Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 3 down vote favorite

Let's \begin{align} f(T)= &f(0)+ \int_0^T f' (t)dt\\ f(T)=&(0)+f' (T)T-\int_0^T f'' (t)tdt\\ f(T)=&(0)+f' (T)t-f'' (T) \frac{T^2}{2}+\int_0^Tf''' (t) \frac{t^2}{2} dt\\ f(T)=&f(0)+f' (T)T-f'' (T) \frac{T^2}{2}+ f'''(T)\frac{T^3}{6}- \int_0^Tf''''(t)\frac{t^3}{6} dt\\ \end{align}

If you compare this to the taylor expansion for $$f(T)=f(0)+f' (0)T+f'' (0) \frac{T^2}{2}+ f'''(0)\frac{T^3}{6}+...$$ and equate coefficients,$f'(T)=f'(0)$, $-f''(T)=f''(0)$,$f'''(T)=f'''(0)$, $-f''''(T)=f''''(0)$: the minus signs seem to indicate the function can only be trivial (for what non-trivial function would behave like this?), but that's not true so my 'derivation' must have a hole in it.

share|improve this question

1 Answer

up vote 2 down vote accepted

After choosing $u = f'(t)$ and $dv=dt$, take $v = t-T$ instead.

share|improve this answer
I've encountered this before, but is this personal preference or 'more correct' than the above version? – Alyosha Jan 22 at 21:30
It's only "more correct" if the "correct" result you want is the Taylor series. You'll get something different if you do something different. Note that $dv=dt$ only defines $u$ up to translation; $v = t+C$ satisfies this for any constant $C$. – Antonio Vargas Jan 22 at 21:32
Well, that thing, an infinite sum of powers of $t$, is equal to $f(t)$, and so is equal to its Maclaurin expansion, an infinite sum of powers of $t$, therefore all the coefficients of the powers of $t$ should be equal in my thing to the Maclaurin expansion. So if it's not incorrect, I must be seeing problems where there are none. Is it that am I making a fuss about nothing that $f^{2n}(0)= f^{2n}(t)$, and this is quite achievable? – Alyosha Jan 22 at 21:43
1  
@Alyosha No, that thing is not an infinite sum in powers of $T$ precisely because its "coefficients" are functions of $T$ -- they are not constants. In short, it is not a power series. You cannot then compare it to a power series and assert that the "coefficients" are equal, because that property is only true when comparing power series. What you have found is an entirely different type of series representation for $f$. – Antonio Vargas Jan 22 at 22:49
1  
@Alyosha Further, it is not clear that your series actually converges to $f$ somewhere, or what kinds of sets it converges on. For instance, power series converge on open intervals, but your series may not. There are a lot of interesting questions you could investigate, if you are so inclined. – Antonio Vargas Jan 22 at 23:04
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.