Mathematical induction.

• Feb 22nd 2011, 11:15 PM
mariama
Mathematical induction.
How can we show this by using mathematical induction

(d^n\dx^n )x^n = n!
• Feb 22nd 2011, 11:45 PM
FernandoRevilla
Quote:

Originally Posted by mariama
How can we show this by using mathematical induction

(d^n\dx^n )x^n = n!

First step . We have

$\dfrac{d^1}{dx^1}x^1=1=1!$

so, the equality is true for $n=1$

What difficulties have you had in the second step?

Remark . If the question is for $n\geq 0$ then, take into account that:

$\dfrac{d^0}{dx^0}f(x)=f(x)$ and $0!=1$

Fernando Revilla
• Feb 22nd 2011, 11:50 PM
mariama
so , Should I plug n+1 instead of n in the next step ?? or how can i make it
• Feb 23rd 2011, 12:05 AM
FernandoRevilla
Quote:

Originally Posted by mariama
so , Should I plug n+1 instead of n in the next step ?? or how can i make it

Suppose:

$\dfrac{d}{dx^n}x^n=n!$

Then, prove:

$\dfrac{d}{dx^{n+1}}x^{n+1}=(n+1)!$

Fernando Revilla
• Feb 23rd 2011, 02:13 AM
deleted
• Feb 23rd 2011, 11:08 AM
mariama
Quote:

Originally Posted by FernandoRevilla
Suppose:

$\dfrac{d}{dx^n}x^n=n!$

Then, prove:

$\dfrac{d}{dx^{n+1}}x^{n+1}=(n+1)!$

Fernando Revilla

Sorry , but i could not prove this
$\dfrac{d}{dx^{n+1}}x^{n+1}=(n+1)!$
Do you have a hint? or how can i start ?
• Feb 23rd 2011, 11:15 AM
TheEmptySet
Quote:

Originally Posted by mariama
Sorry , but i could not prove this
$\dfrac{d}{dx^{n+1}}x^{n+1}=(n+1)!$
Do you have a hint? or how can i start ?

FernandoRevilla has outlined it for you. You need to show that $n \implies n+1$

$\displaystyle \frac{d^{n+1}}{dx^{n+1}}x^{n+1}=\frac{d^n}{dx^n}\l eft( \frac{d}{dx}x^{n+1}\right)$
• Feb 23rd 2011, 12:48 PM
mariama
Quote:

Originally Posted by TheEmptySet
FernandoRevilla has outlined it for you. You need to show that $n \implies n+1$

$\displaystyle \frac{d^{n+1}}{dx^{n+1}}x^{n+1}=\frac{d^n}{dx^n}\l eft( \frac{d}{dx}x^{n+1}\right)$