A practice problem from India

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[@goodplacetostop2973]

0:01 In before comment section invaded by Indians 😂 1:05 y = 0 ? 6:11 Homework 10:11 रुकने के लिए यह एक अच्छी जगह है

[@metablaze3523]

Translation op👌🏻

[@thatkindcoder7510]

@SNEHANSH MAHARAJ Unfortunately, this is a trade secret that cannot be shared

[@goodplacetostop2973]

@SNEHANSH MAHARAJ I have written a truly marvelous explanation of that trick, which this margin is too narrow to contain

[@a_llama]

bruh. invaded by indians?

[@blazedinfernape886]

Perfect translation

[@cosimodamianotavoletti3513]

as an alternative method, after finding f(0)=1 substitute y=-x into the original equation: 1=f(x)f(-x)+2sin²x; f(x)f(-x)=cos(2x), setting x=π/4: f(π/4)=0 or f(-π/4)=0. Substituting the root into the original equation gives f(x±π/4)=-2sinx*sin(±π/4)=∓√2sinx; f(x)=∓√2sin(x∓π/4), so f(x)=-√2sin(x-π/4)=cosx-sinx or f(x)=√2sin(x+π/4)=cosx+sinx, both valid solutions.

[@rudradevaroy1233]

Yes bro I have also applied a differential way.

[@rudradevaroy1233]

In the equation we see that if x and y are inter changed then also the equation is still same

[@rudradevaroy1233]

So we can partially difference wrt x and put x=0 then we get f'(x)=f'(0)f(x)-2sinx

[@rudradevaroy1233]

Then solving this we get f(x)=e^ax (a^2-1)/(a^2+1) + 2/(a^2+1) * (asinx+cosx) . Then after playing with the original and unidifferentiated equation we will get f'(0) ( stands for a) =f(π/2). Then putting it in the solved equation we get a^2 =1. So f(x) = cosx ± sinx

[@yatharthgupta6468]

@@rudradevaroy1233 bro I did the same and arrive at the same ans But how did you find f'(0) to substitute for the constant?

[@henrymarkson3758]

I am Indian JEE aspirant and this is an easy problem, in India we do these problems at age 6, blindfolded, one hand tied behind our back while executing a grade 5.13d rock climb.

[@anshul.singhs]

🤡

[@divyanshjoshi6260]

Kuch bhi

[@yash8055]

😂 lul

[@srijanbhowmick9570]

You nailed it !!

[@ilickcatnip]

Lmfao 😂

[@Grizzly01]

2:46 love the look and pause after the pun. Classic. 😂

[@sohamghosh1865]

This can also be done by forming a differential equation , f'(x)=lim(h->0) (f(x+h)-f(x))/h or( f(x)(f(h)-1) - 2sinxsinh)/h ,since lim(h->0) (f(h)-1)/h = f'(0) and lim(h ->0) sinh/h=1 ,the equation reduces to f'(x)=f(x)f'(0)-2sinx, which can be now solved.

[@bb2fiddler]

Homework is proven by setting x=y=pi/2 Very interesting problem! Thanks for the upload

[@dicksonchang6647]

set x=0 for the fourth equation

[@kairostimeYT]

Same result; x = 0 if fourth eqn; x =pi/2 for the general equation. (Note that we shifted x by pi/2 in third equation to get the fourth which is what caused the value of x to be different)

[@_judge_me_not]

10:11 এটি থামার জন্য ভালো জায়গা 😁😁 You summon us Indians by making such videos

[@of8155]

🤓

[@cyberman362]

f(π/2) can be -1, 0 ,+1 . So there will be 3 solutions. For value 0, you need to put x=0 in final equation carrying f(π/2).

[@agamanbanerjee9048]

For the homework : Just substitute in x=pi/2 in that equation f(x+pi/2)=f(pi/2).f(x)-2sin x Btw really liked the solution. Can you do some more functional equation videos from R->R which involves limiting concepts?

[@edwardlulofs444]

I haven't had a lot of experience with functional equation problems. I found this video very helpful.

