Question

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

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

Express binomial coefficient in terms of Gamma function

The binomial coefficient \(C_{k}^{p}\) can be expressed using gamma functions as: \(C_{k}^{p}=\frac{\Gamma(p+1)}{\Gamma(k+1)\Gamma(p-k+1)}\)

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 \(C_{k}^{1/2}=(-1)^k \frac{(2^{2k})}{2(k!)^2}\).

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.