Kalkulator serii Taylor i Maclaurin (Power)
Znajdź serię Taylora/Maclaurina krok po kroku
Kalkulator znajdzie rozwinięcie szeregu Taylora (lub potęgowego) danej funkcji wokół podanego punktu, z pokazanymi krokami. Można określić rząd wielomianu Taylora. Aby uzyskać wielomian Maclaurina, wystarczy ustawić punkt na 0.
Solution
Your input: calculate the Taylor (Maclaurin) series of x3−3x2 up to n=5
A Maclaurin series is given by f(x)=∞∑k=0f(k)(a)k!xk
In our case, f(x)≈P(x)=n∑k=0f(k)(a)k!xk=5∑k=0f(k)(a)k!xk
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)(x)=f(x)=x3−3x2
Evaluate the function at the point: f(0)=0
Find the 1st derivative: f(1)(x)=(f(0)(x))′=(x3−3x2)′=3x(x−2) (steps can be seen here).
Evaluate the 1st derivative at the given point: (f(0))′=0
Find the 2nd derivative: f(2)(x)=(f(1)(x))′=(3x(x−2))′=6x−6 (steps can be seen here).
Evaluate the 2nd derivative at the given point: (f(0))′′=−6
Find the 3rd derivative: f(3)(x)=(f(2)(x))′=(6x−6)′=6 (steps can be seen here).
Evaluate the 3rd derivative at the given point: (f(0))′′′=6
Find the 4th derivative: f(4)(x)=(f(3)(x))′=(6)′=0 (steps can be seen here).
Evaluate the 4th derivative at the given point: (f(0))′′′′=0
Find the 5th derivative: f(5)(x)=(f(4)(x))′=(0)′=0 (steps can be seen here).
Evaluate the 5th derivative at the given point: (f(0))(5)=0
Now, use the calculated values to get a polynomial:
f(x)≈00!x0+01!x1+−62!x2+63!x3+04!x4+05!x5
Finally, after simplifying we get the final answer:
f(x)≈P(x)=−3x2+x3
Answer: the Taylor (Maclaurin) series of x3−3x2 up to n=5 is x3−3x2≈P(x)=−3x2+x3