Putnam Exam | 2009: A2

[@blackpenredpen]

Such a nice solution!

[@DanGRV]

I noticed you were saying "great" in a suspiciously familiar way on one of your recent videos, haha

[@mortenrimmen1215]

Answer contains 12th root of 2 and 6th root of a ‘triggy’ fraction. “Nice” was not the first word that came to my mind 🤣

[@cycklist]

Definitely the first time I've ever seen the twelfth root of something being the correct answer!

[@joallen5750]

It shows up in music, specifically the western 12-tone equal temperament system. An octave doubles frequency, and there are twelve notes in between increasing geometrically, so the ratio of the frequencies of consecutive notes is twelfth root of 2.

[@cycklist]

@@joallen5750 Cool. Would never have imagined it would have a real world application.

[@jeromesnail]

@@joallen5750 I was about to say that! I remember calculating it myself when I was in high school, to know the ration of one fret to another on a guitar neck. It was really satisfying!

[@stevewarner1188]

Less satisfying is when you realise that the 12 tone system only works approximately. So, octaves sound nice/similar because the frequency of one note is exactly double the other. Chords sound nice when the frequencies of the notes are multiples or nice fractions of each other (so the peaks and troughs keep lining up). On the 12 tone system, notes 4 semitones apart (a natural third) sound nice together (eg like C and E). It turns out that these sound good because they are very nearly in the ratio 5:4. BUT IT’S ONLY APPROXIMATE! The ratio is actually 5.0397:4. And they only sound good because the cube root of 2 just happens to be within about 0.8% of a nice interval of 5/4. Blew my mind when I first realised this. Frankly ... nightmare!

[@leofisher1280]

@@stevewarner1188 Its actually not bad, and you can calculate that the 12TET system is one of the most accurate when it comes to approximating integer ratios. You can solve it by using Just Intonation, where every interval is tuned to the correct ratio, but then your instrument will only work in one key.

[@JB-ym4up]

f(0)=1 Find f(x) when x near 0. Its 1, I found it up there, about 3 lines above the question.

[@akashpremrajan9285]

I dont think that's what "near" means. In calculus I think it means "to first order" or something like that. Not sure.

[@golvanontheroad]

@@akashpremrajan9285 Pretty sure that was a joke.

[@bndrcr82a08e349g]

You win all

[@mcwulf25]

Umm.... yes.

[@comma_thingy]

@@akashpremrajan9285 Near means on some open set containing 0. Since we are talking about R here, it's means we need to know f on at least some open interval containing 0

[@HagenvonEitzen]

The same with a different scenic route: Divide the original equations by f, g, and h to obtain lograrithmic derivatives of f, g, h on the left sides (which will come out handy later at least for f). Now recognize that fgh and 1/(fgh) are ubiquitious on the right hand side, so introduce y=fgh as in the video. By the nature of logarithmic derivatives, recall that adding them produces the logarithmic derivative of the product, i.e., here of y. Then we're at y'/y = 6y + 6/y and quickly at the differential equation for y ...

[@iamtrash288]

looked hard but had a very elegant solution! Enjoyed it immensely!

[@kriswillems5661]

It looks so easy when you explain it and so hard when I've to find the answer myself.

[@agfd5659]

Exactly!

[@looney1023]

Funny that the 12-th root of 2 appears in the answer here, as that's such an important constant in music theory. One question: Why is the question phrased "find f(x) 'near' x=0". To me, 'near' x=0 means to find an approximation, usually a Taylor series, for the function around some point. What you did seemed to be an explicit answer for all x, no? Just not sure what the point of that phrasing is

[@sjoerdo6988]

It essentially meant find the function on this open interval. The function did not have to (and didn't) exist on the whole real axis

[@snowfloofcathug]

I read it as “find lim x -> 0” because it’d be discontinuous. Turns out my interpretation was wrong /shrug. Per how someone else explained it: elsewhere on the interval we’d have picked a different tan(x) = 1 value?

[@reeeeeplease1178]

As Sjoerdo mentioned, generally speaking differential equations only have local solutions This means that your function only solves the equation on some interval If you want a solution on a different interval, you may get a different function This is what "near 0" refers to. They want the function that solves the equation on an interval containing 0

[@minwithoutintroduction]

رائع كالعادة. لا أمل من شروحاتكم. واصل. تحية بنكهة الرياضيات من تنغير في الجنوب الشرقي المغربي

[@markregev1651]

This problem using infinite descent is fun: find all pairs of positive integers (x,y) such that (1+x)(1+y)-1 divides x^2+y^2+1

[@md2perpe]

At 9:50 it would be more correct with ln |f(x)| instead of ln (f(x)).

[@SouthOfHeaven97]

true but note that we are asked to find f(x) near x = 0, and we know f(0) = 1 and f is continuous so it is positive near x = 0

[@mcwulf25]

I have never been comfortable with that formulation. It's like fudging your negative number to make the ln work. Shouldn't be integrating through f= zero anyway.

[@md2perpe]

@@mcwulf25 It's easy to show that if a < b < 0 then ∫_a^b (1/x) dx equals ln (-b) - ln (-a) = [ ln(-x) ]_a^b. We can therefore write a primitive function on (-∞, 0) ∪ (0, ∞) as ln |x|. Integration through 0 should only be done if you know what you are doing, i.e. if you really want the principal value.

[@aljuvialle]

The same with sin and cos under ln

[@davidseed2939]

another version f the product rule is y’/y =f’/f + g’/g + h’/h . which saves a step

[@MightOO]

idea for merch: tshirts with OK GREAT printed on them 😂👍

[@MSCCA]

Also “that’s a good place to stop”

[@HAL-oj4jb]

You were both right lol

[@jimschneider799]

The expression under the radical was just begging for some simplification. After a few false starts, I got it as far as sqrt(2) (sin(6 x) + cos(6 x))/(1 - sin(12 x)), which is nice because the sqrt(2) cancels the 2^(-1/12) factor, giving f(x) = (sin(6 x) + cos(6 x))/(1 - sin(12 x))^(1/6). I haven't verified this in the original equation, but I have confirmed that this gives the same answer for several points near x=0. (edited to add missing 6th root).

[@jimschneider799]

Also, it's easy to see that f(x) is defined and real for [ - pi/24, pi/24 ).

[@hassanalihusseini1717]

Thank you! Always when I watch your explanation of number theoretical related problems I feel so stupid (yes, and at this point I must admit I have a MA in physics, makes me feel stupid even more :-)). But in this problem I could follow you all the way (even I must admit that probably I could not have saved it on my own). Thank you for posting these problems.

[@noway2831]

You have stereo audio on, but I think it's backwards. Personally, I don't really like stereo audio, but if you do that's fine.

[@arthgupta1113]

The audio is messing with my head. Feels like he's circling me.

[@Mephisto707]

This is one beautiful problem with a beautiful solution.

[@rishavmondal8045]

Wonderful solution . Bring more Olympiad and Putnam problems.

[@alejandrolagunes5697]

I love his differencial equiation videos so much!

[@MathyMahdi]

May your channel grow Penn sir! this stuff is currently above my grade but hopefully I will watch them in future!

[@ZaqZiemba]

Oh wow, great problem. The symmetry of the problem is pretty easy to see, but wow did I miss the obvious chain rule applicability!! Regardless, thanks for this and any idea why the problem specifies finding the function “near 0”? Seems unnecessary here...

[@rgqwerty63]

The Picard-Lindelof Theorem only ensures that a solution exists in some neighbourhood of 0

[@easymathematik]

The "near zero" is for the constant in the tan-function. The pi/4.

[@mcwulf25]

That had a bit of everything in it! Other than writing the trig as sec*tan rather than sin/cos^2 I would probably have come up with a similar answer. But not in 15 minutes!!!

[@malawigw]

Nice one! Would be fun to see some kind of discussion after each of these math contest problems about some kind of variation of the problem, how important the actual numbers were in the given problem etc. BTW no need of u-substitution in that integral over tan and cot, one can use the chain rule backwards on that one as well.

[@thayanithirk1784]

Please do the Romanian maths Olympiad 1994 number theory problem.thank u

[@pascallaw5909]

Finally a problem that I can claim is easy enough for all calculus learners..

[@eduardomalacarne9024]

one of many ways to check the beauty of the matematics

[@krabkrabkrab]

I actually don't "think it's a good place to stop". If you expand the trig arguments the 12th root of 2 disappears and what remains is 6th root of [(cos + sin)/(cos - sin)^2]. All trig arguments are 6x.

[@adandap]

If all Putnam problems were calculus I coulda been a contender. Nice problem. Thanks for posting it. BTW if you're in Australia some time hit me up and I'll take you to some of the nice local climbing spots.

[@nikitakipriyanov7260]

Why you've forgot about the period when solving for y? I mean, shouldn't be y = tan(6x+pi/4+pi n)? This can somewhat augment the final solution, because sin and cos change sign when shifted by pi. Or, this sign could have been absorbed by the absolute value, which would in this case stand around sine and cosine. Then, why you've ommited it in logarithms? I mean, antiderivative of 1/x is usually written as ln|x|+C. I feel I forgot a theory of differential equations slightly (this is partially why I watching these videos anyway). Nevertheless I've managed to solve this problem completely to the same answer as yours, except for this absolute value around cos and sin.

[@danielduranloosli]

I noticed my mathematical mind has been irreversibly degenerated by my work as an engineer. From the expressions of (f',g',h') in terms of (f,g,h), you can evaluate [f'(0),g'(0),h'(0)]. Then take the derivatives again to obtain expressions of (f'',g'',h'') in terms of (f,g,h,f',g',h'). Evaluate [f''(0),g''(0),h''(0)]. You can keep expressing higher derivatives and evaluating at x=0 to get all the values you want of [f(0),f'(0),f''(0),f'''(0),f''''(0),...]. With that, you can express the nth-order Taylor series expansion of f(x) around x=0. Ta-dah, "solved" for all intents and purposes imaginable by a limited, terrestrial and practical mind that is totally content in its world of mere Platonian shadows of reality.

