Let #F(x)=int_0^x e^(-5t^4)dt#. Find the MacLaurin polynomial of degree 5 for F(x)?

Question

Ezekiel Alexander · Accepted Answer

# F(x)= x - x^5#

Applying the Calculus Fundamental Theorem:

# F'(x) = e^(-5x^4)#

By definition:

# e^(z)= sum_{k=0}^{oo } z^k/(k!) =1+z + mathbb O(z^2) #

#implies e^(-5x^4)= 1 -5x^4 + mathbb O(x^8) = F'(x)#

Including:

And:

By evaluating an entire series of derivatives and computing as the Maclaurin Series, you can obtain the same result:

# f(x)=f(0)+f^'(0)x+(f^('')(0))/(2!)x^2+(f^((3))(0))/(3!)x^3+...+(f^((n))(0))/(n!)x^n+.... #

However, I wouldn't advise it.

Christopher Agresta · Answer

undefined The MacLaurin polynomial of degree 5 for $ F(x) = \int_0^x e^{-5t^4} dt $ is:

$$ P_5(x) = \sum_{n=0}^{5} \frac{F^{(n)}(0)}{n!}x^n $$

To find the coefficients, $ \frac{F^{(n)}(0)}{n!} $, we need to find the derivatives of $ F(x) $ up to the fifth order evaluated at $ x = 0 $. Then, substitute these values into the polynomial formula.

First, let's find the derivatives of $ F(x) $:
$$ F'(x) = e^{-5x^4} $$
$$ F''(x) = -20x^3 e^{-5x^4} $$
$$ F'''(x) = -60x^2(1-20x^8)e^{-5x^4} $$
$$ F^{(4)}(x) = -240x(1-20x^8)e^{-5x^4} $$
$$ F^{(5)}(x) = -240(1-20x^8-160x^4)e^{-5x^4} $$

Now, evaluate these derivatives at $ x = 0 $:
$$ F(0) = 0 $$
$$ F'(0) = 1 $$
$$ F''(0) = 0 $$
$$ F'''(0) = 0 $$
$$ F^{(4)}(0) = 0 $$
$$ F^{(5)}(0) = -240 $$

Substitute these values into the formula for $ P_5(x) $:
$$ P_5(x) = 0 + \frac{1}{1!}x^1 + 0 + 0 + 0 - \frac{240}{5!}x^5 $$
$$ P_5(x) = x - \frac{4}{3}x^5 $$

Thus, the MacLaurin polynomial of degree 5 for $ F(x) $ is $ P_5(x) = x - \frac{4}{3}x^5 $.