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Problem

Prove all derivatives of: $$ f(x)=\frac{1}{1+x} $$ by induction.

Attempt to solve

I compute few derivatives of $f(x)$ so that i can form general expression for induction hypothesis. I compute all derivatives utilizing formula:

$$ \frac{d}{dx}x^n=nx^{n-1} $$

First 4 derivatives are:

$$ f'(x)=(-1)\cdot(1+x)^{-2}\cdot 1 = -\frac{1}{(1+x)^2} $$ $$ f''(x)=(-1)(-2)(1+x)^{-3}\cdot 1 = \frac{2}{(1+x)^3} $$ $$ f'''(x)=(-1)(-2)(-3)(1+x)^{-4}\cdot 1 = -\frac{6}{(1+x)^4} $$ $$ f''''(x)=(-1)(-2)(-3)(-4)(1+x)^{-5} \cdot 1 = \frac{24}{(1+x)^5} $$

Observe that $(-1)(-2)(-3)(-4)\dots (-n)$ can be generalized with:

$$ (-1)(-2)(-3)(-4)\dots(-n) = (-1)^n\cdot n! $$

Expression follows factorial of $n$ except every other value is positive and every other is negative. If i multiply it by $(-1)^n$ it is positive when $n \mod 2 = 0$ and negative when $n \mod 2 \neq 0$.

Rest of the expression can be generalized as:

$$ (1+x)^{-n-1} = (1+x)^{-(n+1)}=\frac{1}{(1+x)^{n+1}} $$

Combining these gives formula in analytic form:

$$ f(n)= \frac{(-1)^n \cdot n!}{(1+x)^{n+1}} $$

I can form induction hypothesis such that:

$$ \frac{d^n}{dx^n}\frac{1}{1+x} = \frac{(-1)^n\cdot n!}{(1+x)^{n+1}} $$

Induction proof

Base case

Base case when $n=0$:

$$ \frac{d^0}{dx^0}\frac{1}{1+x}=\frac{1}{1+x}=\frac{(-1)^0\cdot 0!}{(1+x)^{0+1}} $$

Induction step

$$ \frac{d^n}{dx^n}\frac{1}{1+x} =_{\text{ind.hyp}} \frac{(-1)^n\cdot n!}{(1+x)^{n+1+1}} $$

$$ \frac{d^n}{dx^n}\frac{1}{1+x} = \frac{(-1)^n\cdot n!}{(1+x)^{n+2}} $$

Now the problem is that formula i used for derivation can only be used recursively. I believe this is correct notation for $n$:th derivative but computing one is only defined recursively with formula i used:

$$ \frac{d}{dx} x^n = nx^{n-1} $$

Which is not defined for case:

$$ \frac{d^n}{dx^n}x^n = \text{ undefined} $$

The idea is to show that this recursion can be expressed in analytical form and it is valid for all $n\in \mathbb{Z}+$ by induction. Problem is i don't know how do you express this in recursive form and how do you get from recursion formula to the analytical one.

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  • $\begingroup$ All of your work seems fine. I'm not sure why the formula for iterated derivatives of $x^n$ is causing you anxiety, as you're only using it for negative values of $n$ (thus you'll never be in a position of needing the value of $\frac{d^n}{dx^n} x^n$). $\endgroup$ – Connor Harris Oct 4 '18 at 16:40
  • $\begingroup$ I am not sure why you say that $\frac{d^n}{dx^n}x^n$ is undefined. Simply applying it to a couple of values from $\mathbb{Z}^+$ leads to the intuitive expression $\frac{d^n}{dx^n}x^n=n!$ $\endgroup$ – mrtaurho Oct 4 '18 at 16:40
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Your hypothesis is

$$\frac{d^n}{dx^n}\frac1{1+x}=\frac{(-1)^nn!}{(1+x)^{n+1}}$$

We want to show that

$$\frac{d^{n+1}}{dx^{n+1}}\frac1{1+x}=\frac{(-1)^{n+1}(n+1)!}{(1+x)^{n+2}}$$

Let's verify:

\begin{align} \frac{d^{n+1}}{dx^{n+1}}\frac1{1+x} &=\frac{d}{dx}\left(\frac{d^n}{dx^n}\frac1{1+x} \right)\\ &=\frac{d}{dx}\left(\frac{(-1)^nn!}{(1+x)^{n+1}} \right) \\ &=(-1)^n(n!) \frac{d}{dx}(1+x)^{-(n+1)} \\ &= (-1)^n(n!) (-(n+1)) (1+x)^{-(n+2)}\\ &=\frac{(-1)^{n+1}(n+1)!}{(1+x)^{n+2}} \end{align}

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Your hypothesis should be

$$\frac{\mathrm d^n}{\mathrm dx^n}\frac{1}{1+x} = \frac{(-1)^n n!}{(1+x)^{n+1}}$$

and what you need for your induction step is:

$$\frac{\mathrm d^{n+1}}{\mathrm dx^{n+1}}f(x)= \frac{\mathrm d}{\mathrm dx}\left( \frac{\mathrm d^n}{\mathrm dx^n}f(x)\right) \quad \forall n \ge 0$$

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You inductive hypothesis is that $$ \frac{d^nf}{dx^n}=\frac{(-1)^nn!}{(1+x)^{n+1}}. $$

Your inductive step would then be $$ \frac{d^{n+1}f}{dx^{n+1}}=\frac d{dx}\,\frac{d^nf}{dx^n}=\frac d{dx}\left(\frac{(-1)^nn!}{(1+x)^{n+1}}\right) =\frac{-(n+1)(-1)^nn!}{(1+x)^{n+2}}=\frac{(-1)^{n+1}(n+1)!}{(1+x)^{n+2}}, $$ which shows that the formula holds.

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