Which usually means "I know it's right, I can justify most of the logic. There are a few obscure leaps of logic, but I kind of have to get back to life. Besides. this .tex document barely compiles as it is"
This approach has a rigorous basis, started by Risch (the Risch algorithm) for Liouvillian functions and extended by Davenport to include algebraic functions. It's the method used by computer algebra systems to integrate.
Finally someone shows us how to create an educated guess for a solution. Usually teacher say, let's guess some (crazy ass shit we just dreamed of last night) expression and see that it works. Thanks Michael.
To be fair, guessing gets easier with mathematical maturity and can seem both obvious yet impossible to explain. Doesn't make it any nicer for the students at that moment though!
@@jaakkovuori9616 Experience is the luxury of the wizened mathematician chewing through a single problem over the course of many weeks. Panicked guessing is the luxury of the novice student rushing a hundred problems in two hours on the midterm exam.
Brilliant. A trick that takes a minute to solve the problem in some cases is definitely a worthy one. I like how you introduced it, the rationale behind the form of the solution.
Another method is u-substitution of u=x^2+1 so that du=2xdx and then x^2=u-1 so the integrand becomes (u-1)/(2sqrt(u))= (1/2))u^(1/2) - (1/2)u^(-1/2) which is easily evaluated via the power rule. The answer obtained is (1/3)*(u-3)*sqrt(u)=(1/3)*(x^2-2)*sqrt(x^2+1)
This is in general what I remember the most from differential equations: Guess that there's some solution involving parts of the function you start with, then figure out what all the constants have to be for it to work. It's not a very universal technique, but there are so few differential equation solving methods that work on any wide variety of function it makes sense that you'd use something like this as the standard way for solving them.
1:38 I would say that you don’t need the quotes around the equals sign. The “degree function” as you put it is a well defined function from the domain of radical/rational functions to the codomain of integers.
Dear Michael in an old scottish maths text book I came across a neat substitution which arrives at the same answer viz; set u=1+x^2 giving du=2xdx thus changing the integrand to 1/2(u-1)du/sqrt(u) from where the answer can be easily obtained. Hope you find this of interest. from Nick Glasgow Scotland.
Clever approach, but the rationalizing substitution you mention, u=x^2+1, is so quick and easy. Honestly, it is much simpler. It would be interesting to see an integral where this new (to me and you) approach really help an otherwise hopeless integral
I agree. I'm guessing this toy example made for a shorter video, but it would be cool to at least see a challenge problem at the end that the viewer can solve.
In my understanding, this is how Wolfram Mathematica does integration: for each given integrand, it builds a parametrization of the likely anti-derivatives, then takes the derivative of it and matches to the integrand until it has extracted the parameters.
this integral can also be done by parts, with u=x², dv=the rest of the stuff. when you see sqrt(x²+1) in the numerator, you can substitute it with (x²+1)/sqrt(x²+1) and continue. piece of cake.
In General, ∫Pn (x)dx/√(ax2 +bx+c) = Qn-1(x)*√(ax2 +bx+c)+λ∫dx/√(ax2 +bx+c) , where Pn(x) is a polynomial of degree n. Qn-1(x) is a polynomial of degree n-1 with uncertain coefficients that have to find, as well as the number λ. Differentiating the said assumed identity and bringing the result to a common denominator, we obtain the equality of two polynomials from which we find these numbers.
You could replace the 1 with a^2 and make the entire integral perfectly consistent, say if you allocated units of length to both x and a. it got me thinking suddenly that this is why inverse trig functions appear. 1/(x^2 + a^2) would have degree -2, so one might expect a degree -1 integral, which is precisely what happens as its 1/a * arctan(x/a) and arctan(x/a) is function of degree 0 combination and so is unitless. 1/sqrt(a^2 - x^2) has degree -1, so one may expect a degree 0 result, which is what happens with the unitless arcsin(x/a).
I usually do this trick like without noticing. Always when i am solving any i tegral or trying to differentiate some functions i try to make some conjectures on what the answer will be for example the degree, the function itself, even sometimes domain of the function.
Another way to do the integral that you didn’t mention is a hyperbolic trig substitution. Letting x=sinh(t), you’ll get sqrt(x^2+1)=sqrt(sinh^2(t)+1)=cosh(t), which will cancel with the dx=cosh(t)dt, leaving you with the integral of sinh^3(t)dt. If you know your hyperbolic trig identities (which are all very similar to regular trig identities) that integral is fairly easy.
Interesting method. It looks like Risch's algorithm where instead of proposing an n-degree polynom as here, one proposes a generic p(x) polynom to be calculated as the result of solving a diferential equation. Specifically, here we get x^3/Sqrt[x^2+1]=d(p(x)*Sqrt[x^2+1])/dx then x*p(x)+p(x)'+x^2*p(x)'=x^3 and by guessing p(x)=a*x^2+b*x+c we get the desired solution.
@@xCorvus7x "This equation is bronstein-solvable": "You can look up the solution in Bronstein" (Bronstein and Semendyayev: Handbook of Mathematics, Moscow 1945).
@@xCorvus7x Bronstein and Semendyayew lists the limits of lots of sequences and series and the solution to large swats of differential equations without actually explaining how to get to those solutions. A Bronstein solution is thus a quick way to attack a problem without the need to understand it.
It looks like the method of undetermined coefficients for solving differential equations. It's useful if you prefer derivatives over integrals, but can be a little time consuming.
@@nikitakipriyanov7260 I'm assuming by "based on" you mean how they were originally derived, but I don't see how that affects what method someone prefers.
