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Math Help - Another Taylor Polynomials problem

  1. #1
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    Another Taylor Polynomials problem

    I still not getting how even start to answer a question related to Taylor Polynomials so please I need some strong help and advice in how to tackle such questions as this ones above:

    i) Use Taylor polynomials p1,p3,p5,... about 0, successively, to evaluate sin(pi/10) to four decimal place showing all your working.


    ii) Use multiplication of Taylor series to find the quartic Taylor polynomial about 0 for the function f(x) = e^x - 1 / sqr of 1+x
    evaluating the coefficients


    I don't need just an answer I also need I follow trough with explanations of what is happening.

    xxx
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  2. #2
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    Re: Another Taylor Polynomials problem

    Well what is the MacLaurin Series for sin(x)? What do you get when you truncate it to 1, 3, 5 places? What do you get when you substitute pi/10?
    Thanks from topsquark and michele
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  3. #3
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    Re: Another Taylor Polynomials problem

    Hi Prove It,

    Following your instruction:

    First I got the Taylor series for sin x ( why this series does not use the even terms?)

    Then I started to calculate :

    p1 = x - 1/3! x^3

    p3 = p1 + 1/5! x^5

    p5 = p3 - 1/7! x^7

    p7 = p5 + 1/9! x^9

    when x = pi/10

    It is correct?
    Last edited by michele; June 24th 2014 at 07:22 AM.
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  4. #4
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    Re: Another Taylor Polynomials problem

    What about the second problem?

    ii) Use multiplication of Taylor series to find the quartic Taylor polynomial about 0 for the function f(x) = e^x - 1 / sqr of 1+x
    evaluating the coefficients

    I'm not sure if I have done it right

    First I obtained the Taylor series for e^x and the subtracted -1 from each term and ended up with the following series:

    e^x - 1 = x - 1 + (1/2)x^2 - (5/6)x^3 - (23/24)x^4 - ....

    Second I used the binomial series to find the taylor serie about 0 for the function 1/sqr 1+x = 1/(1+x)^1/2 = (1+x)^-1/2

    1/sqr 1+x = 1 - (1/2)x + (3/8)x^2 - (5/16)x^3 + (35/128)x^4 - ...

    Third I started to multiply both series term by term:

    e^x - 1/sqr 1+x = (x - 1 + (1/2)x^2 - (5/6)x^3 - (23/24)x^4 - ....) x (1 - (1/2)x + (3/8)x^2 - (5/16)x^3 + (35/128)x^4 - ...)

    =x(1 - (1/2)x + (3/8)x^2 - (5/16)x^3 + (35/128)x^4 - ...) - 1(1 - (1/2)x + (3/8)x^2 - (5/16)x^3 + (35/128)x^4 - ...) +((1/2)x^2)(1 - (1/2)x + (3/8)x^2 - (5/16)x^3 + (35/128)x^4 - ...) - ((5/6)x^3)(1 - (1/2)x + (3/8)x^2 - (5/16)x^3 + (35/128)x^4 - ...) - ((23/24)x^4)(1 - (1/2)x + (3/8)x^2 - (5/16)x^3 + (35/128)x^4 - ...) - ...

    Does it makes any sense or I totally wrong?
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    Re: Another Taylor Polynomials problem

    Quote Originally Posted by michele View Post
    Hi Prove It,

    Following your instruction:

    First I got the Taylor series for sin x ( why this series does not use the even terms?)

    Then I started to calculate :

    p1 = x - 1/3! x^3

    p3 = p1 + 1/5! x^5

    p5 = p3 - 1/7! x^7

    p7 = p5 + 1/9! x^9

    when x = pi/10

    It is correct?
    So what do you get when you sub in x = pi/10?
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  6. #6
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    Re: Another Taylor Polynomials problem

    Quote Originally Posted by Prove It View Post
    So what do you get when you sub in x = pi/10?
    I got:
    p1=0.30899155
    p3=0.30901705
    p5=0.30901699
    p7=0.30901699

    How p5 and p7 agree to 6 decimal places sin(pi/10)=0.3090 to 4 decimal places.

    Right?
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