Simple Taylor Series

**momomath** · August 6th 2010, 05:09 PM

I just wanted to know if my answer to this problem is accurate:

2 degree taylor approximation around (0,0) for the following function:

e^x*ln(1+y)

f(0,0) + fx(0,0)x + fy(0,0)y + 1/2 fxx(0,0)x^2 + 1/2 fyy(0,0)y^2 + fxy(0,0)xy

Also, can someone post the general form for a multivariable taylor expansion?

Thank you.

**chiph588@** · August 6th 2010, 05:25 PM

Taylor series - Wikipedia, the free encyclopedia

**HallsofIvy** · August 7th 2010, 04:46 AM

The Taylor series for the function of two variables, f(x,y), around the point [tex](x_0,y_0), is:
$f(x_0,y_0)+ f_x(x_0,y_0)(x- x_0)+ f_y(x_0, y_0)(y- y_0)+ \frac{f_{xx}(x_0, y_0)}{2}(x- x_0)^2+ \frac{f_{xy}(x_0,y_0)}{2}(x- x_0)(y- y_0)+ \frac{f_{yy}(x_0,y_0)}{2}(y- y_0)^2$ $+ \frac{f_{xxx}(x_0,y_0)}{3!}(x- x_0)^3+ \frac{f_{xxy}(x_0,y_0)}{3!}(x-x_0)^2(y- y_0)+ \frac{f_{xyy}(x_0, y_0)}{3!}(x- x_0)^2(y- y_0)+ \frac{f_{yyy}(x_0,y_0)}{3!}(y- y_0)^3+ \cdot\cdot\cdot$ .

For positive integer n, there are n+ 1 "nth" derivatives to be evaluated at $(x_0, y_0)$ . Each will involve f differentiated with respect to x i times and with respect to y n- i times for i= 0 to n. Each is divided by n! and multiplied by $(x- x_0)^i(y- y_0)^{n-i}$ .

**momomath** · August 7th 2010, 11:44 AM

Thank you. I was able to finally figure it out. I also found a version of the multivariable expansion that uses summation to make it shorter.

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