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The function $f(x)$ is approximated near $x=0$, by the 3rd dgeree Taylor polynomial

$T_3(x)=4-3x+\frac{1}{5}x^2+4x^3$. Give the values of $f(0)$, $f'(0)$, $f''(0)$, $f'''(0)$

To find this, do I take the antiderivatives starting at the 3rd degree Taylor polynomial, and go back? I'm not sure how to approach this problem.

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  • $\begingroup$ Do you recall how to get a Taylor polynomial from a function? $\endgroup$ – Simply Beautiful Art Oct 19 '16 at 20:39
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    $\begingroup$ Write down the Taylor formula and compare the terms. $\endgroup$ – Yves Daoust Oct 19 '16 at 20:42
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We have

$$T_3(x)=f(0)+xf'(0)+\frac{x^2}{2} f''(0)+\frac{x^3}{3!}f'''(0)$$

so by identification of the coefficients of $x^0,x,x^2$ and $x^3$, we get

$f(0)=4$

$f'(0)=-3$

$f''(0)=\frac{2}{5}$

and

$f'''(0)=24$.

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$f(0)=T_3(0)=4$

$f'(0)=T_3'(0)=-3$

$f''(0)=T_3''(0)={2\over 5}$

$f'''(0)=T_3'''(0)=24$

We can't say nothing for highter degree derivatives.

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  • $\begingroup$ Hm, similar can be done to approximately find $f(1),f'(1),f''(1)$ etc. with this idea. :) Intuitive answer. $\endgroup$ – Simply Beautiful Art Oct 19 '16 at 22:13

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