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Problem Involving Two-variable Taylor's Series

  1. Jul 7, 2017 #1

    s3a

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    1. The problem statement, all variables and given/known data
    Write the Taylor's series expansion of the function f(x,y) = e-x2 sin(y) about the point (1,3). (The question doesn't specify how accurate it wants the answer to be, but based on the answer I have, it seems to me that the Taylor's polynomial should be of degree 3.)

    Apparently, the correct answer is:
    e-x2 sin(y) = e-1 sin(3) - 2(x-1) e-1 sin(3) + (y-3) e-1 cos(3) + 1/2 [ 2(x-1)2 e-1 sin(3) - 4(x-1)(y-3) e-1 cos(3) - (y-3)2 e-1 sin(3) ] + . . .

    2. Relevant equations
    I'm trying to do what the guy in this video is saying to do.:


    3. The attempt at a solution
    f(x,y) = e-x2 sin(y)

    fx = -2x e-x2 sin(y)
    fy = e-x2 cos(y)
    fxx = -2 sin(y) e-x2 + 4x2 sin(y) e-x2
    fyy = -e-x2 sin(y)
    fxy = -2x e-x2 cos(y)

    fx(1,3) = -2 e-1 sin(3)
    fy(1,3) = e-1 cos(3)
    fxx(1,3) = 2 sin(3) e-1
    fyy(1,3) = -e-1 sin(3)
    fxy(1,3) = -2 * e-1 cos(3)

    =====================================

    f(1,3):
    e-1 sin(3)

    fx(1,3):
    1/1! * -2 e-1 sin(3) x1
    = -2 e-1 sin(3) x

    fy(1,3):
    1/1! * e(-1) cos(3) y1
    =e-1 cos(3) y

    fxx(1,3):
    1/2! sin(3) e-1 x2
    =1/2 sin(3) e-1 x2

    fyy(1,3):
    1/2! * -e-1 sin(3) y2
    =1/2 * -e-1 sin(3) y2

    fxy(1,3):
    1/1! * -2 * e-1 cos(3) x1 y1
    =-2 * e-1 cos(3) x y

    f(x,y) = sum of the above parts + ...
    f(x,y) = [e-1 sin(3)] + [-2 e-1 sin(3) x] + [e-1 cos(3) y] + [1/2 sin(3) e-1 x2] + [1/2 * -e-1 sin(3) y2] + [-2 * e-1 cos(3) x y] + ...

    However, checking with software, my answer is not the same as the answer that's apparently the correct answer.

    Could someone please help me figure out what it is that I am doing wrong?
     
  2. jcsd
  3. Jul 7, 2017 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    Your series should be about ##x=1## and ##y=3##, so must be expressed in powers and products of ##x-1## and ##y-3##.
     
  4. Jul 7, 2017 #3

    s3a

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    Oh, right!

    That was the biggest problem I had; for what it's worth, another smaller mistake I made was that I had forgotten to bring the factor of 2 from fxx(x,y) to fxx(1,3), and then to the final answer.

    Thanks!
     
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