Question: Show that

for any positive number \(\nu\) and \(x>0\), where \(W\) is the Wronskian function. Hint: First prove the following identity, which states that if \(f\) and \(g\) are solutions of

then \(W[f, g](x)=C e^{-\int p(x) d x}\). Idea: Write down two d.e.'s, one for \(f\) and the same one for \(g\). Write down the definition of \(W[f, g](x)=C e^{-\int p(x) d x}\) and differentiate it. By simplifying, you will get a first order d.e. just for \(W\) which you can easily integrate. Finally, once you have established the result, use it to prove what is required above.

1. Do problem 14 from section 5.3 of the reader.
2. Using the result from (a), prove that \(\sum_{n=0}^{\infty} \frac{1}{2^{n+1}} P_{n}\left(\frac{1}{2}\right)=\frac{1}{\sqrt{3}}\).
3. Prove that \(P_{n}(x)\) has no repeated roots. Hint: suppose \(P_{n}(x)\) has a repeated root at \(x_{0}\), then \(P_{n}\left(x_{0}\right)=P_{n}^{\prime}\left(x_{0}\right)\) (why? explain.). Thus \(P_{n}(x)\) is a solution of the initial value problem \(\left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\) \(n(n+1) y=0 ; \quad y\left(x_{0}\right)=0, \quad y^{\prime}\left(x_{0}\right)=0\). Continue by using the uniqueness theorem for solutions of linear initial value problems.

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