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Error of Lin Approx with Taylors Formula

  • Thread starter glog
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  • #1
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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.
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hi glog! Welcome to PF! :smile:

Show us your mess, and we'll try to unscramble it! :wink:
 
  • #3
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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
 
  • #4
HallsofIvy
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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.
 

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