is this integration trick TOO POWERFUL?
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It seems a lot easier to multiply the second term by e^(1/x)/e^(1/x) rather than multiplying both terms by their conjugates.
Yes, that's much quicker. It requires you to spot that that's going to work, though. The method he used is the natural thing to do if you are just following your nose and don't already know what the answer is going to be.
@@thomasdalton1508 It's a fairly standard trick when you have something of the form 1/(y + 1) + 1/(1/y + 1), which is effectively what the terms in the video are, with y = e^(1/x)
Another way: Split the original integral into one from -1 to 0 and one from 0 to 1. In the first one, change x to -x. Then combine the integrals again; the same simplification will happen.
I'm pretty sure this method and the method splitting into odd/even functions are exactly equivalent (at least for all "nice" functions). The 2 and the 1/2 will always cancel.
Yes! It also generalises: if you have an integral from 1/A to A, you can split it into two integrals; one from 1/A to 1 and one from 1 to A. Then do a substitution of x -> 1/u in either one, and recombine. An example of this is the integral from 0 to infinity of ln(x)/(x^2+1) dx (though you need to worry a little about convergence when you limit A to infinity)
Man.. I sure wish I didnt have to go to other classes and just self study math full time.
If I had to self study math I would SUCK. Classes make me do the work
"I have never let schooling interfere with my education." - Mark Twain
@@user-ct1ns6zw4z but evariste galois was a certified GENIUS - ARE YOU ? not trying to be a smart alleck, just saying ...
i am 14 years and I already know calc pretty decantly
Oh boy let me tell you about math graduate school
Mad respect to Michael for introducing this awesome concept.
Poo in the loo !
@@maalikserebryakov what do you mean?
That construction of f0+f1 seems to fall into that 'just stupid enough to work' category... It's like, duh, of course you can multiply by 1 and add zero, but the effect is unexpected.
I would say it's overpowered by I haven't yet been able to use it for any integral not built for this technique.
This techinuqe is critically important in analytic number theory and in combinatorics. Also, I dont believe you :p, there's just no way that's true. Maybe if you're just out of calc 2 or something
@@MrMctastics I read an entire two volume book on complicated integrals and this technique didn't show up. I'll take your word for this because I've yet to take analytic number theory though.
Nicely done :) Learned something new 👍
I teach a signals and systems course, and instruct my students they should always look at the function first without just jumping into the math, in the hopes of simplifying their approach to the solution. When it comes to the Fourier series and transforms, this can make a very crazy looking function so much easier. Students get amazed how something so simple can save them hours of work. I'm very glad you cover this approach here. Will show my students this video!
Wow, beautiful trick. Thanks Michael
10:07
Wow. Didn't see that coming. Very nice.
I'll start to teach integrals in two weeks... This definitely will be part of my classes
Looks like this method also applies to decompose the square matrix into symmetric and antisymmetric matrices, but in a slightly different way.
Oh neato, the even-odd decomp; Just learned this! This is onna those tricks that i wish was taught like the kings property, Weierstrauss sub, and complex subs.
I have a couple of Paul Nahin's books. One is about i and the other is about Euler's Fabulous Formula. If I recall, he is a retired professor of electrical engineering from the University of New Hampshire
Please, more vídeos on functions decomposition!
Wow! wacky concept, a good place to click the thumbs up 👍
This works for any bounds of integration, as long as they're symmetric around 0. So, one can write:
sin(x) = integral between -x and +x of cos(t)/(e¹ᐟᵗ+1) dt.
seeing the f0 function:
(f(x)+f(-x))/2
obviously reminds me of the definition of cosine with complex exponentials:
(e^(ix)+e^(-ix))/2
but also reminds me of Binet's Formula for the fibonacci numbers:
(phi^n+(-phi)^(-n))/sqrt(5)
which motivates me to try and see if binet's formula can be written in a form involving a complex argument for cosine (real argument for a hyperbolic cosine?). that would be a really neat little identity; the fibonacci numbers in terms of complex trigonometry
You can write the Fibonacci numbers as follows:
(e^(n*ln(phi)) + e^(-n*ln(phi))/sqrt(5)
= 2cos(i*n*ln(phi))/sqrt(5)
=2cosh(n*ln(phi))/sqrt(5)
Thus we get from the definition of the Fibonacci numbers:
cosh(n*ln(phi)) = cosh((n-1)*ln(phi)) + cosh((n-2)*ln(phi))
Hope it helps!
done with this paul j nahin book and i think you should make a video on optical integral which uses riemann sum approach
Huh, I'd learned you could explicitly decompose a function into even+odd from a linear algebra exercise, but I didn't realize you could use it to help solve y-axis-symmetric integrals since you can both ignore the odd part and only worry about half of the even part.
Reminded me of Fourier series from my PDE class
4:32 A Good Place to Start
students here in india learn this in 12th grade, in fact we learn even more of these integral "tricks" to solve questions as quickly as possible, they are used very oftenly in competitive exams aswell.
The words of a true jee aspirant 😂
They are used in symbolic solvers too, anything to speed up that descent of pattern matching
everywhere i go i see these comments bhai yaar 😭
Man, if I had a penny for every indian person saying they learn a math concept or trick when they were in the womb I could retire at 23
@@jeanveramorocho9307 no bro ,he is right this integrals can be solved in less than 30 sec
Did I miss where the improper integral was addressed?
Flammable Maths did this too some years ago if I recall right.
There was an Irish Maths Olympiad (I think?) question that used this trick: the integral from -1 to 1 of 1/(e^x + 1)(x^2 + 1) dx :)
This reminds me of of cosh and sinh in the sense that if f is e^x then f0 and f1 are coshx and sinhx respectively.
How can he even decompose a function into 0 and 1!? I've never seen anything like that before?
@@oom_boudewijns6920 I dont think he does. If you are referring to f0 and f1, that is just notation for a function. Could be replaced with g and h. The Idea is just to use notation similar to f to denote hes splitting f into two similar functions.
Will this trick end the gnarliness of all integrals?
It makes me giddy.
Thank you, professor.
Did I miss where he dealt with the improperness of the integral?
No, but at least heuristically, the integrand has only a jump discontinuity at 0, so it can be integrated over the interval: As x→0−, the denominator approaches 2, so the integrand approaches ½, while as x→0+, the denominator approaches +∞, so the integrand approaches 0+.
Crazy stuff! I can follow it all (except that even and odd thingy, which I suppose one could look up), but I could never find a solution like this.
I like how you guys solve problems, not that I can solve them
Mind properly blown
Interesting how any even function e(x) can be integrated with 1/(1+exp(1/x)) over a symmetric interval to yield half the integral of e(x) over that interval. Actually this can be generalized even more: replacing 1/(1+exp(1/x)) with 1/(1+exp(o(x)) has the same effect, where o(x) is any odd function
And you can use the above to derive the property that the integral from -a to a of 1/(1+exp(o(x)) is equal to a, for any odd function o(x)
Mathematicians hate this one simple trick!
What are you talking about? No Mathematician hates tricks. Mathematicians want students to know when and why the trick works. The trick is fine. However, it only works for symmetric integrals when f(x)+f(-x) is easier to integrate than f(x) which is rarely the case except for functions that were odd in the first place or were easy enough to integrate without symmetry.
It's sarcasm@@nickharland9207
Hi Dr. Penn!
The ad was a little long.
It's a long story but I do appreciate and take on-board this feedback. trust me when I say it's not on purpose or what I'm aiming for, just how it shook out this round.
-Stephanie
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I covered this same topic on my channel a few weeks ago!
A slightly easier simplification involves removing negative exponents from the right term: Then you would have 1/(e^(1/x)+1)+e^(1/x)/(1+e^(1/x)), which more obviously simplifies to 1.
Frequently used in iit adv
This looks viable for climbing the ladder in the new math meta wonder if we'll see it permabanned next season
It is indeed OP!
There is a discontinuity at x=0. Doesn't this make this integral divergent?
The integrand is undefined at x=0, but since both left and right limits exist and are finite, there's no issue for the integral.
Betteridge's law of headlines.
So is the integral from -1 to +1 of (1/x) just 0, then, by similar logic?
In the sense of a principal value, it is, but to be more careful, you do need to show that the integrand is sufficiently well-behaved near singularities.
Since exp(1/x) is not defined at x=0, that's an improper integral - should have been a little more careful there...
Can you show how to use hyper real numbers / infinitesimals to integrate?
hyperreal numbers (or other infinitesimals) are IMO extremely interesting but I don't think they'll help one bit in computing antiderivatives of elementary functions...
What about segregational calculus?
Too bad the AP calculus exam was yesterday
Interesting trick!
I watched one and three quarters times 😂
Here in Brazil we learn this in middle school this is ez
I hope anyone who reads this get that this is a joke
@@Felipe-sw8wp did you travel back in time? why does the reply says 50 minutes ago and the comment itself 47 minutes ago!?!??
@@mcgamescompany I didn't but maybe my future self did.
Looks like an application of Papa Flammy’s int even(x) / (1+a^odd(x)) dx result
Don’t we need to be a bit more careful since this function isn’t continuous at 0?
I was think the same thing. Especially since the limit as x tends to zero does not exist.
Limit doesn't exist, but both left and right-sided limits exist and are finite, so you can split this integral into two convergent parts, which makes it fine.
why would anyone say this integration trick is "too powerful", this technique only applies to a relatively small number of definite integrals, and most aren't going to work out this nicely
JEE mains favorite category of integration
The computation relies on 1/(1+X)+1/(1+1/X)=1/(1+X)+X/(1+X)=(1+X)/(1+X)=1
Michael, Can you please tell me which book is best to learn Calculus?
Jon Rogawski Calculus gives you all the basic tools with plenty of practice problems. A little light on multivariable, but that's ok. I would then recommend a book like schaums outline of calculus to keep you fresh and give you more practice.
This channel takes alot of inspiration from interesting integrals by nahin. A little advanced though
edit: oh wait he recommends it in the video😂
@@MrMctastics Thank you very much.
Is it possible to learn this power?
he just explained it lol
"Not from a Jedi" :)
( exp^(1/x) + 1 )( exp^(-1/x) + 1 ) = ( exp^(1/x) )( exp^(-1/x) ) + exp^(1/x) + exp^(-1/x) + 1 = exp^(1/x) + exp^(-1/x) + 2 right ???
I have the Nahin book you mentioned; it is indeed one you can use for self-teaching and I have recommended it to several students and colleagues.
Does anyone want to try this as a general method in similar integrals?
Substitute,
x+1/x = u
x-1/x = v
Then, use Arithmetic Mean - Harmonic Mean inequality on x and 1/x. Then, calculate u^2 + v^2 in terms of powers of x and replace the powers of x in the inequality with powers of u and v. You will get something like u^2 >= 4-v^2.
Then you calculate the integrals only in terms of u, followed by calculating only in terms of v and compare the results. You should get integrals that are greater than or less than the original integrals. Sometimes, it's a good idea to discuss what they mean physically. (Take care of the 2nπ periodicity).
Cool
You have not shown that the integral EXISTS. E^(1/x) is UNDEFINED at x=0. You simply ignored that issue. I am not saying that your solution is incorrect, but that you never demonstrated that it existed.
When x goes to 0 1/x is inf from the right and -inf from the left and e^1/x goes to inf from right and 0 from left so cosx/(e^1/x+1) goes to 1 from left and 0 from right very rapidly. So it is easy to see the integral exists. There are also tests to see if the integral converges,and michael did it before.i do not think he should show this evry time.the goal of this videos is to solve nice tricky integrals this is not math lecture.
2:20 Stop watchin pron.
Deepak JATAV DJ
Your computation is too complicated. 1/(exp(1/x)+1)+1/(exp(-1/x)+1)=1/(exp(1/x)+1)+exp(x)/(1+exp(1/x))=(1+exp(x))/(1+exp(x))=1
We learn this is 12th grade...
not in the US
@@LucasDimoveo in which grade u guys study it ?
@@integration8274 never, at least until university
@@LucasDimoveo ooh nice, it's actually good to not put that much pressure on a student, here we have to study this and much more in order to clear entrance of college to get admission
@@integration8274 you mean university right?
My Dad taught this to me before I was 3 yrs old.
I taught it to my dad when I was 2 yrs old He probably showed it to your dad
@@silver6054 Well I was considering odd and even functions in the womb. Born prematurely through boredom, I fine tuned and improved my incubator.