Results 1 to 4 of 4

Math Help - Taylor Probs

  1. #1
    Newbie
    Joined
    Mar 2008
    Posts
    19

    Taylor Probs

    Hi I got problems regarding Taylor and need your help on this please

    1.

    p(x) = 2x + 4x^3 - 5x^4

    Find the polynomial i.e


    and prove that


    Using the fact that two polynomials of degree k are equal if and only if they
    have the same coefficients.

    AND

    2.

    Find the following:


    Whichever help I can get would be gr8
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    For the first one, calculate the successive derivatives of p(x) at x=2.

    p(x)=2x + 4x^3 - 5x^4
    p(2)=-44

    p'(x)=2+12x^2-20x^3
    p'(2)=-110

    p''(x)=24x-60x^2
    p''(2)=...

    \begin{aligned} T_{p,2}^{(4)} (x) & =\frac{p(2)}{0!} (x-2)^0+\frac{p'(2)}{1!} (x-2)^1 +\frac{p''(2)}{2!} (x-2)^2+\dots \\<br />
& =\frac{-44}{1} \cdot 1+\frac{-110}{1} \cdot (x-2)+\dots \end{aligned}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2008
    Posts
    19
    Thx for tip Moo I did the following based on your idea...

    p(x)=2x + 4x^3 - 5x^4
    p(2)=-44

    p'(x)=2+12x^2-20x^3
    p'(2)=-110

    p''(x)=24x-60x^2
    p''(2)=-456

    p'''(x)=24 - 120x
    p'''(2)= -216

    p''''(x)=-120

    Taylor Polynomial = [(P(2)/0!)(x-2)^0] + [(P'(2)/1!)(x-2)^1] + [(P''(2)/2!)(x-2)^2] + [(P'''(2)/3!)(x-2)^3] + [(P''''(2)/4!)(x-2)^4]

    After all substitutions and simplifying I got the following:

    = -5x^4 + 4x^3 - 168x^2n+ 530x -528

    which does look slightly similar to P(x).
    Did I go wrong somewhere or is there a few steps to perform?

    Any hints on part 2?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2008
    Posts
    19
    Following the method revamped I ended up with the following:
    P(4)p,2(x)=-44-110x+220-96x^2+384x-384-36x^3+216x^2-432x+288-5x^4+4-x^3-120x^2+160x-80

    which cancelled down to....-5x^4+4x^3+2x which confirms the proof.

    As for part 2 I got the following:

    f(x) = arctan
    f'(x)= 1/(1+x^2)
    f''(x)= -2x/(1+x^2)2
    f'''(x)= [8x^2/(1+x^2)^3] + [2/(1+x^2)^2]

    f(x) = 0
    f'(x)= 1
    f''(x)= 0
    f'''(x)= -2
    0+x+ (0x^2) (-2/3!)x^3

    Any miscalculations?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. charge probs...
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: January 10th 2011, 08:30 PM
  2. complex no probs
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 22nd 2010, 08:42 AM
  3. Help with these 2 calc probs
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 6th 2010, 07:26 AM
  4. Quantum Probs
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: April 8th 2008, 01:01 PM
  5. 3D Trig. Probs
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 16th 2008, 02:46 AM

Search Tags


/mathhelpforum @mathhelpforum