Alright so I am having some difficulty with an inductive proof. I am attempting to prove the following:
Given that the Euler Method is described by the given recursive formula:
$y_{n} = y_{n-1} + h f(x_{n-1}, y_{n-1}), n = 1, 2, 3, ….$
Show that the Euler Approximation $y_{n}$ can be represented by the following formula:
$y_{n} = ( 1 + \frac{1}{n})^{n} , n = 1, 2, 3, ….$
Given that:
$y' = y$ , $y(0) = 1$, and $h = \frac{1}{n}$
Proposed Solution:
Base Case (first re-write the recursive definition as $y_{n} = y_{n-1} (1 + \frac{1}{n})$ to increase clarity)
Let n = 1
$y_{0} (1 + \frac{1}{1}) = (1 + \frac{1}{1})^{1}$
$1 \cdot (1 + \frac{1}{1}) = (1 + \frac{1}{1})$
2 = 2
So the base case checks out.
Inductive Hypothesis
Now assume the following:
$y_{k – 1} (1 + \frac{1}{k}) \Rightarrow (1 + \frac{1}{k})^k \space \space \forall k \in Z^{+}$
Show that: $y_{k} (1 + \frac{1}{1+k}) \Rightarrow (1 + \frac{1}{k+1}) ^{k+1}$
And here is my problem because if I replace $y_{k}$ with $(1 + \frac{1}{k})^k$ I get the following:
$(1 + \frac{1}{k})^k \cdot (1 + \frac{1}{k+1}) ^{k+1} \nRightarrow (1 + \frac{1}{k+1}) ^{k+1}$
Can someone spot what I’m doing wrong?
