series as shown in Fig. p.1-4 with the similar characteristic as in (3)p.5-5, but with the double real root $$ \displaystyle \lambda=-3\ $$, i.e., find values for parameters k,c,m.

Solution
First we put the double root into the characteristic equation: $$ \displaystyle (\lambda+3)^2 =0 $$ $$ \displaystyle \lambda^2 + 6\lambda + 9 = 0 $$ The corresponding L2-ODE is: $$ \displaystyle {y}'' + 6{y}' + 9y = 0 $$ The equation of motion of the spring-dashpot-mass system is: $$ \displaystyle m \left(y_k'' + \frac{k}{c}y_k' \right) + ky_k = 0 $$ or: $$ \displaystyle my_k'' + m \frac{k}{c}y_k' +ky_k = 0 $$ Solving for the parameters k,c,m: $$ \displaystyle m = 1 $$ $$ \displaystyle m \frac{k}{c} = 6 $$ $$ \displaystyle k = 9 $$ We get: $$ \displaystyle m = 1 $$ $$ \displaystyle k = 9 $$ $$ \displaystyle c = \frac{3}{2} $$

Author
This problem was solved and uploaded by: (Radina Dikova)

This problem was proofread by: Mike Wallace

Question
Develop the McLauren series (Taylor series at t=0) for e^t, cos(t), and sin(t).

Solution
The general form of the Taylor series expansion at point a is
 * $$ f(a) + \frac{f'(a)} {1!} (x-a) + \frac{f(a)} {2!} (x-a)^2 + \frac{f'(a)} {3!} (x-a)^3 + ... $$

for $$ e^t $$ at t=0:
 * $$ e^0 + \frac{e^0} {1!} (x) + \frac{e^0} {2!} (x)^2 + \frac{e^0} {3!} (x)^3 + ... $$

$$ = 1 + x + \frac{x^2} {2!} + \frac{x^3} {3!} + ... $$ for $$ \cos(t) $$ at t=0:
 * $$ \cos(0) + \frac{-\sin(0)} {1!} (x) + \frac{-\cos(0)} {2!} (x)^2 + \frac{sin(0)} {3!} (x)^3 + \frac{cos(0)} {4!} (x)^4 + ... $$

$$ = 1 - \frac{x^2} {2!} + \frac{x^4} {4!} + ... $$ for $$ \sin(t) $$ at t=0:
 * $$ \sin(0) + \frac{cos(0)} {1!} (x) + \frac{-\sin(0)} {2!} (x)^2 + \frac{-cos(0)} {3!} (x)^3 + \frac{\sin(x)} {4!} (x)^4 + \frac{\cos(0)} {5!} (x)^5 + ... $$

$$ = x - \frac{x^3} {3!} + \frac{x^5} {5!} + ... $$

Author
This problem was solved and uploaded by: (John North) This problem was checked by: William Knapper

Question
Find a general solution. Check your answer by substitution.

8. $$\displaystyle y^{''}+y^{'}+3.25y=0 $$

15. $$\displaystyle y^{''}+0.54y^{'}+(0.0729+\pi)y=0 $$

Solution
8. $$ \displaystyle y^{''}+y^{'}+3.25y =0 $$

Writing the characteristic equation:
 * $$ \displaystyle \lambda^{2}+\lambda+3.25 =0 $$

Which is now in the form:
 * $$ \displaystyle \lambda ^{2}+a\lambda +b=0 $$

Solving for the two roots:
 * $$ \displaystyle \lambda _{1} = \frac{1}{2}(-a+\sqrt{a^{2}-4b}), \lambda _{2}=\frac{1}{2}(-a-\sqrt{a^{2}-4b}) $$

We will find the discriminant to be less than 0, leading to complex conjugate roots:
 * $$\displaystyle a^{2}-4b=1^{2}-4(3.25)=-12 $$

Which leads to a solution of the form:
 * $$ \displaystyle y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x)) $$

Where $$ \displaystyle \omega^{2} = b-\frac{1}{4}a^{2} $$, therefore:
 * $$ \displaystyle \omega^{2} = 3.25-\frac{1}{4}(1)^{2} =3 \rightarrow \omega=\sqrt{3} $$