Another method of solving this would be adding and subtracting x in the numerator. You'll get two different integrals which are pretty easy solve. You will get the solution in barely 10 lines. But I loved this method too, got to learn something new❤❤❤
So elegant, loved it! Then you just drop a u=x^2+1 and it almost solves itself! We called that kind of move "adding a smart zero" in college, we also "multiplied by smart ones" when needed, it's amazing how much easier your job can become with some well crafted 1's and 0's.
What would happen if our guess was incorrect and the integral wasn't of the form of (ax^2+bx+c)*sqrt(x^2+1) ? Would the system of equation of the coefficients have no solution, or is it just that deriving it wouldn't work?
4:44 The sign between the two terms in this derivative should be minus (-) instead of plus (+) if my derivation is correct. The derivative of sqrt(x^2 + 1) is -x/sqrt(x^2 + 1) with a minus sign, not x/sqrt(x^2 + 1). This leads a, b, and c to be 1, 0, and 2 respectively, which gives us (x^2 + 2) * sqrt(x^2 + 1) as the primitive we were looking for. I also verified this by derivating that expression with respect to x, and I did in fact obtain x^3 / sqrt(x^2 + 1).
@@jompasv1 For a function f, the derivative of sqrt(f) is -f'/2sqrt(f). In this case, f is the function where f(x) = x^2 + 1. Again, you can check that the solution I gave, namely (x^2 + 2) * sqrt(x^2 + 1), is correct by derivating it and obtaining x^3 / sqrt(x^2 + 1).
@@HectaSpyrit The derivative of sqrt(f) is not negative: (x^n)' = n x^(n-1) ...so derivative √x = x^(1/2) is 1/2*x^(1-1/2) = 1/(2√x) for f(x)=x^2+1 you get (√f(x))' = f(x)'/(2√f(x)) = 2x/(2√(x^2+1)) => x/sqrt(x^2 + 1) (equal to original)
Don't forget by chain rule the power of the "outer" function of root(x^2+ 1) is a half. So bringing that down it cancels the 2 from differentiating 2x. This leaves just x.
Maybe the "pseudo-degree" could be expressed in a less fuzzy way: the limit, as x goes to infinity, of the expression in question divided by x to that degree is a nonzero real number.
The method ("trick") of calculating such integrals using indefinite coefficients has a strictly defined algorithm. I wrote about this (essentially quoting one tutorial) in a comment a 21 hours ago. The author was lucky with the chosen example . For example, ∫(x^3 +x^2 )dx/√x^2 +1 = (x^2/3+x/2 -2/3)√x^2 +1 -(1/2) *ln(x+√x^2 +1 ) +C. and, if the first term is found in the way described, then it is hardly possible to guess that there is still a second term and it has the form of a logarithm. Sorry for the repetition. In General, ∫Pn (x)dx/√(ax^2 +bx+c) = Qn-1(x)*√(ax^2 +bx+c)+λ∫dx/√(ax^2 +bx+c) , where Pn(x) is a polynomial of degree n with given coefficients;the numbers a, в.c are also known . Qn-1(x) is a polynomial of degree n-1 with uncertain coefficients that have to find, as well as the number λ. Differentiating the said assumed identity and bringing the result to a common denominator, we obtain the equality of two polynomials from which we find these numbers.
👋I'm sorry, but I don't understand where that λ∫dx/√(ax^2 +bx+c) in the general case you wrote comes from With the example in the video, but using that formula, I get [the same (correct) result] + λ∫dx/(sqrt(x² + 1)), so in this case that would be λ · arsinh(x), but I don't understand how I'm supposed to find that λ = 0
@@giuseppenonna2148 This is a standard calculation method. The integral written in the left part (1) is searched for in the form ∫Pn (x)dx/√(ax^2 +bx+c) = Qn-1(x)*√(ax^2 +bx+c)+λ∫dx/√(ax^2 +bx+c) . (1) If we find all n-indefinite coefficients of the (n-1)-th degree polynomial Qn-1(x), as well as an unknown factor λ, then the task of calculating the original integral will be reduced to calculating a simpler integral ∫dx/√(ax^2+bx+c) . Let me remind you that all the coefficients of the polynomial of the nth degree Pn (x), as well as the numbers a,b,c, are considered known. Differentiate (1) by x, Pn(x)/√(ax^2+bx+c) = d/dx{Qn-1(x)*√(ax^2+bx+c)}+λ/√(ax^2+bx+c) . Pn(x)= d/dx[Qn-1(x)]*(ax^2 +bx+c)+ Qn-1(x)*(2*a*x+b) + λ. (2) From (2) we find the values of the indefinite coefficients, equating the coefficients with the same powers of x.
It was expected that a=1/3. Since you integrate a function that is equivalent to x^2, the result "should" be equivalent to x^3*1/3. You could add this to your guess to have one unknown less.
This is an interesting trick! I wonder if there's a way to make it work with rational functions. It seems like it could be difficult to guess the form of the antiderivative just using degrees.
You kind of sort of use a similar method to this when dealing with rational functions. There is a technique called partial fraction decomposition where you have a rational function of polynomials and then you make it "equal" to a much simpler sum of rational functions and you do some algebra to figure out what those functions are exactly and then you do the integral cause it will be a lot simpler.
It's not only restricted to polynomials. It also applies to ratio functions of polynomials, and roots that undo the greatest degree. This function will still look like a quadratic if you graph it, and study its behavior far away from its poles and zeros.
I like how the German word "ansatz" means both: finding a way to solve a problem or setting something up for fermenting like mixing the ingredients to make liquor or using flour, yeast, milk and water to create a starter for bread.
