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How to prove that $f(x)=\sum_{i=0}^{M-1}(-1)^i\binom{M}{i+1}\frac{1}{1+ix\Delta}$, where $\Delta>0,M>1$, increases with $x$?

Dave
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2 Answers2

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Let us define $\varphi$ by

$$ \varphi(u) = \sum_{i=0}^{M-1} (-1)^i \binom{M}{i+1} u^i. $$

By a simple computation

$$ \varphi(u) = \sum_{i=0}^{M-1} (-1)^i \binom{M-1}{i} \frac{M}{i+1} u^i = \int_{0}^{1} M(1 - su)^{M-1} \, ds, $$

we find that $\varphi(u)$ decreases on $[0, 1]$. And this function is related to our $f$ by

$$ f(x) = \sum_{i=0}^{M-1} (-1)^i \binom{M}{i+1} \frac{1}{1+ix\Delta} = \sum_{i=0}^{M-1} (-1)^i \binom{M}{i+1} \int_{0}^{1} t^{ix\Delta} \, dt = \int_{0}^{1} \varphi(t^{x\Delta}) \, dt. $$

Since $x \mapsto t^{x\Delta}$ decreases for each $t \in (0, 1)$, it follows that $x \mapsto \varphi(t^{x\Delta})$ increases. Therefore the claim follows.

Proof of the 1st equality. Notice that

$$ \binom{M}{i+1} = \binom{M-1}{i}\frac{M}{i+1} \quad \text{and} \quad \int_{0}^{1} s^{i} \, ds = \frac{1}{i+1}. $$

Then

\begin{align*} \varphi(u) &= \sum_{i=0}^{M-1} (-1)^i \binom{M-1}{i} \frac{M}{i+1} u^i \\ &= \sum_{i=0}^{M-1} (-1)^i \binom{M-1}{i} M u^i \int_{0}^{1} s^{i} \, ds \\ &= M \int_{0}^{1} \left( \sum_{i=0}^{M-1} (-1)^i \binom{M-1}{i} u^i s^{i} \right) \, ds \\ &= M \int_{0}^{1} (1 - su)^{M-1} \, ds, \end{align*}

where in the last line we utilized the binomial theorem to simply the sum.

Proof of the 2nd equality. Again, we know that

$$ \int_{0}^{1} t^{ix\Delta} \, dt = \frac{1}{1+ix\Delta}. $$

So we have

\begin{align*} f(x) &= \sum_{i=0}^{M-1} (-1)^i \binom{M}{i+1} \frac{1}{1+ix\Delta} \\ &= \sum_{i=0}^{M-1} (-1)^i \binom{M}{i+1} \int_{0}^{1} t^{ix\Delta} \, dt \\ &= \int_{0}^{1} \left( \sum_{i=0}^{M-1} (-1)^i \binom{M}{i+1} t^{ix\Delta} \right) \, dt \\ &= \int_{0}^{1} \varphi(t^{x\Delta}) \, dt. \end{align*}

Sangchul Lee
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  • Interesting! But why this equality holds? $$ \varphi(u) = \sum_{i=0}^{M-1} (-1)^i \binom{M-1}{i} \frac{M}{i+1} u^i = \int_{0}^{1} M(1 - su)^{M-1} , ds, $$ – Dave Sep 26 '17 at 04:57
  • @Dave, The first equality follows from $$ \binom{M}{i+1} = \frac{M}{i+1}\binom{M-1}{i}, $$ and the second identity follows from the identity $$\int_{0}^{1} s^{i} , ds = \frac{1}{1+i} $$ combined with the binomial theorem. :) – Sangchul Lee Sep 26 '17 at 04:59
  • Dear @ Sangchul Lee, I check the equality you give using matlab, and it is correct from the numerical results. – Dave Sep 26 '17 at 05:13
  • @ Sangchul Lee But I am still not very clear that where is the $\sum$ in the second identity? – Dave Sep 26 '17 at 05:19
  • @Dave, I added some extra explanation. – Sangchul Lee Sep 26 '17 at 05:29
  • @ Sangchul Lee, Thank you very much for your kindly comments on my question. – Dave Sep 26 '17 at 06:06
  • Dear @ Sangchul Lee, if $\frac{1}{1+ix\Delta}$ is replaced by some more general form such as $\frac{x}{a_ix+b_i}$ with $a_i,b_i>0$, can you have some method to handle this? Thank you! – Dave Sep 27 '17 at 03:09
  • @Dave, My understanding on this particular case is that it is related to the iterated forward difference applied to a completely monotone function. For general $a_i$ and $b_i$, though, there is no reason for the same thing to hold. For instance, you can play with $a_i$ and $b_i$ to replicate marty cohen's mistaken example. – Sangchul Lee Sep 27 '17 at 06:36
  • Thank you, @ Sangchul Lee. Would you please explain more on how can I apply the iterated forward difference into the general form function. – Dave Sep 27 '17 at 06:59
  • @Dave, I know not much about that topic. But a general idea is that if your sequence came from some nice function, then the $n$-fold finite difference will approximate the $n$-th derivative. Of course, this is rather a naive intuition rather than an established fact, so any specific claim should be justified case by case. This is almost all I know about it, so you may google it for more information. – Sangchul Lee Sep 27 '17 at 17:18
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I don't think that it's true.

For $m=3$ with $\delta = 1$, $f(x) =1-\dfrac{3}{1+x}+\dfrac{3}{1+2x} =1-3\dfrac{1+2x-(1+x)}{(1+x)(1+2x)} =1-3\dfrac{x}{(1+x)(1+2x)} $.

If $g(x) =\dfrac{x}{(1+x)(1+2x)} $, then, according to Wolfy, $g'(x) =\dfrac{1 - 2 x^2}{(x + 1)^2 (2 x + 1)^2} $ and this changes sign at $1/\sqrt{2}$ so it is not monotonic.

marty cohen
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  • If $M = 3$ and $\Delta = 1$, then $f(x)$ should be $$ f(x) = 3 - \frac{3}{1+x} + \frac{1}{1+2x}. $$ And this function is monotone increasing. – Sangchul Lee Sep 26 '17 at 04:18