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This didn't go right the first time, so I'm going to drastically rephrase the query.

As per this previous question, I am wondering if the two series $$\frac{f(a)+f(b)}{2}\frac{(b-a)}{1!}+\frac{f'(a)-f'(b)}{2}\frac{(b-a)^2}{2!}+\frac{f''(a)+f''(b)}{2}\frac{(b-a)^3}{3!}+\cdots$$

and $$\frac{f(a)+f(b)}{2}\frac{(b-a)}{1!}+\frac{f'(a)-f'(b)}{2^2}\frac{(b-a)^2}{2!}+\frac{f''(a)+f''(b)}{2^3}\frac{(b-a)^3}{3!}+\cdots$$ could possibly be equal. The only difference is the powers of $2$ in the denominator.

NOTE: the numerator is $f^{(k)}(a)+(-1)^kf^{(k)}(b)$ in general.

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    $\begingroup$ Yes if all derivatives are 0's (f is constant)... $\endgroup$
    – user140541
    Apr 16, 2014 at 3:03
  • $\begingroup$ @d.k.o. I mean for (basically) arbitrary functions $f$. $\endgroup$
    – user142299
    Apr 16, 2014 at 3:04
  • $\begingroup$ Is this a different question from the cited one? math.stackexchange.com/questions/755565/estimation-of-integral $\endgroup$ Apr 21, 2014 at 9:23

3 Answers 3

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The identity you give is quite pretty! Yes they are equal.

Let $$\phi(t) = \sum_{n \ge 0}\frac{(b-a)^{n+1}}{(n+1)!} \left( f^{(n)}(a) \cdot t^{n+1} + f^{(n)}(b) \cdot (-1)^n (1-t)^{n+1}\right).$$ The first series in question is $\frac12(\phi(1)+\phi(0))$ and the second is $\phi(\frac12)$. The "why" is that these are equal because, in fact, $\phi$ is constant: $$\phi(t) = \int_a^bf(x) \, dx$$ for all $0 \le t \le 1$. The proof is by splitting the integral into two as $$\int_a^bf(x) \, dx = \int_a^{a(1-t)+bt}f(x) \, dx + \int_{a(1-t)+bt}^bf(x) \, dx$$ and evaluating the first by expanding $f$ as a power series about $a$ and the second by expanding $f$ as a power series about $b$. You will get $\phi(t)$ as the result.

Specifically, the "when" is the condition that the power series expansions $$f(x) = f(a)+f'(a)(x-a)+\frac{1}{2!}f''(a)(x-a)^2+\cdots$$ and $$f(x) = f(b)+f'(b)(x-b)+\frac{1}{2!}f''(b)(x-b)^2+\cdots$$ are valid for all $x \in [a,b]$.

At the risk of sounding totally banal, I've basically just briefly rehashed (and slightly generalized) the derivation you linked to $-$ that post in itself already gives the proof.

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  • $\begingroup$ Well done - clear and concise explanation. Upvote. $\endgroup$ Apr 20, 2014 at 6:58
  • $\begingroup$ I appreciate the post, but frankly I'm not entirely sure why it is receiving so many upvotes. As you said, you have just copied the proof from the original post and slightly modified it. I was really looking for another, different explanation which showed why (intuitively, that is) the two series are equal. $\endgroup$
    – user142299
    Apr 21, 2014 at 3:35
  • $\begingroup$ I guess there's no 'magic' intuition beyond the ideas sketched in the original post. Thank you for the two answers, you have done excellent work and I now understand the identity much better than I did before. Unless I get an answer to my comment on the "super short" proof below, I will certainly award you the bounty. Cheers! And +1 on both. $\endgroup$
    – user142299
    Apr 21, 2014 at 16:36
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Alright, here's another explanation.

The idea is that when you're dealing with an analytic function, if you know $f(a), f'(a), f''(a), ...$, then you already have all the information about $f$. If you also happen to know $f(b), f'(b), f''(b), ...$, well that information is totally redundant. Since $$f(x) = f(a)+f'(a)\cdot(x-a)+\frac{1}{2!}f''(a)\cdot(x-a)^2+\cdots = \sum_n \frac{f^{(n)}(a)}{n!}(x-a)^n$$ and $$f^{(n)}(x) = f^{(n)}(a) + f^{(n+1)}(a)\cdot(x-a) + \frac{1}{2!}f^{(n+2)}(a)\cdot(x-a)^2 + \cdots = \sum_m \frac{f^{(m)}(a)}{(m-n)!}(x-a)^{m-n},$$ just go into the series in question and substitute away all instances of $f^{(n)}(b)$. Then find the coefficient of $f^{(N)}(a)$ for each $N$; you'll find that you get $\frac{(b-a)^{N+1}}{(N+1)!}$ as the coefficient in both cases.


