# Limit Formula For Second Derivative

• Sep 29th 2010, 04:26 PM
mathematicalbagpiper
Limit Formula For Second Derivative
Show that $f''(x)=\displaystyle\lim_{h\to 0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}$.

OK, so I'm trying to think through this but getting something completely different. I'm starting with knowing that:

$f''(x)=\displaystyle\lim_{h\to 0}\frac{f'(x+h)-f'(x)}{h}$

But also that:

$f'(x+h)=\displaystyle\lim_{h\to 0}\frac{f(x+2h)-f(x+h)}{h}$

$f'(x)=\displaystyle\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

So when I substitute these in, I'm instead getting the following:

$\displaystyle\lim_{h\to 0}\frac{\frac{f(x+2h)-f(x+h)}{h}-\frac{f(x+h)-f(x)}{h}}{h}$

$=\displaystyle\lim_{h\to 0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$

Which I know is just off by an $h$. It seems to me that they get the second derivative formula like this:

$f''(x)=\displaystyle\lim_{h\to 0}\frac{f'(x)-f'(x-h)}{h}$

But I don't really understand why it's that way. That seems like $f''(x-h)$ to me. Why is it done that way?
• Oct 7th 2010, 04:42 AM
Pandevil1990
Look here,

Attachment 19214
• Oct 7th 2010, 04:47 AM
Also sprach Zarathustra
Quote:

Originally Posted by mathematicalbagpiper
Show that $f''(x)=\displaystyle\lim_{h\to 0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}$.

OK, so I'm trying to think through this but getting something completely different. I'm starting with knowing that:

$f''(x)=\displaystyle\lim_{h\to 0}\frac{f'(x+h)-f'(x)}{h}$

But also that:

$f'(x+h)=\displaystyle\lim_{h\to 0}\frac{f(x+2h)-f(x+h)}{h}$

$f'(x)=\displaystyle\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

So when I substitute these in, I'm instead getting the following:

$\displaystyle\lim_{h\to 0}\frac{\frac{f(x+2h)-f(x+h)}{h}-\frac{f(x+h)-f(x)}{h}}{h}$

$=\displaystyle\lim_{h\to 0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$

Which I know is just off by an $h$. It seems to me that they get the second derivative formula like this:

$f''(x)=\displaystyle\lim_{h\to 0}\frac{f'(x)-f'(x-h)}{h}$

But I don't really understand why it's that way. That seems like $f''(x-h)$ to me. Why is it done that way?

It is all true! Now just do substitution: x=y-h. (Why you allowed to do this kind substitution?)
• Oct 7th 2010, 04:57 AM
Pandevil1990
Quote:

Originally Posted by Also sprach Zarathustra
It is all true! Now just do substitution: x=y-h. (Why you allowed to do this kind substitution?)

You mean x=y+h...
• Oct 7th 2010, 05:08 AM
Also sprach Zarathustra
Quote:

Originally Posted by Pandevil1990
You mean x=y+h...

No... x=y-h will work.
• Oct 7th 2010, 05:14 AM
Pandevil1990
where exactly are you going to do this substitution?
• Oct 7th 2010, 05:17 AM
Also sprach Zarathustra
Quote:

Originally Posted by Pandevil1990
where exactly are you going to do this substitution?

Here:

$\displaystyle\lim_{h\to 0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$
• Oct 7th 2010, 05:30 AM
Pandevil1990
It's ok...But i think this is a more difficult way than L'hospital 's rule.Thanks...