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Clock and Data Recovery/Structures and types of CDRs

At the receiving end of a data transmission link, the received signal is amplified, filtered and equalized. Then a “slicer” circuit reshapes it and retains just the level transitions and the levels that represent the nominal values of the received signal in that interval between two consecutive transitions. The CDR processes the “sliced” signal
 * to extract the clock signal embedded in its transitions (clock recovery) and
 * to sample and retime the pulses of the “sliced” signal (data recovery).

Clock recover circuits include:
 * the phase locked loop architecture (PLL) -- the most common method of clock recovery
 * the synchronous oscillator—much less common, but which may claim, in certain cases, some advantages over PLL
 * the gated oscillator, that resembles the synchronous oscillator, and that is useful when very fast acquisition is needed and other performances can be traded off in return for it.

Many good books are available, with theory and practical examples, like.

It is useful however, in order to get the best out of them, to have the ability to recognize in each case the fundamental architecture, that is the "Control Systems" block diagram that corresponds to the actual circuit.

Importance of good reference models
The function of a CDR is a relatively simple one. The architectures possible for it are, accordingly, just a few and simple ones (actually just three!). It is nonetheless important to have a good knowledge of those architectures and a good understanding of their mathematical descriptions because these models are the best tools for the engineer that must deal with CDRs.

 * Starting from the definition and specification of (a communication system and of its) CDR(s),
 * and all the way through all the different engineering tasks that logically follow
 * (like design, verification, validation, manufacturing tests, failure analysis, system operation and maintenance),
 * those models can be an invaluable reference for the engineer.
 * He will need them to imagine, specify, design, check, measure and interpret the behavior of a CDR.

The actual implementation of the CDR may differ from the neat, simple structure that the model depicts. Complex digital blocks, DLLs, DSPs may render the analogy difficult to detect, but the fundamental signals and operation of the CDR can not differ. Yielding to the temptation of forgetting the models is a very risky and error prone short cut.

The essential signals and blocks of the architecture must be clearly identified in the actual system. The use of the reference model will then be the best way to make sure that all aspects of the CDR operation are identified and taken into consideration.

The CDR is always designed with the architecture of a PLL (with the obvious addition of the regeneration block, where the received pulses are re-sampled with the local clock). Let’s study the PLL, that is the essential part. It should be kept in mind that such PLL will be specialized for application inside a CDR. The reference model of a PLL in a CDR: unity feedback, first or second order, type 1 or 2. The PLLs inside CDRs are in all cases of the unity feedback kind. The input of the circuit is the phase of a reference signal (a clock or a serial data signal) and the output is the phase of a signal (a serial data stream or a simple clock). The output is locked -as much as the circuit can- to the input signal. The input signal is contrasted with the output signal in a phase comparator, whose output is the error signal. The error signal is processed and then used to control another circuit block that produces the output clock signal.

It may be remarked that the output clock is always phase locked to the input signal, and that it is used to regenerate the input signal in the slave CDRs. In the phase aligner CDRs it is the local clock that regenerates (a phase aligned version of) the received signal.

Essential parts of a CDR
It is important to identify the essential parts (listed below) of the CDR system but also to identify where the received signal and the local clock fit in the architecture. Either one (the received signal or the local clock) may act as input, while the other would simply act as an internal input signal inside the block called “Controlled element” (the block that generates the output signal of the PLL).

The received signal acts as reference input for the PLL when the PLL function is to generate a clock slaved to the received signal itself. The local clock acts instead as reference signal for the PLL when the PLL function is simply to “phase align” the received signal to it, in the cases where the received signal timing is derived from the local clock itself (following a short loop inside a unique clock domain where the local clock is master).

The list of the parts that shall be clearly identified in the CDR are:
 * 1) the phase information (carried by the level transitions) of the received signal
 * 2) the phase information (carried by the level transitions) of the local clock
 * 3) the phase comparator that measures the relative phase of the local clock with respect to the phase of (a signal related to) the received signal
 * 4) an integration block (1/s in the language of Control Systems) or an accumulator if the implementation is of the discrete time type. It makes the control loop able to squeeze down to zero the steady state frequency error (and to a negligible value the corresponding phase error)
 * 5) the regeneration of (a signal related to) the received signal by the local clock.

Order
The order of a control loop (causal, linear and time invariant in our models of CDRs) is the order of the differential equation that describes it. In the language of control systems, the order is the number of poles of the (open or closed loop) system transfer function.
 * 1st order systems are unconditionally stable, are characterised by one parameter only and represent a good reference model for the simplest CDR PLLs.
 * 2nd order systems are unconditionally stable, are characterised by two parameters and represent a good reference model for the other, more complex, CDR PLLs.
 * 3rd and higher order systems are not of practical use in the study of CDRs. They may be unstable in some conditions and, moreover, they are more complex to use but do not offer a better behavior for CDR use than the 2nd order systems. ( Some complexity beyond the simple 2nd order system can be useful to model more accurately some parasitic effects of a CDR circuit, but not to design nor to model performances within the range of functionality of the CDR).

Type
The type of a control loop is the number of poles of the open loop transfer function in the origin (that is, how many times the factor 1/s appears in the open loop transfer function).

The type of a loop tells how well the loop itself is able to track a deviation of its input signal from the nominal value.

(A CDR can operate with a small phase (= sampling time) error without deterioration of performances, provided the error is small enough: a few degrees of jitter around the optimum sampling time do not significantly deteriorate the bit error rate!) (As type 2 implies by definition at least 2nd order or higher, just the type 2 of 2nd order is to consider). In practice the 2nd order type 2 architecture is used because a constant frequency mismatch can be tracked with zero error independently of the loop gain. This is unfortunately accompanied by a softer rejection of the input jitter (= input signal noise) at high frequencies. When the loop gain is variable in a very wide range (because of a bang-bang phase detector) and/or especially when a VCO with poor accuracy must be used, this architecture is preferred.
 * Type 0 systems are able to track a step function ( a phase deviation of the input from the nominal phase expected by the circuit itself) with a small, but finite error. A type 0 PLL can not track at all a linear ramp of phase variation in the received signal (that is a step change of the frequency of the received signal with respect to the frequency of the -free running- local oscillator!). Therefore they are not of large use for CDRs.
 * Type 1 systems are able to track signals that exhibit a step change of phase, without steady-state error. They are not able to track unit ramp inputs without a finite error, though. CDRs are very often type 1 systems, because they can be designed to phase lock:
 * with zero phase error when there is no frequency (just phase) difference
 * with a very small phase error when the master clock embedded in a received signal stream and the free running frequency of the local oscillator are close together.
 * Type 2 systems are able to track signals with a frequency different from the their free running frequency, without steady-state error.

General case
A unity feedback system (all PLL for CDRs are unity feedback) exhibits a finite long term error ( steady state error ) between input and output when the input (besides sinusoids of any frequency and magnitude), includes a constant, or a step function, or a ramp function, etc., depending on the type of the system.

The steady state error, when it exists and is finite and non-zero, can be computed from the open loop transfer function: GH = K $$\tfrac{\prod_{i=1}^m (s-z_i)}{\prod_{i=1}^n (s-p_i)}$$ using the following table, that resumes the most important formulas for the systems representing CDR architectures:

1st order loop
The open loop transfer function (this loop and its equations are described further down in Examples of CDR structures: 1st order, type 1, Oscillator controlled in frequency ("first order slave CDR")) : GH = $$\tfrac{ forward DC gain}{s} = \tfrac{G_{fDC}}{s} $$
 * For a type 1 system let's define the auxiliary quantity K as : K =  $$\lim_{s \to 0}sGH$$

In the case of our 1st order system which is a type 1 system: K =  $$\lim_{s \to 0}s\frac{G_{fDC}}{s}$$  = $$G_{fDC}$$ [sec−1] The steady state error in case of a unit ramp of the input phase is: 1 / K = $$\tfrac{1}{G_{fDC}}$$ [sec]
 * The unit phase ramp function is:
 * 0 [rad]   when t < 0
 * (1 [rad/sec] ) * t [sec] = t [rad]  when t >= 0

The steady state error is generated in practice by the frequency offset between the frequency of the incoming line pulses ($$f_p$$ [2π rad/sec]) and the frequency of the VCO when the latter is driven by the indifferent (e.g. 0 volt) level of the filter output, ($$f_{fr}$$ [2π rad/sec]).  ( $$f_{fr}$$ is also called the VCO "free-running" frequency )  The steady state error due to the frequency mismatch is: $$\tfrac{f_p - f_{fr}}{2\pi\cdot G_{fDC}}$$ [$$rad$$] = $$\tfrac{\omega_p - \omega_{fr}}{G_{fDC}}$$ [rad] This steady state error appears as an offset of the sampling instant away from its optimum point at the center of the received eye.

2nd order loop
For a type 1 system, the open loop transfer function (this loop and its equations are described further down in The CDR based on a second order PLL: Fundamental equations) : GH = $$\tfrac{G_{\phi} G_fG_{VCO}}{s(1+s{\tau_f})} $$
 * The auxiliary quantity K is defined as as :

K =  $$\lim_{s \to 0}sGH$$

In the case of our 2nd order system which is a type 1 system: K =  $$\lim_{s \to 0}s\tfrac{G_{\phi} G_fG_{VCO}}{s(1+s{\tau f})}$$  =  G [sec−1]

The steady state error in case of a unit ramp of the input phase is : 1 / K = 1/ G [sec]

 * The unit phase ramp function is:
 * 0 [rad]    when t < 0
 * (1 [rad/sec] ) * t [sec] = t [rad]  when t >= 0

The steady state error is generated in practice by the frequency offset between the frequency of the incoming line pulses ($$f_p$$ [2π rad/sec] ) and the frequency of the VCO when the latter is driven by the indifferent (e.g. 0 volt) level of the filter output, ($$f_{fr}$$ [2π rad/sec] ).  ( $$f_{fr}$$ is also called the VCO "free-running" frequency )

The steady state error due to the frequency mismatch is: $$\tfrac{f_p - f_{fr}}{2\pi\cdot G}$$ [rad] = $$\tfrac{\omega_p - \omega_{fr}}{G}$$ [rad] This steady state error appears as an offset of the sampling instant away from its optimum point at the center of the received eye.

For a type 2 system the steady state error generated by a frequency offset between the frequency of the incoming line pulses ($$f_p$$ [2π rad/sec]) and the free-running frequency of the VCO, ($$f_{fr}$$ [2π rad/sec]), is exactly 0, as can easily be derived in analogy with the example of the type 1 system.

The right architectural model to study a CDR
No model is perfect. But modeling is essential to human knowledge. The simplest model, amongst the fitting ones, shall be chosen (Occam's razor).

The basic architectural model of a CDR PLL is made up of three forward blocks and a unity feedback.

This section and the following two offer some considerations on the identification of the right architectural model.

 * Flat gain block. In a first order loop a flat gain stage often precedes the VCO. The designer can choose the overall loop gain (an important CDR parameter) without the constraint of the VCO gain value, that is forced by technological choices. Sometimes this block is clearly identified, sometimes it is represented as one with the VCO. This is decided according to what has to be emphasized, but does not change the loop order at all. In a second order loop, shifting the gain curve of the filter block does not change the architecture.
 * High frequency rejection. There is always some sort of filtering at the output of a phase comparator, to reject high signal frequencies (around fp and its harmonics) as well as noise. Both comparator and VCO often have limited bandwidth. But only the phase signal at (much) lower frequencies is of interest for the CDR operation. These low-pass characteristics can almost always be neglected within the scope of our study, as they do not turn a first order loop into a second order loop, nor a second into a third.
 * Second order order loop (type 1) with high ζ value. When ζ is greater than 1, there are two real poles in the loop and the one that creates the jitter cut-off frequency is more important. When ζ is significantly greater than 1, the pole that generates the jitter cut-off is dominant for the loop behavior. A loop with ζ significantly greater than 1 is  very similar to a first order loop whose pole coincides with its dominant pole. Second order loops with  ζ greater than 5 are of no interest for the CDR theory and applications.
 * Second order order loop (type 2) with high ζ value. The closed loop function of this loop exhibits one zero and a dual pole pair at a higher frequency. When ζ is greater than 1, there are two real poles in the loop and the one that creates the jitter cut-off frequency is more important This pole is at a higher frequency, while its companion is located very close to the zero, although at a slightly higher frequency. The pole that generates the jitter cut-off is dominant for the loop behavior, while the effects of the remaining pole and zero tend to counterbalance each other. Such loop is very similar to a first order loop whose pole coincides with its dominant pole. Second order loops with  ζ greater than 5 are of no interest for the CDR theory and applications.

The Architecture number pair is the same, even if Comparator and/or VCO are non-linear
The non-linearities (that are found in practical applications of a PLL for CDR) alter the loop behaviour much less than a change of architecture would.

 * The following pages will show that those non-linearities, when present in real case applications, make the loop behaviour diverge from the linear case less than a change of architecture.

The blocks that may be implemented in a CDR PLL with significantly non-linear characteristic are the phase comparator and sometimes also the VCO.

 * A linear model remains useful for understanding loop behavior, but computer simulations are necessary for a detailed study of performance.

The loop architectures of interest for PLLs used inside CDRs are identified by the same two numbers, both if all their blocks can be modelled as linear or if one or two cannot :

Why only these three?
There are plenty of possible architectures, and for each pair order/type there are several different architectures.