Are you prepared for gory details? Are you sure you really want to see gory details?

So the first series in question is $$\sum_{n \ge 0}\frac{(b-a)^{n+1}}{(n+1)!} \left( f^{(n)}(a) \cdot \frac12 + f^{(n)}(b) \cdot (-1)^n \cdot\frac12\right).$$ Now do the substitution to get $$\sum_{n \ge 0}\frac{(b-a)^{n+1}}{(n+1)!} \left( \frac12 f^{(n)}(a) + \frac12 (-1)^n \sum_m \frac{f^{(m)}(a)}{(m-n)!}(b-a)^{m-n} \right).$$ The coefficient of $f^{(N)}(a)$ in this $$ \frac12 \frac{(b-a)^{N+1}}{(N+1)!} + \frac12 (-1)^0 \frac{(b-a)^{0+1}}{(0+1)!} \frac{(b-a)^{N-0}}{(N-0)!} + \frac12 (-1)^1 \frac{(b-a)^{1+1}}{(1+1)!} \frac{(b-a)^{N-1}}{(N-1)!} + \cdots + \frac12 (-1)^N \frac{(b-a)^{N+1}}{(N+1)!} \frac{(b-a)^{N-N}}{(N-N)!} $$ which can be rewritten as $$ \frac12 \frac{(b-a)^{N+1}}{(N+1)!} \left( 1 + (-1)^0 \binom{N+1}{1} + \cdots + (-1)^N \binom{N+1}{N+1} \right) .$$ Apply the binomial coefficient identity (i.e. just the binomial theorem) $$ (-1)^0 \binom{c}{0} + (-1)^1 \binom{c}{1} + \cdots + (-1)^c \binom{c}{c} = (1-1)^c = 0 $$ to get $\frac{(b-a)^{N+1}}{(N+1)!}$ as promised.


I'll leave out the details of the other verification; you my dear reader can fill them in as an exercise. To briefly sketch it: at the last step you'll get $$ \frac{(b-a)^{N+1}}{(N+1)!} \left( \frac{1}{2^{N+1}} + \frac{(-1)^0}{2^1}\binom{N+1}{1} + \frac{(-1)^1}{2^2} \binom{N+1}{2} + \cdots + \frac{(-1)^N}{2^{N+1}} \binom{N+1}{N+1} \right) $$ and another binomial coefficient identity will bring you to $\frac{(b-a)^{N+1}}{(N+1)!}$.


So there it is. Hey, it wasn't actually so bad. I'd say the other proof is more intuitive, since it actually explains what the two series in question represent. This derivation is more straightforward, in the sense that it requires no real creative leaps.

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Denote a primitive of $f$ by $F$ and add ${1\over2}\bigl(F(a)-F(b)\bigr)$ to both series. Then the first series becomes $$\eqalign{{1\over2}\sum_{k=0}^\infty \bigl(F^{(k)}(a)-(-1)^kF^{(k)}(b)\bigr){(b-a)^k\over k!}&={1\over2}\sum_{k=0}^\infty F^{(k)}(a){(b-a)^k\over k!} -{1\over2}\sum_{k=0}^\infty F^{(k)}(b){(a-b)^k\over k!}\cr &={1\over2}\bigl(F(b)-F(a)\bigr)\ ,\cr}$$ and similarly the second series becomes $$-{1\over2}\bigl(F(a)-F(b)\bigr)+\sum_{k=0}^\infty \bigl(F^{(k)}(a)-(-1)^k F^{(k)}(b)\bigr)\>{\bigl({b-a\over2}\bigr)^k\over k!}$$ $$={1\over2}\bigl(F(b)-F(a)\bigr)+F\left({a+b\over2}\right)-F\left({a+b\over2}\right)={1\over2}\bigl(F(b)-F(a)\bigr)\ .$$

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  • $\begingroup$ Sorry, I don't understand how you're getting from the series to the expression on the right... are you skipping steps or is there some obvious reason that I'm missing? $\endgroup$
    – user142299
    Apr 21, 2014 at 16:29
  • $\begingroup$ @NotNotLogical: There was an obvious reason, see the edit. $\endgroup$ Apr 22, 2014 at 8:18

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