@@peterkrauel7237 Usually the tz is characteristic of a German word, written for English speakers. The z in German makes a "ts" sound, like the z's in pizza. It does the same in Italian as well. For German words that end in a z, they often pair it with a t that isn't there in the original spelling, to remind us how to say the z as a ts sound. The spelling of Leibnitz's name is another example, where Leibniz is the original spelling.
Yep, but “tsu”, which sounds very similar, is found in the romanizations of a lot of Japanese words, in place of the character つ (tsu), though usually followed by a u. For example, in the case of 暗殺 (あんさつ), it’s usually romanized as “ansatsu”. However, since there’s no way to spell “ansatz” precisely in Japanese characters, it would likely be spelled the same (though in katakana: アンサツ).
@@carultch In this case, the German word has a tz though. At the end of words, following a short vowel, Germans normally write -tz. Gottfried Wilhelm Leibniz is an exception from the rule, as he was living before the German spelling was codified, and the spelling of names is usually not changed. The Austrian city of Leibnitz in contrast is written with -tz. PS: The spelling rule is: after l, n and r no tz and no ck. It is Schwatz (chat), but schwarz (black).
After 6:42 we get a system of four equations with three unknowns. There might be a fourth unknown (which in this case is equal to zero), but I have no idea how to "introduce" it in the guess. xD
Serious recommendation. 1) First write expression and only then 2) square root symbol over it. Otherwise it looks not professional, when u constantly write small square root and then prolong the upper line.
Isn't it somehow equivalent to integrating by parts? I see {1/x² + 1}·⁵ x dx and I recognize the derivative of {x² + 1}·⁵ , so I use x² as the part to derivate. Then the new integral is x {x² + 1}·⁵ and I substitute U= x²+1. Eesentially, when integrating by parts, I unconsciously measure the degrees of the possible parts.
Ansatz is roughly translated as a guess solution. So you could say we have an ansatz for this equation, for example. I might be wrong, English is not my first language.
@@azzteke “On occasion, English has pursued other languages down alleyways to beat them unconscious and rifle their pockets for new vocabulary.” - Terry Pratchett
"der Ansatz" is literally "the approach", which makes a lot of sense. ("Welchen Ansatz verfolgen wir heute?" == Which approach are we using/following today?)
In this context, it's no issue to me. You are done solving for the unknown coefficient of C, by the time you add the arbitrary constant of integration. You simply write "reassign C", if you need to be more specific. I think what students would hate is being marked down for recycling a letter, without indicating a mark to tell it apart. It's very common in Differential equations, where your solution takes the form of e^(-t + C), and you reassign C, so you can rewrite it as C*e^(-t). Some instructors require you put a diacritical mark, or subscript to tell them apart, even though there's no chance of confusing them.
This type of thinking is beyond most students but it is the core of real mathematics. Imagine your are Terry Tao and Eli Stein is testing you on your dissertation (Terry has already posted on how that went!! ) and he sets you up for some harmonic analysis( in which of course he is an expert). At that level you need to demonstrate that you understand the fundamental relationships (Terry thought he blew it) and at lot of this is ansatz driven: this looks like this and hence you should get something that looks like this etc. It is at a more sophisticated level than this but it is friggin isomorphic! So Mike, you have kicked some mathematical arse here.
I think that the problem really is the mmr because it caps off too soon so you only need to be a semi good killer to reach high mmr and the same is true for survivors
I think the main reason students dislike this trick is because it is unfair to test students based on their ability to structure a guess, even if at a research level, its a useful skill.
We did not put square root term because it is in the integral, we put it because its integral exists in the integral. And other parts follows from integration by parts/or multiplication property of derivative. If you are experinced in math there is no way around. None of the real mathematicians do want to loss time with these 'easy algebra'
I believe this "trick" is highly unfair. It works only in certain cases. For example, it will work for integrals of kind x^(nm-1)/rootn(x^n+1). Some special polynomial over root of polynomial integral can be solved, like (243x^5+11x+34)/cubicroot(x^3+2x^2+2x+1), but if you change polynomial slightly it won't work.
Я уже написал тут тройку комментариев, больше наверное неприлично)) Тем не менее. Прочитайте их, если нетрудно. Этот метод неопределенных коэффициентов предполагает, что должно быть ещё одно слагаемое. Но, в примере автора оно равно нулю, и он не знает о его возможном существовании. Что касается случая, когда в знаменателе - кубический корень из квадратного трех- члена, т.е. ∫Pn (x)dx/(ax^2 +bx+c)^1/3 , здесь Pn(x) -полином степени n с известными коэффициентами , а, в,с - тоже заданы) ,то ответ ищут в виде Qn-1(x)*(ax^2+bx+c)^(2/3) + λ∫dx/(ax^2 +bx+c) ^1/3. Qn-1(x) - полином степени n-1 c неопределёнными коэффициентами и неизвестной λ. Все они определяются этим методом. Другое дело ∫dx/(ax^2 +bx+c) ^1/3 может не иметь аналитического выражения. Кстати, похоже, что под радикалом может стоять не обязательно квадратный трёхчлен, а полином любой степени.
@@Vladimir_Pavlov Я как раз пробовал на случаях когда степень корня совпадает со степенью полинома под корнем. То есть под кубическим корнем брал кубический полином. Я думаю можно пробовать при любой степени корня и полинома, но в общем случае получается, что число соотношений на коэффициенты в предполагаемом ответе больше, чем число самих коэффициентов, поэтому предполагаемая форма ответа будет работать только при специально подобранном числителе.
The problem of elementary antiderivatives has been entirely resolved by Liouville. I think instead of presenting 'methods', you can just present the algorithm, so that nobody will ever work computing antiderivatives again.
