students hate this trick, but mathematicians use it all of the time!

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Комментарии : 304

@klausolekristiansen2960 2 года назад

A trick students hate, but mathematicians use it all the time: "We leave this as an exercise for the reader."

@douglasmagowan2709 2 года назад

My most hated trick is "It is obvious that..."

@rogerlie4176 2 года назад

@@douglasmagowan2709 There is also "One can see that..." and "A simple calculation shows that..."

@mickschilder3633 2 года назад

Or the infamous ‘recall that’

@andychow5509 2 года назад

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"

@arnabchatterjee4847 2 года назад

And the ugliest one,note that...

@nathanisbored 2 года назад

This reminds me of using dimensional analysis in physics to guess an equation

@SaniFakhouri 2 года назад

Focus on the units! It works out every single time (until it doesn't).

@MrCmon113 2 года назад

E = mv^2

@gpgp 2 года назад

Of course that's a brilliant idea!

@FumblkruschLP 2 года назад

Combining Einstein and Pythagoras, we can clearly see that E = m * (a² + b²).

@BurgoYT 2 года назад

I’ve done this a lot lmao

@jimskea224 2 года назад

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.

@crp2035 2 года назад

Thank you for sharing your vast knowledge! I learned something cool today.

@SuhaibAndrabi 2 года назад

I know some of these words.

@lolerie 2 года назад

Yeah, nobody did Risch algorithm in full yet though in code! LOL.

@jimskea224 2 года назад

@@lolerie It was coded in Reduce and Maple I believe.

@lolerie 2 года назад

@@jimskea224 and in mathematica and in fricas. Neither one is complete, though the last is almost complete.

@goodplacetostop2973 2 года назад

0:58 Homework 7:50 +c 7:55 Homework 8:00 Good Place To Stop

@JonathanMandrake 2 года назад

I love how you're the first to comment xD

@goodplacetostop2973 2 года назад

@@JonathanMandrake I have to, otherwise people won’t know if there is a good place to stop 😂

@anshumanagrawal346 2 года назад

@@goodplacetostop2973 Lol 😂

@Ahmed-Youcef1959 2 года назад

@@goodplacetostop2973 Hahaha

@wontpower 2 года назад

The best place to stop is before I start my homework

@elaadt 2 года назад

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.

@GroundThing 2 года назад

Reminds me of that citation from The Divine and The Human: "This was once revealed to me in a dream"

@jaakkovuori9616 2 года назад

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!

@PegasusTenma1 2 года назад

@@jaakkovuori9616 I agree. With lots of practice and training you tend to “see” some things that others who don’t practice simply cant.

@General12th Год назад

@@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.

@Markgraf_zu_Brandenburg 2 года назад

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..."

@RayaneS 2 года назад

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.

@stevenglowacki8576 2 года назад

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.

@barnabecroizat8641 2 года назад

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!

@RobertoRezende 2 года назад

Never seen an integral solved this way, that's a very useful tool!

@SaveSoilSaveSoil 2 года назад

I have never seen this method before. Thank you for sharing!

@cameronspalding9792 2 года назад

It’s a technique that’s often used when solving pdes

@DrBillDoesMath Год назад

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)

@khaledchatah3425 2 года назад

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.

@pacolibre5411 2 года назад

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.

@gravitas802 2 года назад

What does your well-defined function from radical functions to integers take sqrt(x) to?

@pacolibre5411 2 года назад

@@gravitas802 Good point, I suppose it would be to the rational numbers, not the integers.

@dudono1744 2 года назад

maybe even real numbers

@willyh.r.1216 2 года назад

Nice one, really new to me. Thanks Michael.

@Nikolas_Davis 2 года назад

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.

@kevinkgillette 2 года назад

Brilliantly done, Michael!

@pi_xi 2 года назад

You can use the Landau notation to show the “degree” of that function. (x³ / sqrt(x² + 1)) ∈ O(x²)

@beeble2003 2 года назад

Though you want big-Theta, in this case. Big-O is just an upper bound and we need more than that.

@pi_xi 2 года назад

@@beeble2003 Sure, that would be more correct.

@knutthompson7879 2 года назад

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

@westronic 2 года назад

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.

@GeldarionTFS 2 года назад

Like most powerful mathematical tools, they are harder to use on easy problems and much more scalable to difficult problems.

@Evan490BC 2 года назад

@@GeldarionTFS That's a nice insight.

@blakemaths2469 2 года назад

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

@Evan490BC 2 года назад

Aka Ansätze.

@dirkbruere 2 года назад

"Let us start by making a correct guess as to the solution..."