Giving us a solution of: $$ \displaystyle y=e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)) $$

Checking our solution:
 * $$\displaystyle y=e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)) $$
 * $$\displaystyle y^{'}=-\frac{1}{2}e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))+e^{-x/2}(-\sqrt{3}sin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x)) $$
 * $$\displaystyle y^{''}=-\frac{1}{4}e^{-x/2}(Acos(\sqrt{3}x)+Bsin(\sqrt{3}x))-\frac{1}{2}e^{-x/2}(-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x))-\frac{1}{2}e^{-x/2}(-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x))+e^{-x/2}(-3Acos(\sqrt{3}x)-3Bsin(\sqrt{3}x)) $$

Plugging all of this into the original differential equation and combining like terms gives us:
 * $$\displaystyle [\frac{1}{4}-3-\frac{1}{2}+3.25]e^{-x/2}Acos(\sqrt{3}x) $$
 * $$\displaystyle [\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}-\sqrt{3}]e^{-x/2}Asin(\sqrt{3}x) $$
 * $$\displaystyle [-\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}+\sqrt{3}]e^{-x/2}Bcos(\sqrt{3}x) $$
 * $$\displaystyle [\frac{1}{4}-3-\frac{1}{2}+3.25]e^{-x/2}Bsin(\sqrt{3}x) $$

The terms in the brackets all add up to zero, thus verifying we have a correct solution to our problem.

15. $$ \displaystyle y^{''}+0.54y{'}+(0.0729+\pi )y =0 $$

Writing the characteristic equation:
 * $$ \displaystyle \lambda^{2}+0.54\lambda+(0.0729+\pi ) =0 $$

Which is now in the form:
 * $$ \displaystyle \lambda ^{2}+a\lambda +b=0 $$

Solving for the two roots:
 * $$ \displaystyle \lambda _{1} = \frac{1}{2}(-a+\sqrt{a^{2}-4b}), \lambda _{2}=\frac{1}{2}(-a-\sqrt{a^{2}-4b}) $$

We will find the discriminant to be less than 0, leading to complex conjugate roots:
 * $$\displaystyle a^{2}-4b=0.54^{2}-4(0.0729+\pi)=-4\pi $$

Which leads to a solution of the form:
 * $$ \displaystyle y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x)) $$

Where $$ \displaystyle \omega^{2} = b-\frac{1}{4}a^{2} $$, therefore:
 * $$ \displaystyle \omega^{2} = (0.0729-\pi)-\frac{1}{4}(0.54)^{2} =\pi \rightarrow \omega=\sqrt{\pi} $$

Giving us a solution of: $$ \displaystyle y=e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x)) $$

Checking our solution:
 * $$\displaystyle y=e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x)) $$
 * $$\displaystyle y^{'}=-0.27e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x))+e^{-0.27x}(-\sqrt{\pi}Asin(\sqrt{\pi}x)+\sqrt{\pi}Bcos(\sqrt{\pi}x)) $$
 * $$\displaystyle y^{''}=0.0729e^{-0.27x}(Acos(\sqrt{\pi}x)+Bsin(\sqrt{\pi}x))-0.27e^{-0.27x}(-\sqrt{\pi}Asin(\sqrt{\pi}x)+\sqrt{\pi}Bcos(\sqrt{\pi}x))-0.27e^{-0.27x}(-\sqrt{\pi}Asin(\sqrt{\pi}x)+\sqrt{\pi}Bcos(\sqrt{\pi}x))+e^{-0.27x}(-\pi Acos(\sqrt{\pi}x)-\pi Bsin(\sqrt{\pi}x)) $$

Plugging all of this into the original differential equation and combining like terms gives us:
 * $$\displaystyle [0.0729-\pi-0.1458+0.0729+\pi]e^{-0.27x}Acos\sqrt{\pi}x $$
 * $$\displaystyle [0.27\sqrt{\pi}+0.27\sqrt{\pi}-0.54\sqrt{\pi}]e^{-0.27x}Asin\sqrt{\pi}x $$
 * $$\displaystyle [-0.27\sqrt{\pi}-0.27\sqrt{\pi}+0.54\sqrt{\pi}]e^{-0.27x}Bcos\sqrt{\pi}x $$
 * $$\displaystyle [0.0729-\pi-0.1458+0.0729+\pi]e^{-0.27x}Bsin\sqrt{\pi}x $$

The terms in the brackets all add up to zero, thus verifying we have a correct solution to our problem.

