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# /sci/ - Science & Math

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File: 3 KB, 325x156, integrate.jpg [View same] [iqdb] [saucenao] [google] [report]

Can /sci/ integrate this?

 >> Anonymous Fri Feb 17 13:24:51 2017 No.8682146 >Can /sci/ type this into WolframAlpha?Yes.
 >> Anonymous Fri Feb 17 13:25:23 2017 No.8682148 >>8682139No. I can prove that it is integrable and approximate it if you want.That said, finding the antiderivative is not trivial as you need to factorize the denominator into something no one would even imagine and could only be found via brute force. And the result is a clusterfuck of fractions involving powers, roots, logarithms and even inverse trig functions. As my president would say: SAD!
 >> Anonymous Fri Feb 17 13:27:15 2017 No.8682153 >>8682146>>8682148>these plebs using Wolfram Alphaso it's a no then?
 >> Anonymous Fri Feb 17 13:28:21 2017 No.8682156 >>8682153I literally said 'No.'Can you read? I guess that's a no aswell.
 >> Anonymous Fri Feb 17 13:31:15 2017 No.8682159 >>8682153>Wasting your time manually solving something that a computer can do in a second.You're probably also impressed by people who can recite Pi to a stupid number of digits.
 >> Anonymous Fri Feb 17 13:32:32 2017 No.8682163 >>8682156If you're not smart enough to integrate it give me your best approximation then.
 >> Anonymous Fri Feb 17 13:34:34 2017 No.8682170 >>8682163Give me an interval and I can approximate it through partial riemann sums.
 >> Anonymous Fri Feb 17 13:35:03 2017 No.8682172 >>8682170You pick whatever you want.
 >> Anonymous Fri Feb 17 13:36:09 2017 No.8682177 >>8682159Why wouldn't you be impressed by someone having an amazing memory?
 >> Anonymous Fri Feb 17 13:36:21 2017 No.8682178 >>8682170From 0 to 1
 >> Anonymous Fri Feb 17 13:37:41 2017 No.8682185 >>8682177Only intelligent people impress me.
 >> Anonymous Fri Feb 17 13:39:34 2017 No.8682190 >>8682185Having an amazing memory makes understanding difficult concepts easier.
 >> Anonymous Fri Feb 17 13:45:54 2017 No.8682209 >>8682172From 0 to 1, using 5 partial sums, evaluating at the midpoint of each partition and using only 4 decimal places of each sum I get 0.07616It is pretty off but by hand I won't bother to get something better.
 >> Anonymous Fri Feb 17 13:51:53 2017 No.8682224 >>8682209It is correct to 3 decimal places and from what I remember in high school physicists only use 2 decimal places so I guess you can call me the master integrator from now on.http://www.wolframalpha.com/input/?i=Integral+from+0+to+1+of+x%5E2%2F(x%5E5+%2B+4)
 >> Anonymous Fri Feb 17 13:53:31 2017 No.8682233 File: 24 KB, 608x494, plottttt.jpg [View same] [iqdb] [saucenao] [google] [report] You should be able to figure out the integral from this picture, unless of course you're not mathematically gifted.
 >> Anonymous Fri Feb 17 13:54:59 2017 No.8682236 >>8682224>>8682224>from what I remember in high school physicists only use 2 decimal placesWhat?
 >> Anonymous Fri Feb 17 13:58:14 2017 No.8682244 (1)/(5x^2)*(ln(x^5+4))You're welcome
 >> Anonymous Fri Feb 17 14:00:57 2017 No.8682247 http://www.wolframalpha.com/input/?i=Derivative+of+1%2F(5x%5E2)*(ln(x%5E5%2B4))>>8682244>You're welcomeWhat did he mean by this?
