1
$\begingroup$

This question was written because I have come across a few arc length functions which cannot be evaluated. This also has happened to me when trying to solve some problems of my own. Too often an arc length integral is non elementary causing a conflict with the integration of the function. The only way to take an arc length integral for any function would just be to use the Riemann Sum definition of the integral and maybe painstakingly put it into an alternate form. For example: $$\mathrm{Arclength(f(x), a\le x\le b)=\int_a^b \sqrt{\bigg(\frac{d}{dx}f(x)\bigg)^2+1}\ dx}$$$$\mathrm{=\lim_{N\to \infty}\frac{b-a}{N}\sum_{n=1}^N \sqrt{\bigg[\frac{d}{dx}f\bigg(x+n\frac{b-a}{N}\bigg)\bigg]^2+1}}$$

Let: $$\mathrm{\bigg[\frac{d}{dx}f(x)\bigg]^2+1=T(f(x))=T(f)} $$ This substitution naturally leaves the question of:

$$\mathrm{\sqrt{T(f)}=\sum_{n=u}^vt_n(x)}$$ In conclusion, what is the value of $\mathrm{u,v, and \ t_n(x)}$? One could always try using binomial series, but that would require |T(f)-1|<1 or simply 0<T(f)<2.

My question is what would would be a way to evaluate any arc length function such that we get rid of the square root perhaps through a series expansion? Easily, a substitution may be done to cancel the square root term, but would probably make the integrand more complicated. I have given a well written question, context and some attempts, so please correct me and give me feedback!

$\endgroup$
2
  • $\begingroup$ Is it normal that you last sum starts with $ n = a ?$ $\endgroup$ May 9, 2021 at 23:17
  • $\begingroup$ @hamam_Abdallah sorry, that was supposed to be independent from the integration bounds. Thanks. $\endgroup$ May 9, 2021 at 23:21

2 Answers 2

1
$\begingroup$

You can make series expansions around $x=c$ such that $(a \leq c \leq b)$.

Let $$g(x)=f'(x)=\sum_{n=0}^\infty \alpha_n (x-c)^n$$ For conveniency, let $y=x-c$ and we shall need to truncate the expansion to some order $$g(x)=\sum_{n=0}^p \alpha_n y^n+O(y^{p+1})$$ this would give $$\sqrt{1+g(x)^2}=\sum_{n=0}^p \beta_n y^n+O(y^{p+1})$$ but the relation between the coefficients become quite complex.

For example, using $p=3$,this would give $$\beta_0=\sqrt{1+\alpha_0^2}\qquad \beta_1=\frac {\alpha_0 \alpha_1}{\beta_0}\qquad \beta_2=\frac {\alpha_1^2+2\alpha_0\alpha_2\beta_0^2}{2\beta_0^3}$$ $$\beta_3=\frac {-\alpha_0 \alpha_1^3+2 \beta_ 0^2 \alpha_2 \alpha_1+2 \alpha_0 \beta_ 0^4 \alpha_3 }{2\beta_0^5}$$

Let us try for $$\int_1^3 \sqrt{1+e^{2x}}\,dx$$ making the expansion around $x=2$. It will gives for the integrand $$\sqrt{1+e^4}+\frac{e^4 (x-2)}{\sqrt{1+e^4}}+\frac{e^4 \left(2+e^4\right) (x-2)^2}{2 \left(1+e^4\right)^{3/2}}+\frac{e^4 \left(4+2 e^4+e^8\right) (x-2)^3}{6 \left(1+e^4\right)^{5/2}}+O\left((x-2)^4\right)$$ and the integration will lead to $$L=\frac{6+14 e^4+7 e^8}{3 \left(1+e^4\right)^{3/2}}\sim 17.3975$$ while the exact result is

$$L=-\sqrt{1+e^2}+\sqrt{1+e^6}+\tanh ^{-1}\left(\sqrt{1+e^2}\right)-\tanh ^{-1}\left(\sqrt{1+e^6}\right)\sim 17.5243$$ This make a relative error of $0.72$%.

Adding one term to the expansion would give $$L=\frac{40+176 e^4+272 e^8+188 e^{12}+47 e^{16}}{20 \left(1+e^4\right)^{7/2}}\sim 17.5214$$ reducing the relative error to $0.017$% ($43.4$ times lower than the previous)

Edit

As asked in comments, I had a look at the problem of the arc length of function $$f(x)=\tan[\log(x)]$$ asked in this question and I chose for the inteval of integration $\left(\frac e2,\frac {3e}2\right)$. As anyone could expect, the expansion of $$g(x)=\frac{\sec ^2[\log (x)]}{x}$$ was done around $x=e$ to order $(2n+1)$. Below are given the results $$\left( \begin{array}{cc} n & \text{result} \\ 0 & 4.37301 \\ 1 & 5.59580 \\ 2 & 6.15811 \\ 3 & 6.37893 \\ 4 & 6.47506 \\ 5 & 6.51552 \\ 6 & 6.53216 \\ 7 & 6.53943 \\ 8 & 6.54242 \\ 9 & 6.54368 \end{array} \right)$$

while the "exact" value is $6.54461$.

$\endgroup$
7
  • $\begingroup$ Hello, this is another great answer. Does this work for non-elementary functions, it probably does? Also taking a limit as n goes to $\infty$, would this make the summation the exact answer in exact form? As a side note, is there a closed for solution for the coefficients, without restriction, unless you already defined them within the sum? $\endgroup$ May 10, 2021 at 12:14
  • $\begingroup$ @TymaGaidash.Send me a function of your choice and the range for the integration; I should post the results with some numerical analysis. If you use a CAS, no problem; the formulae becomes enormous but who cares ? By hand, mmmmh ! $\endgroup$ May 10, 2021 at 12:17
  • $\begingroup$ Well there is one interesting question. Anyways, is there a pattern for the coefficients, or it is all based on the taylor series coefficients? Please answer the coefficient question and exact form question, as n goes to infinity maybe, unless I misunderstood something, and I will give the “check mark”. Thank you for your help! $\endgroup$ May 10, 2021 at 12:22
  • $\begingroup$ @TymaGaidash. Just based on Taylor coefficients. No hope for any pattern in most cases (say $99.999$%, I presume). Problem done and answer edited. Do you like it ? Cheers :-) $\endgroup$ May 10, 2021 at 13:04
  • $\begingroup$ How did you derive the general $β_n$ coefficients, or at least your n=1,2,3,4 terms? Yes I like your answer, although now I understand what you mean by “no pattern”. Thanks again. $\endgroup$ May 10, 2021 at 14:13
1
$\begingroup$

You could think instead about where the derivation comes from: $$l=\int dl$$ and we find $dl$ by approximating it as small right angled triangles with two sides of length $dx$ and $df$ giving $dl=\sqrt{(df)^2+(dx)^2}$

So you could take: $$\int\limits_{x=a}^{x=b}\sqrt{(df)^2+(dx)^2}=\lim_{N\to\infty}\sum_{i=1}^N\sqrt{(df_i)^2+(dx_i)^2}$$ where: $$df_i=\frac{f(\delta(i+1))-f(\delta i)}{\delta}$$ $$dx_i=\delta$$ $$\delta=\frac{b-a}{n}$$ and see if you can get anything from that?

$\endgroup$
1
  • $\begingroup$ This is essentially a rewritten parametric arc length Riemann Sum which is kind of copied from the question. It was looked for the square root to be removed through an expansion or another technique. I will see if any other answers come through, but this technically works. Also, rewriting this using algebraic manipulation creates the original definition of the arc length integral... $\endgroup$ May 9, 2021 at 23:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .