Calculadora de series de Taylor y Maclaurin (potencia)

Encontrar la serie Taylor/Maclaurin paso a paso

La calculadora hallará la expansión en serie de Taylor (o potencia) de la función dada alrededor del punto dado, con los pasos mostrados. Puede especificar el orden del polinomio de Taylor. Si desea el polinomio de Maclaurin, simplemente establezca el punto en $0$ .

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: $f\left(0\right)=0$

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: $\left(f\left(0\right)\right)^{\prime }=0$
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: $\left(f\left(0\right)\right)^{\prime \prime }=-6$
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: $\left(f\left(0\right)\right)^{\prime \prime \prime }=6$
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: $\left(f\left(0\right)\right)^{\prime \prime \prime \prime }=0$
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 $x^{3} - 3 x^{2}\approx P\left(x\right)=-3x^{2}+x^{3}$