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Let $f(x) = e^x\cos x$. Approximate the function $f$ through the taylor polynomial at point $x_0 = 0$ and find the degree of the polynomial so that the remainder in the interval $[-\frac{1}{100},\frac{1}{100}]$ is smaller than $10^{-8}$

Approach:

$f(x)= e^x\cos x$

$f'(x)= e^x(\cos x-\sin x)$

$f''(x) = -2e^x\sin x$

$f'''(x)=-2e^x(\sin x+\cos x)$

I don't know what to do next

A.Γ.
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2 Answers2

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$$e^x=1+x+x^2/2+x^3/6+....$$ $$cosx=1-x^/2-x^4/24+......$$ Multiply to get $$e^x.cosx=1+x/2-x^3/12+.....$$

The formula for the error term involves the derivative of $ e^x.cosx$ and $n!$.

With these 3 terms the error condition is met. Thus the degree is 3

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If you are to determine $n$-th derivative [ because the coefficient you are searching for the function $h : x \mapsto e^x\cos(x)$ ] is given for $n \in \mathbb{N}$ by $$ a_n=\frac{h^{(n)}\left(0\right)}{n!} $$

If $f$ et $g$ are differenciable enough, you can use for $n \in \mathbb{N}$ $$ \left(fg\right)^{\left(n\right)}=\sum_{k=0}^{n}\binom{n}{k}f^{\left(k\right)}g^{\left(n-k\right)} $$

What are cosine and exponential $n$-th derivative ?

Atmos
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