Author
This problem was solved and uploaded by: (Josh House) This problem was proofread by: (Radina Dikova)

Question
For the given ODE, find and plot the solution for the L2-ODE-CC corresponding to

$$ \displaystyle \ \lambda^2 + 4\lambda + 13 = 0 \ $$

In another figure superpose this figure, Fig. from R2.6 p.5-6, and the Fig. from R2-1 p.3-7.

This corresponds to the L2-ODE-CC in standard form

$$ \displaystyle \ y'' + 4y' + 13y = r(x) \ $$

Given
Initial conditions

$$ \displaystyle \ y(0) = 1, y'(0) = 0 \ $$

Where there is no excitation

$$ \displaystyle \ r(x) = 0 \ $$

Solution
Solving the quadratic equation for $$ \displaystyle \ \lambda \ $$ determines the roots of the ODE

$$ \displaystyle \ \lambda_1 = \frac{-b + \sqrt{b^2-(4ac)}}{2a} \! = \frac{-4 + \sqrt{(4)^2-4(1)(13)}}{2(1)} \!$$

$$ \displaystyle \ \lambda_2 = \frac{-b - \sqrt{b^2-(4ac)}}{2a} \! = \frac{-4 - \sqrt{(4)^2-4(1)(13)}}{2(1)} \!$$

The roots of the quadratic equation come out to be:

$$ \displaystyle \ \lambda = (-2 \pm 3i) \ $$

The general homogeneous solution of the ODE will resemble the following form.

$$ \displaystyle \ y = C_1e^{a_1x}cosb_1x + C_2e^{a_2x}sinb_2x \ $$

Plugging in the roots into the equation we get

$$ \displaystyle \ y = C_1e^{-2x}cos3x + C_2e^{-2x}sin3x \ $$

Implementing initial conditions where: $$ \displaystyle \ y(0) = 1,   y'(0) = 0 \ $$

$$ \displaystyle \ y(0) = C_1e^{-2(0)}cos3(0) + C_2e^{-2(0)}sin3(0)= 1 \ $$

$$ \displaystyle \ y'(0)= -2C_1e^{-2x}cos3x - 3C_1e^{-2x}sin3x - 2C_2e^{-2x}sin3x + 3C_2e^{-2x}cos3x = 0 \ $$

$$ \displaystyle \ y'(0)= -2(1)e^{-2(0)}cos3(0) + 3C_2e^{-2(0)}cos3(0) = -2 + 3C_2 = 0; \ $$

So from our initial conditions the constants are determined to be

$$ \displaystyle \ C_1 = 1, C_2 = 2/3 \ $$

The final solution becomes

$$ \displaystyle \ y(x) = e^{-2x}cos3x + \frac{2}{3}e^{-2x}sin3x \ $$

Author
This problem was solved and uploaded by: Mike Wallace This problem was proofread by John North

Contribution Summary
Problem 2 was solved and Problems 1, 3, and 4 were proofread by David Herrick 15:56, 6 February 2012 (UTC)

Problem 8 was solved and Problem 5 was proofread by Josh House 23:41, 7 February 2012 (UTC)

Problem 6 was solved and Problem 8 was proofread by Radina Dikova 16:26, 7 February 2012 (UTC)

Problem 9 was solved and Problem 6 was proofread by Mike Wallace 23:35, 7 February 2012 (UTC)

Problems 3 and 5 were solved and Problems 2 and 7 were proofread by William Knapper 05:04, 8 February 2012 (UTC)

Problem 1 was solved by Derik Bell 15:44, 8 February 2012 (UTC)

Problems 4 and 7 were solved and Problem 9 was proofread by John North 15:47, 8 February 2012 (UTC)


text:
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text:
Computer Architecture Lab/SS2012

Project page for the course in summer 2012.

