Showing that for a continuously thrice differentiable function $f(x)$ the following is true

Question

Show that for a continuously thrice differentiable function $f(x)$ :

$$f(x)=f(0)+xf'(0)+\frac{f''(0)x^2}{2}+\frac{1}{2}\int_{0}^{x}f'''(t)(x-t)^2dt$$

I tried by assuming that $f(x)$ can be represented as a sum of powers of $x$.

Say, $$f(x)=a_0+a_1x+a_2x^2+a_3x^3+...$$

By simple differentiation (twice) we conclude that $$a_0=f(0),a_1=f'(0) ,a_2=\frac{f''(0)}{2}$$

So first three terms have been proved equal.

• But how to prove that the sum of the rest of the terms is $\frac{1}{2}\int_{0}^{x}f'''(t)(x-t)^2dt$ ?

• Also, is assuming that $f(x)$ can be represented as a sum of powers of $x$ correct ? Why or why not ?

P.S:

• I don't know Taylor series or Maclaurin series. Please don't use them.

• I have just started learning integral calculus a few days back. So do not use very advanced concepts other than basic differential and integral calculus.

Answer 1

First we use the fundamental theorem of calculus to write $$ f(x) = f(0) + \int_0^x f'(t) dt. $$ Then we integrate by parts: $$ \int_0^x f'(t)dt = \int_0^x f'(t) (t-x)' dt = (t-x) f'(t) \vert_{t=0}^{t=x} + \int_0^x f''(t) (x-t) dt \\ = xf'(0) + \int_0^x f''(t) (x-t) dt. $$ Now we integrate by parts again: $$ \int_0^x f''(t) (x-t) dt = -\int_0^x f''(t) \frac{d}{dt} \frac{(x-t)^2}{2} dt = -\frac{(x-t)^2}{2} f''(t) \vert_{t=0}^{t=x} + \int_0^x f'''(t) \frac{(x-t)^2}{2} dt \\ = \frac{x^2}{2} f''(0) + \int_0^x f'''(t) \frac{(x-t)^2}{2} dt . $$ Chain these all together and we get $$ f(x) = f(0) + x f'(0) + \frac{x^2}{2} f''(0) + \int_0^x f'''(t) \frac{(x-t)^2}{2} dt $$ as desired.