anonymous
  • anonymous
The Taylor polynomial of degree 100 for the function f about x=3 is given by \[p(x)= (x-3)^2 - ((x-3)^4)/2! +... + [(-1)^n+1] [(x-3)^n2]/n! +... - ((x-3)^100)/50!]/ What is the value of f^30 (3)?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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watchmath
  • watchmath
Plug in \(n=15\) to the expression \((-1)^{n+1}/n!\)
watchmath
  • watchmath
Because \(f^{30}(3)\) is the coefficient of \(x^{30}\)
anonymous
  • anonymous
So each term is\[((-1)^{n+1}/n!)(x-3)^{2n}\] and \[(f^{(n)}(3)/n!)(x-3)^n\] so can I say \[(f^{(30)}(3)/30!)(x-3)^{30}=((-1)^{30+1}/30!)(x-3)^{2*30}\] I don't really get where the 15 comes from

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watchmath
  • watchmath
remember that the exponent on \((x-3)\) is \(2n\). And we want this \(2n=30\). So we need to take \(n=15\). Since we are looking for the coefficient of \((x-3)^{30}\)
anonymous
  • anonymous
so \[1/15! = f^{(30)}(3)/30!\]
watchmath
  • watchmath
Ok, let me make this more clear. We want fo select the \(n\) so that we know the coefficient of \((x-3)^{30}\). We know that the coefficient of \((x-3)^{2n}\) is \((-1)^{n+1}/n!\). So in order to find the coefficient of \((x-3)^{30}\) we need to choolse \(n=15\). In that case the coefficient would be \((-1)^{15+1}/15!=1/15!\). So \(f^{(30)}(3)=1/15!\)
anonymous
  • anonymous
Why is the coefficient of \((x-3)^{n}\), \(f^{n}(3)\) and not \(f^{n}(3)/n!\) Isn't each term \((1/n!)(f^{(n)}(3))(x-3)^n\)?
watchmath
  • watchmath
Ah you are right! :)
anonymous
  • anonymous
Thank you so much, I never would have gotten there in the first place :P

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