[@ThainaYu]

Sound is more and more quiet for every clip

[@stefanalecu9532]

The sound is good enough

[@reneszeywerth8352]

Hmmm. I had a different approach (which might lead to one solution not all possible solutions): - set y=0 -> f(0)=1 - set y=-x -> f(0)=f(x)f(-x) +2 sin²x then follows f(x)f(-x)=1-2sin²x = cos² x- sin² x =(cos x + sin x) (cos x - sin x) So f(x)f(-x) = (cos x + sin x) (cos x - sin x) which gives cos(x)+sin(x) and cos(x)-sin(x) as possible solutions

[@shalvagang951]

WHAT ABOUT THE F(-X) DO YOU KNOW THE VALUE WE HAVE TO GET IT ONLY FOR F(X)

[@Reliquancy]

Functional equations are pretty cool. They sort of make think they’re like differential eq’s without the derivatives. 0th order differential eqs maybe, idk lol.

[@mattwoodphd]

Nice to see one of these where the answer isn't f(x)=1 or 0

[@shalvagang951]

MY ANSWER IS ALSO COMING 1 AND HENCE IT IS REAL NUMBER SO IT CAN BE

[@omrizemer6323]

Another method is to expand f(x+y+z) in two ways.

[@abhijeetm29]

Lmao!, looks like India has discovered Michael Penn. Now that's indeed a Good place to stop!

[@vishvajeetkumarbatule5]

Great explanation👍

[@Noam_.Menashe]

I solved this without substituting any number. Was harder than it was supposed to, but also more satisfying.

[@Xoretre]

0=y=1 so 0=1 QED, I successfully managed to break maths! At long last!

[@ashwinraj2033]

Great Video!

[@ashwinraj2033]

"That Is A Good Place To Stop".

[@thedarkknight1865]

Brilliantly framed question. Thanks for this Question.

[@Baruch785]

Well you proved that cos-sin and cos+sin are the only two possible solutions, you didn't prove these are actual solutions

[@ImaginaryMdA]

True. It's probably homework. XD

[@tenayefujaga6341]

I think we must rewrite π as π/2+π/2 then plug it in f(x+y): f(π/2+π/2)

[@user-vg1qo5gi3l]

I think that you should check your answer

[@elardenbergsousa3836]

I used y = 0, then y = -x from the second identity, we have f(0) = f(x)f(-x) - 2(sinx)^2 -> f(x)f(-x) = 1 -2(sinx)^2 Then i factored it. First i tried 1 +- sqrt(2)sinx but it didn't work Then i remembered pythagoeran theorem and it turned f(x)f(-x) = (cosx + sinx)(cosx - sinx) Well it works and we have f(x) cosx +- sinx, but i don't think it ensures it is the only solutions. Am I missing something? Thank You

[@shengliu4596]

This problem can be solved by using ordinary differential equation method. First, we get f(0) = 1 and f'(0) = A, then we do a Taylor expansion around x, that is f(x+dx) = f(x)*[1+f'(0)*dx] - 2*sin(x)*dx. Now, we need to solve this f'(x) = A*f(x) - 2*sin(x). After we get the solution, we can use f(x+pi) = f(x)*f(pi) to determine what A is (turns out A^2 = 1)

[@anshumanagrawal346]

Can you be sure that both the solutions will work? (I don't have a lot of experience with functional equations and it's probably obvious, but if anyone cares to explain, I appreciate it)

[@tanaysodha2229]

Veryfying the Solutions: put x → x+y in f(x) = sinx + cos x, we get: f(x+y) = sin(x+y) + cos (x+y); expanding using trigonometric formulae we get: f(x+y) = sinxcosy + cosxsiny + cosxcosy - sinxsiny; adding and subtracting sinxsiny we get: f(x+y) = sinxcosy + cosxsiny + cosxcosy + sinxsiny - 2sinxsiny; simplifying further: f(x+y) = sinxcosy + cosxcosy + cosxsiny + sinxsiny - 2sinxsiny; f(x+y) = cosy(sinx + cosx) + siny(sinx+cosx) - 2sinxsiny; f(x+y) = (sinx + cosx)(siny + cosy) - 2sinxsiny; f(x+y) = f(x) f(y) - 2sinxsiny. Hence verified. You can do a similar analysis for f(x) = sinx - cosx too!

[@anshumanagrawal346]

@@tanaysodha2229Yeah, I know you can verify it, but I'm curious about is how he was sure the steps he had done were enough to conclude the solutions worked without verification

[@vrj97]

@@anshumanagrawal346 Yeah I don't think so, I think technically he would've had to verify it as well.