Group 1 JDR
Klaas Kelchtermans

Jose Bosca Moya

Diego Panadero Roser

Juan Ricardo Rios Rivas

Group 2
Jinghua Wang

Elakkiya Ellavarason

Nis Schmidt

Group 3
Graeme Best

Oscar Rahnama

Wojciech Pawlak

Matt Birman

Network on a Chip
Project Proposal

Open-source project

Group 4
Andreas B. Svendsen

Attila Sükösd

Kasper Grue Understrup

Rasmus T. Karstensen

title: Computer Architecture Lab/SS2012


text:
User:Egm4313.s12.team11.sheider/R1.5: K 2011 p.59 pb. 4

Problem: Find a general solution to the given ODE. Check your answer by substituting into the original equation.

Given:

$${y}''+ 4{y}'+(\pi^2+4)y = 0\!$$

Solution:

The characteristic equation of this ODE is therefore:

$$\lambda^2 + a\lambda + b = \lambda^2 + 4\lambda  + (\pi^2+4 ) = 0\!$$

Evaluating the discriminant:

$$ a^2 - 4b = 4^2 - 4(\pi^2 + 4) = -4\pi^2 < 0\!$$

Therefore the equation has two complex conjugate roots and a general homogenous solution of the form:

$$ y = e^{-ax/2}(Acos(\omega x) + Bsin(\omega x))\!$$

Where: $$ \omega = \sqrt{b - \frac{1}{4}a^2} = \sqrt{\pi^2 + 4 - \frac{1}{4}(4)^2} = \sqrt{\pi^2} = \pi\!$$

And finally we find the general homogenous solution:

$$ y = e^{-2x}(Acos(\pi x )+ Bsin(\pi x))\!$$

Checking:

We found that:

$$ y = e^{-2x}(Acos(\pi x )+ Bsin(\pi x))\!$$

Differentiating $$ y\!$$ to obtain $${y}'\!$$ and $${y}''\!$$ respectively:

$$ {y}' = -2e^{-2x}(Acos(\pi x)+Bsin(\pi x)) + e^{-2x}(\pi Acos(\pi x)-\pi Bsin(\pi x))\!$$

$$ {y}' = -2e^{-2x}(Acos(\pi x)+Bsin(\pi x)) + \pi e^{-2x}( Acos(\pi x)- Bsin(\pi x))\!$$

$$ {y}'' = 4e^{-2x}(Acos(\pi x)+Bsin(\pi x)) -2 \pi e^{-2x}(Acos(\pi x) - Bsin(\pi x))- 2 \pi e^{-2x}(Acos(\pi x) - Bsin(\pi x)) -\pi^2 e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!$$

$$ {y}'' = 4e^{-2x}(Acos(\pi x)+Bsin(\pi x)) -4 \pi e^{-2x}(Acos(\pi x) - Bsin(\pi x)) -\pi^2 e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!$$

Substituting these equations into the original ODE yields:

$$ 4e^{-2x}(Acos(\pi x)+Bsin(\pi x)) -4 \pi e^{-2x}(Acos(\pi x) - Bsin(\pi x)) -\pi^2 e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!$$

$$ + 4(-2e^{-2x}(Acos(\pi x)+Bsin(\pi x)) + \pi e^{-2x}( Acos(\pi x)- Bsin(\pi x))) + (\pi^2 + 4)(e^{-2x}(Acos(\pi x )+ Bsin(\pi x))) = 0\!$$

$$ 4e^{-2x}(Acos(\pi x)+Bsin(\pi x)) -4 \pi e^{-2x}(Acos(\pi x) - Bsin(\pi x)) -\pi^2 e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!$$

$$ -8e^{-2x}(Acos(\pi x)+Bsin(\pi x)) + 4\pi e^{-2x}( Acos(\pi x)- Bsin(\pi x)) + \pi^2 e^{-2x}(Acos(\pi x )+ Bsin(\pi x)) + 4e^{-2x}(Acos(\pi x )+ Bsin(\pi x)) = 0\!$$

$$ (4-8+4)e^{-2x}(Acos(\pi x)+Bsin(\pi x)) + (-4+4) \pi e^{-2x}(Acos(\pi x) - Bsin(\pi x)) + (\pi^2-\pi^2) e^{-2x}(Acos(\pi x)+Bsin(\pi x)) = 0\!$$

$$ 0 \equiv 0 \!$$

Therefore this solution is correct.


text:
User:Egm4313.s12.team11.sheider/R1.5: K 2011 p.59 pb. 5

Problem: Find a general solution to the given ODE. Check your answer by substituting into the original equation.

