Find the Maclaurin series of [ F(x)=integral dt/(1+t^3) from 0 to x ] and its interval of convergence and ...?

Question

Use the Maclaurin series of [ F(x)=integral dt/(1+t^3) from 0 to x ] to evaluate an approximate value for F(1) =integral dt/(1+t^3) from 0 to 1

what is the maclaurin series of F(x)? I found it Sigma[0 to infinity] ((-1)^n) * (x^(3n+1)) / (3n+1) but I am not sure

Answer 1

1/(1+t^3) is the sum of a geometric serie with r=-t^3
1/(1+x^3) = 1-x^3 +x^6-x^9+x^12++++(-1)^n-1 *x^3(n-1)+++
Its interval of convergence is I xI <1
Integrating term by term for IxI<1

Int = x -x^4/4 +x^7/7 ++++(-1)^n-1*x^(3n-2)/(3n-2)++
At x=1 this is an alternate series

1-1/4+1/7- 1/10 ++++(-1)^n-1 * 1/(3n-2)++ convergent by
the Leibnitz criteria.
The error taking a partial sum of an alternate series is in absolute value less then the first term left out.
So if you want
error <1/00 1/(3n+2)<1/100 so 3n+2>100 and
n>33 so you have to sum till n=33
You can take F(1) =0.8 with error less than 1/13

Answer 2

The general Maclaurin series is

SUM[k=0 to inf] (x^k/k!) f(kth derivative)(0)

Since F(x) = int[0,x][dt/(1+t^3)], by the Fundamental Theorem of Calculus, F'(x) = 1/(1+x^3) and F(0) = 0 and F'(0) = 1.

F''(x) = -3x^2/(1+x^3)^2 so F''(0) = 0

An approximation for F(1) is

F(0) + 1/1! F'(0) + 1/2! F''(0) = 0 + 1 + 0 = 1.

The actual value of F(1) is 0.835648848.

F'''(0) = 0, but F''''(0) = -6, so we can add the term -6/4! = -1/4 to our approximation, making it 3/4. A little bit closer, but still not too accurate.

Answer 3

f(x)= 8 e^9x f(0)= 8 f'(x) = seventy 2 e^9x f'(0)= seventy 2 f''(x) = 8(9)^2 e^9x f''(0) = 8(9)^2 f'''(x) = 8(9)^3 e^9x f'''(0) = 8(9)^3 f(x) = 8 + 8(9)^2 x /a million! + 8(9)^2 x^2 /2! + 8(9)^3 x^3/3! +..... keep on with the Ratio attempt: a(n+a million)/a(n) = 8(9)^(n+a million)x^(n+a million) n! /(n+a million)! 8 (9)^n x^(n) = 9 x /(n+a million) As n strategies infinity 9/(n+a million) -->0 |x| = 0