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Homework Help: Taylor Series and error

  1. Jun 24, 2010 #1

    zcd

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    I'm doing some review over summer before starting college, and one of the practice exams has a question pertaining to the remainder of a taylor series

    1. The problem statement, all variables and given/known data
    Show that [tex]\left|\cos{(1+x)}-\{\cos{(1)}(1-\frac{x^2}{2})-\sin{(1)}(x-\frac{x^3}{3!})\}\right|<\frac{1}{15000}[/tex] for [tex]|x|<0.2[/tex]


    2. Relevant equations
    The two equations I somewhat remember are
    [tex]R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{(n+1)}[/tex]
    and
    [tex]|R_n(x)|\leq{}M_n\frac{r^{n+1}}{(n+1)!}[/tex]
    where M (iirc) is the largest value of the nth derivative in the area of convergence?

    3. The attempt at a solution
    Here's where I'm stuck; my class in high school never fully went over error of taylor approximation because it wasn't part of the curriculum, so we were given the equations without any actual application. To make matters worse, I last took AP calculus BC over a year ago, so this is still very rusty. Can anyone point me in the right direction?
     
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  3. Jun 24, 2010 #2
  4. Jun 24, 2010 #3

    LCKurtz

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    If you expand cos(1+x) in a Taylor series about x = 0 up through the term containing x3 you will have:

    cos(1+x) = (terms from constant through x3) + R3

    where R3 is the error term if you stop with x3. So you will have

    |cos(1+x) - (terms from constant through x3)| = |R3|, where

    [tex]
    R_3(x)=\frac{f^{(4)}(c)}{(4)!}(x-0)^{(4)}
    [/tex]

    If you write out the series you will see it is exactly what is given. All you have to do is estimate the remainder. It tells you that |x| ≤ .2 and you will find it easy to overestimate f(4)(x). Just be brave and work it out. Don't use any decimals, stick with fractions.
     
  5. Jun 24, 2010 #4

    zcd

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    Still getting my feet wet again so correct me if I'm wrong :P

    [tex]\left|\cos{(1+x)}-\{\cos{(1)}(1-\frac{x^2}{2})-\sin{(1)}(x-\frac{x^3}{3!})\}\right|=\left|\cos{(1)}((1-\frac{x^2}{2}+R_3(x))-(1-\frac{x^2}{2}))-\sin{(1)}((x-\frac{x^3}{3!}+R_3(x))-(x-\frac{x^3}{3!}))\right|[/tex]
    [tex]=\left|\cos{(1)}R_3(x)-\sin{(1)}R_3(x)\right|[/tex]

    since [tex]|R_n(x)|\leq{}M_n\frac{r^{n+1}}{(n+1)!}[/tex] and Mn=1 for sin and cos, [tex]|R_3(x)|\leq{}\frac{(\frac{1}{5})^4}{4!}=\frac{1}{15000}[/tex]

    plugging it all back in,
    [tex]\left|\cos{(1)}\frac{1}{15000}-\sin{(1)}\frac{1}{15000}\right|<\frac{1}{15000}[/tex]

    is that right?
     
  6. Jun 25, 2010 #5

    LCKurtz

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    The only thing right is your estimate for R3. The rest is just jumbled confusion.

    Just write the Taylor series for cos(1+x) using the terms up through x3 like I suggested. You know, calculate f(0), f'(0), f''(0) etc. and write the series. This doesn't involve the remainder. Rearrange it to get the expression you are given. Write it as suggested in my previous post.
     
  7. Jun 25, 2010 #6

    zcd

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    I think I get it

    [tex]\cos{(1+x)}=\cos{(1)}-x\sin{(1)}-\frac{x^2}{2}\cos{(1)}+\frac{x^3}{3!}\sin{(1)}+...[/tex]
    [tex]\cos{(1+x)}=\cos{(1)}(1-\frac{x^2}{2})-\sin{(1)}(x-\frac{x^3}{3!})+R_3(x)[/tex]
    [tex]\cos{(1+x)}-\left[\cos{(1)}(1-\frac{x^2}{2})-\sin{(1)}(x-\frac{x^3}{3!})\right]=R_3(x)[/tex]
    [tex]\left|\cos{(1+x)}-\left[\cos{(1)}(1-\frac{x^2}{2})-\sin{(1)}(x-\frac{x^3}{3!})\right]\right|=\left|R_3(x)\right|[/tex]
    [tex]\left|\cos{(1+x)}-\left[\cos{(1)}(1-\frac{x^2}{2})-\sin{(1)}(x-\frac{x^3}{3!})\right]\right|\leq\frac{1}{15000}[/tex]

    how does the =< turn into <?
     
  8. Jun 25, 2010 #7

    LCKurtz

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    Good job. Look at the limits on x. Are you given strict inequality in your estimate?
     
  9. Jun 25, 2010 #8

    zcd

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    Ahh I see. Thanks for all the help :)
     
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