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Assuming $c\ne0$ $$\frac{\rm d}{{\rm d}y} c^{xy} = xc^{xy}\ln(c)$$ $$\frac{\rm d}{{\rm d}x} \left(xc^{xy}\ln(c)\right) = xy\ln^2(c)c^{xy}+c^{xy}\ln(c) =c^{xy}$$ Cancelling $c^{xy}$ $$xy\ln^2(c)+\ln(c)=1\implies xy=\frac{1-\ln(c)}{\ln^2(c)}$$assuming $c\ne1$ Clearly this cannot hold for all $x$ and $y$ since the left expression is constant. Now we ...

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The Champernowne constant $C_{10} = 0.12345678910111213141516…$ can be expressed exactly as an infinite series: $$C_{10}=\sum_{n=1}^\infty\sum_{k=10^{n-1}}^{10^n-1}\frac{k}{10^{n(k-10^{n-1}+1)+9\sum_{l=1}^{n-1}10^{l-1}l}}$$

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(a) I would use the limit test. By showing $\lim_{n \to \infty} \frac{f(n)}{1.9n^{4}} \to \infty$, you get $f(n) \in \Omega(n^{1.4})$. Alternatively, you can prove this by induction. (b) Again, the limit test is your friend.

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