But just three are valid for CDR PLLs. Why no other?

It has been anticipated that in a network phase aligners and slaves are needed:

Here are the reasons in detail for the 3 architectures  (note that there are 3 network functions and 3 architectures, but that they do not necessarily correspond exactly one to one ):
 * 1) All CDR PLLs are unity feedback loops.  The phase of the output signal (= of the local clock that locks into the incoming signal level transitions) must be directly compared with the phase (the level transitions) of the input signal.
 * This definition holds good for all CDR PLLs, including the phase aligners. In phase aligners the output signal of the PLL is not the independent clock that the input signal must be aligned to, but the clock generated by the PLL and used to write into the elastic buffer.
 * Contrary to CDR PLLs where input and output frequencies are the same, frequency synthesizer PLLs often have circuitry that makes the output frequency an m/n multiple of the input, and part of it is located in the loop feedback path.
 * 1) The order of the loop can only be one or two. Order 0 cannot lock, while order 3 and higher are prone to unstability or do not add any advantage.
 * 2) The type must be at least 1 (which limits the possible choices to the three cases), to achieve phase lock. Therefore a loop of 1st order can only be used if of  type 1, while a loop of 2nd order can be either type 1 or type 2 (the type cannot be higher than the order, by definition).
 * 3) For the 2 loops of order 2, the filter block (that must be of order 1) must be a low pass ( ωp  < ωz  ), because the jitter and noise of higher frequencies must be more, and not less, rejected. Considering all types of order 1 filters:

In conclusion:
 * for a phase aligner, that only needs a good jitter tolerance, the 1 – 1 is chosen, because of its fast acquisition, good tolerance, and robustness against both non-linearity and drift of characteristic in the circuit components.
 * for slave CDRs, the (slave) PLL architectures 2-1 and 2-2 are chosen.
 * 2 – 1 is preferred for regenerators, in particular when components with good linearity and characteristics are available
 * 2 - 2 is preferred for monolythic implementation, where components have stronger non-linearities and variability of characteristics

The following pages of this book shall show that the best matching of applications and architectures is:


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AQA Business Studies/Ratio Analysis

Ratios in a business, are important in order to compare the efficiency with either another company in the same or a similar market or previous years.

There are 5 types of Ratios we will look at:

 * Shareholders Ratios
 * Profitability Ratios
 * Liquidity Ratios
 * Activity Ratios
 * Financial Ratios

Dividend Per Share
The amount of the overall dividend that each share will get - what essentially shareholders are investing for.. The Dividend is announced each year and can be found in the Trading, Profit and Loss Account after Tax has been taken off.

$$Dividend\ Per\ Share=\frac{Overall\ Dividend}{Number\ of\ Shares}$$

Dividend Yield
The percentage of the Market Price that the Shareholders will get in dividend. This shows potential investors how much a company values it's shareholders investment.

$$Dividend\ Yield=\frac{Dividend\ Per\ Share}{Market\ Price\ Per\ Share}$$

Gross Profit Margin
Gross Profit is the Revenue minus the cost of making the products.

Gross Profit Margin is the measure of how efficiently a company is making the products to sell and turn into revenue (remember - Sales revenue and sales turnover are the same things). The higher the percentage, the better for the business because it means that the Profit Margin on each product is high - for a 'company', a higher dividend could be given.

To increase the Gross Profit Margin:
 * Supplier costs could be decreased if a cheaper one is available
 * Price per unit can be increased
 * Sales turnover can be increased (marketing or price decrease - but has an adverse effect on overall profit)

$$Gross\ Profit\ Margin=\frac{Gross\ Profit}{Sales\ Turnover} \times 100$$

Net Profit Margin
Net Profit is Revenue minus all the costs of the day to day running of the business - Costs of making the products, Overheads, Vehicles, Machinery, etc.

Net Profit Margin is the measure of how efficiently a company is turning Revenue into Net Profit. The higher the Net Profit Margin, is again the better. It shows what percentage of the revenue is available to cover taxes to be paid to inland revenue and how healthy the dividend will be.

To increase the Net Profit Margin:
 * (Everything used to increase Gross Profit Margin)
 * Carefully monitor Overhead usage (turning lights off, turning heating off)
 * Budget on expenditure carefully

$$\mbox{Net Profit  Margin}=\frac{\mbox{Net Profit}}{\mbox{Sales Turnover}} \times 100$$

Return on Capital Employed (ROCE)
ROCE is also known as the primary accounting ratio. It measures how efficiently a business is investing capital in a business in order to make profit. Sales revenue can again be found on the Trading, Profit and Loss account and the Capital Employed on the balance sheet. Capital Employed can be worked out by adding the Shareholders capital and Long-term liabilities.

The higher the figure, the more attractive investment the business would be to investors (More returns on their capital!).It can be compared to previous years (show levels of previous investment and expected returns and whether it is satisfactory or not) or other businesses ROCE. It should also be compared to bank interest rates - what's the point in the risk of investing it in a business if you could invest it in a risk-free bank and earn more in interest rates than you would on ROCE?

ROCE can be adjusted by:
 * Earning more revenue with the same amount of capital employed
 * Earning the same amount of revenue but decreasing the Capital used to generate the revenue.

$$ROCE=\frac{Sales\ Turnover}{Capital\ Employed} \times 100$$

Liquidity Ratios
Liquidity ratios measure how effectively a business can sell it's assets to cover it's liabilities.

Acid Test Ratio
Acid test is the measure of how effectively a business can liquidate (pay off creditors) without considering the stock. Stock is taken off because it is the most illiquid asset that a business owns. In order to sell it immediately the stock becomes very cheap and so loses its value.

It is recommended a business keeps this at about 0.8 as the stock will always be worth something even if a very small proportion to what it's actual value is. If the number becomes lower, the business may experience difficulties covering it's short term debts. If the number increases, the business is probably using the cash in an unprofitable or unproductive form.

$$Acid\ Test=\frac{Current\ Assets\ -\ Stock}{Current\ Liabilities}$$

Current Ratio
Not actually needed (it's not in the syllabus) but useful to know as the theory for acid test links in.

$$Current\ Ratio=\frac{Current\ Assets}{Current\ Liabilities}$$

Gearing Ratio
Shows how reliant a business is on External sources of finance. It is used by banks and lenders to see how risky a business is before they give them the loan. Highly geared companies are known as the riskiest type of business because they are reliant on loans for the day to day running of assets, which they haven't yet paid off.

$$Gearing=\frac{Long\ term\ Liabilities}{Capital\ Employed}$$

Asset Turnover
$$Asset\ Turnover=\frac{Sales\ Turnover}{Net\ Assets}$$

Stock Turnover
$$Stock\ Turnover=\frac{Overall\ Cost\ of\ Goods\ Sold}{Average\ Number\ of\ Stock}$$

Debtor Days
$$Debtor\ Days=\frac{Debtors}{Sales\ Turnover}$$

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

title: AQA Business Studies/Ratio Analysis


text: User talk:83.131.146.88

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User talk:Hubbak~enwikibooks

Welcome, !

Come introduce yourself at the new users page. If you have any questions, you can ask there or contact me personally. – Mike.lifeguard  | talk 21:56, 19 March 2008 (UTC)

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Please upload a version which is not defaced like that. We generally don't allow even a watermark, much less a glaring banner proclaiming ownership. Thanks. – Mike.lifeguard  | talk 21:56, 19 March 2008 (UTC)

Your account will be renamed
Hello,

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User talk:DavidPickett

Welcome!
Welcome, !

Come introduce yourself at the new users page. If you have any questions, you can ask there or contact me personally.-- – Mike.lifeguard  | talk 21:57, 19 March 2008 (UTC)

title: User talk:DavidPickett


text: User talk:68.5.8.17

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Talk:Bambara

>>i like in India

Untitled
Is this like the first I in India, or like the second i in India? Cilantrohead (talk) 22:52, 19 March 2008 (UTC)
 * Like the second "i"; this is not an ideal choice of word as example.--A12n (discuss • contribs) 17:54, 21 September 2017 (UTC)

title: Talk:Bambara


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Prealgebra for Two-Year Colleges/Workbook AIE/Circle and bar graphs

Circle Graphs and Bar Graphs

Why
Circle graphs and bar graphs are often used rather than tables to display data. Graphs can give you a quick feeling for the data.

Learning Objectives

 * 1) Construct clear, accurate circle graphs from tables.
 * 2) Construct clear, accurate bar graphs from tables.

Activity
1)	What does it mean for a circle graph to be “clear”?

 * Graph has title. Slices are labeled with categories (words) and data (numbers).

2)	Construct a circle graph for the table below.

a)	First Look on the last page of this handout. Cut out the protractor. You will use the protractor to measure angles in your circle graph. b)	Fill in the last column of the table (degrees). i)	Show your calculations in the table. ii)	What is a reasonable place to round off to? Why?

 * Round off to the ones, because I cannot estimate any better than that from the protractor.

c)	Now draw the circle graph.

 * Title: Absence from Work Due to Carpal Tunnel Syndrome

3)	Construct a circle graph for the table below.

a)	What is the total number of students in the survey?

 * 700

b)	Use proportions to fill in the last column of the table (degrees). Show your proportions in the table. Round off to a reasonable place.

 * See table above

c)	Now draw the circle graph.

 * Title ="Where students live while attending college"

4)	What does it mean for a bar graph to be “clear”?

 * Graph has title. Axes are labeled. Axes are marked with values and category names.

5)	Create a vertical bar graph for the table to the right.

a)	What is the smallest number?
 * 50

What is the largest number?

 * 105

What is the difference?

 * 55

b)	How many “tic marks” are there on the graph?

 * 14

Leading U.S Passenger Airlines Airline	# of Passengers (in millions) Delta	105 United	87 American	81 Southwest	59 U.S. Airways	58 Northwest	50 c)	If you put the smallest number at the bottom, and the largest number at the top, how much would each tic mark be worth? Show your calculation.

 * 55 / 14 ≈ 3.928

d)	Round the answer in Part (c) up to a “nice” number.

 * 5

e)	What does a “zig-zag” on the axis mean? Will you use one? Why?

 * A zig-zag means that you have skipped part of the axis; the numbers have jumped. I will use one because I am starting the numbers at 50, but each tic mark is only 5.

f)	Make the bar graph.


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Chess Opening Theory/1. d4/1...Nf6/2. Nf3/2...e6/3. Bf4

= London System Indian Setup with a Early 2. Nf3 = The London System has evolved significantly since its showcase in the 1922 London tournament. A prominent shift in modern play is the preference for 2.Bf4 instead of the traditional 2.Nf3. Both sides have seen strategic advancements in the London System.

The move order with 2. Nf3 before 3. Bf4 in the London System can be seen as a more flexible way to reach the London System and has its specific advantages and drawbacks.

For a general introduction to the London System, take a closer look at 1. d4 d5 2. Bf4

Advantages:

 * 1) Flexibility: By playing 2. Nf3 first, White keeps more options open. This can be particularly useful if Black opts for setups that may not be best met with the traditional London System. For instance, White can divert into other openings or setups based on Black's moves.
 * 2) Control over e5: Central control, particularly over the e5 and e4 squares, is vital in the London System. The move 2. Nf3 immediately prevents any potential pawn break with ...e5 by Black.
 * 3) Avoiding Immediate Pressure: Certain move orders by Black might put immediate pressure on the d4 pawn (like ...c5 and ...Qb6 setups). The knight on f3 provides additional support to the d4 pawn.

Drawbacks:

 * 1) Delaying the London System's Set-up: The very essence of the London System is rapid and harmonious development, particularly the dark-squared bishop. By playing 2. Nf3 first, White is essentially delaying the typical London setup. By shifting the knight from f3 to d4, White diminishes their hold on the e5 square.
 * 2) When White plays 2.Nf3, it can prompt Black to opt for 2...c5 because the subsequent 3.Bf4 can be met by 3...cxd4 4.Nxd4 Nd7.
 * 3) Allowing Black's Equalizing Attempts: Black might employ certain systems that challenge White's center more directly before White can establish the typical London System setup.
 * 4) Potential Transpositions: The move 2. Nf3 allows various transpositions into other openings, which could be a disadvantage if the White player is specifically aiming to play the London System and avoid other lines.

Theory table
'''1. d4 Nf6 2. Nf3 e6 3. Bf4'''


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User:Beamer

Hi. I live in California.--Beamer (talk) 00:39, 20 March 2008 (UTC)

title: User:Beamer


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Ordinary Differential Equations/Frobenius Solution to the Hypergeometric Equation

In the following we solve the second-order differential equation called the hypergeometric differential equation using Frobenius method, named after Ferdinand Georg Frobenius. This is a method that uses the series solution for a differential equation, where we assume the solution takes the form of a series. This is usually the method we use for complicated ordinary differential equations.

The solution of the hypergeometric differential equation is very important. For instance, Legendre's differential equation can be shown to be a special case of the hypergeometric differential equation. Hence, by solving the hypergeometric differential equation, one may directly compare its solutions to get the solutions of Legendre's differential equation, after making the necessary substitutions. For more details, please check the hypergeometric differential equation

We shall prove that this equation has three singularities, namely at x = 0, x = 1 and around infinity. However, as these will turn out to be regular singular points, we will be able to assume a solution on the form of a series. Since this is a second-order differential equation, we must have two linearly independent solutions.

