How can we show this by using mathematical induction

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

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- Feb 22nd 2011, 11:15 PMmariamaMathematical induction.
How can we show this by using mathematical induction

(d^n\dx^n )x^n = n! - Feb 22nd 2011, 11:45 PMFernandoRevilla

. We have*First step*

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

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

What difficulties have you had in the second step?

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

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

Fernando Revilla - Feb 22nd 2011, 11:50 PMmariama
so , Should I plug n+1 instead of n in the next step ?? or how can i make it

- Feb 23rd 2011, 12:05 AMFernandoRevilla
Suppose:

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

Then, prove:

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

Fernando Revilla - Feb 23rd 2011, 02:13 AMArchie Meade
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- Feb 23rd 2011, 11:08 AMmariama
- Feb 23rd 2011, 11:15 AMTheEmptySet
- Feb 23rd 2011, 12:48 PMmariama