@@lolerie if no one has implemented it in full, then why teach students to use it if it's not practical. I mean, they should learn about it, but still.
@@bcthoburn because people are working on it still even now. In 2012 Miller wrote a very important dissertation on it which was a breakthrough. It is very practical and fast, it is the only algorithm that is really used for this. It is just the implementation of it is not complete still 70 years late, because it is 100 pages long of crazy math!
Spoiler alert... For those wondering how the trig. sub. and rationalization work here it is. For trig sub, use x=Tan \theta, then realize that the integral is essentially f(sec \theta) d(sec \theta) resulting in integral of a polynomial ie. substitution on sec For rationalization, first multiply and divide by the denominator to mke denominator rational. Now split the x^3 /(1+x^2) as x -x/(1+x^2) (in other words, divide the polynomials). Now use the fact that all terms are of the form f(1+x^2)d(1+x^2) ie. substitution on 1+x^2
Not a math related question, but when I'm trying to view this video, or any of your videos for that matter, they are all fuzzy and look like a blocked cable channel. Anyone know why this is?
Your connection was poor when you viewed this video, so YT's auto quality setting gave you a low resolution? Either that or you've consistantly seen this person's videos close enough to their upload that they don't have all quality options yet. Those are my guesses.
A trick students hate, but mathematicians use it all the time: "We leave this as an exercise for the reader."
My most hated trick is "It is obvious that..."
@@douglasmagowan2709 There is also "One can see that..." and "A simple calculation shows that..."
Or the infamous ‘recall that’
Which usually means "I know it's right, I can justify most of the logic. There are a few obscure leaps of logic, but I kind of have to get back to life. Besides. this .tex document barely compiles as it is"
And the ugliest one,note that...
This reminds me of using dimensional analysis in physics to guess an equation
Focus on the units! It works out every single time (until it doesn't).
E = mv^2
Of course that's a brilliant idea!
Combining Einstein and Pythagoras, we can clearly see that E = m * (a² + b²).
I’ve done this a lot lmao
This approach has a rigorous basis, started by Risch (the Risch algorithm) for Liouvillian functions and extended by Davenport to include algebraic functions. It's the method used by computer algebra systems to integrate.
Thank you for sharing your vast knowledge! I learned something cool today.
I know some of these words.
Yeah, nobody did Risch algorithm in full yet though in code! LOL.
@@lolerie It was coded in Reduce and Maple I believe.
@@jimskea224 and in mathematica and in fricas. Neither one is complete, though the last is almost complete.
0:58 Homework
7:50 +c
7:55 Homework
8:00 Good Place To Stop
I love how you're the first to comment xD
@@JonathanMandrake I have to, otherwise people won’t know if there is a good place to stop 😂
@@goodplacetostop2973 Lol 😂
@@goodplacetostop2973
Hahaha
The best place to stop is before I start my homework
Finally someone shows us how to create an educated guess for a solution.
Usually teacher say, let's guess some (crazy ass shit we just dreamed of last night) expression and see that it works.
Thanks Michael.
Reminds me of that citation from The Divine and The Human: "This was once revealed to me in a dream"
To be fair, guessing gets easier with mathematical maturity and can seem both obvious yet impossible to explain. Doesn't make it any nicer for the students at that moment though!
@@jaakkovuori9616 I agree. With lots of practice and training you tend to “see” some things that others who don’t practice simply cant.
@@jaakkovuori9616 Experience is the luxury of the wizened mathematician chewing through a single problem over the course of many weeks.
Panicked guessing is the luxury of the novice student rushing a hundred problems in two hours on the midterm exam.
Brilliant. A trick that takes a minute to solve the problem in some cases is definitely a worthy one. I like how you introduced it, the rationale behind the form of the solution.
This method reminds me a typical phrase in some math books(especially about differential equations): "Let's search for a solution in the form of..."
Another method is u-substitution of u=x^2+1 so that du=2xdx and then x^2=u-1 so the integrand becomes (u-1)/(2sqrt(u))= (1/2))u^(1/2) - (1/2)u^(-1/2) which is easily evaluated via the power rule.
The answer obtained is (1/3)*(u-3)*sqrt(u)=(1/3)*(x^2-2)*sqrt(x^2+1)
This is in general what I remember the most from differential equations: Guess that there's some solution involving parts of the function you start with, then figure out what all the constants have to be for it to work. It's not a very universal technique, but there are so few differential equation solving methods that work on any wide variety of function it makes sense that you'd use something like this as the standard way for solving them.
1:38 I would say that you don’t need the quotes around the equals sign. The “degree function” as you put it is a well defined function from the domain of radical/rational functions to the codomain of integers.
What does your well-defined function from radical functions to integers take sqrt(x) to?
@@gravitas802 Good point, I suppose it would be to the rational numbers, not the integers.
maybe even real numbers
Dear Michael in an old scottish maths text book I came across a neat substitution which arrives at the same answer viz; set u=1+x^2 giving du=2xdx thus changing the integrand to
1/2(u-1)du/sqrt(u) from where the answer can be easily obtained.
Hope you find this of interest.
from Nick Glasgow Scotland.
Clever approach, but the rationalizing substitution you mention, u=x^2+1, is so quick and easy. Honestly, it is much simpler. It would be interesting to see an integral where this new (to me and you) approach really help an otherwise hopeless integral
I agree. I'm guessing this toy example made for a shorter video, but it would be cool to at least see a challenge problem at the end that the viewer can solve.
Like most powerful mathematical tools, they are harder to use on easy problems and much more scalable to difficult problems.
@@GeldarionTFS That's a nice insight.
