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It's easier than you think. $$x(t)= A\sin(t)+B\cos(t)$$ Now differentiate to find $x'(t)$ $$x'(t)= A\cos(t) - B\sin(t) $$ Sub this into your original equation and solve for A and B $$A\cos(t) - B\sin(t) - 3A\sin(t) - 3B\cos(t) = \frac{\cos(t)}{2}$$ Simplifying to $$\cos(t)[A - 3B - \frac12] - \sin(t)[B + 3A] = 0$$ Since there is no $\sin(t)$ or ...


You can do this if and only if $c \geq 0$ (assuming that the max exists and is finite, otherwise you need $c>0$). If $c<0$ then $\max \{ cf \}=c \min \{ f \}$.


Let $g=p_{n+1}-p_n$ and write $p=p_n$ for simplicity. Then you're asking for $\log\log(p+g)-\log\log p$ which is $g/p\log p+O(g^2/p^2\log p).$ This Of course if you want $\sum_{k=1}^n\left(\frac{1}{f(k)}-\log\log p_k\right)=O(1)$ then $\sum_{k=1}^n\frac{1}{f(k)}\sim\sum_{k=1}^n\log\log p_k\sim n\log\log n.$ In fact $\log\log p_k=\log\log k+\log\log k/\log ...


If you define $S$ to be the disconnected set above and the height of the line above a horizontal line to be the value of $f(x)$, you can see how it is false. This can be extended to higher dimensions in an analogous way.


Since the directional derivative along any direction $\hat s$ is $\hat s \cdot \nabla f(\vec x)=0$ in $S$, then $f$ must be a piece-wise constant on $S$.


if $S$ is disconnected, give $f$ a different constant value on each component of $S$ and we see that the conjecture is false.


Assuming I understand where you're trying to put your parentheses (since you have $\frac1{p_n}$ on the LHS I presume it's trying to be 'captured' by the sum), then the sum should be easily relatable to the Mertens constant : $\sum(\log\log p_{n+1}-\log\log p_n)$ telescopes, so if you sum from $n=1$ to $n=K$ your partial sum will be just $\log\log ...

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