Given:

$${y}''+ 2\pi{y}'+\pi^2y = 0\!$$

Solution:

The characteristic equation of this ODE is therefore:

$$\lambda^2 + a\lambda + b = \lambda^2 + 2\pi\lambda  + \pi^2= 0\!$$

Evaluating the discriminant:

$$ a^2 - 4b = (2\pi)^2 - 4(\pi^2) = 4\pi^2 - 4\pi^2 = 0\!$$

Therefore the equation has a real double root and a general homogenous solution of the form:

$$ y = (c_{1}+c_{2}x)e^{-ax/2}\!$$

And finally we find the general homogenous solution:

$$ y = (c_{1}+c_{2}x)e^{-\pi x}\!$$

Checking:

We found that:

$$ y = (c_{1}+c_{2}x)e^{-\pi x}\!$$

Differentiating $$ y\!$$ to obtain $${y}'\!$$ and $${y}''\!$$ respectively:

$$ {y}' = c_{2}e^{-\pi x} - \pi (c_{1}+c_{2}x)e^{-\pi x}\!$$

And,

$$ {y}'' = -\pi c_{2} e^{-\pi x}- \pi c_{2}e^{-\pi x} + \pi^2(c_{1}+c_{2}x)e^{-\pi x}\!$$

$$ {y}'' = -2\pi c_{2} e^{-\pi x} + \pi^2(c_{1}+c_{2}x)e^{-\pi x}\!$$

Substituting these equations into the original ODE yields:

$$ -2\pi c_{2} e^{-\pi x} + \pi^2(c_{1}+c_{2}x)e^{-\pi x} + 2\pi(c_{2}e^{-\pi x} - \pi (c_{1}+c_{2}x)e^{-\pi x}) + \pi^2((c_{1}+c_{2}x)e^{-\pi x}) = 0\!$$

$$ -2\pi c_{2} e^{-\pi x} + \pi^2(c_{1}+c_{2}x)e^{-\pi x} + 2\pi c_{2}e^{-\pi x} - 2\pi^2 (c_{1}+c_{2}x)e^{-\pi x} + \pi^2(c_{1}+c_{2}x)e^{-\pi x} = 0\!$$

$$ (-2\pi + 2\pi) c_{2} e^{-\pi x} + (\pi^2 - 2\pi^2 + \pi^2)(c_{1}+c_{2}x)e^{-\pi x} = 0\!$$

$$ 0 \equiv 0 \!$$

Therefore this solution is correct.


text:
User talk:Bratzjoy

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title: User talk:Bratzjoy


text:
User:Egm4313.s12.team18.roberts.jr

= Josh Roberts =

Useful Links
Course Wiki Course Website Team 18 Writing Tools Good model reports Report Table Exam Tips Email Archive

Problem
Find a general solution. Check your answer by substitution.

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$$  \displaystyle y''+4y'+(\pi^2+4)y=0 $$     (1)
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Solution
Note This is a homogenous linear 2nd order ordinary differential equation (ODE) with constant coefficients. Trial Solution:
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$$  \displaystyle y=e^{\lambda x} $$ (2)
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$$  \displaystyle y'=\lambda e^{\lambda x} $$ (3)
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$$  \displaystyle y''=\lambda^2 e^{\lambda x} $$ (4) Plugging (2),(3), and (4) into (1)
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$$  \displaystyle (\lambda^2 + 4\lambda + (\pi^2+4))e^{\lambda x} = 0 $$     (5)
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The characteristic equation is
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$$  \displaystyle \lambda^2 + 4\lambda + (\pi^2+4) = 0 $$     (6)
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Note The coefficients of this quadratic equation are
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$$  \displaystyle a=1 $$
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$$  \displaystyle b=4 $$
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$$  \displaystyle c=\pi^2+4 $$
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The discriminant can be found as follows
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$$  \displaystyle \Delta = b^2 - 4ac $$     (7)
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$$  \displaystyle \Delta = 4^2 - 4(1)(\pi^2+4) $$
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Note The discriminant is negative. Therefore, Eq (6) has complex roots of the form

