Let $f(x) = x^n$. Show that $f^{(n)} = n!$ and $f^{(m)} (x) = 0$ for all $m > n$.

Question

I'm supposed to use mathematical induction to solve this

1. Show that $P(1)$ is true
2. Assume $P(K)$ is true
3. Show that $P(K+1)$ is true

How do I approach this problem?

Answer 1

You approach the problem by following the steps. Keep in mind that $P(n)$ is the statement

$f^{(n)} = n!$ and $f^{(m)}(x)=0$ for $m>n$.

• Now, if this is $P(n)$, what is $P(1)$?
• How would you prove that $P(1)$ is true?

Once that is done, try to write down $P(k)$ (which is what you assume to be true) and $P(k+1)$ (which is what you want to prove). Write them both down and try to see what the connection between them is. The statements will be very similar.

Answer 2

Notice, $n\in\mathbb{Z^+}$:

$$\frac{\text{d}^n}{\text{d}x^n}\left(x^a\right)=\begin{cases}a!\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\text{when}\space n=a\\ 0\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\text{when}\space n>a\\ \frac{a!}{(a-n)!}\cdot x^{a-n}\space\space\space\space\space\space\space\space\text{when}\space n<a \end{cases}$$

Because (example):

$$\frac{\text{d}^3}{\text{d}x^3}\left(x^2\right)=\frac{\text{d}^2}{\text{d}x^2}\left(2x^1\right)=\frac{\text{d}}{\text{d}x}\left(2x^0\right)=0$$ $$\frac{\text{d}^3}{\text{d}x^3}\left(x^3\right)=\frac{\text{d}^2}{\text{d}x^2}\left(3x^2\right)=\frac{\text{d}}{\text{d}x}\left(6x^1\right)=6x^0=6=3!$$

And to proof those, use this:

$$\frac{\text{d}}{\text{d}x}\left(x^a\right)=ax^{a-1}$$ $$\frac{\text{d}^2}{\text{d}x^2}\left(x^a\right)=a(a-1)x^{a-2}$$ $$\frac{\text{d}^3}{\text{d}x^3}\left(x^a\right)=a(a-1)(a-2)x^{a-3}$$ $$\frac{\text{d}^4}{\text{d}x^4}\left(x^a\right)=a(a-1)(a-2)(a-3)x^{a-4}$$ $$\frac{\text{d}^5}{\text{d}x^5}\left(x^a\right)=a(a-1)(a-2)(a-3)(a-4)x^{a-5}$$