I will write a problem with an answer that apparently is wrong. My question would be what is wrong with this solution.

Define $$B(s)=\sum_{i=0}^s{{2\,s-i-1\choose s-1}\frac {i}{s}{5}^{i}{2}^{s-i}}$$ for integers $s>0$. The problem is to find $f(x)$ such that $$\int_{-2\sqrt{2}}^{2\sqrt{2}}x^{2s}f(x)dx=B(s)\qquad\text{and}\qquad \int_{-2\sqrt{2}}^{2\sqrt{2}}x^{2s-1}f(x)dx=0.\tag{$*$}$$

To find $f(x)$ we can use Zeilberg's algorithm and obtain that $B(s)$ satisfy the following recurrence relation $$25\,B \left( s \right) -3\,B \left( s+1 \right) =10\,{\frac {{8}^{s} \Gamma \left( s+1/2 \right) }{\sqrt {\pi }\Gamma \left( s+2 \right) }}.$$

Now, we can write the right hand side as an integral $$10\,{\frac {{8}^{s} \Gamma \left( s+1/2 \right) }{\sqrt {\pi }\Gamma \left( s+2 \right) }}=\int_{-2\sqrt{2}}^{2\sqrt{2}}{\frac {5{x}^{2\,s}\sqrt {8-{x}^{2}}}{2\pi }}dx.\tag{$**$}$$

Now replacing (*) and (**) in the recurrence, we obtain $$25\,\int_{-2\sqrt{2}}^{2\sqrt{2}}x^{2s}f(x)dx -3\,\int_{-2\sqrt{2}}^{2\sqrt{2}}x^{2s+2}f(x)dx=\int_{-2\sqrt{2}}^{2\sqrt{2}}{\frac {5{x}^{2\,s}\sqrt {8-{x}^{2}}}{2\pi }}dx.$$

Taking derivatives and solving for $f(x)$, we obtain $$f(x)={\frac {5\sqrt {8-{x}^{2}}}{2\pi \, \left( 25-3\,{x}^{2} \right) }}.$$

It seems everything is correct to me. However, if we take $s=1$ $$\int_{-2\sqrt{2}}^{2\sqrt{2}}x^{2}f(x)dx=\frac{20}{9},$$ but $$B(1)=5.$$

In fact it is wrong for every $s$ I fix.

Would you be kind and give me some ideas on how to fix my solution to find $f(x)$?


The problem is in your last step "Taking derivatives and solving for etc." Namely, the display before this sentence can be rewritten as $$\int_{-2\sqrt{2}}^{2\sqrt{2}}x^{2s}g(x)dx = 0,\qquad s=1,2,3,\dots,\tag{1}$$ where $$g(x):=(25-3x^2)f(x)-\frac {5\sqrt {8-{x}^{2}}}{2\pi}.$$ From here you conclude that $g(x)$ is identically zero, which is bad logic. For example, any odd function $g:[-2\sqrt{2},2\sqrt{2}]\to\mathbb{R}$, i.e. one that satisfies $g(-x)=-g(x)$, automatically satisfies (1).

P.S. The OP added the condition that $f(x)$ is even, which makes $g(x)$ even as well. In this case $g(x)$ is identically zero as the OP claimed originally (assuming of course that $f(x)$ is continuous). So the conclusion is that $f(x)$ does not exist.

  • $\begingroup$ I see. In my case $f(x)$ is an even function, because the odd moments are zero. How can I recover $f(x)$ then? $\endgroup$ – LuHell Jan 18 '17 at 17:58
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    $\begingroup$ @LuHell: Well, if $f(x)$ is an even function, then so is $g(x)$, and then we get that $g(x)$ is identically zero (at least when $g(x)$ is $x^2$ times a continuous function). So in this case the only candidate for $f(x)$ is the one that you indicate. However, this $f(x)$ does not work, which means that no $f(x)$ satisfies your conditions. $\endgroup$ – GH from MO Jan 18 '17 at 18:02
  • $\begingroup$ No $f(x)$ satisfy that? I fail to see why. $\endgroup$ – LuHell Jan 18 '17 at 18:04
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    $\begingroup$ @LuHell: Because this is what your argument gives. Your initial conditions lead to a contradiction. $\endgroup$ – GH from MO Jan 18 '17 at 18:04
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    $\begingroup$ @LuHell: This theorem is about general measures, while you are looking for absolutely continuous measures of the form $f(x)dx$. For most sequences of moments there is no function $f(x)$ that you are looking for. Example: if $\int_0^1 x^r dE(x)=2^{-r}$ for each $r$, then $dE(x)$ is the Dirac measure at $1/2$. This measure is not absolutely continuous with respect to $dx$ (in fact it is concentrated on a single point), hence there is no function $f:[0,1]\to\mathbb{R}$ such that $\int_0^1 x^r f(x) dx=2^{-r}$ for each $r$. At any rate, this site is not about teaching basic measure theory. $\endgroup$ – GH from MO Jan 18 '17 at 21:23

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