The problem however will be that our assumed solutions may or not be independent, or worse, may not even be defined (depending on the value of the parameters of the equation). This is why we shall study the different cases for the parameters and modify our assumed solution accordingly.

The equation
Solve the hypergeometric equation around all singularities:

x(1-x)y''+\left\{ \gamma -(1+\alpha +\beta )x \right\}y'-\alpha \beta y=0 $$

Solution around x = 0
Let

 * $$\begin{align}

P_0(x) = -\alpha \beta, && P_1(x) = \gamma - (1+\alpha +\beta )x, && P_2(x) = x(1-x) \end{align}$$

Then

 * $$P_2(0) = 0, P_2 (1)=0.\,$$

Hence, x = 0 and x = 1 are singular points. Let's start with x = 0. To see if it is regular, we study the following limits:

 * $$\begin{align}

\lim_{x \to a} \frac{(x - a) P_1(x)}{P_2(x)} &= \lim_{x \to 0} \frac{(x - 0)(\gamma - (1 + \alpha + \beta)x)}{x(1 - x)} = \lim_{x \to 0} \frac{x(\gamma - (1 + \alpha + \beta)x)}{x(1 - x)} = \gamma \\ \lim_{x \to a} \frac{(x - a)^2 P_0(x)}{P_2(x)} &= \lim_{x \to 0} \frac{(x - 0)^2(-\alpha \beta)}{x(1 - x)} = \lim_{x \to 0} \frac{x^2 (-\alpha \beta)}{x(1 - x)} = 0 \end{align}$$

Hence, both limits exist and x = 0 is a regular singular point. Therefore, we assume the solution takes the form

 * $$y = \sum_{r=0}^\infty a_r x^{r + c}$$

with a0 ≠ 0. Hence,

 * $$\begin{align}

y' = \sum_{r = 0}^\infty a_r(r + c) x^{r + c - 1} && y''= \sum_{r = 0}^\infty a_r(r + c)(r + c - 1) x^{r + c - 2}. \end{align}$$

Substituting these into the hypergeometric equation, we get

 * $$\begin{align}

&x \sum_{r = 0}^\infty a_r(r + c)(r + c - 1) x^{r + c - 2} - x^2 \sum_{r = 0}^\infty a_r(r + c)(r + c - 1) x^{r + c - 2} + \gamma \sum_{r = 0}^\infty a_r(r + c) x^{r + c - 1} \\ &\quad -(1 + \alpha + \beta) x\sum_{r = 0}^\infty a_r(r + c) x^{r + c - 1} -\alpha \beta \sum_{r = 0}^\infty a_r x^{r + c} = 0 \end{align}$$ That is,
 * $$\begin{align}

&\sum_{r = 0}^\infty a_r(r + c)(r + c - 1) x^{r + c - 1} -\sum_{r = 0}^\infty a_r(r + c)(r + c - 1) x^{r + c}  +\gamma \sum_{r = 0}^\infty a_r(r + c) x^{r + c - 1} \\ &\quad -(1 + \alpha + \beta) \sum_{r = 0}^\infty a_r(r + c) x^{r + c}  -\alpha \beta \sum_{r = 0}^\infty a_r x^{r + c} =0 \end{align}$$

In order to simplify this equation, we need all powers to be the same, equal to r + c - 1, the smallest power. Hence, we switch the indices as follows:

 * $$\begin{align}

&\sum_{r = 0}^\infty a_r(r + c)(r + c - 1)x^{r + c - 1} -\sum_{r = 1}^\infty a_{r - 1}(r + c - 1)(r + c - 2) x^{r + c - 1} +\gamma \sum_{r = 0}^\infty a_r(r + c) x^{r + c - 1} \\ &\quad -(1 + \alpha + \beta) \sum_{r = 1}^\infty a_{r - 1}(r + c - 1) x^{r + c - 1} -\alpha \beta \sum_{r = 1}^\infty a_{r - 1} x^{r + c - 1} =0 \end{align}$$

Thus, isolating the first term of the sums starting from 0 we get

 * $$\begin{align}

&a_0 (c(c-1) + \gamma c) x^{c - 1} + \sum_{r = 1}^\infty a_r(r + c)(r + c - 1) x^{r + c - 1} -\sum_{r = 1}^\infty a_{r - 1}(r + c - 1)(r + c - 2) x^{r + c - 1} \\ &\quad + \gamma \sum_{r = 1}^\infty a_r(r + c) x^{r + c - 1} -(1 + \alpha + \beta) \sum_{r = 1}^\infty a_{r - 1}(r + c - 1) x^{r + c - 1} -\alpha \beta \sum_{r = 1}^\infty a_{r - 1} x^{r + c - 1} = 0 \end{align}$$

Now, from the linear independence of all powers of x, that is, of the functions 1, x, x2, etc., the coefficients of xk vanish for all k. Hence, from the first term, we have

 * $$ a_{0} (c(c - 1) + \gamma c) = 0\,$$

which is the indicial equation. Since a0 ≠ 0, we have

 * $$c(c - 1 + \gamma) = 0.\,$$

Hence,

 * $$\begin{align}

c_1 = 0 && c_2 = 1 - \gamma \end{align}$$

Also, from the rest of the terms, we have

 * $$\begin{align}

&((r + c)(r + c - 1) + \gamma(r+c)) a_r \\ &\quad + (-(r + c - 1)(r + c - 2) - (1 + \alpha + \beta)(r + c - 1) - \alpha\beta) a_{r - 1} = 0 \end{align}$$

Hence,

 * $$\begin{align}

a_r &= \frac{(r + c - 1)(r + c - 2) + (1 + \alpha + \beta)(r + c - 1) + \alpha\beta} {(r + c)(r + c - 1) + \gamma(r + c)} a_{r - 1} \\ &= \frac{(r + c -1)(r + c + \alpha + \beta - 1) + \alpha\beta} {(r + c)(r + c + \gamma - 1)} a_{r - 1} \end{align}$$

But

 * $$\begin{align}

&(r + c - 1)(r + c + \alpha + \beta - 1) + \alpha\beta \\ &\quad = (r + c - 1)(r + c + \alpha - 1) + (r + c - 1)\beta + \alpha\beta \\ &\quad = (r + c - 1)(r + c + \alpha - 1) + \beta(r + c + \alpha - 1) \end{align}$$ Hence, we get the recurrence relation
 * $$a_r = \frac{(r + c + \alpha - 1)(r + c + \beta - 1)}{(r + c)(r + c + \gamma - 1)} a_{r - 1}, \text{ for } r \geq 1.$$

Let's now simplify this relation by giving ar in terms of a0 instead of ar − 1. From the recurrence relation (note: below, expressions of the form (u)r refer to the Pochhammer symbol).
 * $$\begin{align}

a_1 &= \frac{(c + \alpha)(c + \beta)}{(c + 1)(c + \gamma)} a_0 \\ a_2 &= \frac{(c + \alpha + 1)(c + \beta + 1)}{(c + 2)(c + \gamma + 1)} a_1 = \frac{(c + \alpha + 1)(c + \alpha)(c + \beta + 1)} {(c + 2)(c + 1)(c + \gamma)(c + \gamma + 1)} a_0 = \frac{(c + \alpha)_2 (c + \beta)_2}{(c + 1)_2 (c + \gamma)_2} a_0 \\ a_3 &= \frac{(c + \alpha + 2)(c + \beta + 2)}{(c + 3)(c + \gamma + 2)} a_2 = \frac{(c + \alpha)_2 (c + \alpha + 2)(c + \beta )_2 (c + \beta + 2}            {(c + 1)_2 (c + 3)(c + \gamma)_2 (c + \gamma + 2)} a_0 \\      &= \frac{(c + \alpha)_3 (c + \beta)_3}                    {(c + 1)_3 (c + \gamma)_3} a_0 \end{align}$$

As we can see,

 * $$a_r =\frac{(c + \alpha)_r (c + \beta)_r}{(c + 1)_r(c + \gamma)_r} a_0, \text{ for } r \geq 0$$

Hence, our assumed solution takes the form

 * $$y = a_0 \sum_{r = 0}^\infty \frac{(c + \alpha)_r (c + \beta)_r}{(c + 1)_r (c + \gamma)_r} x^{r + c}. $$

We are now ready to study the solutions corresponding to the different cases for c1 − c2 = γ − 1 (it should be noted that this reduces to study the nature of the parameter γ: whether it is an an integer or not).

γ not an integer
Then y1 = y|undefined and y2 = y|undefined. Since
 * $$y = a_0 \sum_{r = 0}^\infty \frac{(c + \alpha )_r (c + \beta)_r}{(c + 1)_r (c + \gamma)_r} x^{r + c},$$

we have
 * $$\begin{align}

y_1 &= a_0 \sum_{r = 0}^\infty \frac{(\alpha)_r (\beta)_r}{(1)_r (\gamma)_r} x^r \\ &= a_0 \cdot {{}_2 F_1}(\alpha, \beta; \gamma; x) \\ y_2 &= a_0 \sum_{r = 0}^\infty \frac{(\alpha + 1 - \gamma)_r (\beta + 1 - \gamma)_r} {(1 - \gamma + 1)_r (1 - \gamma + \gamma)_r} x^{r + 1 - \gamma} = a_0 x^{1 - \gamma} \sum_{r = 0}^\infty \frac{(\alpha + 1 - \gamma)_r (\beta + 1 - \gamma)_r} {(1)_r (2 - \gamma)_r} x^r \\ &= a_0 x^{1 - \gamma} {{}_2 F_1}(\alpha - \gamma + 1, \beta - \gamma + 1; 2 - \gamma; x) \end{align}$$ Hence, $$y = A' y_1 + B' y_2.$$ Let A′ a0 = a and B′ a0 = B. Then
 * $$y = A {{}_2 F_1}(\alpha, \beta; \gamma; x)

+ B x^{1 - \gamma} {{}_2 F_1}(\alpha - \gamma + 1, \beta - \gamma + 1; 2 - \gamma; x)\,$$

γ = 1
Then y1 = y|undefined. Since γ = 1, we have
 * $$y = a_0 \sum_{r = 0}^\infty \frac{(c + \alpha)_r (c + \beta)_r}{(c + 1)_r^2} x^{r + c}.$$

Hence,
 * $$\begin{align}

y_1 &= a_0 \sum_{r = 0}^\infty \frac{(\alpha)_r (\beta)_r}{(1)_r (1)_r} x^r = a_0 {{}_2 F_1}(\alpha, \beta; 1; x) \\ y_2 &= \left.\frac{\partial y}{\partial c}\right|_{c = 0}. \end{align}$$ To calculate this derivative, let
 * $$M_r = \frac{(c + \alpha)_r (c + \beta)_r}{(c + 1)_r^2}.$$

Then
 * $$\ln(M_r) = \ln\left(\frac{(c + \alpha)_r (c + \beta)_r}{(c + 1)_r^2}\right)

= \ln(c + \alpha)_r + \ln(c + \beta)_r - 2\ln(c + 1)_r$$ But
 * $$\ln(c + \alpha)_r = \ln\bigl((c + \alpha)(c + \alpha + 1) \cdots (c + \alpha + r - 1)\bigr)

= \sum_{k = 0}^{r - 1} \ln(c + \alpha + k).$$ Hence,
 * $$\begin{align}

\ln(M_r) &= \sum_{k = 0}^{r - 1} \ln(c + \alpha + k)            + \sum_{k = 0}^{r - 1} \ln(c + \beta + k)             - 2 \sum_{k = 0}^{r - 1} \ln(c + 1 + k) \\ &= \sum_{k = 0}^{r - 1} \bigl(\ln(c + \alpha + k) + \ln(c + \beta + k) -2 \ln(c + 1 + k)\bigr) \end{align}$$ Differentiating both sides of the equation with respect to c, we get:
 * $$\frac{1}{M_r} \frac{\partial M_r}{\partial c}

= \sum_{k = 0}^{r - 1} \left(\frac{1}{c + \alpha + k} + \frac{1}{c + \beta +k }               - \frac{2}{c + 1 + k}\right).$$ Hence,
 * $$\frac{\partial M_r}{\partial c}

= \frac{(c + \alpha)_r (c + \beta)_r}{(c + 1)_r^2} \sum_{k = 0}^{r - 1} \left(\frac{1}{c + \alpha + k} + \frac{1}{c + \beta + k} - \frac{2}{c + 1 + k}\right). $$ Now,
 * $$y = a_0 x^c \sum_{r = 0}^\infty \frac{(c + \alpha)_r (c + \beta)_r}{(c + 1)_r^2} x^r

= a_0 x^c \sum_{r = 0}^\infty M_r x^r.$$ Hence,
 * $$\begin{align}

\frac{\partial y}{\partial c}  &= a_0 x^c \ln(x) \sum_{r = 0}^\infty \frac{(c + \alpha)_r (c + \beta)_r}{(c + 1)_r^2} x^r \\ &\quad + a_0 x^c \sum_{r = 0}^\infty \left(\frac{(c + \alpha)_r (c + \beta)_r}{(c + 1)_r^2}                    \left\{\sum_{k = 0}^{r - 1} \left(\frac{1}{c + \alpha + k} + \frac{1}{c + \beta + k}                                                       - \frac{2}{c + 1 + k} \right) \right\} \right) x^r \\ &= a_0 x^c \sum_{r = 0}^\infty \frac{(c + \alpha)_r (c + \beta)_r}{(c + 1)_r)^2}                    \left(\ln x + \sum_{k = 0}^{r - 1} \left(\frac{1}{c + \alpha + k}                                     +\frac{1}{c + \beta + k} - \frac{2}{c + 1 + k} \right) \right) x^{r}. \end{align}$$ For c = 0, we get
 * $$y_2 = a_0 \sum_{r = 0}^\infty \frac{(\alpha)_r (\beta)_r}{(1)_r^{2}} \left(\ln x

