Solving for $T$ in an equation

Question

Can anyone please help me solve for $T$ in this equation:

$$0=M\cdot S+ \frac{T}{0.755}-\frac{T}{0.755} \cosh\frac{0.755M}{2T}.$$

Thanks in advance.

Answer 1

If I properly undertand, you want to solve for $T$ the equation $$MS+\frac T{0.755}-\frac T{0.755}\cosh\left(\frac M{2 \frac T {0.755}} \right)=0$$ For more simplicity, let us define $x=\frac T{0.755 M}$ (assuming $M\neq 0$).

This reduces the equation to $$S+x\left(1-\cosh \left(\frac{1}{2 x}\right)\right)=0$$ or $$x\left(\cosh \left(\frac{1}{2 x}\right)-1\right)=S$$ Plot function $$f(x)=x\left(\cosh \left(\frac{1}{2 x}\right)-1\right)$$ and look where, more or less, the function value is $\approx S$. Now, you have all elements to start a numerical method such as bisection or Newton.

In practice, if you consider large values of $x$, a Taylor expansion gives $$f(x)=\frac{1}{8 x}+\frac{1}{384 x^3}+O\left(\frac{1}{x^5}\right)$$ so an inital guess could be $x_0=\frac{1}{8 S}$.

Let us try using $S=2$; Newton iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.062500 \\ 1 & 0.071230 \\ 2 & 0.082445 \\ 3 & 0.096799 \\ 4 & 0.114242 \\ 5 & 0.132214 \\ 6 & 0.144649 \\ 7 & 0.148437 \\ 8 & 0.148689 \\ 9 & 0.148690 \end{array} \right)$$ Doing the same for $S=0.2$ $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.625000 \\ 1 & 0.654210 \\ 2 & 0.655858 \\ 3 & 0.655863 \end{array} \right)$$

Edit

It could even be nicer using $x=\frac 1{4y}$ which makes the equation $$g(y)=\sinh ^2\left({y}\right)-2Sy$$

Answer 2

Setting $x=T/(0.755M)$, we can write your equation as $$0=\underbrace{S+x\left(1-\cosh\left(\frac{1}{2x}\right)\right)}_{:=f(x)}.$$

So basically you are looking for a zero of $f(x)$. However, this equation can probably not be resolved for $x$ in elementary term, i.e using functions that might be familiar to you.

But what you can do is guessing a solution, e.g. $x=1$ (which might be a bad guess, but does not matter) and then iteratively improve this guess by some numerical method, e.g. Newton's method.

Example. Using Newton's method, we can approximate a zero of $f(x)$ in the following way. Guess an initial value for the zero, e.g. $x_0=1$. Now you will need the derivative $f'$. Compute

$$x_1=x_0-\frac{f(x_0)}{f'(x_0)}.$$

This value $x_1$ is a better guess for the zero. Do it again

$$x_2=x_1-\frac{f(x_1)}{f'(x_1)}.$$

Now $x_2$ is an even better guess than $x_1$. You can repeat this over and over until you see that your guess is not really changing anymore (certainly after something like ten steps). You can assume this value to be your solution.

From this approximate solution you can recover $T$ via $T=0.755M\cdot x$.