In my understanding, this is how Wolfram Mathematica does integration: for each given integrand, it builds a parametrization of the likely anti-derivatives, then takes the derivative of it and matches to the integrand until it has extracted the parameters.
Great stuff again :)
I could add that since the integrand is odd, then the antiderivative will be even : this gets rid of the "b" from the beginning!
I have never seen this method before. Thank you for sharing!
You can use the Landau notation to show the “degree” of that function. (x³ / sqrt(x² + 1)) ∈ O(x²)
Though you want big-Theta, in this case. Big-O is just an upper bound and we need more than that.
@@beeble2003 Sure, that would be more correct.
this integral can also be done by parts, with u=x², dv=the rest of the stuff. when you see sqrt(x²+1) in the numerator, you can substitute it with (x²+1)/sqrt(x²+1) and continue. piece of cake.
that's my absolute least favorite method Lol
You could also just add and subtract x from the numerator. You obtain xsqrt(x^2+1)-x/sqrt(x^2+1) in the integrand, which is easily integrable
A lot of maths research involves doing rough approximations like this to get an idea of what to expect, before moving onto more rigorous work
Aka Ansätze.
Never seen an integral solved this way, that's a very useful tool!
In General, ∫Pn (x)dx/√(ax2 +bx+c) = Qn-1(x)*√(ax2 +bx+c)+λ∫dx/√(ax2 +bx+c) ,
where Pn(x) is a polynomial of degree n. Qn-1(x) is a polynomial of degree n-1 with uncertain
coefficients that have to find, as well as the number λ.
Differentiating the said assumed identity and bringing the result to a common denominator, we obtain the equality of two polynomials from which we find these numbers.
You could replace the 1 with a^2 and make the entire integral perfectly consistent, say if you allocated units of length to both x and a. it got me thinking suddenly that this is why inverse trig functions appear. 1/(x^2 + a^2) would have degree -2, so one might expect a degree -1 integral, which is precisely what happens as its 1/a * arctan(x/a) and arctan(x/a) is function of degree 0 combination and so is unitless. 1/sqrt(a^2 - x^2) has degree -1, so one may expect a degree 0 result, which is what happens with the unitless arcsin(x/a).
I usually do this trick like without noticing. Always when i am solving any i tegral or trying to differentiate some functions i try to make some conjectures on what the answer will be for example the degree, the function itself, even sometimes domain of the function.
It’s a technique that’s often used when solving pdes
Another way to do the integral that you didn’t mention is a hyperbolic trig substitution. Letting x=sinh(t), you’ll get sqrt(x^2+1)=sqrt(sinh^2(t)+1)=cosh(t), which will cancel with the dx=cosh(t)dt, leaving you with the integral of sinh^3(t)dt. If you know your hyperbolic trig identities (which are all very similar to regular trig identities) that integral is fairly easy.
I think it's still considered trig substitution
This can also be done using integration by parts if one lets u=x^2 and dv = (x/\sqrt{x^2+1})dx.
Interesting method. It looks like Risch's algorithm where instead of proposing an n-degree polynom as here, one proposes a generic p(x) polynom to be calculated as the result of solving a diferential equation. Specifically, here we get x^3/Sqrt[x^2+1]=d(p(x)*Sqrt[x^2+1])/dx then x*p(x)+p(x)'+x^2*p(x)'=x^3 and by guessing p(x)=a*x^2+b*x+c we get the desired solution.
"Let us start by making a correct guess as to the solution..."
"students hate this trick, but mathematicians use it all of the time!"
- Looking up the result in Bronstein?
"looking up the solution" sounds lik soemthing mathematicians hate but students love to do
Bronstein?
@@xCorvus7x "This equation is bronstein-solvable": "You can look up the solution in Bronstein" (Bronstein and Semendyayev: Handbook of Mathematics, Moscow 1945).
@@SiqueScarface thanks
@@xCorvus7x Bronstein and Semendyayew lists the limits of lots of sequences and series and the solution to large swats of differential equations without actually explaining how to get to those solutions. A Bronstein solution is thus a quick way to attack a problem without the need to understand it.
It looks like the method of undetermined coefficients for solving differential equations. It's useful if you prefer derivatives over integrals, but can be a little time consuming.
Virtually anyone prefers derivatives over intergals. Basic integration techniques are based on it, both integration by parts and substitution.
@@nikitakipriyanov7260 I'm assuming by "based on" you mean how they were originally derived, but I don't see how that affects what method someone prefers.
Another method of solving this would be adding and subtracting x in the numerator. You'll get two different integrals which are pretty easy solve. You will get the solution in barely 10 lines. But I loved this method too, got to learn something new❤❤❤
So elegant, loved it! Then you just drop a u=x^2+1 and it almost solves itself!
We called that kind of move "adding a smart zero" in college, we also "multiplied by smart ones" when needed, it's amazing how much easier your job can become with some well crafted 1's and 0's.
What would happen if our guess was incorrect and the integral wasn't of the form of (ax^2+bx+c)*sqrt(x^2+1) ? Would the system of equation of the coefficients have no solution, or is it just that deriving it wouldn't work?
Probably you either get no solutions or a solution that has a contraddiction in it
@@davidcroft95 yeah but I don't know if it's guaranteed. You could end up with a wrong answer without ever knowing
Brilliantly done, Michael!
I saw this trick in the book A synopsis of pure and applied mathematics.
Integration by parts destroyed this intergal. Split x³ into x² and x. Take the derivative of x² and integrate x/√x²+1
I just change the x^3 into something that makes it easier to integrate: x^3 = x(x^2+1) -x; gives me two terms that are relatively easy to integrate.
