<s>[INST] How would one calculate absolute error under root operations? I know absolutely nothing about error analysis and I was wondering if someone could help me out. I want to approximate the value of $t$ that satisfies this equation: $$\sqrt{(\frac{1}{200}e^{2t}-\frac{1}{200}e^{-2t})^{2}+(\frac{1}{50}e^{2t})^{2}}=1.$$ I want to drop the term with the negative power, to get the following:
\begin{align}
\sqrt{(\frac{1}{200}e^{2t})^{2}+(\frac{1}{50}e^{2t})^{2}}&=1 \\
e^{2t}\sqrt{(\frac{1}{200})^{2}+(\frac{1}{50})^{2}}&=1 \\
2t+\frac{1}{2}\ln{\left((\frac{1}{200})^{2}+(\frac{1}{50})^{2}\right)
}&=0 \\
t&=-\frac{1}{4}\ln\left((\frac{1}{200})^{2}+(\frac{1}{50})^{2}\right)
\end{align}
However, I don't know by how much this approximates my actual value of t. How would one go about calculating the absolute error? Thanks in advance.
 [/INST] I'll not answer the error propagation part, but show you how to calculate the $t$ value explicitly: By squaring the inital equation and setting $X = e^{2t}$, you obtain
$$\frac{1}{200^2}(X-\frac{1}{X})^2 + \frac{1}{50^2}X^2 = 1$$
$$\frac{1}{200}(X^2 - 2 + \frac{1}{X^2}) + \frac{4}{50}X^2 = 200$$
$$\frac{1}{200}(X^4 - 2X^2 + 1) + \frac{4}{50}X^4 = 200 X^2$$
$$(X^4 - 2X^2 +1) + 16 X^4 = 40000 X^2$$
$$17X^4 - 40002X^2 + 1 = 0$$
Now substitute $Y = X^2$, solve the quadratic equation, take the square root to obtain $X$ and then the logarithm to obtain $t$.
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