Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Taylor Polynomial Approximation of log(2.25)

  1. Mar 24, 2010 #1
    1. The problem statement, all variables and given/known data
    Determine the order two Taylor polynomial, p2(x, y), for

    f(x, y) = log e (1 + x2 + y4)

    about point (0, 1)


    ANSWER:
    loge (2) + 2y - 2 + [tex]\frac{1}{2}[/tex] [ x2 - 2y2 + 4y - 2 ]

    Managed that question and should be correct. If not, do let me know =)



    Part 2: Using p2 (x, y), find a quadratic approximation to log e (2.25) to 4 decimal places. Use a calculator to find the value of log e (2.25) to 4 decimal places, and comment on your answer.


    2. Relevant equations

    2nd Order Taylor Polynomial
    p2(x, y) = f(x, y) + (x + a)[tex]\frac{\partial f}{\partial x}[/tex] + (y + b)[tex]\frac{\partial f}{\partial y}[/tex] + [tex]\frac{1}{2}[/tex] [ (x - a)2[tex]\frac{\partial ^2 f}{\partial x^2}[/tex] + (x - a)(y - b)[tex]\frac{\partial ^2 f}{\partial x\partial y}[/tex] + (y - b)2[tex]\frac{\partial ^2 f}{\partial y ^2}[/tex] ]


    3. The attempt at a solution

    I'm not too sure where to go from there...I thought of putting p2(x, y) as log 2 (2.25) but I'm not sure what that does after simplifying...

    Any assistance would be of much help! =)
    Thanks
     
  2. jcsd
  3. Mar 24, 2010 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You got the sign of fyy wrong.
    What do you mean by "putting p2(x, y) as log 2 (2.25)"?

    Just choose values of x and y so that f(x,y)=log 2.25 and plug those values into your expansion.
     
  4. Mar 24, 2010 #3
    I got fyy = -2

    This was from fyy = [tex]\frac{12y^2( 1 + x^2 + y^4 ) - 16y^2}{(1 + x^2 + y^4 ) ^2}[/tex]

    and since at point (0,1),

    fyy = [tex]\frac{12 (1)^2( 1 + (0)^2 + (1)^4 ) - 16 (1)^2}{(1 + (0)^2 + (1)^4 ) ^2}[/tex]
    fyy = -2

    Isn't it? Unless I stuffed up the Taylor polynomial part somewhere...


    Oh, when I meant putting p2(x, y) as log e (2.25) [correction], I meant like:

    loge(2.25) = loge (2) + 2y - 2 + [tex]\frac{1}{2}[/tex] [ x2 - 2y2 + 4y - 2 ]

    But yeah..got me no where =(


    I don't understand how I can choose values for x and y because how can we figure out what is loge (2.25) from something like loge (2) + constant. Seeing that we just choose values for x and y...unless there's the use of a calculator. Could you help shed some light on this? Thanks =)
     
  5. Mar 24, 2010 #4

    Mark44

    Staff: Mentor

    You have f(x, y) = ln (1 + x2 + y4),
    so f(0, 1) = ln(1 + 02 + 12) = ln 2

    You want ln 2.25, which can be represented as f(.5, 1). Use the polynomial you found to approximate f(.5, 1) by p2(.5, 1).
     
  6. Mar 24, 2010 #5

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    No, the top is 12x2-16 = 24-16 = 8, and the bottom is 4.
    The base of the log didn't confuse me; I figured that was a typo. It was your wording, particularly your choice of verb.
     
  7. Mar 24, 2010 #6
    Oh sorry. What would have been a better choice of verb?


    Thanks a lot guys, I believe I managed to get it now.

    Worked it out to be loge (2) + [tex]\frac{1}{8}[/tex] = 0.8181

    Hopefully that's right =)
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook