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Chapter 6: Problem 11

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. \(C_{k}^{p}=\frac{\Gamma(p+1)}{\Gamma(k+1)\Gamma(p-k+1)}\)\n b. \(C_{k}^{1/2}=(-1)^k \frac{(2^{2k})}{2(k!)^2}\)\n c. The Maclaurin series expansion for \((1+x)^{1/2}\) confirms the result from part b: \(\sum_{k=0}^\infty(-1)^k \frac{(2^{2k})}{2(k!)^2}x^k\).

Step by step solution

01

Express binomial coefficient in terms of Gamma function

The binomial coefficient \(C_{k}^{p}\) can be expressed using 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

For \(p=1/2\), the binomial coefficient becomes: \(C_{k}^{1/2}=\frac{\Gamma(1/2+1)}{\Gamma(k+1)\Gamma(1/2-k+1)} =\frac{\sqrt{\pi}}{k!\Gamma(1/2-k+1)}\). Using the duplication formula of gamma function, we can further simplify above expression: \(C_{k}^{1/2}=\frac{(2(-1)^k (-4)^{k}(1)(3)(5)...(2k-1))}{k!(2k)!}\). Simplifying the factorials will give \(C_{k}^{1/2}=(-1)^k \frac{(2k)!(2^{2k})}{2(2k!)(k!)^2}\) which simplifies to \(C_{k}^{1/2}=(-1)^k \frac{(2^{2k})}{2(k!)^2}\).
03

Derive Maclaurin series for \((1+x)^{1/2}\) and confirm the answer

The Maclaurin series expansion for \((1+x)^{1/2}\) is given by \(\sum_{k=0}^\infty C_{k}^{1/2}x^k\). Substituting \(C_{k}^{1/2}\) from above, we get \(\sum_{k=0}^\infty(-1)^k \frac{(2^{2k})}{2(k!)^2}x^k\). This confirms the result from step 2.

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Most popular questions from this chapter

Find the eigenvalues and eigenfunctions of the given Sturm-Liouville lems: a. \(y^{\prime \prime}+\lambda y=0, y^{\prime}(0)=0=y^{\prime}(\pi)\) b. \(\left(x y^{\prime}\right)^{\prime}+\frac{\lambda}{x} y=0, y(1)=y\left(e^{2}\right)=0\).

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. \(y+4 y=x^{2} \quad y(0)=y(1)=0\).

The Hermite polynomials, \(H_{n}(x)\), satisfy the following: i. \(=\int_{-\infty}^{\infty} e^{-x^{2}} H_{n}(x) H_{m}(x) d x=\sqrt{\pi} 2^{n} n ! \delta_{n, m}\) ii. \(H_{n}^{\prime}(x)=2 n H_{n-1}(x)\). iii. \(H_{n+1}(x)=2 x H_{n}(x)-2 n H_{n-1}(x)\) iv. \(H_{n}(x)=(-1)^{n} e^{x^{2}} \frac{d^{n}}{d x^{n}}\left(e^{-x^{2}}\right) .\) ng these, show that a. \(H_{n}^{\prime \prime}-2 x H_{n}^{\prime}+2 n H_{n}=0\). [Use properties ii. and iii.] b. \(\int_{-\infty}^{\infty} x e^{-x^{2}} H_{n}(x) H_{m}(x) d x=\sqrt{\pi} 2^{n-1} n !\left[\delta_{m, n-1}+2(n+1) \delta_{m, n+1}\right] .\) [Use properties i. and iii.]

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Show that a Sturm-Liouville operator with periodic boundary condion \([a, b]\) is self-adjoint if and only if \(p(a)=p(b)\). [Recall that periodic dary conditions are given as \(u(a)=u(b)\) and \(\left.u^{\prime}(a)=u^{\prime}(b) .\right]\).
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