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I am able to do all the parts except the very last. I have been trying to coax the differential equation $\frac{M'}{M}=t$ or something to that effect but I don't see how I can achieve this. Hints would be much appreciated.

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We have $M(2t)=M(t)^3 M(-t)$ and $M(t)=M(-t)$, so $M(2t)=M(t)^4$. Hence we have a functional equation for $m(t)=t^{-2}\log M(t)$ (let $m(0)=\frac12s^2=\frac12$): $$m(2t)=m(t)\quad\text{for all }t\tag{1}$$ But we know $$ m(t)=\frac12+o(1)\text{ for small }t\tag{2} $$ from the expansion of $M(t)$.

Now equations (1) and (2) together implies $m$ is constant $\frac12$. Can you see why?

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  • $\begingroup$ So letting $o(1) = f_1(t)$ we can see that $f_1(t)=f_1(2t) \implies f$ constant and $m(0) = 1/2$ gives the result? Question: how did we get equation (2) $\endgroup$ – user3184807 May 22 at 10:13
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    $\begingroup$ Not quite. By (1), $m(t)=m(t/2)=m(t/4)=\dots$. By (2), we have $m(t/2^n)\to\frac12$. The only way to reconcile these two is if $m(t)=\frac12$, any other value would violate the limit. $\endgroup$ – user10354138 May 22 at 10:16
  • $\begingroup$ oh yes of course $\endgroup$ – user3184807 May 22 at 10:19

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