Pretty nice trick, however adding and substracting x in the numerator seems pretty obvious too!!
i cant belive i just used and discovered this yesterday myself
this video inspired me to stop falling asleep and get back to learning! thanks you Michael !
so.. a and c are the charges of quarks?
Hi,
8:03 : would have been a better place to stop, to avoid being hidden by the splash screens.
For fun:
6:42 : "great".
After a while checking the reasonableness of a result becomes second nature. This was an interesting video.
4:44 The sign between the two terms in this derivative should be minus (-) instead of plus (+) if my derivation is correct. The derivative of sqrt(x^2 + 1) is -x/sqrt(x^2 + 1) with a minus sign, not x/sqrt(x^2 + 1). This leads a, b, and c to be 1, 0, and 2 respectively, which gives us (x^2 + 2) * sqrt(x^2 + 1) as the primitive we were looking for.
I also verified this by derivating that expression with respect to x, and I did in fact obtain x^3 / sqrt(x^2 + 1).
Where are you getting the minus sign from?
@@jompasv1 For a function f, the derivative of sqrt(f) is -f'/2sqrt(f). In this case, f is the function where f(x) = x^2 + 1.
Again, you can check that the solution I gave, namely (x^2 + 2) * sqrt(x^2 + 1), is correct by derivating it and obtaining x^3 / sqrt(x^2 + 1).
@@HectaSpyrit The derivative of sqrt(f) is not negative: (x^n)' = n x^(n-1)
...so derivative √x = x^(1/2) is 1/2*x^(1-1/2) = 1/(2√x)
for f(x)=x^2+1 you get (√f(x))' = f(x)'/(2√f(x)) = 2x/(2√(x^2+1)) => x/sqrt(x^2 + 1) (equal to original)
Just use the substitution u=x^2+1 for this integral. Very simple.
Great presentation as usual! I would be grateful l if you started a series on the basics of funtional analysis, Hilbert spaces etc.
Nice one, really new to me. Thanks Michael.
Sir, please a video on recurrence.
When taking the derivative at 4:58 shouldn't there be a 2 attached to the x as the derivative of x^2 is 2x?
Don't forget by chain rule the power of the "outer" function of root(x^2+ 1) is a half. So bringing that down it cancels the 2 from differentiating 2x. This leaves just x.
@@oberonthefirst8886 thanks for pointing that out oberon
"This is not a method of integration that's universal" Yeah I kinda got that from the fact you weren't sweating and yelling about it
Not to say you're sweaty and yelly, but a trick that applies to all integrals would be something to sweat n yell about
@@fenryrtheshaman wat is another scenario I could sweat and yell about?
@@chairwood proving P = NP
@@fenryrtheshaman u kno wat. that is very true. thx
@@fenryrtheshaman wow I clicked your channel and saw wkukvods in our common subscriptions... :(
Maybe the "pseudo-degree" could be expressed in a less fuzzy way: the limit, as x goes to infinity, of the expression in question divided by x to that degree is a nonzero real number.
The method ("trick") of calculating such integrals using indefinite coefficients has a strictly defined algorithm. I wrote about this (essentially quoting one tutorial) in a comment a 21 hours ago.
The author was lucky with the chosen example . For example,
∫(x^3 +x^2 )dx/√x^2 +1 = (x^2/3+x/2 -2/3)√x^2 +1 -(1/2) *ln(x+√x^2 +1 ) +C. and, if the first term is found in the way described, then it is hardly possible to guess that there is still a second term and it has the form of a logarithm.
Sorry for the repetition.
In General, ∫Pn (x)dx/√(ax^2 +bx+c) = Qn-1(x)*√(ax^2 +bx+c)+λ∫dx/√(ax^2 +bx+c) ,
where Pn(x) is a polynomial of degree n with given coefficients;the numbers a, в.c are also known . Qn-1(x) is a polynomial of degree n-1 with uncertain coefficients that have to find, as well as the number λ.
Differentiating the said assumed identity and bringing the result to a common denominator, we obtain the equality of two polynomials from which we find these numbers.
👋I'm sorry, but I don't understand where that λ∫dx/√(ax^2 +bx+c) in the general case you wrote comes from
With the example in the video, but using that formula, I get [the same (correct) result] + λ∫dx/(sqrt(x² + 1)), so in this case that would be λ · arsinh(x), but I don't understand how I'm supposed to find that λ = 0
@@giuseppenonna2148 This is a standard calculation method.
The integral written in the left part (1) is searched for in the form
∫Pn (x)dx/√(ax^2 +bx+c) = Qn-1(x)*√(ax^2 +bx+c)+λ∫dx/√(ax^2 +bx+c) . (1)
If we find all n-indefinite coefficients of the (n-1)-th degree polynomial
Qn-1(x), as well as an unknown factor λ, then the task of calculating the original integral will be reduced to calculating a simpler integral ∫dx/√(ax^2+bx+c) .
Let me remind you that all the coefficients of the polynomial of the nth degree Pn (x), as well as the numbers a,b,c,
are considered known.
Differentiate (1) by x,
Pn(x)/√(ax^2+bx+c) = d/dx{Qn-1(x)*√(ax^2+bx+c)}+λ/√(ax^2+bx+c) .
Pn(x)= d/dx[Qn-1(x)]*(ax^2 +bx+c)+ Qn-1(x)*(2*a*x+b) + λ. (2)
From (2) we find the values of the indefinite coefficients, equating the coefficients with the same powers of x.
Oh you are right, I didn't notice λ was findable by comparing the coefficients too
Thanks a lot!
6:20 2b ♥
That's cool.
Is there a clear proof of that rule working where the requirements on the functions on which it works?
