MAT102 Lecture 16

• 2
  • a) If $c_1,\dots ,c_n$ are real numbers and $f_1,\dots ,f_n$ are functions, show that $\frac{d}{dx}\left(\sum _{k=1}^n{c}_k{f}_k\right)=\sum _{k=1}^n{c}_k\frac{d}{dx}(f_k)$
    • Base Case
      • This says that the derivative of the $\sum _{k=1}^n{c}_n{f}_n$ is the same as if we just sum the derivative of $f_k$ and pull our the constant.
      • Let $P(n)=D\left(\sum _{k=1}^n{c}_k{f}_k\right)$
      • Let $Q(n)=\sum _{k=1}^n{c}_{k}D(f_k)$
      • For $P(n)$ let's prove that $P(1)$ is $Q(1)$
        • We know that this is just the sum of each differentiation with a constant in it like
        • $f(x)=2x^2$
        • then sum $f(x)$ blah blah
        • It's the same as $2f(x)=x^2$
        • Well through properties of sums and derivatives we already know that $\frac{d}{dx}\left(\sum _{k=1}^n{c}_k{f}_k\right)$ is just the $\sum _{k=1}^n\frac{d}{dx}{c}_k{f}_k$
        • No matter the constant, we can pull it out.
        • $\sum _{k=1}^n{c}_k\frac{d}{dx}{f}_k$
        • This already is $Q(n)$
        • So now we need to show that $Q(n)$ is true for all $n$
    • Induction Hypothesis:
      • Assume that $P(n)=Q(n)$ for an arbitrary $n\in \mathbb{N}$.
    • Induction Step:
      • $P(n)=Q(n)⟹P(n+1)=Q(n+1)$
      • $\frac{d}{dx}\left(\sum _{k=1}^{n+1}{c}_k{f}_k\right)=\sum _{k=1}^{n+1}{c}_k\frac{d}{dx}(f_k)$
      • $\frac{d}{dx}\left(\sum _{k=1}^{n+1}{c}_k{f}_k\right)=\frac{d}{dx}(\left(\sum _{k=1}^n{c}_k{f}_k\right)+c_{n+1}{f}_{n+1})$
      • $\frac{d}{dx}(\left(\sum _{k=1}^n{c}_k{f}_k\right)+c_{n+1}{f}_{n+1})=\frac{d}{dx}\left(\sum _{k=1}^n{c}_k{f}_k\right)+\frac{d}{dx}(c_{n+1}{f}_{n+1})$
      • $\frac{d}{dx}\left(\sum _{k=1}^{n+1}{c}_k{f}_k\right)=(\sum _{k=1}^n{c}_k\frac{d}{dx}(f_k))+c_{n+1}\frac{d}{dx}(f_{n+1})$
      • $(\sum _{k=1}^n{c}_k\frac{d}{dx}(f_k))+c_{n+1}\frac{d}{dx}(f_{n+1})=\sum _{k=1}^{n+1}{c}_k\frac{d}{dx}(f_k)$
      • $\frac{d}{dx}\left(\sum _{k=1}^{n+1}{c}_k{f}_k\right)=\sum _{k=1}^{n+1}{c}_k\frac{d}{dx}(f_k)$
  • b) For $n,k\in \mathbb{N}$, define $F(n,k)=\prod _{i=1}^k\frac{n-i+1}{i}$. Let's say that $F(n,0)=1$ for any choice of $n$. Show that if $n,k\ge 1$ then $F(n,k)=F(n-1,k-1)+F(n-1,k)$.
    • $n,k\in \mathbb{N}:n,k\ge 1$
    • Case 1: $n=1=k$
      • $F(1,1)$
        • $\prod _{i=1}^1\frac{1-i+1}{i}$
        • The product notation $\prod _{i=1}^n{a}_i$ means to multiply together the terms $a_i$ from $i=1$ to $i=n$. That is,
          • $\prod _{i=1}^n{a}_i=a_1\cdot a_2\cdot a_3\cdot \dots \cdot a_n$.
        • In your example, the upper limit of the product is $n=1$, so the product only contains one term. The term is given by the formula $\frac{1-i+1}{i}$. Plugging in $i=1$, we get:
          • $\prod _{i=1}^1\frac{1-i+1}{i}=\frac{1-1+1}{1}=\frac{1}{1}=1$.
      • $F(0,0)$
        • By definition we know this is $1$
      • $F(0,1)$
        • $\prod _{i=1}^1\frac{0-i+1}{i}$
        • $\prod _{i=1}^1\cdots =0$
      • $F(1,1)=F(0,0)+F(0,1)$
      • $1=1+0$
      • yes