# 2022-05-23-Monday Just watched https://www.youtube.com/watch?v=bOXCLR3Wric Thanks for the problems, Grant! ## Problem 1 Take the derivative of both sides of $ \frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n} $ How can you use this to compute the expected number of times you have to roll a fair six-sided die before seeing the number 1? ### Solution Take derivative $\left(\frac{1}{1-x}\right)^2=\sum_{n=1}^{\infty} n x^{n-1}$ Notice that if $x=\frac{5}{6}$ then the right side is proportional to a weighted average of rolls taken to see a one. $ \begin{aligned} \left(\frac{1}{1-\frac{5}{6}}\right)^2&=\sum_{n=1}^{\infty} n \left(\frac{5}{6}\right)^{n-1}\\ \frac16\left(\frac{1}{1-\frac{5}{6}}\right)^2&=\frac16\sum_{n=1}^{\infty} n \left(\frac{5}{6}\right)^{n-1}\\ \end{aligned}$ The $\frac16$ accounts for the $n^\text{th}$ roll, while all rolls before that should be non-ones, having probabilities $\frac56$ each time. When simplified $\frac16\left(\frac{1}{1-\frac{5}{6}}\right)^2=\frac16(6)^2=\boxed6$ ## Problem 2 Compute the sum $ \sum_{k=0}^{n} 2^{k}\left(\begin{array}{l}n \\ k\end{array}\right) $ (Hint, expand the expression $(1+x)^{n}$) ### Solution Take the hint $(1+x)^{n}=\sum_{k=0}^{n}x^k{n\choose k}$ allow $x=2$ $\boxed{3}=(1+2)^{n}=\sum_{k=0}^{n}2^k{n\choose k}$ ## Problem 3 Let $f_{n}$ be the $n$th Fibonacci number, and consider the function $ F(x)=\sum_{n=0}^{\infty} f_{n} \frac{x^{n}}{n !} $ Explain why the differential equation $F^{\prime \prime}(x)=F^{\prime}(x)+F(x)$ is true. For bonus points, solve this differential equation, and use it to find a closed-form expression for $f_{n}$ ### Solution ## Problem 4 Suppose $f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}$ What do the coefficients of $f(x) /(1-x)$ tell you? ### Solution Let $g(x)=\frac{f(x)}{1-x}=\frac{\sum_{n=0}^{\infty} a_{n} x^{n}}{1-x}=\sum_{n=1}^{\infty} b_{n-1} x^{n-1}$ So that $b_nx^n(1-x)=b_nx^n-b_nx^{n-1}$ and $b_{n-1}x^{n-1}(1-x)=b_{n-1}x^{n-1}-b_{n-1}x^{n}$ Meaning $a_nx^n=b_nx^n-b_{n-1}x^{n}$