# Thread: taylor theorem proof - extra term?

1. ## taylor theorem proof - extra term?

I am working through the proof of taylor's theorem going from this step:

$\displaystyle f(a)+f'(a)(x-a)+\int_{a}^{x}(x-t)f''(t)dt$

to this step:

$\displaystyle f(a)+f'(a)(x-a)+\frac{1}{2}(x-a)^2f''(a)+\frac{1}{2}\int_{a}^{x}(x-t)^2f'''(t)dt$

for the integral in the first equation I have
$\displaystyle u=f''(t), u'=f'''(t)$
$\displaystyle v=-\frac{1}{2}(x-t)^2+\frac{x^2}{2}, v'=(x-t)dt$

so for uv I get the 0.5*(x-a)^2*f''(a) just fine.

but for integral of u'v I get:
$\displaystyle \frac{1}{2}\int_{a}^{x}(x-t)^2f'''(t) + \frac{x^2}{2} dt$
and if i integrate and evaluate the extra $\displaystyle \frac{x^2}{2}$ term, it does not seem to help.
Which step/s are incorrect?

2. To show:
$\displaystyle \int_{a}^{x}(x-t)f''(t)dt = \frac{1}{2}(x-a)^2f''(a)+\frac{1}{2}\int_{a}^{x}(x-t)^2f'''(t)dt$

Set u(t) = f''(t)
Set dv = (x-t)dt

Then v = $\displaystyle \frac{-(x-t)^2}{2}$ and
$\displaystyle \int_{a}^{x}(x-t)f''(t)dt$ = $\displaystyle \int_{a}^{x}udv$ = $\displaystyle \frac{1}{2}(x-a)^2f''(a)+\frac{1}{2}\int_{a}^{x}(x-t)^2f'''(t)dt$