[@yatharthgupta6468]

Actually the answer is coming out to be (1/c^2+1)(2c cosx -2sincx + (c^2-1)(c)e^cx) where c is f'(0) Is there any way to find f'(0)?

[@internetbad3575]

There's a nice way of doing this with some calculus and differential equations, but it's not entirely rigorous as I am not that well-versed in calc and diff. eq. Please enlighten me on how to patch this up: As we've seen, for f(0) = 0 the solution is f(x) = 0. The only other choice is f(1) = 1. Remember that f'(x) = lim_h->0 [f(x+h) - f(x)] / h. So: f'(x) = lim_h->0 [f(x)f(h) - 2sinxsinh - f(x)] / h = lim_h->0 f(x) * (f(h) - 1) / h - 2sinx * sinh/h f(x) & sinx are constant, and lim_h->0 of sinh/h = 1, and lim_h->0 (f(h) - 1) / h = A, where A is some constant. This is where we assume for whatever reason that f(x) is differentiable, so that the limit may exist, and as a consequence lim_h->0 f(h) = 1 (hence why I don't entirely like this method. Maybe it can be proven that f(x) is differentiable for all x.) We get the differential equation: f'(x) = (A - 1)f(x) - 2sinx, the solution to which is pretty disgusting: f(x) = 2[(A - 1)sinx + cosx] / (A^2 - 2A + 2) + Be^(Ax - x). Note that if we set A = 0, B = 0 (again, no justification that no other solutions exist, I suck), we get f(x) = cosx - sinx, and if we set A = 2, B = 0, we get cosx + sinx as a solution.

[@yatharthgupta6468]

Did using same method Actually the answer is (1/c^2+1)(2c cosx -2sincx + (c^2-1)(c)e^cx) where c is f'(0) Is there any way to find f'(0)?

[@OscarCunningham]

Liking the intro music!

[@srijanbhowmick9570]

No way did I just solve a functional equation on my own with ABSOLUTELY NO HINTS ????????!!!!!!

[@youcefkenane8973]

At 5:59 plug x=0 in the last equation to get the homework

[@Sush]

Nice!

[@shalvagang951]

@@Sush ARE YOU MAD HOW IT CAN BE IF WE HAVE TO PUT TWO VALUES OF X AND Y AND THAT WILL PIE /2

[@Sush]

@@shalvagang951 Uh... just take the fact that: f(x + y) = f(x)f(y) - 2sin(x)sin(y) Now for x = y = π/2 you get f(π) = f(π/2)f(π/2) - 2sin(π/2)sin(π/2) Which means f(π) = [f(π/2)]² - 2 as required.

[@Sush]

The thing is the sir made a middle step of saying well if f(x + y) = f(x)f(y) - 2sin(x)sin(y) Then f([x + π/2] + [π/2]) = f(x + π/2)f(π/2) - 2sin(x + π/2)sin(π/2) This must be true FOR ALL X, since the function f mas this property. It must then work for x = 0. That is f(π/2 + π/2) = f(π/2)f(π/2) - 2sin(π/2)sin(π/2) Which concludes with the final step I put on my last comment. Now go relax a little.

[@shalvagang951]

You don’t understand what i said I said for that homework Micheal give it should be pie by 2 for both values Now you realx ok bro

[@rikthecuber]

Hw solution x=y=pi/2

[@MrWarlls]

For the first calculation, I think you write a mistake. y=0 and not y=1.

[@sashimanu]

That’s can be thought of a degenerate case of a very thin zero with infinite eccentricity 😅

[@barisdemir7896]

this question was best ... I like more solutions

[@peterquartararo3249]

fascinating.

[@ImaginaryMdA]

This was neat!

[@noumanegaou3227]

There is a countable and compact set in same time?

[@TedHopp]

So in the end, you solved the homework problem for us! 🎉

[@LetsbeHonest97]

So, I did this problem and my doubt was the coefficient of cosx and sinx.

[@wellingtonbalmant5965]

Put zero on x+pi equation.

[@srinjoyganguly3650]

Will use of calculus make the problem easier to solve ?

[@heliocentric1756]

y=1 (zero) And that's a good place to stop.

[@Mohamed.Soltan1991]

Wonderful 💖💖💖

[@yoav613]

I found the solution by guessing but this solution is great!!

