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This is a Question from an Analysis 1 exam. The question is as follows: Decide if the functions $f: \mathbb{R} \longrightarrow \mathbb{R}$ can be written as the difference of two monotonically increasing functions

a) $f(x) = \cos(x)$

b) $f(x) = x^2$

For the moment I’m working on a) my first thought would be to use the MVT and receive something in the form of $\cos(x)+2x = -\sin(x)-2x$ but as we see -$\sin(x)$ is not monotonically increasing. Obviously one could also answer with $\cos(x) = (\cos(x)+2x) - 2x$ but I fear this answer would not be accepted by my professor. If you have any tips or answers for either a) or b) id be grateful

Nik
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fluffy
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2 Answers2

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This should work $$ \cos(x) = \sum_{k=0}^\infty (-1)^{k} \frac{x^{2k}}{(2k)!} = \sum_{k=0}^\infty \frac{x^{2(2k)}}{(2(2k))!} - \sum_{k=0}^\infty \frac{x^{2(2k+1)}}{(2(2k+1))!} $$

Leaves
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  • Those functions are increasing for $x \ge 0$, but not for $x < 0$. – Martin R Jan 11 '23 at 10:48
  • Great Example ! [ +1 ] Suggestion 1 : I think , you should include some content to Prove that the two terms are indeed monotonic. Suggestion 2 : You might include some Image Plot of the two terms. – Prem Jan 11 '23 at 10:59
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Because it's an Analysis 1 exam, you can use differentiation to help construct such functions. We note the desired subtractor as $g(x)$ (then $f=(f+g)-g$ should be our answer), and to be convenient, just suppose $g(x)$ is continuouslly differentiable.

For a), we need $\begin{cases} -\sin(x)+g'(x)>0 \\ g'(x)>0 \end{cases}\Rightarrow g'(x)>\max\left\{\sin(x), 0\right\}$, and letting $g(x)=2x$ (just as the construction in your answer) helps;

For b), we need $\begin{cases} 2x+g'(x)>0 \\ g'(x)>0 \end{cases}$. We put $g'(x)=x^2+2\Rightarrow g(x)=\dfrac{1}{3}x^3+2x$ gives the solution.

erpxyr2001
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