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Problem: Show that $P^{'}_{2n+1}(0)= \frac{(-1)^n (2n+1){^{2n}}C_n}{2^n}$.

My attempt: Given: $P_{n}(x)=\frac{1}{2^n}\sum_{k=0}^{\frac{1}{2}n}(-1)^{k} {{^n}}C_k{^{2n-2k}C_n}x^{n-2k}$

Let $n = 2n+1$. We have $P_{2n+1}(x)=\frac{1}{2^{2n+1}}\sum_{k=0}^{{n}+\frac{1}{2}}(-1)^{k} {{^{2n+1}}}C_k{^{2(2n+1)-2k}C_{2n+1}}x^{2n+1-2k}$

Now, let $k=n$.

$P_{2n+1}(x)=\frac{1}{2^{2n+1}}\sum_{n=0}^{{n}+\frac{1}{2}}(-1)^{n} {{^{2n+1}}}C_n{^{2(2n+1)-2n}C_{2n+1}}x^{2n+1-2n}$

$P_{2n+1}(x)=\frac{1}{2^{2n+1}}\sum_{n=0}^{{n}+\frac{1}{2}}(-1)^{n} {{^{2n+1}}}C_n{^{2n+2}C_{2n+1}}x^{1}$

$P_{2n+1}^{'}(x)=\frac{1}{2^{2n+1}}\sum_{n=0}^{{n}+\frac{1}{2}}(-1)^{n} {{^{2n+1}}}C_n{^{2n+2}C_{2n+1}}$

Upon simplifying, we have

$P^{'}_{2n+1}(0)= \frac{(-1)^n (2n+1){^{2n}}C_n}{2^{2n}}$

This result is different from the one given. I have missed out something, please help!

user107224
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Lawrence Orewa
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1 Answers1

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Your result is correct although it is difficult to read the derivation: I strongly suggest you use the binomial coefficient notation $\binom{a}{b}$ instead of $^aC_b$. My own derivation is as follows: The recurrence relation $$(x^2-1)P_n'=n(xP_n-P_{n-1})$$ implies that $$P_n'(0)=nP_{n-1}(0)$$ so it remains to find the value of $P_n(0)$. As $$P_n(x)=\frac{1}{2^n}\sum_{j=0}^{n}\binom{n}{j}^2(x+1)^{n-j}(x-1)^j$$ we deduce that $$P_n(0)=\frac{1}{2^n}\sum_{j=0}^{n}\binom{n}{j}^2(-1)^j$$ Now from the fact that $(1-x)^m(1+x)^m=(1-x^2)^m$, equating the coefficients for $x^m$, we derive that $$ \begin{equation} \sum_{j=0}^{m}\binom{m}{j}^2(-1)^j=\sum_{j=0}^{m}\binom{m}{j}(-1)^j\binom{m}{m-j}\cdot (+1)^j= \begin{cases}0 &\text{if m is odd}\\ \binom{m}{m/2}(-1)^{m/2} & \text{if m is even} \end{cases} \end{equation}$$ And thus $$P_{2n+1}'(0)=(2n+1)P_{2n}(0)=\frac{2n+1}{2^{2n}}\binom{2n}{n}(-1)^n$$

Maestro13
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