Answer:
We want to find f(x) where f'(x) = cos(x)/x
Then we need to integrate f'(x) over x.
[tex]f(x) = \int\limits {f'(x)} \, dx = \int\limits {\frac{cos(x)}{x} } \, dx[/tex]
We must use a Taylor/McLauren polynomial to solve it.
Cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! .....
then the integral is:
[tex]\int\limits {cos(x)/x \, dx = \int\limits {\frac{1}{x} - \frac{x}{2!} + \frac{x^3}{4!} .... \, dx[/tex]
[tex]= lnIxI - \frac{x^2}{2*2!} + \frac{x^4}{4*4!} + .... \frac{x^{2n} }{(2n)*(2n)!} + C[/tex]
And we can write this as:
f(x) = lnIxI + ∑(-1)^n*(x^2n/(2n*(2n)!)) + C
fron n = 1 to n = ∞
and f(1) = 3.
ln(1) + ∑(-1)^n*(1/(2n*(2n)!)) + C = 3
∑(-1)^n*(1/(2n*(2n)!)) + C = 3
The sum, that is an alternating series and the terms are decreasing, so it converges, and the value can be obtained with a calculator, it is something around -0.24
-0.24 + C = 3
C = 3 - 0.24 = 2.76
f(x) = lnIxI + ∑(-1)^n*(x^2n/(2n*(2n)!)) + 2.76