+ \sum_{k = 0}^{r - 1} \left(\frac{1}{\alpha + k} + \frac{1}{\beta + k}                      - \frac{2}{1 + k} \right) \right)x^{r}.$$ Hence, y = C′ y1 + D′ y2.  Let C′ a0 = C and D′ a0 = D.  Then
 * $$y = C {{}_2 F_1}(\alpha, \beta; 1; x)

+ D \sum_{r = 0}^\infty \frac{(\alpha)_r (\beta)_r}{(1)_r^2} \left(\ln x                 + \sum_{k = 0}^{r - 1} \left(\frac{1}{\alpha + k} + \frac{1}{\beta + k}                 - \frac{2}{1 + k} \right) \right) x^r$$

γ ≤ 0
From the recurrence relation
 * $$a_r = \frac{(r + c + \alpha - 1)(r + c + \beta - 1)}{(r + c)(r + c + \gamma - 1)} a_{r - 1},$$

we see that when c = 0 (the smaller root), a1 − γ → ∞. Hence, we must make the substitution a0 = b0 (c - ci), where ci is the root for which our solution is infinite. Hence, we take a0 = b0 c and our assumed solution takes the new form
 * $$y_b = b_0 x^c \sum_{r = 0}^\infty \frac{c(c + \alpha)_r (c + \beta)_r}{(c + 1)_r (c + \gamma)_r} x^r $$

Then y1 = yb|undefined. As we can see, all terms before
 * $$\frac{c(c + \alpha)_{1 - \gamma}(c + \beta)_{1 - \gamma}}

{(c + 1)_{1 - \gamma}(c + \gamma)_{1 - \gamma}} x^{1 - \gamma} $$ vanish because of the c in the numerator. Starting from this term however, the c in the numerator vanishes. To see this, note that
 * $$(c + \gamma)_{1 - \gamma} = (c + \gamma)(c + \gamma + 1) \cdots c.$$

Hence, our solution takes the form
 * $$\begin{align}

y_1 &= b_0 \left(             \frac{(\alpha)_{1 - \gamma} (\beta)_{1 - \gamma}}{(1)_{1 - \gamma} (\gamma)_{-\gamma}} x^{1 - \gamma}            + \frac{(\alpha)_{2 - \gamma} (\beta)_{2 - \gamma}}{(1)_{2 - \gamma} (\gamma)_{-\gamma}(1)} x^{2 - \gamma}            + \frac{(\alpha)_{3 - \gamma} (\beta)_{3 - \gamma}}{(1)_{3 - \gamma} (\gamma)_{-\gamma}(1)(2)} x^{3 - \gamma}            + \cdots \right) \\ &= \frac{b_0}{(\gamma)_{-\gamma}} \sum_{r = 1 - \gamma}^\infty \frac{(\alpha)_r (\beta)_r}{(1)_r (1)_{r + \gamma - 1}} x^r. \end{align}$$ Now,
 * $$y_2 = \left.\frac{\partial y_b}{\partial c}\right|_{c = 1 - \gamma}.$$

To calculate this derivative, let
 * $$M_r = \frac{c(c + \alpha)_r (c + \beta)_r}{(c + 1)_r(c + \gamma)_r}.$$

Then following the method in the previous case, we get
 * $$\frac{\partial M_r}{\partial c}

= \frac{c(c + \alpha)_r (c + \beta )_r}{(c + 1)_r (c + \gamma)_r} \left\{ \frac{1}{c} + \sum_{k = 0}^{r - 1} \left(\frac{1}{c + \alpha + k} + \frac{1}{c + \beta + k}                                 - \frac{1}{c  + 1 + k} -\frac{1}{c + \gamma + k} \right) \right\}. $$ Now,
 * $$y_b = b_0 \sum{r = 0}^\infty \frac{c(c + \alpha)_r (c + \beta)_r}{(c + 1)_r (c + \gamma)_r} x^{r + c}

= b_0 x^c \sum_{r = 0}^\infty M_r x^r.$$ Hence,
 * $$\begin{align}

\frac{\partial y}{\partial c}  &= b_0 x^c \ln(x) \sum_{r = 0}^\infty \frac{c(c + \alpha)_r (c + \beta)_r}{(c + 1)_r (c + \gamma)_r} x^r \\ &\quad + b_0 x^c \sum_{r = 0}^\infty \frac{c(c + \alpha)_r (c + \beta)_r}{(c + 1)_r (c + \gamma)_r} \left\{ \frac{1}{c} + \sum_{k = 0}^{r - 1} \left(\frac{1}{c + \alpha + k} + \frac{1}{c + \beta + k}                              - \frac{1}{c + 1 + k} - \frac{1}{c + \gamma + k} \right) \right\} x^r \end{align}$$ Hence,
 * $$\begin{align}

\frac{\partial y}{\partial c}  = b_0 x^c \sum_{r = 0}^\infty &\frac{c(c + \alpha)_r (c + \beta)_r}{(c + 1)_r (c + \gamma)_r} \Biggl(\ln x + \frac{1}{c} + \\  &\quad + \sum_{k = 0}^{r - 1} \left(\frac{1}{c + \alpha + k} + \frac{1}{c + \beta + k}               - \frac{1}{c + 1 + k} - \frac{1}{c + \gamma + k} \right) \Biggr) x^r. \end{align}$$ At c = 1- γ, we get y2. Hence, y = E′ y1 + F′ y2. Let E′ b0 = E and F′ b0 = F. Then
 * $$\begin{align}

y &= \frac{E}{(\gamma)_{-\gamma}} \sum_{r = 1 - \gamma}^\infty \frac{(\alpha)_r (\beta)_r} {(1)_r (1)_{r + \gamma - 1}} x^r \\ &\quad\begin{align} {}+ F x^{1 - \gamma} \sum_{r = 0}^\infty &\frac{(1 - \gamma) (\alpha + 1 - \gamma)_r (\beta + 1 - \gamma)_r} {(2 - \gamma)_r (1)_r} \Biggl(\ln x + \frac{1}{1 - \gamma} + \\  &+ \sum_{k = 0}^{r - 1} \left(\frac{1}{\alpha + k + 1 - \gamma} + \frac{1}{\beta + k + 1 - \gamma} - \frac{1}{2 + k - \gamma} - \frac{1}{1 + k} \right) \Biggr) x^r. \end{align} \end{align}$$

γ > 1
From the recurrence relation
 * $$a_r = \frac{(r + c + \alpha - 1)(r + c + \beta - 1)}{(r + c)(r + c + \gamma - 1)} a_{r - 1},$$

we see that when c = 1 - γ (the smaller root), aγ − 1 → ∞. Hence, we must make the substitution a0 = b0(c − ci), where ci is the root for which our solution is infinite. Hence, we take a0 = b0(c + γ - 1) and our assumed solution takes the new form:
 * $$y_b = b_0 x^c \sum_{r = 0}^\infty \frac{(c + \gamma -1 )(c + \alpha )_r (c + \beta)_r}

{(c + 1)_r (c + \gamma)_r} x^r.$$ Then y1 = yb|undefined. All terms before
 * $$ \frac{(c + \gamma - 1)(c + \alpha)_{\gamma - 1} (c + \beta)_{\gamma - 1}}

{(c + 1)_{\gamma - 1}(c + \gamma)_{\gamma - 1}} x^{\gamma - 1}$$ vanish because of the c + γ - 1 in the numerator. Starting from this term, however, the c + γ - 1 in the numerator vanishes. To see this, note that
 * $$(c + 1)_{\gamma - 1} = (c + 1)(c + 2)\cdots(c + \gamma - 1).$$

Hence, our solution takes the form
 * $$\begin{align}

y_1 &= b_0 x^{1 - \gamma} \left(\frac{(\alpha + 1 - \gamma)_{\gamma - 1} (\beta + 1 - \gamma)_{\gamma - 1}}                                     {(2 - \gamma)_{\gamma - 2} (1)_{\gamma - 1}} x^{\gamma - 1}                               + \frac{(\alpha + 1 - \gamma)_{\gamma} (\beta + 1 - \gamma)_{\gamma}}                                      {(2 - \gamma)_{\gamma - 2} (1) (1)_{\gamma}} x^{\gamma}                               + \cdots \right) \\ &= \frac{b_0}{(2 - \gamma)_{\gamma - 2}} x^{1 - \gamma} \sum_{r = \gamma - 1}^\infty \frac{(\alpha + 1 - \gamma)_r (\beta + 1 - \gamma)_r} {(1)_r (1)_{r + 1 - \gamma}} x^r. \end{align}$$ Now,
 * $$y_2 = \left.\frac{\partial y_b}{\partial c}\right|_{c = 0}.$$

To calculate this derivative, let
 * $$M_r = \frac{(c + \gamma - 1)(c + \alpha)_r (c + \beta)_r}{(c + 1)_r (c + \gamma)_r}.$$

Then following the method in the second case above,
 * $$\begin{align}

\frac{\partial M_r}{\partial c}  &= \frac{(c + \gamma - 1) (c + \alpha)_r (c + \beta)_r}{(c + 1)_r (c + \gamma)_r} \Biggl(\frac{1}{c + \gamma - 1} + \\   &\qquad + \sum_{k = 0}^{r - 1} \left(\frac{1}{c + \alpha + k} + \frac{1}{c + \beta + k}                                      - \frac{1}{c + 1 + k} - \frac{1}{c + \gamma + k} \right) \Biggr) \end{align}$$ Now,
 * $$y_b = b_0 \sum_{r = 0}^\infty \frac{(c + \gamma - 1) (c + \alpha)_r (c + \beta)_r}

{(c + 1)_r (c + \gamma)_r} x^{r + c}          = b_0 x^c \sum_{r = 0}^\infty M_r x^r.$$ Hence,
 * $$\begin{align}

\frac{\partial y}{\partial c}  &= b_0 x^c \ln(x) \sum_{r = 0}^\infty \frac{(c + \gamma - 1) (c + \alpha)_r (c + \beta)_r} {(c + 1)_r (c + \gamma)_r} x^r +\\ &\qquad\begin{align} {}+ b_0 x^c \sum_{r = 0}^\infty &\frac{(c + \gamma - 1) (c + \alpha)_r (c + \beta)_r} {(c + 1)_r (c + \gamma)_r} \Biggl(\frac{1}{c + \gamma - 1} + \\            &+ \sum_{k = 0}^{r - 1} \left(\frac{1}{c + \alpha + k} + \frac{1}{c + \beta + k} - \frac{1}{c + 1 + k}                                          - \frac{1}{c + \gamma + k} \right) \Biggr)x^r \end{align} \\ &\begin{align} {}= b_0 x^c \sum_{r = 0}^\infty &\frac{(c + \gamma - 1) (c + \alpha)_r (c + \beta)_r} {(c + 1)_r (c + \gamma)_r} \Biggl(\ln x + \frac{1}{c + \gamma - 1} + \\                  &+ \sum_{k = 0}^{r - 1} \left(\frac{1}{c + \alpha + k} + \frac{1}{c +\beta + k}                                                 - \frac{1}{c + 1 + k} - \frac{1}{c + \gamma + k}\right) \Biggr) x^r. \end{align} \end{align}$$ At c = 0 we get y2. Hence, y = G′y1 + H′y2. Let G′b0 = E and H′b0 = F. Then
 * $$\begin{align}

y = \frac{G}{(2 - \gamma)_{\gamma - 2}} &x^{1 - \gamma} \sum_{r = \gamma - 1}^\infty \frac{(\alpha + 1 - \gamma)_r (\beta + 1 - \gamma)_r} {(1)_r (1)_{r + 1 - \gamma}} x^r \\ &\begin{align} {} + H \sum_{r = 0}^\infty &\frac{(1 - \gamma) (\alpha + 1 - \gamma)_r (\beta + 1 - \gamma)_r} {(2 - \gamma)_r (1)_r} \Biggl(\ln x + \frac{1}{\gamma - 1} + \\                               &+ \sum_{k = 0}^{r - 1} \left(\frac{1}{\alpha + k} + \frac{1}{\beta + k}                                           - \frac{1}{1 + k} - \frac{1}{\gamma + k} \right) \Biggr) x^r. \end{align} \end{align}$$

Solution around x = 1
Let us now study the singular point x = 1. To see if it is regular,
 * $$\begin{align}

