Solving $f(x) = \frac{x^2}2 +x - \int_0^x f(t)dt, x\in[0,1] $ with Iteration Method

Question

I have problem solving the following integral equation : $$f(x) = \frac{x^2}2 +x - \int_0^x f(t)dt, x\in[0,1] $$ using the iteration method with initial approach $f_0(x)= \frac{x^2}2 +x$

I applied Picard Iteration and then I found the following results :

$f_0(x) = x+\frac{x^2}2 $

$f_1(x) = x-\frac{x^3}6 $

$f_2(x) = x +\frac{x^4}{24}$

$f_3(x) = x-\frac{x^5}{120}$

$f_4(x) =x +\frac{x^6}{720}$

then I found the pattern:

$$f_n(x) = \sum_{j=0}^n(-1)^n(\frac{x^{n+1}}{(n+1)!} + \frac{x^{n+2}}{(n+2)!} ) $$

I don't know if it is right. If it is right, how should I continue ? Any help would be greatly appreciated.

Answer 1

Your $f_0(x)$ , $f_1(x)$ ,$f_2(x)$ , $f_3(x) , f_4(x)$ are correct.

HINT :

The pattern is :
$$f_n(x)=x+(-1)^n\frac{x^{n+2}}{(n+2)!}$$

You have to prove it. Use the above $f_n(x)$ to compute $f_{n+1}(x)$ according to the iterative method. Check that $f_{n+1}(x)$ agrees with the above pattern.

Finally, you can show that $\frac{x^{n+2}}{(n+2)!}\to 0$ when $n\to\infty$. Thus $$f(x)=x$$ Of course, this can be directly found from the integral equation.