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.

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Solution

Your input: calculate the Taylor (Maclaurin) series of $$$x^{3} - 3 x^{2}$$$ 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.

$$$f^{(0)}\left(x\right)=f\left(x\right)=x^{3} - 3 x^{2}$$$

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

  1. Find the 1st derivative: $$$f^{(1)}\left(x\right)=\left(f^{(0)}\left(x\right)\right)^{\prime}=\left(x^{3} - 3 x^{2}\right)^{\prime}=3 x \left(x - 2\right)$$$ (steps can be seen here).

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

  2. Find the 2nd derivative: $$$f^{(2)}\left(x\right)=\left(f^{(1)}\left(x\right)\right)^{\prime}=\left(3 x \left(x - 2\right)\right)^{\prime}=6 x - 6$$$ (steps can be seen here).

    Evaluate the 2nd derivative at the given point: $$$\left(f\left(0\right)\right)^{\prime \prime }=-6$$$

  3. Find the 3rd derivative: $$$f^{(3)}\left(x\right)=\left(f^{(2)}\left(x\right)\right)^{\prime}=\left(6 x - 6\right)^{\prime}=6$$$ (steps can be seen here).

    Evaluate the 3rd derivative at the given point: $$$\left(f\left(0\right)\right)^{\prime \prime \prime }=6$$$

  4. Find the 4th derivative: $$$f^{(4)}\left(x\right)=\left(f^{(3)}\left(x\right)\right)^{\prime}=\left(6\right)^{\prime}=0$$$ (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: $$$f^{(5)}\left(x\right)=\left(f^{(4)}\left(x\right)\right)^{\prime}=\left(0\right)^{\prime}=0$$$ (steps can be seen here).

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

Now, use the calculated values to get a polynomial:

$$$f\left(x\right)\approx\frac{0}{0!}x^{0}+\frac{0}{1!}x^{1}+\frac{-6}{2!}x^{2}+\frac{6}{3!}x^{3}+\frac{0}{4!}x^{4}+\frac{0}{5!}x^{5}$$$

Finally, after simplifying we get the final answer:

$$$f\left(x\right)\approx P\left(x\right) = -3x^{2}+x^{3}$$$

Answer: the Taylor (Maclaurin) series of $$$x^{3} - 3 x^{2}$$$ up to $$$n=5$$$ is $$$x^{3} - 3 x^{2}\approx P\left(x\right)=-3x^{2}+x^{3}$$$