&\lim_{x \to a} \frac{(x - a) P_1(x)}{P_2(x)} = \lim_{x \to 1} \frac{(x - 1) (\gamma - (1 + \alpha + \beta)x)}{x(1 - x)} \\ &\quad = \lim_{x \to 1} \frac{-(\gamma - (1 + \alpha + \beta)x)}{x} = 1 + \alpha + \beta - \gamma \\ &\lim_{x \to a} \frac{(x - a)^2 P_0(x)}{P_2(x)} = \lim_{x \to 1} \frac{(x - 1)^2 (-\alpha\beta)}{x(1 - x)} = \lim_{x \to 1} \frac{(x - 1) \alpha \beta}{x} = 0 \end{align}$$ Hence, both limits exist and x = 1 is a regular singular point. Now, instead of assuming a solution on the form
 * $$y = \sum_{r = 0}^\infty a_r (x - 1)^{r + c},$$

we will try to express the solutions of this case in terms of the solutions for the point x = 0. We proceed as follows: we had the hypergeometric equation
 * $$x(1 - x)y'' + (\gamma - (1 + \alpha + \beta)x)y' - \alpha\beta y = 0.\,$$

Let z = 1 - x. Then
 * $$\begin{align}

&\frac{dy}{dx} = \frac{dy}{dz} \times \frac{dz}{dx} = -\frac{dy}{dz} = -y' \\ &\frac{d^2 y}{dx^2} = \frac{d}{dx}\left( \frac{dy}{dx} \right) = \frac{d}{dx}\left( -\frac{dy}{dz} \right) = \frac{d}{dz}\left( -\frac{dy}{dz} \right) \times \frac{dz}{dx} =\frac{d^{2}y}{dz^{2}} = y'' \end{align}$$ Hence, the equation takes the form
 * $$z(1 - z) y'' + (\alpha + \beta - \gamma + 1 - (1 + \alpha + \beta)z) y' - \alpha\beta y = 0.\,$$

Since z = 1 - x, the solution of the hypergeometric equation at x = 1 is the same as the solution for this equation at z = 0. But the solution at z = 0 is identical to the solution we obtained for the point x = 0, if we replace each γ by α + β - γ + 1. Hence, to get the solutions, we just make this substitution in the previous results. Note also that for x = 0, c1 = 0 and c2 = 1 - γ. Hence, in our case, c1 = 0 while c2 = γ - α - β. Let us now write the solutions. It should be noted in the following we replaced each z by 1 - x.

γ − α − β not an integer
$$\begin{align} y &= A \cdot {{}_2 F_1}(\alpha, \beta; \alpha + \beta - \gamma + 1; 1 - x) \\ &\quad + B (1 - x)^{\gamma - \alpha - \beta} {{}_2 F_1}(\gamma - \alpha, \gamma - \beta;                                                          \gamma - \alpha - \beta + 1; 1 - x) \end{align}$$

γ − α − β = 0
$$\begin{align} y &= C \cdot {{}_2 F_1}(\alpha, \beta; 1; 1 - x) \\ &\quad + D \sum_{r = 0}^\infty \frac{(\alpha)_r (\beta)_r}{(1)_r^2} \left(\ln(1 - x)        + \sum_{k = 0}^{r - 1} \left(\frac{1}{\alpha + k} + \frac{1}{\beta + k} - \frac{2}{1 + k}\right)\right) (1 - x)^r \end{align}$$

γ − α − β > 0
$$\begin{align} y &= \frac{E}{(\alpha + \beta - \gamma + 1)_{\gamma - \alpha - \beta - 1}} \sum_{r = 1 - \gamma}^\infty \frac{(\alpha)_r (\beta)_r}{(1)_r (1)_{r + \alpha + \beta - \gamma}} (1 - x)^r + {}\\ &\quad\begin{align} {} + F(1 - x)^{\gamma - \alpha - \beta} \sum_{r = 0}^\infty & \frac{(\gamma - \alpha - \beta)(\gamma - \beta)_r (\gamma - \alpha)_r} {(1 + \gamma - \alpha - \beta)_r (1)_r} \Biggl(\ln(1 - x) + \frac{1}{\gamma - \alpha - \beta} + {} \\          &+ \sum_{k = 0}^{r - 1} \left(\frac{1}{k + \gamma - \beta} + \frac{1}{k + \gamma - \alpha} - \frac{1}{1 + k + \gamma - \alpha - \beta} - \frac{1}{1 + k} \right) \Biggr) (1 - x)^r \end{align} \end{align}$$

γ − α − β < 0
$$\begin{align} y &= \frac{G}{(1 + \gamma - \alpha - \beta)_{\alpha + \beta - \gamma - 1}} (1 - x)^{\gamma - \alpha - \beta} \sum_{r = \alpha + \beta - \gamma}^\infty \frac{(\gamma - \beta )_r (\gamma - \alpha)_r} {(1)_r (1)_{r + \gamma - \alpha - \beta}} (1 - x)^r + {}\\ &\quad\begin{align} {} + H \sum_{r = 0}^\infty & \frac{(\gamma - \alpha - \beta)(\gamma - \beta)_r (\gamma - \alpha)_r} {(1 + \gamma - \alpha - \beta)_r (1)_r} \Biggl(\ln(1 - x) + \frac{1}{\alpha + \beta - \gamma} + {} \\               & + \sum_{k = 0}^{r - 1} \left(\frac{1}{\alpha + k} + \frac{1}{\beta + k}                                           - \frac{1}{1 + k} - \frac{1}{\alpha + \beta - \gamma + 1 + k} \right)                  \Biggr) (1 - x)^r \end{align} \end{align}$$

Solution around infinity
Finally, we study the singularity as x → ∞. Since we can't study this directly, we let x = s−1. Then the solution of the equation as x → ∞ is identical to the solution of the modified equation when s = 0. We had

 * $$\begin{align}

& x(1-x)y''+\left\{ \gamma -(1+\alpha +\beta )x \right\}y'-\alpha \beta y=0 \\ & \frac{dy}{dx}=\frac{dy}{ds}\times \frac{ds}{dx}=-s^{^{2}}\times \frac{dy}{ds}=-s^{^{2}}y' \\ & \frac{d^{2}y}{dx^{2}}=\frac{d}{dx}\left( \frac{dy}{dx} \right)=\frac{d}{dx}\left( -s^{^{2}}\times \frac{dy}{ds} \right)=\frac{d}{ds}\left( -s^{^{2}}\times \frac{dy}{ds} \right)\times \frac{ds}{dx} \\ & \left( (-2s)\times \frac{dy}{ds}+(-s^{2})\frac{d^{2}y}{ds^{2}} \right) \times (-s^{2})=2s^{3}y'+s^{4}y'' \end{align} $$ Hence, the equation takes the new form

 * $$ \frac{1}{s} \left(1 - \frac{1}{s}\right) \left(2 s^3 y' + s^4 y''\right)

+ \left(\gamma - (1 + \alpha + \beta)\frac{1}{s} \right) (-s^2 y') - \alpha \beta y = 0$$

which reduces to

 * $$(s^{3}-s^{2})y''+ \bigl((2-\gamma )s^{2}+(\alpha +\beta -1)s\bigr)y'-\alpha \beta y = 0.$$

Let

 * $$ P_{0}(s)=-\alpha \beta, \qquad

P_{1}(s)=((2-\gamma )s^{2}+(\alpha +\beta -1)s), \qquad P_{2}(s)=(s^{3}-s^{2}).$$

As we said, we shall only study the solution when s = 0. As we can see, this is a singular point since P2(0) = 0. To see if it's regular,

 * $$\begin{align}

& \underset{s\to a}{\mathop{\lim }}\,\frac{\left( s-a \right)P_{1}(s)}{P_{2}(s)}=\underset{s\to 0}{\mathop{\lim }}\,\frac{\left( s-0 \right)((2-\gamma )s^{2}+(\alpha +\beta -1)s)}{(s^{3}-s^{2})}=\underset{s\to 0}{\mathop{\lim }}\,\frac{((2-\gamma )s^{2}+(\alpha +\beta -1)s)}{s^{2}-s} \\ & =\underset{s\to 0}{\mathop{\lim }}\,\frac{((2-\gamma )s+(\alpha +\beta -1))}{s-1}=1-\alpha -\beta \text{ }\text{. } \\ & \underset{s\to a}{\mathop{\lim }}\,\frac{\left( s-a \right)^{2}P_{0}(s)}{P_{2}(s)}=\underset{s\to 0}{\mathop{\lim }}\,\frac{\left( s-0 \right)^{2}\left( -\alpha \beta  \right)}{(s^{3}-s^{2})}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( -\alpha \beta  \right)}{s-1}=\alpha \beta \text{ }\text{.} \end{align}$$

Hence, both limits exist and s = 0 is a regular singular point. Therefore, we assume the solution takes the form

y=\sum\limits_{r=0}^{\infty }{a_{r}s^{r+c}} $$ with a0 ≠ 0.

Hence,

 * $$y'=\sum\limits_{r=0}^{\infty }{a_{r}(r+c)s^{r+c-1}}$$ and $$y''=\sum\limits_{r=0}^{\infty }{a_{r}(r+c)(r+c-1)s^{r+c-2}}.

$$

Substituting in the modified hypergeometric equation we get

 * $$\begin{align}

& (s^{3}-s^{2})y''+((2-\gamma )s^{2}+(\alpha +\beta -1)s)y'-\alpha \beta y=0 \\ & s^{3}\sum\limits_{r=0}^{\infty }{a_{r}(r+c)(r+c-1)s^{r+c-2}}-s^{2}\sum\limits_{r=0}^{\infty }{a_{r}(r+c)(r+c-1)x^{r+c-2}} \\ & +(2-\gamma )s^{2}\sum\limits_{r=0}^{\infty }{a_{r}(r+c)s^{r+c-1}}+(\alpha +\beta -1)s\sum\limits_{r=0}^{\infty }{a_{r}(r+c)s^{r+c-1}}-\alpha \beta \sum\limits_{r=0}^{\infty }{a_{r}s^{r+c}}=0 \end{align}$$

i.e.,

 * $$\begin{align}

\sum\limits_{r=0}^{\infty }&{a_{r}(r+c)(r+c-1)s^{r+c+1}}-\sum\limits_{r=0}^{\infty }{a_{r}(r+c)(r+c-1)x^{r+c}} \\ & +(2-\gamma )\sum\limits_{r=0}^{\infty }{a_{r}(r+c)s^{r+c+1}}+(\alpha +\beta -1)\sum\limits_{r=0}^{\infty }{a_{r}(r+c)s^{r+c}}-\alpha \beta \sum\limits_{r=0}^{\infty }{a_{r}s^{r+c}}=0 \end{align}$$

In order to simplify this equation, we need all powers to be the same, equal to r + c, the smallest power. Hence, we switch the indices as follows
 * $$\begin{align}

& \sum\limits_{r=1}^{\infty }{a_{r-1}(r+c-1)(r+c-2)s^{r+c}}-\sum\limits_{r=0}^{\infty }{a_{r}(r+c)(r+c-1)x^{r+c}} \\ & +(2-\gamma )\sum\limits_{r=1}^{\infty }{a_{r-1}(r+c-1)s^{r+c}}+(\alpha +\beta -1)\sum\limits_{r=0}^{\infty }{a_{r}(r+c)s^{r+c}}-\alpha \beta \sum\limits_{r=0}^{\infty }{a_{r}s^{r+c}}=0 \end{align}$$

Thus, isolating the first term of the sums starting from 0 we get

\begin{align} & a_{0}\left\{ -(c)(c-1)+(\alpha +\beta -1)(c)-\alpha \beta \right\}s^{c}+\sum\limits_{r=1}^{\infty }{a_{r-1}(r+c-1)(r+c-2)s^{r+c}} \\ & -\sum\limits_{r=1}^{\infty }{a_{r}(r+c)(r+c-1)x^{r+c}}+(2-\gamma )\sum\limits_{r=1}^{\infty }{a_{r-1}(r+c-1)s^{r+c}} \\ & +(\alpha +\beta -1)\sum\limits_{r=1}^{\infty }{a_{r}(r+c)s^{r+c}}-\alpha \beta \sum\limits_{r=1}^{\infty }{a_{r}s^{r+c}}=0 \\ \end{align}$$

Now, from the linear independence of all powers of s (i.e., of the functions 1, s, s2, ..., the coefficients of sk vanish for all k. Hence, from the first term we have

 * $$a_{0}\left\{ -(c)(c-1)+(\alpha +\beta -1)(c)-\alpha \beta \right\}=0$$

which is the indicial equation. Since a0 ≠ 0, we have

 * $$(c)(-c+1+\alpha +\beta -1)-\alpha \beta )=0.\,$$

Hence, c1 = α and c2 = β.