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$$  \displaystyle \lambda = -1/2a \pm i\omega $$     (8) where
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$$  \displaystyle \omega=\sqrt{c- b^2/4} $$     (9)  Now The general solution to the ODE is
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$$  \displaystyle y(x)=e^{-bx/2} (Acos\omega x + Bsin\omega x) $$ (10) Plugging known quantities into (9) and (10)
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$$  \displaystyle y(x)=e^{-2x} (Acos\sqrt{(\pi^2+4)-4}\,\,x + Bsin\sqrt{(\pi^2+4)-4}\,\,x) $$
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$$  \displaystyle y(x)=e^{-2x} (Acos\pi\,x + Bsin\pi\,x) $$
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Problem
Given the two roots and the initial conditions:
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$$  \displaystyle \lambda_1=-2\qquad\qquad\lambda_2 = +5 $$
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$$  \displaystyle y(0)=1\qquad\qquad y'(0)=0 $$ Find the non-homogeneous L2-ODE-CC in standard form and the solution in terms of the initial conditions and the general excitation $$r(x)=0$$.
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Consider no excitation:
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$$  \displaystyle r(x)=0 $$
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Plot the solution.

Generate 3 non-standard (and non-homogeneous) L2-ODE-CC that admit the 2 values in (3a) p.3-7 as the 2 roots of the corresponding characteristic equation.

Solution

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$$  \displaystyle (\lambda-\lambda_1)(\lambda-\lambda_2) $$     (2.1.1)
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Now Insert the given roots into (1)
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$$  \displaystyle (\lambda-(-2))(\lambda-5) $$     (2.1.2) By the FOIL method,  (2) gives the characteristic equation for the unknown L2-ODE-CC.
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$$  \displaystyle \lambda^2-3\lambda-10=0 $$     (2.1.3)
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From this characteristic equation, (3), we can find the non-homogeneous L2-ODE-CC.
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$$  \displaystyle y''-3y'-10=r(x) $$     (2.1.4)
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Now It is known from the given roots that the homogeneous L2-ODE-CC has the solution
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$$  \displaystyle y_H(x)=C_1e^{-2x} + C_2e^{5x} $$     (2.1.5)
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Therefore, the overall solution is
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$$  \displaystyle y(x)=y_H(x)+y_P(x) $$     (2.1.6)
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$$  \displaystyle y(x)=C_1e^{-2x}+C_2e^{5x}+y_P(x) $$     (2.1.7) where $$y_P(x)$$ is the particular solution.
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Now Using the given initial conditions
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$$  \displaystyle y(0)=1\qquad y'(0)=0 $$
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$$  \displaystyle y(0)=C_1+C_2+y_P(0)=1 $$     (2.1.8) Differentiating $$y(x)$$ and then inserting the initial condition.
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$$  \displaystyle y'(x)=-2C_1e^{-2x}+5C_2e^{5x}+y_P'(x) $$     (2.1.9)
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$$  \displaystyle y'(0)=-2C_1+5C_2+y_P'(0)=0 $$     (2.1.10)  Now Multiplying (2.1.8) by 2 and adding it to (2.1.10)
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$$  \displaystyle (2C_1+2_C2+2y_P(0)=2) + (-2C_1+5C_2+y_P'(0)=0) $$
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$$  \displaystyle C_2=\frac{2}{7}-\frac{2}{7}y_P(0)-\frac{1}{7}y_P'(0) $$     (2.1.11) By inserting (2.1.11) into (2.1.8) and combining like terms $$C_1$$ is also found.
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$$  \displaystyle C_1=\frac{5}{7}-\frac{5}{7}y_P(0)+\frac{1}{7}y_P'(0) $$     (2.1.12)  Now The overall solution $$y(x)$$ is given in terms of the initial conditions.
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$$  \displaystyle y(x)=\frac{1}{7}[(5-5y_P(0)+y_P'(0))e^{-2x}+(2-2y_P(0)-y_P'(0))e^{5x}] $$     (2.1.13)
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Now Considering no excitation

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$$  \displaystyle r(x)=0 $$ Which causes
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$$  \displaystyle y_P(x)=0\,\,,\,\,y_P'(x)=0 $$ So
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$$  \displaystyle C_1=\frac{5}{7} $$
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 (2.1.14)
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$$  \displaystyle C_2=\frac{2}{7} $$ The homogeneous solution to the L2-ODE-CC is therefore
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 (2.1.15)
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$$  \displaystyle y_H(x)=\frac{5}{7}e^{-2x}+\frac{2}{7}e^{5x} $$
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 (2.1.16)
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This solution will now be plotted using MATLAB.

