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Math Help - Taylor Polynomials

  1. #1
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    Taylor Polynomials

    Question given:

    Taylor Polynomials-taylors-polynomials.png
    (I think you can click to enlarge it..)

    Both the question and solution are highlighter.. But I Don't unnderstand.. My answer would be:

    P5(x) = (x-1) + [(x2/2!)(x-1)2] + [(x3/3!)(X-1)3] + [(x4/4!)(X-1)4 + [(x5/5)(x-1)5

    Where are all them zeros coming from, and where did the 120 come from D: !

    thank you
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  2. #2
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    Re: Taylor Polynomials

    Isn't 5! = 120?
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  3. #3
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    Re: Taylor Polynomials

    yeah - I'm not understanding why they would put 120/5!
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  4. #4
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    Re: Taylor Polynomials

    f(1)=1^5-5*1^4+10*1^3-10*1^2+6*1-2=0
    Let's calculate first five derivatives:
    f'(x)=5x^4-20x^3+30x^2-20x+6, so f'(1)=5*1^4-20*1^3+30*1^2-20*1+6=1
    f''(x)=20x^3-60x^2+60x-20, so f''(1)=20*1^3-60*1^2+60*1-20=0
    f'''(x)=60x^2-120x+60, so f'''(1)=60*1^2-120*1+60=0
    f^{(4)}(x)=120x-120, so f^{(4)}(1)=120*1-120=0
    f^{(5)}(x)=120, so f^{(5)}(1)=120
    Now, by definition
    P_5(1)=f(1)+\frac{f'(1)}{1!}(x-1)+\frac{f''(1)}{2!}(x-1)^2+\frac{f'''(1)}{3!}(x-1)^3+\frac{f^{(4)}(x)}{4!}(x-1)^4+\frac{f^{(5)}(x)}{5!}(x-1)^5
    Or
    P_5(1)=0+\frac{1}{1!}(x-1)+\frac{0}{2!}(x-1)^2+\frac{0}{3!}(x-1)^3+\frac{0}{4!}(x-1)^4+\frac{120}{5!}(x-1)^5=\frac{1}{1!}(x-1)+\frac{120}{5!}(x-1)^5
    Since 1!=1 and 5!=120 then P_5(1)=(x-1)+(x-1)^5
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