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Chapter 6: Problem 10

The coefficients \(C_{k}^{p}\) in the binomial expansion for \((1+x)^{p}\) are given by $$ C_{k}^{p}=\frac{p(p-1) \cdots(p-k+1)}{k !} $$ a. Write \(C_{k}^{p}\) in terms of Gamma functions. b. For \(p=1 / 2\), use the properties of Gamma functions to write \(C_{k}^{1 / 2}\) in terms of factorials. c. Confirm your answer in part b. by deriving the Maclaurin series expansion of \((1+x)^{1 / 2}\)

Short Answer

Expert verified
a) \(C_{k}^{p}=\frac{\Gamma(p+1)}{\Gamma(k+1)\Gamma(p-k+1)}\)\nb) \(C_{k}^{1 / 2}= \frac{(2k-1)!!}{2^{2k} } \cdot \frac{1 }{k!}\)\nc) The Maclaurin series expansion of \((1+x)^{1 / 2}\) is \(\displaystyle\sum_{k=0}^\infty \frac{(2k-1)!!}{2^{2k} } \cdot \frac{1 }{k!} x^{k}\)

Step by step solution

01

Write \(C_{k}^{p}\) in terms of Gamma functions

The coefficient \(C_{k}^{p}\) in the binomial expansion for \((1+x)^{p}\) is given by \(C_{k}^{p}=\frac{p(p-1)...(p-k+1)}{k !}\). The Gamma function \(\Gamma(n)\) is given by \(\Gamma(n)=(n-1)!\). Therefore, we can write \(C_{k}^{p}\) in terms of Gamma functions as: \(C_{k}^{p}=\frac{\Gamma(p+1)}{\Gamma(k+1)\Gamma(p-k+1)}\)
02

Write \(C_{k}^{1 / 2}\) in terms of factorials

We are given \(p= 1 / 2\), Exchange \(p\) with \(1/2\) in the expression of \(C_{k}^{p}\) we get: \(C_{k}^{1 / 2}=\frac{\Gamma(3/2)}{\Gamma(k+1)\Gamma(1/2-k)}\). Using the property of the Gamma function for half-integer values, \(\Gamma(1/2+n)=(2n-1)!!/2^n\), where \(!!\) denotes the double factorial, we simplify the equation to get: \(C_{k}^{1 / 2}= \frac{(2k-1)!!}{2^{2k} } \cdot \frac{1 }{k!}\)
03

Confirm the result by deriving the Maclaurin series

The Maclaurin series expansion of \((1+x)^{1 / 2}\) is \(\displaystyle\sum_{k=0}^\infty C_{k}^{1 / 2} x^{k}\) After substituting \(C_{k}^{1 / 2}\) from Step 2, we get the series \(\displaystyle\sum_{k=0}^\infty \frac{(2k-1)!!}{2^{2k} } \cdot \frac{1 }{k!} x^{k}\) which confirms our result from part b

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Most popular questions from this chapter

Determine the solvability conditions for the nonhomogeneous boundralue problem: \(u^{\prime}(\pi / 4)=\beta\).

A solution of Bessel's equation, \(x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-n^{2}\right) y=0\), can be found using the guess \(y(x)=\sum_{j=0}^{\infty} a_{j} x^{j+n} .\) One obtains the recurrence relation \(a_{j}=\frac{-1}{j(2 n+j)} a_{j-2} .\) Show that for \(a_{0}=\left(n ! 2^{n}\right)^{-1}\), we get the Bessel function of the first kind of order \(n\) from the even values \(j=2 k\) : $$ J_{n}(x)=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k !(n+k) !}\left(\frac{x}{2}\right)^{n+2 k} $$

Use the method of eigenfunction expansions to solve the problems: a. \(y^{\prime \prime}+4 y=x^{2}, \quad y^{\prime}(0)=y^{\prime}(1)=0\). b. \(y+4 y=x^{2} \quad y(0)=y(1)=0\).

Determine the solvability conditions for the nonhomogeneous boundralue problem: \(u^{\prime}(\pi / 4)=\beta\).

Bessel functions \(J_{p}(\lambda x)\) are solutions of \(x^{2} y^{\prime \prime}+x y^{\prime}+\left(\lambda^{2} x^{2}-p^{2}\right) y=\) 0. Assume that \(x \in(0,1)\) and that \(J_{p}(\lambda)=0\) and \(J_{p}(0)\) is finite.s a. Show that this equation can be written in the form $$ \frac{d}{d x}\left(x \frac{d y}{d x}\right)+\left(\lambda^{2} x-\frac{p^{2}}{x}\right) y=0 $$ This is the standard Sturm-Liouville form for Bessel's equation. b. Prove that $$ \int_{0}^{1} x J_{p}(\lambda x) J_{p}(\mu x) d x=0, \quad \lambda \neq \mu $$ by considering $$ \int_{0}^{1}\left[J_{p}(\mu x) \frac{d}{d x}\left(x \frac{d}{d x} J_{p}(\lambda x)\right)-J_{p}(\lambda x) \frac{d}{d x}\left(x \frac{d}{d x} J_{p}(\mu x)\right)\right] d x $$ Thus, the solutions corresponding to different eigenvalues \((\lambda, \mu)\) are orthogonal. c. Prove that $$ \int_{0}^{1} x\left[J_{p}(\lambda x)\right]^{2} d x=\frac{1}{2} J_{p+1}^{2}(\lambda)=\frac{1}{2} J_{p}^{\prime 2}(\lambda) $$
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