Personal epiphany about Taylor theorem, true?

[Reverberant]

Hello,

I'm taking my first calculus course right now, and something struck me regarding the remainder in integral form of a Taylor series expansion:

Let's say we have a Taylor expansion of the (n-1):th order, which has a remainder of the form

Now, my claim is that if we integrate by parts n-1 times (I think), then every term of the integrated remainder, except for the last one, will cancel out a term of the Taylor polynomial; leaving only f(x). Thus, all we have left is the equation f(x)=f(x).

Is this true, and if so, can you confirm that the form of integration by parts to be used is

and not

?

Thanks in advance!

 

[SammyS]

Hello,

I'm taking my first calculus course right now, and something struck me regarding the remainder in integral form of a Taylor series expansion:

Let's say we have a Taylor expansion of the (n-1):th order, which has a remainder of the form

Now, my claim is that if we integrate by parts n-1 times (I think), then every term of the integrated remainder, except for the last one, will cancel out a term of the Taylor polynomial; leaving only f(x). Thus, all we have left is the equation f(x)=f(x).

Is this true, and ...

Thanks in advance!

Just do it.

$$\displaystyle \text{We use the notation, } f^{(n)}(t)={{d^n}\over{dt^n}}f(t)\,.$$

$$\displaystyle \text{Let } u= (x-t)^{n-1}\ \to \ {{d}\over{dt}}u=-(n-1)(x-t)^{n-2}\,.\ \ \text{Let } dv=f^{(n)}(t)}\, dt\ \to \ v=f^{(n-1)}(t)(n-1)(x-t)^{n-2}\,.$$

$$\displaystyle {{1}\over{n!}}\int_a^x{n(x-t)^{n-1}\,f^{(n)}(t)}\, dt$$ $$\displaystyle = {{1}\over{(n-1)!}}\left(\left[ (x-t)^{n-1}f^{(n-1)}(t)\right]_a^x - \int_a^x{-(n-1)(x-t)^{n-2}\,f^{(n-1)}(t)}\, dt \right)$$

$$\displaystyle = {{0\cdot f^{(n-1)}(x)-(x-a)^{n-1}f^{(n-1)}(a)}\over{(n-1)!}}\,+\,{{1}\over{(n-1)!}}\left( \int_a^x{(n-1)(x-t)^{n-2}\,f^{(n-1)}(t)}\, dt \right)$$

$$\displaystyle = \,-\,{{(x-a)^{n-1}f^{(n-1)}(a)}\over{(n-1)!}}\,+\,{{1}\over{(n-1)!}}\left( \int_a^x{(n-1)(x-t)^{n-2}\,f^{(n-1)}(t)}\, dt \right)$$
​

This will subtract the (n-1)^th order term of the Taylor expansion of f(x) and leave a Taylor expansion of the (n-2)^th order plus the integral which is the remainder.

So, yes your claim is correct!

 

[Reverberant]

Sweet, thanks!