 >> Anonymous Fri Feb 17 14:01:41 2017 No.8682249 This is really just partial fractions. The bottom is quintic but has one obvious real root. You then factor the quartic and it will be ugly most likely but you seriously can just partial fractions this bitch
 >> Anonymous Fri Feb 17 14:04:07 2017 No.8682256 >>8682236Back in high school for physics class they would tell us to just round to 2 decimal places most of the time.
 >> Anonymous Fri Feb 17 14:06:14 2017 No.8682260 >>8682256(−√5−1)ln(∣∣x(2x+5√4(√5−1))+295∣∣)+(√5−1)ln(∣∣x(2x+5√4(−√5−1))+295∣∣)20⋅5√16+ln(∣∣x+225∣∣)5⋅5√16−(−8⋅√5−2135⋅5√4+8)arctan(4x−5√4(√5+1)2910√−√5+295⋅435−3)5⋅252√−√5+295⋅435−3⋅5√16+(−8⋅√5+2135⋅5√4−8)arctan(4x+5√4(√5−1)2910√√5+295⋅435−3)5⋅252√√5+295⋅435−3⋅5√16ugly doesn't even begin to describe this
 >> Anonymous Fri Feb 17 14:12:37 2017 No.8682268 >>8682256I see, just thought you meant that actual physicists only round to two decimal values
 >> Anonymous Fri Feb 17 14:23:31 2017 No.8682286 How can you guys claim that you're so smart yet you cant integrate that?
 >> Anonymous Fri Feb 17 14:28:07 2017 No.8682297 >>8682249>>8682260my bad
 >> Anonymous Fri Feb 17 14:28:35 2017 No.8682299 >>8682190>>8682177an amazing long term memory used to be valuable, but supply and demand bruhnow we have cheap memories that easily outclass humanswhat matters for intelligence is that size of your working memory, which is completely different from the memory needed for reciting digits of pi
 >> Anonymous Fri Feb 17 14:30:57 2017 No.8682302 >>8682297>my badYou aren't wrong.You are supposed to factorize the denominator and then partial fractions.But lets just say, some fractions are more partial than others.
 >> Anonymous Fri Feb 17 14:44:04 2017 No.8682326 If you can't integrate it,then just leave /sci/
 >> Anonymous Fri Feb 17 14:58:14 2017 No.8682358 File: 583 KB, 800x900, Marisa is confused.png [View same] [iqdb] [saucenao] [google] [report] >>8682139I'm shit at this and not much of a /sci/ person in all honestly. I would Imagine you could do one of the following:Rewrite x^2/x^5+4 as log|x^5+4|, then, when integrating turn it into an exponential e^x^5+4.Alternatively multiply by bottom brackets and integrate the solution. I need help on this sorta stuff desu though so help would be appreciated.
 >> Anonymous Fri Feb 17 15:02:40 2017 No.8682364 >>8682358this dumb weeaboo had been shitting up the board in search for attention for days now.just letting you guys know so that you don't reply to him and he might just leave
 >> Anonymous Fri Feb 17 15:08:18 2017 No.8682374 >>8682364This is my first time on the board in months, I've been doing further maths for while and this shit has been causing some grievance, sorry for the misunderstanding.
 >> Anonymous Fri Feb 17 15:08:20 2017 No.8682375 >>8682364I was about to tell him how wrong he was but I see, he is just pretending to be retarded.Or maybe he is retarded but he also knows that he is retarded so he is roleplaying as a retarded person who doesn't know he is retarded.Really makes you think
 >> Anonymous Fri Feb 17 15:09:21 2017 No.8682378 >>8682375No, I'm probably retarded, what am I doing wrong here? I've never been that great at maths, I just want a quick solution.
 >> Anonymous Fri Feb 17 15:10:50 2017 No.8682379 >>8682378You can't>Rewrite x^2/x^5+4 as log|x^5+4|That doesn't even make sense.
 >> Anonymous Fri Feb 17 15:13:38 2017 No.8682389 File: 185 KB, 714x337, Capture.png [View same] [iqdb] [saucenao] [google] [report] >>8682379Would it have to be in the context shown here then? (where you are integrating the exponent of an exponential).