@shubhodeepde3927 2 года назад

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❤❤❤

@orti1283 2 года назад

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.

@Pklrs 2 года назад

Great presentation as usual! I would be grateful l if you started a series on the basics of funtional analysis, Hilbert spaces etc.

@emersonschmidt7882 2 года назад

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.

@celadon2048 2 года назад

Slick. I shoulda thought of this.

@Vladimir_Pavlov 2 года назад

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.

@nickkilpatrick4999 2 года назад

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.

@dr.rahulgupta7573 2 года назад

Excellent presentation. vow !!

@schrodingerbracat2927 2 года назад

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.

@CarlyDayDay 2 года назад

that's my absolute least favorite method Lol

@aqeel6842 Год назад

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

@jamiewalker329 2 года назад

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).

@michaelanisimov7896 2 года назад

this video inspired me to stop falling asleep and get back to learning! thanks you Michael !

@reneszeywerth8352 2 года назад

"students hate this trick, but mathematicians use it all of the time!" - Looking up the result in Bronstein?

@zairaner1489 2 года назад

"looking up the solution" sounds lik soemthing mathematicians hate but students love to do

@xCorvus7x 2 года назад

Bronstein?

@SiqueScarface 2 года назад

@@xCorvus7x "This equation is bronstein-solvable": "You can look up the solution in Bronstein" (Bronstein and Semendyayev: Handbook of Mathematics, Moscow 1945).

@xCorvus7x 2 года назад

@@SiqueScarface thanks

@SiqueScarface 2 года назад

@@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.

@padraiggluck2980 2 года назад

After a while checking the reasonableness of a result becomes second nature. This was an interesting video.

@fariesz6786 2 года назад

so.. a and c are the charges of quarks?

@jaewok5G 2 года назад

holy lurking schist!! I _love_ this!

@captainsnake8515 2 года назад

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.

@CarlyDayDay 2 года назад

I think it's still considered trig substitution

@777NCM 2 года назад

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.

@TheHellBoy05 6 месяцев назад

Integration by parts destroyed this intergal. Split x³ into x² and x. Take the derivative of x² and integrate x/√x²+1

@farfa2937 2 года назад

6:20 2b ♥

@waitotong9590 2 года назад

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”

@CM63_France 2 года назад

Hi, 8:03 : would have been a better place to stop, to avoid being hidden by the splash screens. For fun: 6:42 : "great".

@cernejr 2 года назад

Clever! I like it a lot.

@housamkak646 2 года назад

It really is such a nice new method!

@samegawa_sharkskin 2 года назад

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.

@samegawa_sharkskin 2 года назад

*mechanical pencil for real this is an amazing trick i feel like trying this out on all sorts of other integrals

@fenryrtheshaman 2 года назад

"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

@fenryrtheshaman 2 года назад

Not to say you're sweaty and yelly, but a trick that applies to all integrals would be something to sweat n yell about

@chairwood 2 года назад

@@fenryrtheshaman wat is another scenario I could sweat and yell about?

@fenryrtheshaman 2 года назад

@@chairwood proving P = NP

@chairwood 2 года назад

@@fenryrtheshaman u kno wat. that is very true. thx

@chairwood 2 года назад

@@fenryrtheshaman wow I clicked your channel and saw wkukvods in our common subscriptions... :(

@fartoxedm5638 2 года назад

Pretty nice trick, however adding and substracting x in the numerator seems pretty obvious too!!

@Anishkrsinghiit-jee 2 года назад

Sir, please a video on recurrence.

@MrRyanroberson1 2 года назад

definitely a very nice trick, and it feels quite good whenever it pays soff

@markcomerford3208 2 года назад

This can also be done using integration by parts if one lets u=x^2 and dv = (x/\sqrt{x^2+1})dx.

@cenaalan5825 2 года назад

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.

@peterhall6656 2 года назад

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.

@masoomladkaaproudvegan2876 2 года назад

i cant belive i just used and discovered this yesterday myself

@ichwillfrieden1635 Год назад

I saw this trick in the book A synopsis of pure and applied mathematics.

@CarlyDayDay 2 года назад

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 2 года назад

Virtually anyone prefers derivatives over intergals. Basic integration techniques are based on it, both integration by parts and substitution.

@CarlyDayDay 2 года назад

@@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.

@lam2506 2 года назад

Just use the substitution u=x^2+1 for this integral. Very simple.

@Metodones 2 года назад

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?

@davidcroft95 2 года назад

Probably you either get no solutions or a solution that has a contraddiction in it

@l1mbo69 2 года назад

@@davidcroft95 yeah but I don't know if it's guaranteed. You could end up with a wrong answer without ever knowing

@ciderfan823 2 года назад

I remember nothing about integrals; it's been over 20 years since I learned about them in high school.

@JCCyC Год назад

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.

@as-qh1qq 2 года назад

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

@galgrunfeld9954 2 года назад

That's cool. Is there a clear proof of that rule working where the requirements on the functions on which it works?

@gkatoch1982 2 года назад

Very interesting trick.

@Vladimir_Pavlov 2 года назад

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.

@giuseppenonna2148 Год назад

👋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

@Vladimir_Pavlov Год назад

@@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.

@giuseppenonna2148 Год назад

Oh you are right, I didn't notice λ was findable by comparing the coefficients too Thanks a lot!

@michaelleue7594 2 года назад

Partial Fraction Decomposition: Integral Edition.

@orzhovthief 2 года назад

This tricks resembles dimensional analysis in physics

@peceed 2 года назад

"Guessing solution" is the most powerful and universal mathematical trick.

@bobfr4806 2 года назад

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.

@alicewyan 2 года назад

Nice!

@minwithoutintroduction 2 года назад

عمل رائع واصل.

@supermanifold 2 года назад

Curious to know how this generalizes!

@danidiaz9510 2 года назад

Let's not pretend like mathematicians solve integrals manually

@SiqueScarface 2 года назад

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 2 года назад

I thought it was the Japanese word for assassination (暗殺) !

@SiqueScarface 2 года назад

@@peterkrauel7237 Lets kill the anti-derivative!

@carultch 10 месяцев назад

@@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.

@peterkrauel7237 10 месяцев назад

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).

@SiqueScarface 10 месяцев назад

@@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).

@adrianschmidt3583 2 года назад

Wo kommt das deutsche Thumbnail plötzlich her?

@TJStellmach 2 года назад

"Ansatz" in mathematics is the method of pursuing a solution via a guess as to the form it must take.

@NotSoObvious 2 года назад

It's like he's speaking a foreign language, I haven't done actual maths in 10 years, why am I here?

@MathFromAlphaToOmega 2 года назад

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.

@jimmykitty 2 года назад

Right!! ❤🌿🌿

@insignia201 2 года назад

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.

@adityaekbote8498 2 года назад

4:32 Chen Lu

@marcofrigerio2217 2 года назад

You can try the substitution x=sinh(t).

@carultch 10 месяцев назад

Sinchy

@rafakrecichwost3666 2 года назад

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

@drewmandan 2 года назад

If I've learned anything from studying differential equations, it's this: if the solution needs another term, it's always a ln.

@MarcusCactus 2 года назад

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.

@phnml8440 2 года назад

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

@adamwho9801 Год назад

This feels like a variation on partial fractions

@egillandersson1780 2 года назад

A great trick ! I like it

@phee4174 2 года назад

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?

@oberonthefirst8886 2 года назад

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.

@phee4174 2 года назад

@@oberonthefirst8886 thanks for pointing that out oberon

@avirajhanspal8167 2 года назад

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

@bryancrompton7238 2 года назад

A Differential Equations expert has entered the chat.

@pacolibre5411 2 года назад

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.

@pietergeerkens6324 2 года назад

Yep. That stuff should only be on Putnam exams, to be absolutely sure students writing have never seen it before.

@whyyat3470 2 года назад

at 5:00, the derivative should be multiplied by 2x (derivative of the inside) not just x, right?

@davidgould9431 2 года назад

The 2 from 2x is cancelled by the ½ from the square root.

@whyyat3470 2 года назад

@@davidgould9431 Yep. Duh. Thanks

@eugenet453 2 года назад

It made me nervous when he didn't add the constant in the first place

@harshshah6937 2 года назад

I thought I was ahead in modularity but i was

@aidanmargarson8910 2 года назад

It all comes out in the wash!

@leif_p 2 года назад

That is *One Weird Trick* !

@HectaSpyrit 2 года назад

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 2 года назад

Where are you getting the minus sign from?

@HectaSpyrit 2 года назад

@@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).

@panPetr0ff 2 года назад

@@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)

@Maniclout 2 года назад

My first year professor is from germany and he always said Ansatz, for us it became more of a meme xD

@StefaanHimpe 2 года назад

What students hate is to use the same variable C for two completely different things (-2/3 and constant after taking antiderivative).

@alysdexia 2 года назад

but c != C, if only professionals knew how to say those properly.

@carultch 10 месяцев назад

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.