You can try the substitution x=sinh(t).
Sinchy
at 5:00, the derivative should be multiplied by 2x (derivative of the inside) not just x, right?
The 2 from 2x is cancelled by the ½ from the square root.
@@davidgould9431 Yep. Duh. Thanks
Excellent presentation. vow !!
It was expected that a=1/3.
Since you integrate a function that is equivalent to x^2, the result "should" be equivalent to x^3*1/3.
You could add this to your guess to have one unknown less.
Could you do a few more examples of this technique? Also, what is the name of that method? I'd like to search for it, if possible.
I would say the name of the method is "guess and check" :)
It is called the "ansatz" method.
@@TJStellmach I've already googled that name. "ansatz" is "approach" in german.
@@raystinger6261I prefer the name "prototype solution" .
I had to read the title twice as I thought it was one of those “mathematicians hate this trick, but students use it all the time”
I remember nothing about integrals; it's been over 20 years since I learned about them in high school.
This is an interesting trick! I wonder if there's a way to make it work with rational functions. It seems like it could be difficult to guess the form of the antiderivative just using degrees.
Right!! ❤🌿🌿
You kind of sort of use a similar method to this when dealing with rational functions. There is a technique called partial fraction decomposition where you have a rational function of polynomials and then you make it "equal" to a much simpler sum of rational functions and you do some algebra to figure out what those functions are exactly and then you do the integral cause it will be a lot simpler.
"Guessing solution" is the most powerful and universal mathematical trick.
Is the hesitancy regarding the proposed definition of degree the fact that the degree is typically restricted to polynomials?
It's not only restricted to polynomials. It also applies to ratio functions of polynomials, and roots that undo the greatest degree. This function will still look like a quadratic if you graph it, and study its behavior far away from its poles and zeros.
holy lurking schist!! I _love_ this!
Partial Fraction Decomposition: Integral Edition.
It really is such a nice new method!
I like how the German word "ansatz" means both: finding a way to solve a problem or setting something up for fermenting like mixing the ingredients to make liquor or using flour, yeast, milk and water to create a starter for bread.
I thought it was the Japanese word for assassination (暗殺) !
@@peterkrauel7237 Lets kill the anti-derivative!
@@peterkrauel7237 Usually the tz is characteristic of a German word, written for English speakers. The z in German makes a "ts" sound, like the z's in pizza. It does the same in Italian as well. For German words that end in a z, they often pair it with a t that isn't there in the original spelling, to remind us how to say the z as a ts sound. The spelling of Leibnitz's name is another example, where Leibniz is the original spelling.
Yep, but “tsu”, which sounds very similar, is found in the romanizations of a lot of Japanese words, in place of the character つ (tsu), though usually followed by a u. For example, in the case of 暗殺 (あんさつ), it’s usually romanized as “ansatsu”. However, since there’s no way to spell “ansatz” precisely in Japanese characters, it would likely be spelled the same (though in katakana: アンサツ).
@@carultch In this case, the German word has a tz though. At the end of words, following a short vowel, Germans normally write -tz. Gottfried Wilhelm Leibniz is an exception from the rule, as he was living before the German spelling was codified, and the spelling of names is usually not changed. The Austrian city of Leibnitz in contrast is written with -tz.
PS: The spelling rule is: after l, n and r no tz and no ck. It is Schwatz (chat), but schwarz (black).
After 6:42 we get a system of four equations with three unknowns. There might be a fourth unknown (which in this case is equal to zero), but I have no idea how to "introduce" it in the guess. xD
If I've learned anything from studying differential equations, it's this: if the solution needs another term, it's always a ln.
Serious recommendation. 1) First write expression and only then 2) square root symbol over it. Otherwise it looks not professional, when u constantly write small square root and then prolong the upper line.
Slick. I shoulda thought of this.
Isn't it somehow equivalent to integrating by parts? I see {1/x² + 1}·⁵ x dx and I recognize the derivative of {x² + 1}·⁵ , so I use x² as the part to derivate. Then the new integral is x {x² + 1}·⁵ and I substitute U= x²+1.
Eesentially, when integrating by parts, I unconsciously measure the degrees of the possible parts.
Clever! I like it a lot.
definitely a very nice trick, and it feels quite good whenever it pays soff
This feels like a variation on partial fractions
Curious to know how this generalizes!
How is ansatz used in English?
Ansatz is roughly translated as a guess solution. So you could say we have an ansatz for this equation, for example.
I might be wrong, English is not my first language.
@@mrswats But "Ansatz" is German!
@@azzteke indeed :)
@@azzteke “On occasion, English has pursued other languages down alleyways to beat them unconscious and rifle their pockets for new vocabulary.” - Terry Pratchett
"der Ansatz" is literally "the approach", which makes a lot of sense. ("Welchen Ansatz verfolgen wir heute?" == Which approach are we using/following today?)
This tricks resembles dimensional analysis in physics
What students hate is to use the same variable C for two completely different things (-2/3 and constant after taking antiderivative).
but c != C, if only professionals knew how to say those properly.
In this context, it's no issue to me. You are done solving for the unknown coefficient of C, by the time you add the arbitrary constant of integration. You simply write "reassign C", if you need to be more specific.
I think what students would hate is being marked down for recycling a letter, without indicating a mark to tell it apart. It's very common in Differential equations, where your solution takes the form of e^(-t + C), and you reassign C, so you can rewrite it as C*e^(-t). Some instructors require you put a diacritical mark, or subscript to tell them apart, even though there's no chance of confusing them.
if the frame of your board doesn't fit into the frame of your camera , then maybe its because of not using the right lens
What does ansatz mean?
