Results 1 to 5 of 5

Math Help - integral and taylor.

  1. #1
    Member i_zz_y_ill's Avatar
    Joined
    Mar 2008
    Posts
    140

    integral and taylor.

    part (iv) h is small

    k part (i) find integral of (1)/(4x^2+9)^0.5 dx between limits 0 and 2 DONE
    (ii) given f(x)=arsinh(4/3+x) find f'(x) f''(x) bladeblah..
    (iii) find maclaurin series for arsinh(4/3+x) as far as x^2
    (iv) hence show that, when h is small integral of xarsinh(4/3+x)dx
    roughly = (2/3)(h)^3

    i got part (iii) as; ln3 + (4+3x)/5 -(54x^2 +144x +96)/125 +....
    presuming this is wright in (iv) i get some utter rubbish -ve no. i screwed it up in rage! grr ???
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Oct 2007
    From
    London / Cambridge
    Posts
    591
    A taylor series centred about a is given by.

    f(a+x) = f(a) + x f'(a) + \frac{f''(a)}{2}x^2

    so we have arcsinh \left( \frac{4}{3} + x \right) = arcsinh \left( \frac{4}{3} \right) + \frac{1}{\sqrt{\frac{4}{3}^2+1}} x - \frac{\frac{4}{3}}{\left( \frac{4}{3}^2+1 \right)^\frac{3}{2}} <br />
\frac{x^2}{2}

    so the simplified series is

    arcsinh \left( \frac{4}{3} + x \right) = \ln3 + \frac{3}{5} x - \frac{18}{125} x^2

    Your wording for part vi) makes little sense.

    do you mean \int_{0}^{h}{x\mbox{arcsinh}\left( \frac{4}{3}+x \right)dx} \approx <br />
 \frac{2}{3}h^{3} ? for small h

    Bobak
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member i_zz_y_ill's Avatar
    Joined
    Mar 2008
    Posts
    140

    err yeh

    yeh i did thnx.... and the last bit??......is the taylor series the same as maclaurin??? not supposed to use taylor yet!?

    maclaurin is; f(x) = f'(0) + f'(0)x + (f''(0)x^2)/2! etc sorry about this,,,,,,,i think i must've gone wrong in thw maclaurin/taylir series when i did
    (f'(o)(4/3+x)^2)/2! someone on another post for that q told me the latter was right ,,am i wrong,,well fro ur anwer i mustve been,,,erm k let me give it a go
    Last edited by i_zz_y_ill; April 12th 2008 at 03:55 PM. Reason: >
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member i_zz_y_ill's Avatar
    Joined
    Mar 2008
    Posts
    140

    got it cheers buddy!!!!

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Oct 2007
    From
    London / Cambridge
    Posts
    591
    Quote Originally Posted by i_zz_y_ill View Post
    you got \int_{0}^{h}{x\mbox{arcsinh}\left( \frac{4}{3}+x \right)dx} \approx <br />
 \frac{2}{3}h^{3} ?

    I am fairly sure that result is incorrect. could you post your working.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Test functions as a taylor series with integral remainder
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: May 13th 2011, 08:53 AM
  2. Help with integral taylor polynomials
    Posted in the Calculus Forum
    Replies: 5
    Last Post: November 29th 2008, 07:30 PM
  3. help on taylor series (integral)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 20th 2008, 05:07 AM
  4. Estimate an integral via Taylor's polynomials
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 21st 2008, 11:05 AM
  5. Replies: 9
    Last Post: April 3rd 2008, 05:50 PM

Search Tags


/mathhelpforum @mathhelpforum