Take 2: When/Why are these equal?

Question

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.

Answer 1

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.

Answer 2

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.

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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.

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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 $\frac{(b-a)^{N+1}}{(N+1)!}$.

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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.

Answer 3

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 $$-{1\over2}\bigl(F(a)-F(b)\bigr)+\sum_{k=0}^\infty \bigl(F^{(k)}(a)-(-1)^k F^{(k)}(b)\bigr)\>{\bigl({b-a\over2}\bigr)^k\over k!}$$ $$={1\over2}\bigl(F(b)-F(a)\bigr)+F\left({a+b\over2}\right)-F\left({a+b\over2}\right)={1\over2}\bigl(F(b)-F(a)\bigr)\ .$$