Calculating banking rates - Equation [Note: I have no mathematical background at all and as this is my first time posting here, let me know if I forget anything in my post, especially regarding the tags as I can't just pick up "equation" and I don't understand the popular tags' meaning.]
I try to calculate a rate for banking purpose.
This formula gives me the monthly payment I'll be entitled to reimburse each month :
$M = \frac{C\cdot\frac{t}{12}}{1-\left(1+\frac{t}{12}\right)^{-12n}}$
Given :

C = 200000 (if I loan 200k€)
t = 0,02 (at a rate of 2%)
n = 20 (over 20 years)
M = ? (I'll have to pay M €/months)

$M = \frac{200000*\frac{0,02}{12}}{1-(1+\frac{0,02}{12})^-12*20} = 1011.77 \approx 1012€/\text{month}$
Now my problem is : if I change the $C$ value to $199000$ how can I get the new $t$ from the following equation ?
$1012 = \frac{199000\cdot\frac{t}{12}}{1-(1+\frac{t}{12})^{-12\cdot20}}$
I know the answer is $t = 2,05%$ but I have no idea how to calculate it so I can't integrate it into my code. Thanks for your help.
 A: HINT:
We can develop an approx formula by logarithmic differentiation
$$ \text{Let}  \; t/12= x,m= (-12 n)\;$$
$$ \dfrac{Cx}{1-(1+x)^m} =M$$
$$ \log C + \log x - \log (1-(1+x)^m) = \log M$$
$$\dfrac{ \Delta C}{C} + \dfrac{ \Delta x}{x}+ \dfrac{  m (1+x)^{m-1} \cdot \Delta x}{(1-(1+x)^m)}=0$$
$$\dfrac{ \Delta C}{C} + \Delta x \cdot \left( \dfrac{1 }{x}+ \dfrac{  m (1+x)^{m-1} }{1-(1+x)^m}\right) =0$$
It does not depend on $M$
$$ \Delta C= 0.01* 10^5 , \Delta t = 12 \Delta x $$
which relation allows to you to find $\Delta x$ and so the new $ \Delta t.$
A: Since you are very close to the first situation, use it for starting Newton method.
Its iterates will be
$$\left(
\begin{array}{cc}
 0 & 0.02000000000 \\
 1 & 0.02056153090 \\
 2 & 0.02056062632
\end{array}
\right)$$
Edit
Since, in comments, you confirmed that $M$ is exactly the same is both cases, then you want to solve for $i_2$ the more general problem
$$C_1 \frac {i_1} {1-(1+i_1)^{-m}}=C_2 \frac {i_2} {1-(1+i_2)^{-m}}$$ where $i_k=\frac {t_k}{12}$ and $m=n t$.
Using Taylor expansion for the rhs around $i_2=i_1$ we have
$$\frac {i_2} {1-(1+i_2)^{-m}}=\frac{i_1}{1-(i_1+1)^{-m}}+(i_2-i_1) \left(\frac{1}{1-(i_1+1)^{-m}}-\frac{i_1 m
   (i_1+1)^{m-1}}{\left((i_1+1)^m-1\right)^2}\right)+O\left((i_2-i_1)^2\right)$$
Ignoring the higher order terms, we just need to solve for $i_2$ the equation
$$0=-\frac{i_1 (i_1+1)^m (C_1-C_2)}{(i_1+1)^m-1}+C_2\frac{ (i_1+1)^{m-1} \left((i_1+1)^{m+1}-
   (i_1( m+1)+1)\right)}{\left((i_1+1)^m-1\right)^2}(i_2-i_1)$$
which gives, as an approximation,
$$i_2\sim i_1 \frac {C_1-C_2}{C_2}\frac {i_1 (i_1+1) \left((i_1+1)^m-1\right) } {(i_1+1)^{m+1}-i_1(m+1)-1 }$$
Applied to your numbers, this gives $t_2=0.0205368$ which does not look too bad.
