-1

I have tried to solve the following equation in the title and resulted to solving it via plotting the graphs (not a very good solution I know) so I would like to know what is the proper way of solving this equation with just basic maths (no calculus) Thanks

Daniel P
  • 2,710
Nav Bhatthal
  • 1,057
  • Just like to point out that the $-7x$ is an exponent. – Nav Bhatthal Dec 12 '22 at 16:34
  • It seems to me that the equation doesn't simplify further (other than subtracting $7$ from each side), so the best way is to approximate the solution using the Taylor series of $\cos$ and $\exp$. – Daniel P Dec 12 '22 at 16:37
  • In that case, how is it that we can have an equation that is in a single variable $x$ and is still "unsolvable" via normal means, what is the limitation that makes this happen? – Nav Bhatthal Dec 12 '22 at 16:39
  • Most equations are not solvable anatically, it's not surprising. You have to use some numerical algorithms. – Lelouch Dec 12 '22 at 16:43
  • 4
    Most equations are "unsolvable via normal means". Consider $e^x = 2$. Now you might think to yourself, "but I can solve that!", but can you really? Yes, we have "invented" the inverse function of $f(x) = e^x$, and called it $\ln(x)$, however calculating $\ln(2)$ comes down to evaluating its Taylor series with a good enough accuracy. In your case, we "simply" need to find the inverse of $f(x) = \cos(3x)+22e^{-7x}$ (in an interval where it's invertible), which we could call $g(x)$ or something. Then evaluate the Taylor series of $g(x)$ at $22-7=15$ to find the solution for the original equation. – Daniel P Dec 12 '22 at 16:43
  • This series solution and Newton’s method could work, but they both require calculus to derive them. – Тyma Gaidash Dec 12 '22 at 17:05
  • There is likely not an analytical solution you can find. Plotting the function $\cos(3x)+2e^{-7x}-15$ (which is just the left side minus the right side) reveals that there is only one solution, between $-0.5$ and $0$. So we can find this solution using binary search, which I did, so $x \approx -0.281371$ is the only solution. – Daniel P Dec 12 '22 at 17:35
  • 1
    Or as Moo mentioned, Newton's method is also a really good way of finding solutions, if you can calculate the derivative of $f(x) = \cos(3x)+2e^{-7x}-15$, which is $f'(x) = -3\sin(3x)-14e^{-7x}$. Now start from a point let's say $x=0$ and step to $x-f(x)/f'(x)$. Repeat this a couple $100$ times, and you'll find the solution. There is a great animation on Wikipedia that shows how it works. – Daniel P Dec 12 '22 at 17:35
  • @DanielP I mean tbf by that definition then the only true elementary functions are polynomials. This is kinda a nuanced view idk – Max0815 Dec 12 '22 at 19:11
  • @navop I did not downvote, but please add some more detail on how you tried solving the problem in addition to what you have already. Otherwise, more people may downvote – Тyma Gaidash Dec 12 '22 at 19:26

2 Answers2

2

Here is an analytic solution using Lagrange reversion:

$$2e^{-7x}+\cos(3x)=15\implies x=-\frac17\ln\left(\frac{15}2+\sum_{n=1}^\infty\frac1{(-2)^nn!}\left.\left(\frac{d^{n-1}}{da^{n-1}}\cos^n\left(\frac37\ln(a)\right)\right)\right|_{a=\frac{15}2}\right)$$

Now use complex definitions, expand as a binomial series, and take the $n$th derivatives using factorial power $u^{(v)}$:

$$\left.\left(\frac{d^{n-1}}{da^{n-1}}\cos^n\left(\frac37\ln(a)\right)\right)\right|_{a=\frac{15}2}= \sum_{k=0}^n\frac{\binom nk}{2^n} \left.\left(\frac{d^{n-1}}{da^{n-1}} a^{\frac{3i}7(2k-n)} \right)\right|_{a=\frac{15}2}=2^{-n}\sum_{k=0}^n\binom nk \left(\frac{15}2\right)^{1+\frac{6 i k}7-\left(1+\frac37i\right)n}\left(\frac{3i}7(2k-n)\right)^{(n-1)}$$

Therefore we expand and simplify to get:

$$\boxed{2e^{-7x}+\cos(3x)=15\implies x=-\frac17\ln\left(\frac{15}2+\sum_{n=1}^\infty\sum_{k=0}^n\frac{\left(\frac{3i}7(2k-n)!\right)\left(\frac{15}2\right)^{\frac{3i}7(2k-n)-n+1}}{\left(\frac{3i}7(2k-n)-n+1\right)!(n-k)!k!(-4)^n}\right)}$$

shown here. From numerical testing, the upper limit may be $n$ or $\infty$ too. Based on this result for $e^x+a\sin(bx)=c$, there is no closed form for the inner sum. Maybe there is a simplification of this double hypergeometric series?

Тyma Gaidash
  • 12,081
1

We can get the range for the value of $x$ as: $$\cos (3x) +7=-2e^{-7x}+22 \\\Rightarrow \cos y=-2e^{-\frac73y}+15$$ For range: $$-1\leq-2e^{-\frac73y}+15\leq1\\\Rightarrow7\leq e^{-\frac73y}\leq8\\\Rightarrow8^{-3/7}\leq e^y\leq7^{-3/7}\\\Rightarrow-\frac37\ln8\leq y\leq-\frac37\ln7\\\Rightarrow-0.297\leq x\leq-0.278$$

This range leaves us quite close to the answer, which is $-0.281$.

Also, it's still better to draw a graph just to be sure on the number of solutions part, which, as you have already drawn, looks somewhat like:

enter image description here

So, now we know that the number of solutions for our equation is $1$.

From here, we may still further narrow our range using the following "elementary" approach:
We are going to shorten our range for $x$ in accordance to our LHS value.

So, from above, we know that for the equation: $$\cos (3x)=-2e^{-7x}+15$$ the RHS traverses from $-1$ to $1$ for $x$ from $-0.278$ to $-0.297$.
Just a rough check for $x=-0.3$, we have LHS $= \cos(-0.9^\mathcal{c}) \sim\cos51.6^\circ$ and $\cos51.6^\circ>\cos 60^\circ$ that is, $\cos51.6^\circ>0.5$.
For $x>-0.3$ we will still have $\cos(3x)>0.5$.
Also, since RHS varies from $1$ to $-1$ which is a magnitude of $2$, and we know our value is greater than $0.5$ so, we require only $1/4^{th}$ of that, that is, $0.5$ or from $1$ to $0.5$ (instead of $-1$).
So, similarly for $x$ which is from $-0.278$ to $-0.297$, difference of $\sim0.02$, and take $1/4^{th}$ of it, that is, a difference of $0.05$.
Thus, we can say that our required answer, $x$, should lie in the range $-0.278$ to $-0.283$ $(= 0.278 + 0.05)$.

Happy Learning!

InanimateBeing
  • 3,074