[@TheLuciano13579]

The question ask for f(x) near 0. Couldn't we just use a linear aproximation? Its easy to see that f'(0)=3 in the first equation, so the linear aproximation is 3x+ 1 that is really close to f(x) near 0

[@nim64]

It asks for f(x) near 0, not something that's really close to f(x) near 0

[@TheLuciano13579]

If that is true than the condition "near 0" is useless sinse he doesn't use It to calculate f(x)

[@ShapelessMonstrosity]

@@TheLuciano13579 He used the "near 0" constraint to choose which solution for c to use when solving arctan y = 6x + c. Otherwise there would be infinitely many solutions for c.

[@yitongbig589]

Would you recommend me to watch this video tin order to prepare the Oxford's MAT and interview for undergraduate program?

[@firemaniac100]

a kind of where is wally problem!!!

[@erickherrerapena8981]

Increíble, ¡¡buen vídeo!!

[@brucelavoie7333]

Using this same method I found that the results for g(x) and h(x) are the following. Almost by inspection h(x) = [1/(6th root of 2)]{6th root of [sin()/cos^3()]} and with a little more work I found that g(x) = [1/(12th root of 8)]{6th root of [sin^4()/cos()]}. Someone want to confirm this or tell me where I messed up?

[@biswarupsaha2495]

Sir your previous swiss mathematical Olympiad question about number theory solution is very hard process,pls try this solve a little bit easy way!!!!!!?????

[@lucasm.b.4390]

I love your videos.

[@DepressionVarietyVlog]

how do you get your solutions? Do you just work them out yourself?

[@easymathematik]

Probably yes. If you are experienced enough you have a focused view for hidden hints (symmetry, derivatives, ...) Like here. The product rule is kind of obvious to consider if you focus on the coefficients.

[@pikkutonttu2697]

I used the same method to solve the problem, I even solved g and h. However I recalled that partial differential equations are sometimes solved by method of separation of variables. It is actually possible to try to solve partial equation of form nabla dot f = 6f^2 +6. However my abilites end there, I cant solve it any other way than by separation of variables, which leads to where I started. Maybe someone can solve it neatly and then use its solution to solve the actual problem.

[@madhavvarshney9210]

Solves a hard problem Finds the answer “That’s a good place to stop”

[@dp121273]

Sounds so weird with that opposite-side sound. Why not just use a mono microphone?

[@osianrhys1257]

Boss man at math

[@НикитаАрляпов]

7:00

[@O_Capivara]

This guy is awesome. I'd never think about in this way

[@aakashchakraborty3673]

Love ir videos .

[@mathbbn2676]

Teachers have started to teach the lessons learned during the COVID 19, and students must learn to understand and prepare for the COVID 19 exam.

[@xshortguy]

i got the first part of solution right as a taylor polynomial. f(0) = 1. The rest was trivial :)

[@holyshit922]

Systems of nonlinear ODE'a are as difficult as first order PDE

[@amirb715]

beautiful !

[@geeache1891]

I came to INT(1/(y2+1)dy)=INT(1/2 (1/(y-1)-1/(y+1))dy=ln(y-1)-ln(y+1) rather than arctan y

[@Notthatkindofdr]

Careful: your second expression does not equal the first expression. You "factored" y^2+1 as if it was y^2-1.

[@polyhistorphilomath]

It should still work out using y-i and y+i

[@JJJ-gc2mi]

so nice!

[@RobotProctor]

12th root..... Is this a music theory question :)

[@TheTar05]

Nice problem

[@psy7669]

awesome

[@shokan7178]

The audio panning changes whenever you rotate your head and it's very disorienting.

[@schweinmachtbree1013]

I'm triggered that you didn't bring the twelfth root of 2 inside the main radical 😠 😠 😠

[@romaing.1510]

You could deduce it at first glance

[@enalaxable]

I smell a math music connection here, how uncannily ;)

[@josephquinto5812]

It frightens me that you consider certain mathematics hard...this shit is already rocket science to me 😂

[@thefranklin6463]

Mic upgrade!

[@gautambansal8952]

Wow, I just solved this question and this video came out what are the chances lol Edit: just watched the whole video I just had to find f(a)g(a)h(a)

[@ashrafmohamed5102]

🌹

[@adityaverma1676]

I am the only one who is impressed by the stereo sound?

[@CauchyIntegralFormula]

This is a cute problem

[@abderrahmanenedjadi7475]

Man, I hope you to gimme your email

[@rajanihatti]

First view

[@maosatela]

definite integral , DEFINITE INTEGRAL

[@sunitaagrawal2751]

Its a pretty eazy problem