Also, from the rest of the terms we have
 * $$\begin{align}

& \left\{ (r+c-1)(r+c-2)+(2-\gamma )(r+c-1) \right\}a_{r-1} \\ & +\left\{ -(r+c)(r+c-1)+(\alpha +\beta -1)(r+c)-\alpha \beta \right\}a_{r}=0 \end{align}$$

Hence,
 * $$\begin{align}

& a_{r}=-\frac{\left\{ (r+c-1)(r+c-2)+(2-\gamma )(r+c-1) \right\}}{\left\{ -(r+c)(r+c-1)+(\alpha +\beta -1)(r+c)-\alpha \beta \right\}}a_{r-1} \\ & \text{  }=\frac{\left\{ (r+c-1)(r+c-\gamma ) \right\}}{\left\{ (r+c)(r+c-\alpha -\beta )+\alpha \beta  \right\}}a_{r-1} \end{align}$$ But
 * $$\begin{align}

(r+c)(r+c-\alpha -\beta )+\alpha \beta &=(r+c-\alpha )(r+c)-\beta (r+c)+\alpha \beta \\ & =(r+c-\alpha )(r+c)-\beta (r+c-\alpha ). \end{align}$$

Hence, we get the recurrence relation

a_{r}=\frac{(r+c-1)(r+c-\gamma )}{(r+c-\alpha )(r+c-\beta )}a_{r-1},\,\forall r \ge 1 $$

Let's now simplify this relation by giving ar in terms of a0 instead of ar − 1. From the recurrence relation,
 * $$\begin{align}

& a_{1}=\frac{(c)(c+1-\gamma )}{(c+1-\alpha )(c+1-\beta )}a_{0} \\ & a_{2}=\frac{(c+1)(c+2-\gamma )}{(c+2-\alpha )(c+2-\beta )}a_{1}=\frac{(c+1)(c)(c+2-\gamma )(c+1-\gamma )}{(c+2-\alpha )(c+1-\alpha )(c+2-\beta )(c+1-\beta )}a_{0} \\ & =\text{ }\frac{(c)_{2}(c+1-\gamma )_{2}}{(c+1-\alpha )_{2}(c+1-\beta )_{2}}a_{0} \end{align}$$

As we can see,
 * $$a_{r}=\frac{(c)_{r}(c+1-\gamma )_{r}}{(c+1-\alpha )_{r}(c+1-\beta )_{r}}a_{0}\text{   }\forall \text{r}\ge \text{0}

$$

Hence, our assumed solution takes the form

y=a_{0}\sum\limits_{r=0}^{\infty }{\left( \frac{(c)_{r}(c+1-\gamma )_{r}}{(c+1-\alpha )_{r}(c+1-\beta )_{r}}s^{r+c} \right)} $$

We are now ready to study the solutions corresponding to the different cases for c1 − c2 = α − β.

α − β not an integer
Then y1 = y|undefined and y2 = y|undefined. Since

y=a_{0}\sum\limits_{r=0}^{\infty }{\left( \frac{(c)_{r}(c+1-\gamma )_{r}}{(c+1-\alpha )_{r}(c+1-\beta )_{r}}s^{r+c} \right)} $$,

we have

\begin{align} y_{1}&=a_{0}\sum\limits_{r=0}^{\infty }{\left( \frac{(\alpha )_{r}(\alpha +1-\gamma )_{r}}{(1)_{r}(\alpha +1-\beta )_{r}}s^{r+\alpha } \right)}=a_{0}s^{\alpha }_{2}F_{1}(\alpha ,\text{ }\alpha +1-\gamma ;\text{ }\alpha +1-\beta ;\text{ s)} \\ y_{2}&=a_{0}\sum\limits_{r=0}^{\infty }{\left( \frac{(\beta )_{r}(\beta +1-\gamma )_{r}}{(\beta +1-\alpha )_{r}(1)_{r}}s^{r+\beta } \right)=a_{0}s^{\beta }_{2}F_{1}(\beta ,\text{ }\beta +1-\gamma ;\text{ }\beta +1-\alpha ;\text{ s)}} \end{align} $$

Hence, y = A′y1 + B′y2. Let A′a0 = A and B′a0 = B. Then, noting that s = x-1,
 * $$\begin{align}

& y=Ax^{-\alpha }_{2}F_{1}(\alpha ,\text{ }\alpha +1-\gamma ;\text{ }\alpha +1-\beta ;\text{ x}^{-1})+Bx^{-\beta }_{2}F_{1}(\beta ,\text{ }\beta +1-\gamma ;\text{ }\beta +1-\alpha ;\text{ x}^{-1}) \\ \end{align} $$

α − β = 0
Then y1 = y|undefined. Since α = β, we have

y=a_{0}\sum\limits_{r=0}^{\infty }{\left( \frac{(c)_{r}(c+1-\gamma )_{r}}{\left( (c+1-\alpha )_{r} \right)^{2}}s^{r+c} \right)}$$

Hence,
 * $$\begin{align}

& y_{1}=a_{0}\sum\limits_{r=0}^{\infty }{\left( \frac{(\alpha )_{r}(\alpha +1-\gamma )_{r}}{(1)_{r}(1)_{r}}s^{r+\alpha } \right)}=a_{0}s^{\alpha }_{2}F_{1}(\alpha ,\text{ }\alpha +1-\gamma ;\text{ 1; s)} \\ & y_{2}=\frac{\partial y}{\partial c}\text{ at }c=\alpha \end{align}$$

To calculate this derivative, let
 * $$\begin{align}

& M_{r}=\frac{(c)_{r}(c+1-\gamma )_{r}}{\left( (c+1-\alpha )_{r} \right)^{2}} \end{align} $$

Then using the method in the case γ = 1 above, we get
 * $$\begin{align}

& \frac{\partial M_{r}}{\partial c}=\frac{(c)_{r}(c+1-\gamma )_{r}}{\left( (c+1-\alpha )_{r} \right)^{2}}\left\{ \sum\limits_{k=0}^{r-1}{\left( \frac{1}{c+k}+\frac{1}{c+1-\gamma +k}-\frac{2}{c+1-\alpha +k} \right)} \right\} \\ \end{align}$$

Now,
 * $$\begin{align}

y=a_{0}s^{c}\sum\limits_{r=0}^{\infty }{\left( \frac{(c)_{r}(c+1-\gamma )_{r}}{\left( (c+1-\alpha )_{r} \right)^{2}}s^{r} \right)}=a_{0}s^{c}\sum\limits_{r=0}^{\infty }{M_{r}s^{r}} \end{align}$$

Hence
 * $$\begin{align}

& =a_{0}s^{c}\text{ ln(s)}\sum\limits_{r=0}^{\infty }{\left( \frac{(c)_{r}(c+1-\gamma )_{r}}{\left( (c+1-\alpha )_{r} \right)^{2}}s^{r} \right)} \\ & \text{       }+a_{0}s^{c}\sum\limits_{r=0}^{\infty }{\left( \frac{(c)_{r}(c+1-\gamma )_{r}}{\left( (c+1-\alpha )_{r} \right)^{2}}\left\{ \sum\limits_{k=0}^{r-1}{\left( \frac{1}{c+k}+\frac{1}{c+1-\gamma +k}-\frac{2}{c+1-\alpha +k} \right)} \right\}s^{r} \right)} \\ \end{align}$$

Hence,
 * $$\begin{align}

& \frac{\partial y}{\partial c}=a_{0}s^{c}\sum\limits_{r=0}^{\infty }{\left( \left( \frac{(c)_{r}(c+1-\gamma )_{r}}{\left( (c+1-\alpha )_{r} \right)^{2}} \right)\left( \ln s+\sum\limits_{k=0}^{r-1}{\left( \frac{1}{c+k}+\frac{1}{c+1-\gamma +k}-\frac{2}{c+1-\alpha +k} \right)} \right)s^{r} \right)} \\ \end{align}$$

For c = α we get
 * $$\begin{align}

& y_{2}=a_{0}s^{\alpha }\sum\limits_{r=0}^{\infty }{\left( \left( \frac{(\alpha )_{r}(\alpha +1-\gamma )_{r}}{\left( (1)_{r} \right)^{2}} \right)\left( \ln s+\sum\limits_{k=0}^{r-1}{\left( \frac{1}{\alpha +k}+\frac{1}{\alpha +1-\gamma +k}-\frac{2}{1+k} \right)} \right)s^{r} \right)} \\ & \end{align}$$ Hence, y = C′y1 + D′y2. Let C′a0 = C and D′a0 = D. Noting that s = x-1,

y=Cx^{-\alpha }_{2}F_{1}(\alpha ,\alpha +1-\gamma ; 1; x^{-1})$$

 * $$+Dx^{-\alpha }\sum\limits_{r=0}^{\infty }{\left( \left( \frac{(\alpha )_{r}(\alpha +1-\gamma )_{r}}{\left( (1)_{r} \right)^{2}} \right)\left( \ln x^{-1}+\sum\limits_{k=0}^{r-1}{\left( \frac{1}{\alpha +k}+\frac{1}{\alpha +1-\gamma +k}-\frac{2}{1+k} \right)} \right)x^{-r} \right)}$$

α − β > 0
From the recurrence relation

 * $$a_{r}=\frac{(r+c-1)(r+c-\gamma )}{(r+c-\alpha )(r+c-\beta )}a_{r-1}

$$ we see that when c = β (the smaller root), aα - β → ∞. Hence, we must make the substitution a0 = b0(c − ci), where ci is the root for which our solution is infinite. Hence, we take a0 = b0(c − β) and our assumed solution takes the new form

 * $$y_{b}=b_{0}\sum\limits_{r=0}^{\infty }{\left( \frac{(c-\beta )(c)_{r}(c+1-\gamma )_{r}}{(c+1-\alpha )_{r}(c+1-\beta )_{r}}s^{r+c} \right)}

$$

Then y1 = yb|undefined. As we can see, all terms before

 * $$\frac{(c-\beta )(c)_{\alpha -\beta }(c+1-\gamma )_{\alpha -\beta }}{(c+1-\alpha )_{\alpha -\beta }(c+1-\beta )_{\alpha -\beta }}s^{\alpha -\beta }

$$

vanish because of the c − β in the numerator.

But starting from this term, the c − β in the numerator vanishes. To see this, note that

(c+1-\alpha )_{\alpha -\beta } =(c+1-\alpha )(c+2-\alpha )\cdots(c-\beta ). $$

Hence, our solution takes the form
 * $$\begin{align}

& y_{1}=b_{0}\left( \frac{(\beta )_{\alpha -\beta }(\beta +1-\gamma )_{\alpha -\beta }}{(\beta +1-\alpha )_{\alpha -\beta -1}(1)_{\alpha -\beta }}s^{\alpha -\beta }+\frac{(\beta )_{\alpha -\beta +1}(\beta +1-\gamma )_{\alpha -\beta +1}}{(\beta +1-\alpha )_{\alpha -\beta -1}(1)(1)_{\alpha -\beta +1}}s^{\alpha -\beta +1}+... \right) \\ & =\frac{b_{0}}{(\beta +1-\alpha )_{\alpha -\beta -1}}\sum\limits_{r=\alpha -\beta }^{\infty }{\left( \frac{(\beta )_{r}(\beta +1-\gamma )_{r}}{(1)_{r}(1)_{r+\beta -\alpha }}s^{r} \right)} \\ & \\ \end{align}$$

Now,
 * $$y_{2}=\left.\frac{\partial y_{b}}{\partial c}\right|_{c=\alpha}.$$

To calculate this derivative, let
 * $$ M_{r}=\frac{(c-\beta )(c)_{r}(c+1-\gamma )_{r}}{(c+1-\alpha )_{r}(c+1-\beta )_{r}}.$$

Then using the method in the case γ = 1 above we get
 * $$\begin{align}

\frac{\partial M_r}{\partial c}  &= \frac{(c - \beta)(c)_r (c + 1 - \gamma)_r}{(c + 1 - \alpha)_r (c + 1 - \beta)_r}\Biggl(\frac{1}{c - \beta} + \\   &\qquad + \sum_{k = 0}^{r - 1} \left(\frac{1}{c + k} + \frac{1}{c + 1 - \gamma + k}                                       -\frac{1}{c + 1 - \alpha + k} - \frac{1}{c + 1 - \beta + k} \right) \Biggr) \end{align}$$

Now,
 * $$ y_{b}=b_{0}\sum\limits_{r=0}^{\infty }{\left( \frac{(c-\beta )(c)_{r}(c+1-\gamma )_{r}}{(c+1-\alpha )_{r}(c+1-\beta )_{r}}s^{r+c} \right)}=b_{0}x^{c}\sum\limits_{r=0}^{\infty }{M_{r}s^{r}}

$$

Hence,
 * $$\begin{align}

\frac{\partial y}{\partial c} &= b_0 s^c \ln(s) \sum_{r = 0}^\infty \frac{(c - \beta)(c)_r (c + 1 - \gamma)_r} {(c + 1 - \alpha)_r (c + 1 - \beta)_r} s^r \\ &\quad\begin{align} {}+ b_0 s^c \sum_{r = 0}^\infty & \frac{(c - \beta) (c)_r (c + 1 - \gamma)_r} {(c + 1 - \alpha)_r (c + 1 - \beta)_r} \Biggl(\frac{1}{c - \beta} + \\   &+ \sum_{k = 0}^{r - 1} \left(\frac{1}{c + k} + \frac{1}{c + 1 - \gamma + k} - \frac{1}{c + 1 - \alpha + k}                               - \frac{1}{c + 1 - \beta + k} \right) \Biggr) s^r \end{align} \end{align} $$

Hence,
 * $$\begin{align}

\frac{\partial y}{\partial c} = b_0 s^c \sum_{r = 0}^\infty &\frac{(c - \beta)(c)_r (c + 1 - \gamma)_r} {(c + 1 - \alpha)_r (c + 1 - \beta)_r} \Biggl(\ln s + \frac{1}{c - \beta } + \\ &+ \sum_{k = 0}^{r - 1} \left(\frac{1}{c + k} + \frac{1}{c + 1 - \gamma + k}                                    - \frac{1}{c + 1 - \alpha + k} - \frac{1}{c + 1 - \beta + k}\right) \Biggr) s^{r} \end{align} $$

