how can I solve a equation like that $40939=200 \cdot 0.54 \cdot T+\sigma \cdot \varepsilon\cdot T^4$ with a simple scientific calculator?

Question

$\varepsilon = 0.8$ and $\sigma = 5.669 \cdot 10^{-8}$

We are only allowed to use simple scientific calculador during the tests. How can I solve this equation like that? Find T.

$40939=200 \cdot 0.54 \cdot T+\sigma \cdot \varepsilon\cdot T^4$

Answer 1

Newton's method

If you are allowed to use a calculator I'd recommend you use Newton's method:

1. Define the function

$$ f(T) = \sigma \epsilon T^4 + 200 \cdot 0.54 T - 40939 $$

2. Calculate the derivative

$$ f'(T) = 4\sigma \epsilon T^3 + 200 \cdot 0.54 $$

3. Define the recurrence

$$ T_{n + 1} = T_n - \frac{f(T_n)}{f'(T_n)} = T_n - \frac{\sigma \epsilon T_n^4 + 200 \cdot 0.54 T_n - 40939}{4\sigma \epsilon T_n^3 + 200 \cdot 0.54} $$

4. Iterate away! I will start with $T_0 = 400$, from this you can calculate $T_1$, $\cdots$

$$ \begin{array}{ccc}\hline n & T_n & f(T_n) \\ \hline 0 & 400 & 3 422.01 \\ 1 & 371.39 & 33.9676 \\ 2 & 371.10068848 & 0.0031 \\ 3 & 371.10066165 & 2.18\times10^{-11} \\ \hline \end{array} $$

After the fourth iteration you find the solution $T = 371.10066165$

Even simpler

1. Isolate the linear term in your equation

$$ T = \frac{40939 - \sigma \epsilon T^4}{200 \cdot 0.54} $$

2. Define the function

$$ g(T) = \frac{40939 - \sigma \epsilon T^4}{200 \cdot 0.54} $$

3. Iterate the recursion

$$ T_{n + 1} = g(T_n) $$

It converges slower, but at the end you get back the solution above

Answer 2

Obviously, you can compute a base value $T_0=40939/(200⋅0.54)=379.06481481...$ from the linear part of the equation. It remains to be seen if the 4th degree term constitutes a perturbation or has a substantial contribution to $T=T_0⋅x$. $$ 40939=200⋅0.54⋅T_0⋅x+σ⋅ε⋅T_0^4⋅x^4=40939x+936.3761480165862⋅x^4 $$ or reduced $$ 1=x+\delta⋅x^4,~~δ=0.02287247241057637. $$ This indeed counts as perturbation, so that $x=1-δ+c_2δ^2+c_3δ^3+...$. Inserting this perturbation series gives $$ 0=δ⋅((1-δ+c_2δ^2+...)^4-1)+c_2δ^2+c_3δ^3+..., $$ from where by comparing coefficients $c_2=4$ and $6+4c_2+c_3=0$, $$ x\approx 1-δ+4δ^2-22δ^3=0.979\pm 0.002 $$ but the coefficients of this series grow fast, so that the error after $n$ terms is about $0.2^n$.

Or you could apply one or two iterations of Newton's method starting from $x_1=1-δ$, $$ x_2=x_1-\frac{1+δ⋅x_1^3-x_1^{-1}}{3δ⋅x_1^2+x_1^{-2}} =x_1\frac{2-x_1+2δ⋅x_1^4}{1+3δ⋅x_1^4} $$ giving the iterates

0.978986870787873
0.9789899963061174
0.9789899963149125

or $T=371.10066165867$

Answer 3

There are various approaches to numeric root finding, all of which amount to guess and check. They differ in how you choose the next guess based on what information you have gained so far. For this equation I would first find which term on the right is important by solving the equation left after you ignore each one. Often one is large and one is small, so $$40939=200\cdot 0.54T\\T=\frac {40939}{200\cdot 0.54}\approx 379\\40939=0.8\cdot 5.669\cdot 10^{-8}T^4\\T=\left(\frac{40939}{0.8\cdot 5.669\cdot 10^{-8}}\right)^{1/4}\approx974$$ The first term is dominant because the second requires a much higher $T$. Now write your equation in fixed point form $$T=\frac 1{200\cdot 0.54}\left(40939-0.8\cdot 5.669\cdot 10^{-8}T^4\right)$$ start with $379$ and iterate to convergence. Fixed point iteration often converges more slowly than Newton, but I find not needing to take the derivative is easier.