 >> Anonymous Fri Feb 17 15:14:03 2017 No.8682390 >>8682139Just residue theorem it nigga
 >> Anonymous Fri Feb 17 15:18:03 2017 No.8682395 >>8682389What the fuck? That has nothing to do with this.
 >> Anonymous Fri Feb 17 15:21:19 2017 No.8682400 >>8682395I figured that if you can integrate factors in that manner then you would be able to do the same in this case. Sure its something completely different, though I'd assume a link (though obviously false as proven).
 >> Anonymous Fri Feb 17 15:22:42 2017 No.8682404 >>8682400Adding on to this, I'd assume the correct way is using partial fractions then integrating them individually?
 >> Anonymous Fri Feb 17 15:25:14 2017 No.8682414 >>8682400In your example of differential equation that exponent of an integral is the general solution of that kind of DE.You can't really put the log inside the integral and then put an e outside. It is not the same. You could put an e and a log inside but then that would be completely useless.>>8682404Yes, this is what you have to do and there is no other way.
 >> Anonymous Fri Feb 17 15:55:22 2017 No.8682473 >>8682414Great, thanks for the help! Sorry if I came off as retarded, I'm not very good at this.
 >> Anonymous Fri Feb 17 16:41:06 2017 No.8682600 Too lazy to compute the residues, but residue integration seems to work if you choose the contour $C_{R,\epsilon}$ that comprises a small semicircle around $-\sqrt[5](4)$ of radius e, then the segment $[-\sqrt[5](4)+\epsilon, R]$, then a semicircle from -R to R, then the segment $[-R, -\sqrt[5](4)-\epsilon]$.As you let R go to infinity, the integral over the big semicircle goes to 0. Then you let epsilon go to 0 and then you have to compute the limit of the integral over the small semicircle (probably a dominated convergence argument). That tells you that the integral you want plus the limit of the small integral is 2pi i times the sum of the residues, which I won't compute but that wolfram can compute.Still, it works very well.
 >> Anonymous Fri Feb 17 16:51:01 2017 No.8682627 >>8682139Easy, just Taylor expand it and integrate term by term.
 >> Anonymous Sat Feb 18 00:59:04 2017 No.8683665 >>8682358Take your pedophile cartoons back to >>>/a/
 >> Anonymous Sat Feb 18 04:51:24 2017 No.8683954 >>8682139Yes. Any fractions of polynomials are easily solvable.
 >> Anonymous Sat Feb 18 05:10:39 2017 No.8683979 >>8682139why can't i use integration by partial fractions here? im a brainlet and i need an explanation
 >> Anonymous Sat Feb 18 06:09:38 2017 No.8684061 >all these retards who cant't do itYou literally just factor the denominator, decompose into partial fractions, separate the integral and integrate each part as u'/u or the inverse of a first or second degree polynomial which you should kill yourself if you're a math major and can't do it with your eyes closed.
 >> Anonymous Sat Feb 18 06:11:46 2017 No.8684066 >>8684061holy shit im a complete brainlet and i got it >>8683979holy shit
 >> Anonymous Sat Feb 18 06:51:50 2017 No.8684142 >>8683979>>8684061>>8684066that's the solution btw>>8682260
 >> Anonymous Sat Feb 18 06:56:36 2017 No.8684151 >>8684142o well i lied cant belive someone took the time to do that
 >> Anonymous Sat Feb 18 06:59:10 2017 No.8684154
 >> Anonymous Sat Feb 18 07:02:13 2017 No.8684158 >>8684154let me believe what i want to believe
 >> Anonymous Sat Feb 18 07:02:21 2017 No.8684159 >>8684151a computer took the time to do that>>8684154i don't use wolfram cause it's shit shows no stepshttp://www.integral-calculator.com/#just type it in there just to get a scope of how much work is required to get the solution
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