At c = α we get y2. Hence, y = E′y1 + F′y2. Let E′b0 = E and F′b0 = F. Noting that s = x-1 we get
 * $$\begin{align}

y &= \frac{E}{(\beta + 1 - \alpha)_{\alpha - \beta - 1}} \sum_{r = \alpha - \beta}^\infty \frac{(\beta)_r (\beta + 1 - \gamma)_r} {(1)_r (1)_{r + \beta - \alpha}} x^{-r} \\ &\quad\begin{align} {} + F x^{-\alpha} \sum_{r = 0}^\infty & \frac{(\alpha - \beta) (\alpha)_r (\alpha + 1 - \gamma)_r} {(1)_r (\alpha + 1 - \beta)_r} \Biggl(\ln x^{-1} + \frac{1}{\alpha -\beta } \\    &+ \sum_{k = 0}^{r - 1} \left(\frac{1}{\alpha + k} + \frac{1}{\alpha + 1 + k - \gamma} -\frac{1}{1 + k}                                 -\frac{1}{\alpha + 1 + k - \beta} \right) \Biggr) x^{-r} \end{align} \end{align}$$

α − β < 0
From the symmetry of the situation here, we see that

 * $$\begin{align}

y &= \frac{G}{(\alpha + 1 - \beta)_{\beta - \alpha - 1}} \sum_{r = \beta - \alpha}^\infty \frac{(\alpha)_r (\alpha + 1 - \gamma)_r} {(1)_r (1)_{r + \alpha - \beta}} x^{-r} \\ &\quad\begin{align} {} + H x^{-\beta} \sum_{r = 0}^\infty &\frac{(\beta - \alpha) (\beta)_r (\beta + 1 - \gamma)_r} {(1)_r (\beta + 1 - \alpha)_r} \Biggl(\ln x^{-1} + \frac{1}{\beta - \alpha } \\    &+ \sum_{k = 0}^{r - 1} \left(\frac{1}{\beta + k} + \frac{1}{\beta + 1 + k - \gamma} - \frac{1}{1 + k}                                - \frac{1}{\beta + 1 + k - \alpha} \right) \Biggr) x^{-r} \end{align} \end{align}$$


text:
Israeli History

History of Israel In this book you will learn the history of the modern state of Israel and the people who live in it. The aim of this book is to be easy to read, informative, and unbiased. It uses plenty of primary sources to give you a sense of what people thought at the time. If you have something to add to this book, or you'd like to correct an error, just click "edit this page" and fix it yourself!

Table of Contents

 * /The Zionist Movement/
 * /Palestine under the Mandate/
 * /From World War II to Partition/
 * /From Partition to Independence/
 * /War, Ceasefire and Refugees/
 * /Challenges for a New State/
 * /Six-Day War, Yom Kippur War, Settlements/
 * /Peace and Likud Rule/
 * 1980s: First Lebanon War and Infitada
 * 1990-2005: Oslo, Netanyahu, Sharon
 * /Israel Today/

Authors
Daniel Shalom is the pseudonym for a Jewish man living in the United States.

Joanna Kristall works for Yeshiva University.

Related Wikibooks

 * Judaism
 * Hebrew
 * Cookbook:Cuisine of Israel

title: Israeli History


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title: User talk:76.234.115.211


text: User talk:Ultracouturedesign

title: User talk:Ultracouturedesign


text:
Muggles' Guide to Harry Potter/Magic/Tebo

Overview
Tebos are ash-coloured warthogs with the ability to become invisible, making them extremely dangerous creatures.

Extended Description
Tebos are found in Congo and Zaire. Their hide is valuable in the making of protective shields and clothing.


text:
Muggles' Guide to Harry Potter/Magic/Winged horse

Overview
Winged horses are found in a variety of breeds worldwide.

Extended Description
The owner of a winged horse is required by law to regularly enchant the creature with a Disillusionment Charm to deceive any onlooking Muggles.

Fantastic Beasts and Where to Find Them describes the following varieties of winged horse:

 * Abraxan — A powerful and large-sized palomino. In Harry's fourth year at Hogwarts, the Triwizard delegation from Beauxbatons arrives in a carriage drawn by twelve enormous winged palominos that, according to Madame Maxime who breeds them, require "forceful handling" and drink only single-malt whiskey.

 * Aethonan — A chestnut-coated British breed.

 * Granian — A relatively agile breed, grey in color.

 * Thestral — A large, skeletal winged horse, invisible to anyone who has not witnessed death. Thestrals are bred at Hogwarts.

A Hippogriff, while being a winged creature with the body of a horse, is not mentioned as being a variety of winged horse.


text: Muggles' Guide to Harry Potter/Magic/Streeler


text: Muggles' Guide to Harry Potter/Magic/Snidget


text: User talk:193.62.43.1

title: User talk:193.62.43.1


text:
User talk:212.240.143.240

Join Us!
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title: User talk:212.240.143.240


text:
Wikijunior:World War II/Soviet-German War

The Soviet-German War also known in Russia as The Great Patriotic War, was fought between Germany and its allies: Finland, Italy, Hungary, Romania and Bulgaria against the Soviet Union. It officially began on 22 June 22 1941 and ended on 8 May 1945. People of the countries of the former Soviet Union celebrate their victory over Germany on 9 May every year.


text: User talk:12.110.140.45

title: User talk:12.110.140.45


text: User talk:72.4.81.178

title: User talk:72.4.81.178


text:
User talk:Ranjith

Welcome!
Welcome, Ranjith!

Come introduce yourself at the new users page. If you have any questions, you can ask there or contact me personally.-- – Mike.lifeguard  | talk 20:42, 20 March 2008 (UTC)

title: User talk:Ranjith


text: User talk:76.91.104.93

title: User talk:76.91.104.93


text:
Prealgebra for Two-Year Colleges/Workbook AIE/Properties of addition

Materials Needed: None

Why
What does addition of fractions have in common with addition of decimals? What do these have in common with addition of expressions that contain variables? What does finding a perimeter have in common with balancing your checkbook?

Learning Objectives

 * 1) Explain why when we add, we need common denominators or like place values or like terms or common units.
 * 2) Solve problems that involve finding totals, finding perimeters, or combining values.
 * 3) Use the commutative property of addition to aid in performing mental addition.

Warm-up
An child in elementary is asked, "What is the value of 3 dimes and 5 pennies?" The child answers, "3 + 5 = 8. The money is worth 8 cents." Explain to the child why his reasoning is wrong, and how he could find the correct answer.

You can't add dimes and pennies because they have different values.

A correct way to add would be to convert the 3 dimes to 30 cents, and then to add the 8 pennies. The answer would be 38 cents.

Activity
...


text:
Prealgebra for Two-Year Colleges/Workbook AIE/Representations of numbers

Materials Needed: None

Why
What is the connection between fractions, decimals, percents, number lines, and measurements? What are the interpretations and uses for negative numbers?

Learning Objectives

 * 1) Convert between fractions, decimals, mixed numbers, percents, number lines, and two-dimensional or three-dimensional representations of quantities.
 * 2) Interpret negative numbers and use negative numbers to describe situations.

Warm-up
...

Activity
...


text:
Prealgebra for Two-Year Colleges/Workbook AIE/Properties of subtraction

Materials Needed: None

Why
What does subtraction of fractions have in common with subtraction of decimals? What do these have in common with subtraction of expressions that contain variables? How is subtraction related to addition?

Learning Objectives

 * 1) Explain why when we subtract, we need common denominators or like place values or like terms or common units.
 * 2) Solve problems that involve finding changes, differences, or missing pieces.
 * 3) Use the commutative property of addition to aid in performing mental subtraction.

Warm-up
...

Activity
...


text:
Prealgebra for Two-Year Colleges/Workbook AIE/Properties of multiplication

Materials Needed: None

Why
...

Learning Objectives

 * 1) Use the distributive property to estimate products, and then to correct your estimate.
 * 2) Find and use patterns in the multiplication table.

Warm-up
You have $80 in your wallet. You want to buy 3 DVD's for $19.95 each. (Fortunately, you are in Oregon where there is no sales tax.)

 * 1. Approximate the price a DVD to the nearest whole dollar.
 * 2. Estimate the cost of the three DVD's.
 * 3. Estimate how much money you will have in your wallet after the purchase.
 * 4. What do you need to add or subtract from your approximate price of a DVD to get the actual price?
 * 5. What do you need to add or subtract from your estimate of the cost of the three DVD's to get the actual cost?
 * 6. Exactly how much money you will have in your wallet after the purchase?
 * 7. Explain how your estimation and calculation is related to the following application of the distributive property. Be specific about why you are adding or subtracting.
 * $$\begin{align}

80 - \underbrace{3(20 - 0.05)} & = 80 - \underbrace{3(20)} + \underbrace{3(0.05)} \\ & = \underbrace{80 - 60} + 0.15 \\ & = \underbrace{20 + 0.15} \\ & = 20.15 \\ \end{align} $$

Activity
...


text:
User talk:Sathish121

Welcome!
Welcome, Sathish121!

Come introduce yourself at the new users page. If you have any questions, you can ask there or contact me personally.-- – Mike.lifeguard  | talk 21:13, 20 March 2008 (UTC)

title: User talk:Sathish121


text:
User talk:Parth.Satish

Welcome!
Welcome, Parth.Satish!

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title: User talk:Parth.Satish


text:
Talk:Java Programming/Keywords/super

Following use of 'super' is not mentioned:

Which means that GenericClass may only be SomeClass or a superclass/superinterface to SomeClass
 * It is now described. 20:12, 29 November 2013 (UTC)

Used in constructors?
Is super also used in constructors?

public class MyClass2 extends MyClass1 { ...  // constructor public MyClass2(ClassA1 arg1, ClassA2 arg2, ClassA3 arg3) {   super(arg1, arg2) // calls MyClass1 constructor ... } } . Albmont (talk) 19:21, 30 December 2009 (UTC)
 * Yes, as you write it. 20:12, 29 November 2013 (UTC)

title: Talk:Java Programming/Keywords/super


text:
Talk:Java Programming/Keywords/extends

Following use of 'extends' is not mentioned:  Which means that GenericClass may only be SomeClass or a subclass/subinterface to SomeClass

title: Talk:Java Programming/Keywords/extends


text: Template:CategoryList

title: Template:CategoryList


text: Template:CategoryJunction

title: Template:CategoryJunction


text: Template:CategoryIntersection

title: Template:CategoryIntersection


text:
User talk:146.201.93.158

Join Us!
-- – Mike.lifeguard  | talk 23:30, 20 March 2008 (UTC)

title: User talk:146.201.93.158


text:
User talk:Makavich von wiseguy

Welcome!
Welcome, Makavich von wiseguy!

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title: User talk:Makavich von wiseguy


text:
User talk:Osama~enwikibooks

Welcome!
Welcome, Osama!

Come introduce yourself at the new users page. If you have any questions, you can ask there or contact me personally.-- – Mike.lifeguard  | talk 23:51, 20 March 2008 (UTC)

Your account will be renamed
Hello,

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

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

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

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:35, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 05:20, 19 April 2015 (UTC)

title: User talk:Osama~enwikibooks


text:
User talk:Just 0ne man

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Welcome, Just 0ne man!

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title: User talk:Just 0ne man


text:
User talk:Hardwarecompugeek~enwikibooks

Welcome!
Welcome, Hardwarecompugeek!

Come introduce yourself at the new users page. If you have any questions, you can ask there or contact me personally.-- – Mike.lifeguard  | talk 23:51, 20 March 2008 (UTC)

Your account will be renamed
Hello,

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

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

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

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:16, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 05:01, 19 April 2015 (UTC)

title: User talk:Hardwarecompugeek~enwikibooks


text:
Talk:CAT-Tools/DéjàVu X/deal with USB dongle problems

This is the worst tools ever I used with Windows XP no matter what instruction you follow you still get this stupid demo mode every time you run the application even te dongl is there


text:
Prealgebra for Two-Year Colleges/Appendix (procedures)/Adding fractions

Find a common denominator and then add the numerators. Reduce the result, if you can.


text:
Prealgebra for Two-Year Colleges/Appendix (procedures)/Subtracting fractions

Find a common denominator and then subtract the numerators. Reduce the result if you can.


text:
Prealgebra for Two-Year Colleges/Appendix (procedures)/Reducing fractions

Equivalent fractions represent the same number, although they are written differently. For example,
 * $$\frac{2}{4}$$ and $$\frac{1}{2}$$

represent the same amount. The latter fraction has a smaller denominator, so we say the fraction has been reduced.

You can reduce a fraction by dividing the numerator and denominator by the same number. For example,
 * $$\frac{75}{100} = \frac{75 {\color{Red}\div 5}}{100 {\color{Red}\div 5}} = \frac{15}{20}$$

and
 * $$\frac{15}{20} = \frac{15 {\color{Red}\div 5}}{20 {\color{Red}\div 5}} = \frac{3}{4}$$.

When we ask you to simplify a fraction or reduce a fraction such as 75/100, we want you to find the equivalent fraction with the smallest possible denominator. In this case, the fully reduced form is 3/4. The only number that we could divided into both 3 and 4 would be 1, but that would not reduce the fraction any further.
 * $$\frac{3}{4} = \frac{3 {\color{Red}\div 1}}{4 {\color{Red}\div 1}} = \frac{3}{4}$$.

