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Math Help - Taylor polynomials- please check

  1. #1
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    Taylor polynomials- please check

    Use taylor polynomials about 0 to evaluate cos(0.52) to 4 deciamal places, showing all your working.

    What I did;

    Since the taylor polynomial about 0 for cosine function is

    p n (x) = 1-1/2!x^2 + 1/4!x^4- 1/6!x^6+.....1/n!x^n

    Hence (did all working)

    p2(0.52) = 0.8646

    P4(0.52) = 0.867846

    p6(0.52) = 0.867819

    p8(0.52) = 0.867819

    So cos(0.52) = 0.8678
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  2. #2
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    Quote Originally Posted by fair_lady0072002
    Use taylor polynomials about 0 to evaluate cos(0.52) to 4 deciamal places, showing all your working.

    What I did;

    Since the taylor polynomial about 0 for cosine function is

    p n (x) = 1-1/2!x^2 + 1/4!x^4- 1/6!x^6+.....1/n!x^n

    Hence (did all working)

    p2(0.52) = 0.8646

    P4(0.52) = 0.867846

    p6(0.52) = 0.867819

    p8(0.52) = 0.867819

    So cos(0.52) = 0.8678
    But you have to prove that is accurate to 4 decimal places

    By Taylor's theorem,
    \cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}+R_4
    Where, (the remainder term)
    R_4=\frac{f^{(5)}(z_n)(x-c)^{n+1}}{(n+1)!} for some, z_n between zero and x
    But |f^{(5)}(z_n)=-\sin (z_n)|\leq 1
    Thus, what you have is that the remainder term is,
    |R_n|\leq \frac{(.52)^5}{5!}=.00041
    This is correct up to three decimal places.
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  3. #3
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    ThePerfectHacker

    Thanx very much.

    At this point proof of approximations found by evaluating Taylor Polynomials is beyond the scope of my course. I will definitely keep the workings for later ref.

    Thanx once more.
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  4. #4
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    Quote Originally Posted by fair_lady0072002
    Thanx very much.

    At this point proof of approximations found by evaluating Taylor Polynomials is beyond the scope of my course. I will definitely keep the workings for later ref.

    Thanx once more.
    Honestly, the biggest challenge in an approximation problem is not approximation but proof that it is accurate.
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