Kalkulator serii Taylor i Maclaurin (Power)

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