English | Español
  1. Lar
  2. calculadoras
  3. calculadoras: cálculo I
  4. calculadora de cálculo

Calculadora da Série Taylor e Maclaurin (Power)

Encontre a série Taylor/Maclaurin passo a passo

A calculadora encontrará a expansão em 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 de Maclaurin, basta definir 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.

If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below.

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

Notas Relacionadas: Taylor Polynomial , Maclaurin Polynomials of Common Functions