unJon
unJon
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July 29th, 2022 at 12:53:34 PM permalink
If you have wolfram alpha pro you can just type in arc length 1/X and click show step by step solution.
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gordonm888
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July 29th, 2022 at 2:38:04 PM permalink
Recognizing that this was either an unsolvable integral (i.e., no analytic form for the anti-derivative) or at least one for which the solution is not widely known, I was just typing in notes in my posts as I studied it. I realized that my posts were not directly helpful, sho88, I was just trying to come up with a good idea on how to start the analysis.

I did notice that the integral of (1/sqrt(a2+u2) = arsinh (u/a) + C This is the closest form that I have found for a simple integral relevant to your problem.

Also, the integral of(1/(u*sqrt(a2+u2)) = arcsch |(u/a)| + C where I think arcsch is an arc-hyperbolic cosecant.

All of which seems to be in sync with teliot's comment about Wolfram's advice to look at hyperbolic functions. However, hyperbolas have different mathematical properties than the curve you are studying and I'm not sure if they will reduce to your curve by, say, setting the distance between the two foci to 0.

Certainly not an "easy math problem."
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ssho88
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July 29th, 2022 at 6:19:44 PM permalink
Quote: gordonm888

Recognizing that this was either an unsolvable integral (i.e., no analytic form for the anti-derivative) or at least one for which the solution is not widely known, I was just typing in notes in my posts as I studied it. I realized that my posts were not directly helpful, sho88, I was just trying to come up with a good idea on how to start the analysis.

I did notice that the integral of (1/sqrt(a2+u2) = arsinh (u/a) + C This is the closest form that I have found for a simple integral relevant to your problem.

Also, the integral of(1/(u*sqrt(a2+u2)) = arcsch |(u/a)| + C where I think arcsch is an arc-hyperbolic cosecant.

All of which seems to be in sync with teliot's comment about Wolfram's advice to look at hyperbolic functions. However, hyperbolas have different mathematical properties than the curve you are studying and I'm not sure if they will reduce to your curve by, say, setting the distance between the two foci to 0.

Certainly not an "easy math problem."
link to original post




Possible to solve this problem with Taylor series ?

Taylor series, (1+y)^0.5 = 1 +y/2 - y^2/8 + y^3/16 - (5*y^4)/128 + (14*y^5)/256 +. . . . . .

For my case, just replace y with 1/x^4 . . . .then the integration will be very easy !

Any comments ?
Last edited by: ssho88 on Jul 29, 2022
gordonm888
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July 30th, 2022 at 1:20:57 PM permalink
Nice. But with y = x^4, is there any convergence issue when x<1?
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teliot
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July 30th, 2022 at 5:38:26 PM permalink
Quote: ssho88


Possible to solve this problem with Taylor series ?

Taylor series, (1+y)^0.5 = 1 +y/2 - y^2/8 + y^3/16 - (5*y^4)/128 + (14*y^5)/256 +. . . . . .

For my case, just replace y with 1/x^4 . . . .then the integration will be very easy !

Any comments ?
link to original post

For a numerical approximation, this is exactly what I would do.
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ssho88
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July 30th, 2022 at 7:00:38 PM permalink
Quote: gordonm888

Nice. But with y = x^4, is there any convergence issue when x<1?
link to original post



Convergence issue is my concern too. Thanks for your comment.
ssho88
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July 30th, 2022 at 7:01:58 PM permalink
Quote: teliot

Quote: ssho88


Possible to solve this problem with Taylor series ?

Taylor series, (1+y)^0.5 = 1 +y/2 - y^2/8 + y^3/16 - (5*y^4)/128 + (14*y^5)/256 +. . . . . .

For my case, just replace y with 1/x^4 . . . .then the integration will be very easy !

Any comments ?
link to original post

For a numerical approximation, this is exactly what I would do.
link to original post



Thanks for your comment.
gordonm888
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gordonm888
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August 6th, 2022 at 5:14:38 AM permalink
Did the Taylor series converge? Were you able to solve this integral?
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ssho88
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August 6th, 2022 at 6:53:50 AM permalink
Here is the integral solution:-



However, I think that series only converge if -1 < x < 1. So I think still not able to find length of arc 1/x from x = 1 to x = 3. Any comment?
unJon
unJon
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August 6th, 2022 at 7:09:14 AM permalink
Quote: ssho88

Here is the integral solution:-



However, I think that series only converge if -1 < x < 1. So I think still not able to find length of arc 1/x from x = 1 to x = 3. Any comment?
link to original post



Just the same comment I made twice before.

https://www.wolframalpha.com/input?i=arc+length+1%2FX&assumption=%7B%22F%22%2C+%22ArcLength%22%2C+%22a%22%7D+-%3E%221%22&assumption=%7B%22F%22%2C+%22ArcLength%22%2C+%22b%22%7D+-%3E%223%22
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