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I am tasked with the proof that $x =0$ is a unique stationary point and a minimum of $\cos(x) + \cosh(x) = 2\sum_{k=0}^{\infty} \frac{x^{4k}}{(4k)!}$.

What the markscheme does, concerning the "minimum" and "uniqueness" part is incomprehensible to me:

Given that the derivative is the series for $\sinh(x) - \sin(x)$ which is odd and strictly positive for all $x > 0$, we have proven that the point is unique. The fact that the point is a min can be proven in two ways: either by the fact that the series for $\cos(x) + \cosh(x)$ is positive for all $x \neq 0$ or by finding the fourth derivative which equals 2 at $x=0$, hence its a local minimum.

I have no idea why $\sinh(x) - \sin(x)$ being odd and strictly positive for all $x > 0$ helps me prove that the point is unique. Similarly, I dont understand the issue about the fourth derivative or the actual series being positive for all x not equal to 0 implying that the point is a min. Can someone please clarify this for me?

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  • $\begingroup$ Edit: 2 in front of sum $\endgroup$ – Ruslan Mushkaev Jan 16 '17 at 14:43
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Let $f(x)=\cos x+\cosh x$. Then $f(0)=2$ and if $x\ne0$ we have: $$ f(x)=2+2\sum_{k=1}^\infty\frac{x^{4k}}{(4k)!}>2. $$ On the other hand $f'(x)=-\sin x+\sinh x$. We have $f'(0)=0$ and $f'(x)<0$ if $x<0$, $f'(x)>0$ if $x>0$. This implies that $f$ is decreasing on $(-\infty,0)$, increasing on $(0,\infty)$ and that $f$ attains a unique minimum at $x=0$.

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  • $\begingroup$ Apologies, I've made a slight edit to the question: I forgot the 2 in front of the sum $\endgroup$ – Ruslan Mushkaev Jan 16 '17 at 14:46
  • $\begingroup$ I don't get it: given that the powers in the power series are all even and the coefficients are all strictly positive, it is clear that $x=0$ is the unique point of minimum - why write the paragraph about $f'$ anymore? $\endgroup$ – Alex M. Jan 16 '17 at 14:52
  • $\begingroup$ @AlexM. Because in the text quoted by the OP both arguments are used, and the OP asks about both of them. $\endgroup$ – Julián Aguirre Jan 16 '17 at 15:04

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