How would one calculate absolute error under root operations?

Question

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.

Answer 1

You want to solve $f(t_0)=u_0$ and have found an approximation $f(t_1)=u_1$. Lets assume $t_0\le t_1$. I we have $f'(t)\ne 0$ in an closed interval $I$ containing $t_0$ and $t_1$, then $f$ has a differentiable inverse $f^{-1}$ in this interval $I$ and therefore we have $$|t_0-t_1| = |f^{-1}(u_0)-f^{-1}(u_1)| \le \sup_{u\in f(I)}|f^{-1}(u)|\cdot |u_0-u_1|.$$

Further we have $$f^{-1}(f(t))=t$$ and so according to the chain rule $$(f^{-1})'(f(t))\cdot f'(t) = 1$$ and so $$\sup_{u \in f(I)} |(f^{-1})'(u)| = \sup_{t \in I} \frac{1}{|f'(t)|}=\frac{1}{\inf_\limits{t \in I}|f'(t)|}.$$

So finally we have the following

Lemma
If $t_0$ is the solution of $$f(t)=u_0$$ and $t_1$ is an approximative solution of this equation and $f$ has a countinous non vanishing first derivate in an open interval that contains the closed interval $$I=[\min(t_0,t_1), \max(t_0,t_1)]$$ then $$|t_0-t_1|\le \inf_{t \in I}\frac{1}{f'(t)}\cdot |u_0-f(t1)|.$$

If we think that $f$ and $f'$ are well behaving and that $t_1$ is sufficient near to $t_0$ then we will approximate $\inf_{t \in I}\frac{1}{f'(t)}$ by $\frac{1}{f'(t_1)}.$

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In this special situation we have

$$f(t)=\sqrt{(\frac{1}{200}e^{2t}-\frac{1}{200}e^{-2t})^{2}+(\frac{1}{50}e^{2t})^{2}}$$

$$t_1=-\frac{1}{4}\ln\left((\frac{1}{200})^{2}+(\frac{1}{50})^{2}\right)\approx 1.9408553472599643$$ $$f'(t)=\frac{(\frac{1}{200}e^{2t}-\frac{1}{200}e^{-2t}+\frac{1}{50}e^{2t})\cdot {(\frac{1}{100}e^{2t}+\frac{1}{100}e^{2t}+\frac{1}{25}e^{2t})}}{f(t)}$$ We use $$f(e^{2t_0})\approx f(e^{2t_1}) \\ f(e^{-2t_0})\approx 0 \\ f(e^{-2t_1})\approx 0$$ $$f(t_1)\approx f(t_0)=1$$ and get $$f'(t_1)\approx\frac{\frac{1}{40}\frac{3}{50}e^{4t_1}}{1}=3.5294117647058822$$ we take this as an approximation of $\inf_\limits{t \in I}|f'(t)|.$ If I plug in your approximation $t_1$ in the function $f$ I get $0.999975005000125$. So finally we have an absolute error

$$|t_0-t_1|\approx 0.000025 \cdot 3.529 \approx 9\times 10^{-5}$$ and a relative error is $$|\frac{t_0-t_1}{t_1}|\approx \frac{0.000025 \cdot 3.529}{1.941} \approx 4.5\times 10^{-5}.$$

The exact solution is calculated by solving a biquadratic equation $$\frac{\log\left(\frac{\sqrt{76 \sqrt{ 69259}+20001}}{\sqrt{17}}\right)}{2}\approx1.9408678442914902.$$

The difference is $1.9408678442914902-1.9408553472599643=1.2497031525837698\times 10^{-5}.$

You can find an approximation of the other solution by assuming that $e^{-2t}$ is large and then find an estimation of the error in a similar way.

Answer 2

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 $t$.