Now Generate 3 non-standard and non-homogeneous L2-ODE-CC that satisfy the roots given in the problem statement.
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$$  \displaystyle 2y''-6y'-20=r(x) $$
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 (2.1.17)
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$$  \displaystyle \frac{3}{2}y''-\frac{9}{2}y'-15y=r(x) $$
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 (2.1.18)
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$$  \displaystyle 3y''-9y'-30y=r(x) $$
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 (2.1.19)
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Exam 1 Material
Date: Tue, Feb 28, 2012

Study Guide
Sec 2.1 - Homogeneous L2-ODEs
 * A 2nd order ODE can be called linear if it can be written as
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$$  \displaystyle y'' + p(x)y'+q(x)y = r(x) $$     (1)
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 * Linear in y and its derivatives.
 * This equation is in standard form.

If $$r(x)\equiv 0$$, then (1) reduces to
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$$  \displaystyle y'' + p(x)y'+q(x)y = 0 $$     (2)
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 * }
 * which is called a homogeneous L2-ODE.

If $$r(x)\not\equiv 0 $$, then (1) is called non homogeneous.
 * Superposition Principle (Linearity Principle)
 * For a homogeneous linear ODE (2), any linear combination of two solutions on an open interval I is again a solution
 * of (2) on I. In particular, for such an equation, sums and constant multiples of solutions are again solutions.

Sec 2.2 - Homogeneous Linear ODEs with Constant Coefficients These equations take the form
 * Initial Value Problems
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$$  \displaystyle y'' + ay'+by = 0 $$     (3) Homogeneous solutions to these equations take the form
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$$  \displaystyle y_h'' + ay_h'+by_h = 0 $$     (4) and can be solved by the method of trial solution, also known as the method of undetermined coefficients. The trial solution is
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$$  \displaystyle y = e^{\lambda x} $$ (5) By inserting (5) and its 1st and 2nd derivatives into (4), one can obtain the characteristic equation
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$$  \displaystyle \lambda^2 + a\lambda + b = 0 $$     (6) Finding the roots to the characteristic equation will allow for the homogeneous solution to be found. We know that for a quadratic equation such as (6) that there are 3 kinds of roots, depending on the sign of the discriminant $$a^2 - 4b$$,namely.
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 * (Case 1)  Two real roots if $$a^2 - 4b > 0$$
 * (Case 2)  A real double root if $$a^2 - 4b = 0$$
 * (Case 3)  Complex conjugate roots if $$a^2 - 4b < 0$$

There will be two distinct roots $$\lambda_1 and  \lambda_2$$ Therefore, the solution to (4) will be
 * Case 1
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$$  \displaystyle y = C_1e^{\lambda_1 x} + C_2e^{\lambda_2 x} $$ (7) If the discriminant is zero it is easily seen that we only get one root (double root), hence only one solution.
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 * Case 2
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$$  \displaystyle y = (C_1 + C_2x)e^{-ax/2} $$     (8) If the discriminant is negative, we know that the roots will be complex and of the form
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 * Case 3
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$$  \displaystyle \lambda = -(1/2)a \pm i\omega $$ A basis of real solutions can be obtained that is
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$$  \displaystyle y_1 = e^{-ax/2}cos{\omega x} $$
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$$  \displaystyle y_2 = e^{-ax/2}sin{\omega x} $$ where
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$$  \displaystyle \omega^2 = b - (1/4)a^2 $$ Finally, the general solution is
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$$  \displaystyle y = e^{-ax/2}(Acos{\omega x} + Bsin{\omega x}) $$     (9)
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Exam 2 Material
Date: Tue, Apr 24, 2012

title: User:Egm4313.s12.team18.roberts.jr


text:
User:Egm4313.s12.team18.alvinb

Team 18!

title: User:Egm4313.s12.team18.alvinb


text:
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