German for "approach".
Besides, we just witnessed proof by tactical guessing 🤣
This type of thinking is beyond most students but it is the core of real mathematics. Imagine your are Terry Tao and Eli Stein is testing you on your dissertation (Terry has already posted on how that went!! ) and he sets you up for some harmonic analysis( in which of course he is an expert). At that level you need to demonstrate that you understand the fundamental relationships (Terry thought he blew it) and at lot of this is ansatz driven: this looks like this and hence you should get something that looks like this etc. It is at a more sophisticated level than this but it is friggin isomorphic! So Mike, you have kicked some mathematical arse here.
4:32 Chen Lu
hey mikey pencil!!! Great job with this video! Shared with my other math friend
I am going to try it on ALL the integrals! Hehe.
*mechanical pencil
for real this is an amazing trick i feel like trying this out on all sorts of other integrals
I think that the problem really is the mmr because it caps off too soon so you only need to be a semi good killer to reach high mmr and the same is true for survivors
I think the main reason students dislike this trick is because it is unfair to test students based on their ability to structure a guess, even if at a research level, its a useful skill.
Yep. That stuff should only be on Putnam exams, to be absolutely sure students writing have never seen it before.
Let's not pretend like mathematicians solve integrals manually
We did not put square root term because it is in the integral, we put it because its integral exists in the integral. And other parts follows from integration by parts/or multiplication property of derivative. If you are experinced in math there is no way around. None of the real mathematicians do want to loss time with these 'easy algebra'
Very interesting trick.
A Differential Equations expert has entered the chat.
It's like he's speaking a foreign language, I haven't done actual maths in 10 years, why am I here?
Nice, but surely in this case, the u-substitution is just natural and so much easier?
I believe this "trick" is highly unfair. It works only in certain cases. For example, it will work for integrals of kind x^(nm-1)/rootn(x^n+1). Some special polynomial over root of polynomial integral can be solved, like (243x^5+11x+34)/cubicroot(x^3+2x^2+2x+1), but if you change polynomial slightly it won't work.
Я уже написал тут тройку комментариев, больше наверное неприлично)) Тем не менее. Прочитайте их, если нетрудно. Этот метод неопределенных коэффициентов предполагает, что должно быть ещё одно слагаемое. Но, в примере автора оно равно нулю, и он не знает о его возможном существовании.
Что касается случая, когда в знаменателе - кубический корень из квадратного трех-
члена, т.е. ∫Pn (x)dx/(ax^2 +bx+c)^1/3 , здесь Pn(x) -полином степени n с известными коэффициентами , а, в,с - тоже заданы) ,то ответ ищут в виде Qn-1(x)*(ax^2+bx+c)^(2/3) + λ∫dx/(ax^2 +bx+c) ^1/3. Qn-1(x) - полином степени n-1 c неопределёнными коэффициентами и неизвестной λ.
Все они определяются этим методом. Другое дело ∫dx/(ax^2 +bx+c) ^1/3
может не иметь аналитического выражения.
Кстати, похоже, что под радикалом может стоять не обязательно квадратный трёхчлен, а полином любой степени.
@@Vladimir_Pavlov Я как раз пробовал на случаях когда степень корня совпадает со степенью полинома под корнем. То есть под кубическим корнем брал кубический полином. Я думаю можно пробовать при любой степени корня и полинома, но в общем случае получается, что число соотношений на коэффициенты в предполагаемом ответе больше, чем число самих коэффициентов, поэтому предполагаемая форма ответа будет работать только при специально подобранном числителе.
Легко решается через гиперболические функции. Не очень понимаю, какая проблема с этим интегралом
Лол, я вообще решил через t=x² и немного простой арифметики
You can’t read the screen, the point is literally to show a DIFFERENT way of doing it!
@@thetheoreticalnerd7662 Omg thanks cap
It made me nervous when he didn't add the constant in the first place
The problem of elementary antiderivatives has been entirely resolved by Liouville. I think instead of presenting 'methods', you can just present the algorithm, so that nobody will ever work computing antiderivatives again.
What is the algorithm?
Nobody implemented the Risch algorithm in full. Alas.
@@lolerie if no one has implemented it in full, then why teach students to use it if it's not practical. I mean, they should learn about it, but still.
@@bcthoburn because people are working on it still even now. In 2012 Miller wrote a very important dissertation on it which was a breakthrough. It is very practical and fast, it is the only algorithm that is really used for this. It is just the implementation of it is not complete still 70 years late, because it is 100 pages long of crazy math!
@@lolerie Oh I see. Thank you!
A good method to integrate
Spoiler alert...
For those wondering how the trig. sub. and rationalization work here it is.
For trig sub, use x=Tan \theta, then realize that the integral is essentially f(sec \theta) d(sec \theta) resulting in integral of a polynomial ie. substitution on sec
For rationalization, first multiply and divide by the denominator to mke denominator rational. Now split the x^3 /(1+x^2) as x -x/(1+x^2) (in other words, divide the polynomials). Now use the fact that all terms are of the form f(1+x^2)d(1+x^2) ie. substitution on 1+x^2
Would polynomial long division help you here, too, maybe?
Not a math related question, but when I'm trying to view this video, or any of your videos for that matter, they are all fuzzy and look like a blocked cable channel. Anyone know why this is?
Your connection was poor when you viewed this video, so YT's auto quality setting gave you a low resolution? Either that or you've consistantly seen this person's videos close enough to their upload that they don't have all quality options yet. Those are my guesses.
Wo kommt das deutsche Thumbnail plötzlich her?
"Ansatz" in mathematics is the method of pursuing a solution via a guess as to the form it must take.