solving $\frac{d}{dx}(2^x+3^x\cos(\pi x))=0$

Question

is there any "elementary" method to solve this equation

\begin{align} \frac{d}{dx}\big(2^x+3^x\cos(\pi x)\big)=0 \end{align}

I tried to use wolfram alpha for solutions but it couldn't do it.

Answer 1

As said in comments, finding the zeros of function $$f(x)=a^x\log (2)-\pi \sin (\pi x)+\log (3) \cos (\pi x) \qquad \text{where}\qquad a=\frac 23$$ needs numerical methods.

However, you probably noticed already that the solutions are closer and closer to integer values. So, let $x=n+t$ $$f(t)= a^{n+t}\log (2)+(-1)^n (\log (3) \cos (\pi t)-\pi \sin (\pi t))$$

Expand $f(t)$ around $t=0$ $$f(t)=\sum_{k=0}^\infty \frac{b_k}{k!}\,t^k$$ where the first coefficients are $$\left( \begin{array}{cc} k & b_k \\ 0 & \log (2) a^n+(-1)^n \log (3) \\ 1 & \log (2) a^n \log (a)-\pi ^2 (-1)^n \\ 2 & \log (2) a^n \log ^2(a)-\pi ^2 (-1)^n \log (3) \\ 3 & \log (2) a^n \log ^3(a)+\pi ^4 (-1)^n \\ 4 & \log (2) a^n \log ^4(a)+\pi ^4 (-1)^n \log (3) \\ \end{array} \right)$$

Now, use power series reversion to make $$t=-\frac{\log (2) a^n+(-1)^n \log (3)}{\log (2) a^n \log (a)-\pi ^2 (-1)^n}-$$ $$\frac{\left(\log (2) a^n+(-1)^n \log (3)\right)^2 \left(\log (2) a^n \log ^2(a)-\pi ^2 (-1)^n \log (3)\right)}{2 \left(\log (2) a^n \log (a)-\pi ^2 (-1)^n\right)^3}+\cdots$$

Using only the terms given in the above table, some results (do not forget to add $n$ to $t_{(n)}$ in order to have $x_{(n)}$ $$\left( \begin{array}{ccc} n & \text{estimate} & \text{solution} \\ 1 & 0.063864 & 0.063883 \\ 2 & 0.134349 & 0.135012 \\ 3 & 0.088028 & 0.088117 \\ 4 & 0.119182 & 0.119560 \\ 5 & 0.098540 & 0.098693 \\ 6 & 0.112356 & 0.112642 \\ 7 & 0.103170 & 0.103361 \\ 8 & 0.109305 & 0.109556 \\ 9 & 0.105220 & 0.105429 \\ 10 & 0.107945 & 0.108182 \\ \end{array} \right)$$ and we could do much better adding a few more terms.

When $n$ will be very large, the term $a^{n+t}$ will be neglible and $$\log (3) \cos (\pi t)-\pi \sin (\pi t)\sim 0\quad \implies \quad t_{(\infty)}=\frac 1 \pi \tan ^{-1}\left(\frac{\log (3)}{\pi }\right)=0.107082$$

Edit

If you want to have a good explicit approximation, use one single iteration of Newton method starting with $t_0=t_{(\infty)}$. This will give

$$t_1=t_0-\frac{f(t_0)}{f'(t_0)}$$ which not ugly at all since $$f(t_0)=\log (2) a^{n+\frac{\tan ^{-1}\left(\frac{\log (3)}{\pi }\right)}{\pi }}$$ and $$f'(t_0)=\log (2) \log (a) a^{n+\frac{\tan ^{-1}\left(\frac{\log (3)}{\pi }\right)}{\pi }}-\pi (-1)^n \sqrt{\pi ^2+\log ^2(3)}$$

Thsi will give $$\left( \begin{array}{ccc} n & \text{estimate} & \text{solution} \\ 1 & 0.0640247402 & 0.0638833045 \\ 2 & 0.1349745836 & 0.1350117865 \\ 3 & 0.0881290392 & 0.0881171739 \\ 4 & 0.1195568672 & 0.1195602065 \\ 5 & 0.0986940499 & 0.0986930271 \\ 6 & 0.1126417901 & 0.1126420864 \\ 7 & 0.1033608318 & 0.1033607427 \\ 8 & 0.1095558965 & 0.1095559227 \\ 9 & 0.1054293120 & 0.1054293042 \\ 10 & 0.1081819025 & 0.1081819025 \\ \end{array} \right)$$