| Thread Closed |
Taylor series with partial derivatives |
Share Thread | Thread Tools |
| Feb28-04, 04:07 PM | #1 |
|
|
Taylor series with partial derivatives
We were gievn a question in tutorial last week asking us to calculate the Taylor series of the function f(x,y) = e^(x^(2) + y^(2)) to second order in h and k about the point x=0, y=0
I've got the forumla here with all the h's and k's in it and have it written down, but it's actually how to work it out that's confusing me. f(a,b) + 1/1! (hd/dx + kd/dy)f(a'b) etc.... My confusion is do you multiply out the brackets so you'd have- f(a,b) + 1/1! (hdf(a,b)/dx + kdf(a,b)/dy) So you do the derivatives and then sub in the values of x and y Or, do you leave it as it is the first tiem I wrote it and end up with- 1+ (hd/dx + kd/dx) +0.5(hd/dx + kd/dy)^2 etc I know there are more terms but I've so much trouble typing out mathematical terms on this computer! I know this is v.obvious etc but I just want to get this clear in my head cos I have a test this Wednesday at uni and I want to go in with a fighting chance! [:)] Thanks! [:D] |
| Feb28-04, 04:41 PM | #2 |
|
Recognitions:
|
Your first statement is correct. I'm not sure I see what you've done in the second.
(hd/dx +kd/dy) is an operator that acts on f, you then evaluate it at (0,0) in this case. the generic term of degree n in the expansion about (0,0) is [tex]\frac{1}{n!}\sum_{r=0}^n \binom{n}{r}h^rk^{n-r}\frac{\partial^nf}{\partial x^r \partial y^{n-r}}[/tex] with the function understood to be evaluated at (0,0) |
| Feb29-04, 05:22 PM | #3 |
|
|
Thanks very much![:D]
|
| Thread Closed |
| Thread Tools | |
Similar Threads for: Taylor series with partial derivatives
|
||||
| Thread | Forum | Replies | ||
| Partial Derivatives | Calculus & Beyond Homework | 0 | ||
| Partial Derivatives | Calculus | 2 | ||
| Partial Derivatives | Calculus | 11 | ||
| partial derivatives | Calculus | 5 | ||
| partial derivatives | Calculus | 5 | ||