Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Can anyone validate if my understanding regarding numeric differentiation is correct??

$z = f(x,y)$ is an analytic function. $$\frac{\partial z}{\partial x} = \frac{f(x+h,y)-f(x,y)}{h}$$ $$\frac{\partial z}{\partial y} = \frac{f(x,y+h)-f(x,y)}{h}$$

Thanks in advance...

share|improve this question
    
The limit as h goes to 0 is the derivative on continuous functions. The way written above works well for discrete functions (aka x = 0,1,2,3.... and h=1). –  kaine Feb 5 '13 at 14:52

1 Answer 1

up vote 0 down vote accepted

Numerically spoken you have

$$\frac{\partial z}{\partial x} \approx\frac{f(x+h,y)-f(x,y)}{h}$$ $$\frac{\partial z}{\partial y} \approx \frac{f(x,y+h)-f(x,y)}{h}$$

Note that there are different ways to approximate the derivative. The most common ones are forward, backward and central differences. $$\frac{\partial z}{\partial y} \approx \frac{f(x,y+h)-f(x,y)}{h}\approx \frac{f(x,y)-f(x,y-h)}{h}\approx \frac{f(x,y+h)-f(x,y-h)}{2h}$$

And if you take the limit you get $$\frac{\partial z}{\partial x} =\lim_{h\rightarrow 0}\frac{f(x+h,y)-f(x,y)}{h}$$ $$\frac{\partial z}{\partial y}= \lim_{h\rightarrow 0}\frac{f(x,y+h)-f(x,y)}{h}$$

share|improve this answer
    
So... for appropriately small h the equation I stated is correct.... –  user2043404 Feb 5 '13 at 16:30
    
Yes, even though it is rather $\approx$ than $=$. –  sonystarmap Feb 5 '13 at 16:39

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.