# Calculadora das séries Taylor e Maclaurin (potência)

A calculadora encontrará a expansão da série de Taylor (ou potência) da função dada em torno do ponto dado, com as etapas mostradas. Você pode especificar a ordem do polinômio de Taylor. Se você quiser o polinômio Maclaurin, defina o ponto como $0$.

Enter a function:

Enter a point:

For Maclaurin series, set the point to 0.

Order n=

Evaluate the series and find the error at the point

The point is optional.

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## Solution

Your input: calculate the Taylor (Maclaurin) series of $\sin{\left(x \right)}$ up to $n=5$

A Maclaurin series is given by $f\left(x\right)=\sum\limits_{k=0}^{\infty}\frac{f^{(k)}\left(a\right)}{k!}x^k$

In our case, $f\left(x\right) \approx P\left(x\right) = \sum\limits_{k=0}^{n}\frac{f^{(k)}\left(a\right)}{k!}x^k=\sum\limits_{k=0}^{5}\frac{f^{(k)}\left(a\right)}{k!}x^k$

So, what we need to do to get the desired polynomial is to calculate the derivatives, evaluate them at the given point, and plug the results into the given formula.

$f^{(0)}\left(x\right)=f\left(x\right)=\sin{\left(x \right)}$

Evaluate the function at the point: $f\left(0\right)=0$

1. Find the 1st derivative: $f^{(1)}\left(x\right)=\left(f^{(0)}\left(x\right)\right)^{\prime}=\left(\sin{\left(x \right)}\right)^{\prime}=\cos{\left(x \right)}$ (steps can be seen here).

Evaluate the 1st derivative at the given point: $\left(f\left(0\right)\right)^{\prime }=1$

2. Find the 2nd derivative: $f^{(2)}\left(x\right)=\left(f^{(1)}\left(x\right)\right)^{\prime}=\left(\cos{\left(x \right)}\right)^{\prime}=- \sin{\left(x \right)}$ (steps can be seen here).

Evaluate the 2nd derivative at the given point: $\left(f\left(0\right)\right)^{\prime \prime }=0$

3. Find the 3rd derivative: $f^{(3)}\left(x\right)=\left(f^{(2)}\left(x\right)\right)^{\prime}=\left(- \sin{\left(x \right)}\right)^{\prime}=- \cos{\left(x \right)}$ (steps can be seen here).

Evaluate the 3rd derivative at the given point: $\left(f\left(0\right)\right)^{\prime \prime \prime }=-1$

4. Find the 4th derivative: $f^{(4)}\left(x\right)=\left(f^{(3)}\left(x\right)\right)^{\prime}=\left(- \cos{\left(x \right)}\right)^{\prime}=\sin{\left(x \right)}$ (steps can be seen here).

Evaluate the 4th derivative at the given point: $\left(f\left(0\right)\right)^{\prime \prime \prime \prime }=0$

5. Find the 5th derivative: $f^{(5)}\left(x\right)=\left(f^{(4)}\left(x\right)\right)^{\prime}=\left(\sin{\left(x \right)}\right)^{\prime}=\cos{\left(x \right)}$ (steps can be seen here).

Evaluate the 5th derivative at the given point: $\left(f\left(0\right)\right)^{\left(5\right)}=1$

Now, use the calculated values to get a polynomial:

$f\left(x\right)\approx\frac{0}{0!}x^{0}+\frac{1}{1!}x^{1}+\frac{0}{2!}x^{2}+\frac{-1}{3!}x^{3}+\frac{0}{4!}x^{4}+\frac{1}{5!}x^{5}$

Finally, after simplifying we get the final answer:

$f\left(x\right)\approx P\left(x\right) = x- \frac{1}{6}x^{3}+\frac{1}{120}x^{5}$

Answer: the Taylor (Maclaurin) series of $\sin{\left(x \right)}$ up to $n=5$ is $\sin{\left(x \right)}\approx P\left(x\right)=x- \frac{1}{6}x^{3}+\frac{1}{120}x^{5}$