# Error of Lin Approx with Taylors Formula

• glog
Here, the maximum value of fxx is 4e^2, the maximum value of fxy is 4e^2, and the maximum value of fyy is 4e^2. So the error is less than or equal to (1/2)[4e^2(x-1)^2+ 8e^2(x-1)(y-1)+ 4e^2(y-1)^2]= 2e^2[(x-1)^2+ 2(x-1)(y-1)+ (y-1)^2]= 2e^2[(x-1)^2+ 2xy- 2x- 2y+ 1+ (y

#### glog

Let f(x,y) = e^(x^2 + y^2)
Use Taylor's theorem to show that the error in the linear approximation L(1,1)(x,y) is at most 5e^2[(x-1)^2+(y-1)^2] if 0 <= x <= 1, 0 <= y <= 1.

I've taken the partial and second partial derivatives and tried plugging it into this theorem but I get messy algebra and cannot simplify to the above.

Thanks guys.

Welcome to PF!

Hi glog! Welcome to PF!

Show us your mess, and we'll try to unscramble it!

sure, thanks!

so i get the partial derivs:
f(x,y) = e^2
fx(x,y) = 2e^2
fy(x,y) = 2e^2
fxx(x,y) = 4e^2
fxy(x,y) = 4e^2
fyy(x,y) = 4e^2

I get my linear approx:
L1,1(x,y) = 2ex^2+2ey^2-3e^2

so then, by Taylor:
|e^(x^2 + y^2) - (2ex^2+2ey^2-3e^2)| = 1/2[fxx(1,1) x^2 + 2fxy(1,1) xy + fyy(1,1) y^2]

and to be honest, I'm not sure where to go from here. :s

glog said:
sure, thanks!

so i get the partial derivs:
f(x,y) = e^2
fx(x,y) = 2e^2
fy(x,y) = 2e^2
fxx(x,y) = 4e^2
fxy(x,y) = 4e^2
fyy(x,y) = 4e^2
Wrong notation. You mean f(1,1)= e^2 etc.

I get my linear approx:
L1,1(x,y) = 2ex^2+2ey^2-3e^2
That is not "linear"! Do you mean 2e^2 x+ 2e^2 y- 3e^2?

so then, by Taylor:
|e^(x^2 + y^2) - (2ex^2+2ey^2-3e^2)| = 1/2[fxx(1,1) x^2 + 2fxy(1,1) xy + fyy(1,1) y^2]

and to be honest, I'm not sure where to go from here. :s
No, "Taylor" doesn't say that! The error for the 2nd degree Taylor's polynomial function to f(x,y) about (a, b) is (1/2)[fxx(x*,y*)(x-1)^2+ 2fxy(x*,y*)(x-1)(y-1)+ fyy(x*,y*)(y-1)^2] where x* is some number between 1 and x and y* is some number between 1 and y. Of course we don't know what x* and y* are- if we did we could evaluate that "error", add it on and get the exact value! We do know that the error is less than or equal to that with fxx(x*,y*), fxy(x*,y*), and fyy(x*,y*) taken as as at least the maximum possible values between 1 and x, 1 and y.