Question

how to find second partial derivative of u(x,y) by using u(x+h,y) and u(x,y+h)
Please, help

Answer 1

@freckles

Answer 2

\[\dfrac{\partial^2 u}{\partial x^2} +\dfrac{\partial^2u}{\partial y}=\\\dfrac{(x-h,y)-2u(x,y)+u(x+h,y)}{h^2}+\dfrac{(x,y-h)-2u(x,y)+u(x,+hy)}{h^2}\]
But I don't know how to get it.

Answer 3

@IrishBoy123

Answer 4

well loser, the quotient rule looks to be in there

so when i've finished my nightly ablutions, i'm defo up for having a go at this. though i know that @freckles is a much safer bet than i :-)

being pedantic with the red....

\[\dfrac{\partial^2 u}{\partial x^2} +\dfrac{\partial^2u}{\partial y}=\\\dfrac{\color{red}{u}(x-h,y)-2u(x,y)+u(x+h,y)}{h^2}+\dfrac{\color{red}{u}(x,y-h)-2u(x,y)+u(x,+hy)}{h^2}\]

Answer 5

Thanks anyway. I am waiting.

Answer 6

\[\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=f'(x) \]
\[f''(x)=\lim_{h \rightarrow 0} \frac{f'(x+h)-f'(x)}{h} \\ f''(x)=\lim_{h \rightarrow 0} \frac{\lim_{h \rightarrow 0} \frac{f(x+h+h)-f(x+h)}{h}-\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}}{h}\]

Answer 7

\[f''(x)=\lim_{h \rightarrow 0} \frac{f(x+2h)-f(x+h)-f(x+h)+f(x)}{h^2 } \\ f''(x)=\lim_{h \rightarrow 0} \frac{f(x+2h)-2 f(x+h)+f(x)}{h^2}\]

so looks like we need to show this is equivalent to
\[f''(x)=\lim_{h \rightarrow 0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}\]
based on what you have above...

Answer 8

I'm thinking you might want to try a sub

Answer 9

yes a sub seems to work

Answer 10

Thanks a lot. I think I got it.