You are also allowed to multiply the numerator and denominator by the same number. We usually call this finding equivalent fractions.


text:
Prealgebra for Two-Year Colleges/Appendix (procedures)/Multiplying fractions

Multiply the numerators to find the numerator of the result. Multiply the denominators to find the denominators of the result. Then reduce the result. For example,
 * $$\frac{2}{21} \times \frac{35}{22} = \frac{70}{462}$$.

We can then reduce
 * $$\frac{70}{462} = \frac{70 {\color{Red}\div 2}}{462 {\color{Red}\div 2}} = \frac{35}{231} = \frac{35 {\color{Red}\div 7}}{231 {\color{Red}\div 7}} = \frac{5}{33}$$.

An easier way to multiply the same fractions is first find the prime factors of the numerators and denominators, then reduce, then multiply.
 * $$\frac{2}{21} \times \frac{35}{22} = \frac{2}{3 \cdot 7} \times \frac{5 \cdot 7}{2 \cdot 11} = \frac{\color{Red}\not{2}}{3 \cdot {\color{Blue}\not{7}}} \times \frac{5 \cdot {\color{Blue}\not{7}}}{{\color{Red}\not{2}} \cdot 11} = \frac{5}{33}$$.


text:
Prealgebra for Two-Year Colleges/Appendix (procedures)/Dividing fractions

To divide fractions, we change the division problem into a multiplication problem, and we replace the second factor with its reciprocal. Example:
 * $$\frac{2}{21} \div \frac{\color{Red}22}{\color{Blue}35} = \frac{2}{21} \times \frac{\color{Blue}35}{\color{Red}22}$$.

Then we multiply and reduce as usual.


text:
Outdoor Education/Scope of outdoor education

This chapter describes what outdoor education is and isn't, definitions, similar terms and concepts, and how the meaning/interpretation of outdoor education may have varied over time and across cultures.

Tasks

 * 1) Revise Outdoor education
 * 2) Transfer relevant content to here
 * 3) Add refs from Outdoor education to here


text: Wikibooks:Requests for deletion/AAVE

title: Wikibooks:Requests for deletion/AAVE


text:
User:タチコマ robot/Editcounter

Month-Graph:yes

Namespacesum-Graph:yes

Weekday-Graph:yes

Hour-Graph:yes

Usercontributions:yes

title: User:タチコマ robot/Editcounter


text:
Talk:African American Vernacular English

Source?
What is the source of this book? Fredlesaltique (discuss • contribs) 23:12, 20 February 2023 (UTC)


text:
Prealgebra for Two-Year Colleges/Appendix (procedures)/Reading Decimal Numbers

The word and is used only for the decimal point; And is also used to separate a whole number and a fraction. A comma is used at every third place, starting at the decimal point and moving left. Frequently a comma or space is used at every third place moving to the right of the decimal point.

Additionally, no comma is used before and. Also, all decimals end in ths except unitary decimals that end in th.


text:
Prealgebra for Two-Year Colleges/Appendix (procedures)/Adding whole numbers with carrying

Addition
Addition operations are denoted by the + sign and the - (-)= +. The addition operator (plus sign) will take any two numbers, called addends, as operands to work on. The result is called the sum of the two numbers.

The operation of addition is commutative. This means that the addition of two numbers will give the same sum regardless of the order in which the numbers are added.

Example:
 * $$7 + 5 = 12$$ and $$5 + 7 = 12$$

 * $$3 + 2 + 4 = 9$$ and $$4 + 3 + 2 = 9$$

Vertical Addition
To add vertically, stack the numbers on top.

52 19

Now add the numbers in the first or right column. If you get a number with tens in it, put what's in the tens place on the tens place in the problem. You should have:

1 52  19   1

Add the tens column now.

1 52  19  71

Repeat in a similar fashion for each additional column (hundreds, thousands, etc.)until you finish the problem.

Exercises
In Exercises 1–25, find the sum.
 * 1. $$5 + 3$$
 * 2. $$8 + 7$$
 * 3. $$9 + 2$$
 * 4. $$6 + 3$$
 * 5. $$1 + 4$$
 * 6. $$2 + 17$$
 * 7. $$12 + 11$$
 * 8. $$53 + 8$$
 * 9. $$41 + 9$$
 * 10. $$84 + 12$$
 * 11. $$16 + 17$$
 * 12. $$7,576 + 5,345$$
 * 13. $$2,345 + 3,245$$
 * 14. $$8,952 + 9,423$$
 * 15. $$2,783 + 2,389$$
 * 16. $$189,583 + 1,574,822$$
 * 17. $$1.5 + 2.7$$
 * 18. $$5.4 + 3.9$$
 * 19. $$8.3 + 9.2$$
 * 20. $$2.23 + 4.89$$
 * 21. $$534.4 + 34.675$$
 * 22. $$348.904 + 23,498.2$$
 * 23. $$1.673 + 48,210.38$$
 * 24. $$10.4823 + 94.29478$$
 * 25. $$128.52 + 2,070.24$$

Answers

 * 1. $$8$$
 * 2. $$15$$
 * 3. $$11$$
 * 4. $$9$$
 * 5. $$5$$
 * 6. $$19$$
 * 7. $$23$$
 * 8. $$61$$
 * 9. $$50$$
 * 10. $$96$$
 * 11. $$33$$
 * 12. $$12,921$$
 * 13. $$5,590$$
 * 14. $$18,375$$
 * 15. $$5,172$$
 * 16. $$1,764,405$$
 * 17. $$4.2$$
 * 18. $$9.3$$
 * 19. $$17.5$$
 * 20. $$7.12$$
 * 21. $$569.075$$
 * 22. $$23,847.104$$
 * 23. $$48,212.053$$
 * 24. $$104.77708$$
 * 25. $$2,198.76$$

Subtraction
Subtraction, as you probably also have seen, uses the minus (-) sign. The generic subtraction operator will take any two numbers as operands. The result is called the difference of the two numbers.

Subtraction is not a commutative operation. Changing the order of the operands will likely give a different (not the same) result.

Example:
 * $$7 - 5 = 2$$ and $$5 - 7 = -2$$

 * $$3 - 2 - 4 = -3$$ and $$4 - 3 - 2 = -1$$

Exercises
In Exercises 26–50, find the difference.
 * 26. $$5 - 3$$
 * 27. $$8 - 7$$
 * 28. $$9 - 2$$
 * 29. $$6 - 3$$
 * 30. $$1 - 4$$
 * 31. $$2 - 17$$
 * 32. $$12 - 11$$
 * 33. $$53 - 8$$
 * 34. $$41 - 9$$
 * 35. $$84 - 12$$
 * 36. $$16 - 17$$
 * 37. $$7,576 - 5,345$$
 * 38. $$2,345 - 3,245$$
 * 39. $$8,952 - 9,423$$
 * 40. $$2,783 - 2,389$$
 * 41. $$1,574,822 - 189,583$$
 * 42. $$2.7 - 1.5$$
 * 43. $$5.4 - 3.9$$
 * 44. $$8.3 - 9.2$$
 * 45. $$2.23 - 4.89$$
 * 46. $$10.38 - 1.673$$
 * 47. $$534.4 - 34.675$$
 * 48. $$348.904 - 23,498.2$$
 * 49. $$10.4823 - 94.29478$$
 * 50. $$2,070.24 - 128.52$$

Answers

 * 26. $$2$$
 * 27. $$1$$
 * 28. $$7$$
 * 29. $$3$$
 * 30. $$-3$$
 * 31. $$-15$$
 * 32. $$1$$
 * 33. $$45$$
 * 34. $$32$$
 * 35. $$72$$
 * 36. $$-1$$
 * 37. $$2,231$$
 * 38. $$-900$$
 * 39. $$-471$$
 * 40. $$394$$
 * 41. $$1,385,239$$
 * 42. $$1.2$$
 * 43. $$1.5$$
 * 44. $$-0.9$$
 * 45. $$-2.66$$
 * 46. $$8.707$$
 * 47. $$499.725$$
 * 48. $$-23,149.296$$
 * 49. $$-83.81248$$
 * 50. $$1,941.72$$


text:
Prealgebra for Two-Year Colleges/Appendix (procedures)/Subtracting whole numbers with borrowing

Addition
Addition operations are denoted by the + sign and the - (-)= +. The addition operator (plus sign) will take any two numbers, called addends, as operands to work on. The result is called the sum of the two numbers.

The operation of addition is commutative. This means that the addition of two numbers will give the same sum regardless of the order in which the numbers are added.

Example:
 * $$7 + 5 = 12$$ and $$5 + 7 = 12$$

 * $$3 + 2 + 4 = 9$$ and $$4 + 3 + 2 = 9$$

Vertical Addition
To add vertically, stack the numbers on top.

52 19

Now add the numbers in the first or right column. If you get a number with tens in it, put what's in the tens place on the tens place in the problem. You should have:

1 52  19   1

Add the tens column now.

1 52  19  71

Repeat in a similar fashion for each additional column (hundreds, thousands, etc.)until you finish the problem.

Exercises
In Exercises 1–25, find the sum.
 * 1. $$5 + 3$$
 * 2. $$8 + 7$$
 * 3. $$9 + 2$$
 * 4. $$6 + 3$$
 * 5. $$1 + 4$$
 * 6. $$2 + 17$$
 * 7. $$12 + 11$$
 * 8. $$53 + 8$$
 * 9. $$41 + 9$$
 * 10. $$84 + 12$$
 * 11. $$16 + 17$$
 * 12. $$7,576 + 5,345$$
 * 13. $$2,345 + 3,245$$
 * 14. $$8,952 + 9,423$$
 * 15. $$2,783 + 2,389$$
 * 16. $$189,583 + 1,574,822$$
 * 17. $$1.5 + 2.7$$
 * 18. $$5.4 + 3.9$$
 * 19. $$8.3 + 9.2$$
 * 20. $$2.23 + 4.89$$
 * 21. $$534.4 + 34.675$$
 * 22. $$348.904 + 23,498.2$$
 * 23. $$1.673 + 48,210.38$$
 * 24. $$10.4823 + 94.29478$$
 * 25. $$128.52 + 2,070.24$$

Answers

 * 1. $$8$$
 * 2. $$15$$
 * 3. $$11$$
 * 4. $$9$$
 * 5. $$5$$
 * 6. $$19$$
 * 7. $$23$$
 * 8. $$61$$
 * 9. $$50$$
 * 10. $$96$$
 * 11. $$33$$
 * 12. $$12,921$$
 * 13. $$5,590$$
 * 14. $$18,375$$
 * 15. $$5,172$$
 * 16. $$1,764,405$$
 * 17. $$4.2$$
 * 18. $$9.3$$
 * 19. $$17.5$$
 * 20. $$7.12$$
 * 21. $$569.075$$
 * 22. $$23,847.104$$
 * 23. $$48,212.053$$
 * 24. $$104.77708$$
 * 25. $$2,198.76$$

Subtraction
Subtraction, as you probably also have seen, uses the minus (-) sign. The generic subtraction operator will take any two numbers as operands. The result is called the difference of the two numbers.

Subtraction is not a commutative operation. Changing the order of the operands will likely give a different (not the same) result.

Example:
 * $$7 - 5 = 2$$ and $$5 - 7 = -2$$

 * $$3 - 2 - 4 = -3$$ and $$4 - 3 - 2 = -1$$

Exercises
In Exercises 26–50, find the difference.
 * 26. $$5 - 3$$
 * 27. $$8 - 7$$
 * 28. $$9 - 2$$
 * 29. $$6 - 3$$
 * 30. $$1 - 4$$
 * 31. $$2 - 17$$
 * 32. $$12 - 11$$
 * 33. $$53 - 8$$
 * 34. $$41 - 9$$
 * 35. $$84 - 12$$
 * 36. $$16 - 17$$
 * 37. $$7,576 - 5,345$$
 * 38. $$2,345 - 3,245$$
 * 39. $$8,952 - 9,423$$
 * 40. $$2,783 - 2,389$$
 * 41. $$1,574,822 - 189,583$$
 * 42. $$2.7 - 1.5$$
 * 43. $$5.4 - 3.9$$
 * 44. $$8.3 - 9.2$$
 * 45. $$2.23 - 4.89$$
 * 46. $$10.38 - 1.673$$
 * 47. $$534.4 - 34.675$$
 * 48. $$348.904 - 23,498.2$$
 * 49. $$10.4823 - 94.29478$$
 * 50. $$2,070.24 - 128.52$$

Answers

 * 26. $$2$$
 * 27. $$1$$
 * 28. $$7$$
 * 29. $$3$$
 * 30. $$-3$$
 * 31. $$-15$$
 * 32. $$1$$
 * 33. $$45$$
 * 34. $$32$$
 * 35. $$72$$
 * 36. $$-1$$
 * 37. $$2,231$$
 * 38. $$-900$$
 * 39. $$-471$$
 * 40. $$394$$
 * 41. $$1,385,239$$
 * 42. $$1.2$$
 * 43. $$1.5$$
 * 44. $$-0.9$$
 * 45. $$-2.66$$
 * 46. $$8.707$$
 * 47. $$499.725$$
 * 48. $$-23,149.296$$
 * 49. $$-83.81248$$
 * 50. $$1,941.72$$