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Find a 4th degree polynomial approximation for $f(x)=\cos(x)$ near $x = 0$ by producing a polynomial $p(x)=a+bx+cx^2+dx^3+fx^4$ for which $p^j (0)=f^j (0)$ for $0\leq j\leq4$.

This should be solved with calculus at the AB/BC level.

I have no idea where I'm going with this. Help!! Thank you!

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By $f^j$, do you mean the $j$-th derivative? The more standard notation would be $f^{(j)}$. –  user61527 Sep 29 '13 at 5:45
3  
Presumably, you know how to calculate the first few derivatives of $\cos x$ and evaluate them at zero, and you know the same for the polynomial $p(x)$; set them equal to get equations for the coefficients of $p$. –  Gerry Myerson Sep 29 '13 at 5:48
    
@GerryMyerson, why don't you turn that into an answer? –  dfeuer Sep 29 '13 at 5:49
    
@dfeuer, if OP can understand what I've written, and then post it as an answer with all the details worked out, I think that will be better than if I write it up. But thanks for the vote of confidence! –  Gerry Myerson Sep 29 '13 at 5:52
    
@GerryMyerson, I actually was suggesting that you copy the text of the comment, delete the comment, and post that same text as an answer. –  dfeuer Sep 29 '13 at 5:53

2 Answers 2

Start like this :

  1. What should $a = p(0)$ be to satisfy your requirements.
  2. Given that, use the mean value theorem on the interval $[0,x]$ to find what $p'(0) = b$ should be.
  3. Rinse repeat.
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The Mean Value Theorem? Is that really needed? –  Gerry Myerson Sep 29 '13 at 5:51
    
No, it isn't. Doesn't hurt to use it though. –  Prahlad Vaidyanathan Sep 29 '13 at 6:23

Notice that $p(0)=a$ and $\cos(0)=1$, so $a=1$. By taking derivatives, you should be able to find the other coefficients similarly.

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