[@hashimabbas3977]

Hi. From 🇮🇳

[@matteoanoffo1447]

With y=-x i found f(x)=√cos(2x), it's possible?

[@matteoanoffo1447]

I notice that it's R→R so it's not a valido solution

[@olau5478]

@@matteoanoffo1447 why is it not valid? because cos(2x) can be < 0?

[@matteoanoffo1447]

@@olau5478 yes because the Dominio Is R and √cos(2x) doesn't exist for Every number

[@thedarkknight1865]

f(x)+f(-x)+2sin² x=1 How can you get that function, when you don't know the value of f(-x)

[@matteoanoffo1447]

@@thedarkknight1865 1-sin²x=cos(2x) and since cos Is an even function f(-x)=f(x), so f(x)f(-x)=f²(x)=cos(2x)

[@har011]

317th viewer from India And very old subscriber 😁

[@mcwulf25]

Good one

[@abhigyakumar3705]

Great vid

[@fonzi102]

isn't this a solution? 2cos(x+y)=2cosxcosy-2sinxsiny

[@ayoubabid8783]

This is an easy function equation problem, i can solve it just by doing some classic substitution

[@olau5478]

ok

[@shalvagang951]

@@olau5478 SO REPLY

[@alainbarnier1995]

Like it !

[@abdallahal-dalleh6453]

The homework can be done by setting x = y

[@divyanshjoshi6260]

Or putting x=y=π/2

[@davaariantara3704]

nice video, but mainly though, nice camera heh

[@fedorlozben6344]

This Is Impossible!

[@StarsManny]

Careful with the music, Michael. Once you allow the dreaded music to invade your videos it tends to spread out and dominate, like the red weed in war of the worlds.

[@stephensu4371]

y is 0，cool

[@Harshit_Pro]

1:06 you wrote 1 and said 0

[@kodirovsshik]

Nice intro

[@rrr1304]

Hii 🙏🏻🙏🏻🙏🏻

[@Thy_panda_king]

come on fellow Indians... let's invade here

[@thedarkknight1865]

Kuch dimaag se paidal ho kya Har jagah bachabazi Achi nhi lgti

[@ilickcatnip]

@@thedarkknight1865 dude :/

[@Thy_panda_king]

@@thedarkknight1865 ok uncle, jao aap apne bachon ko sambhalo..... ye sab youtube pe majak aapki bas ki nhi

[@thedarkknight1865]

@@Thy_panda_king ye har jgah apni chaap mat chora kro Sabhya log ki tarah seekho aur chalte bano, tum log majak bna dete ho mere desh ka

[@Thy_panda_king]

@@thedarkknight1865 aapka dimag kitna viksit hai wo iss line se dikh gaya..... bhai desh hame bhi pyara hai... lekin me apne ghar ke 4 diwaar se bahar ka life janta hu.. choro kya bole abb

[@flabbypenguin]

Nice thumbnail .

[@harrymattah418]

Vocal fry at its best. Videos may be as excellent without this US vocal habit.

[@divyansh_19]

IndianJEE aspirants!! Assemble!!

[@ilickcatnip]

Lol

[@rikthecuber]

Assembled! Ready for durther command.

[@crazy4hitman755]

I can’t believe I solved a functional equation

[@mithutamang3888]

LIKE A VIKRAM BATRA RIP 😭😭😭!!! 😁😁👍👍

[@Miyamoto_345]

I can understand but it is not really relevant in this math channel

[@ilickcatnip]

@@Miyamoto_345 true

[@yashgupta7995]

What with the flag?? It's a damn math problem.

[@sharathchandra1029]

It's a marketing strategy i think most of his viewers are indian and we watch anything that mentions indian in a video

[@ilickcatnip]

Well he had the Mexico flag in one video and the Canadian Maple leaf in another. What's the problem? A nation's flag is meant to represent it 🤷‍♂️ Stop charging fellow KZbinrs with stupid arguments.

[@blackmuskveetandoor2487]

@@ilickcatnip true

[@juyifan7933]

Thats the source of the problem. Whenever he does a math contest problem he uses the flag of the country where it came from. People are butthurt for everything nowadays. Seriously, complaining about a flag!

[@AmirHossein1986]

sound quality,bad.video quality,bad. bye.

[@adiaphoros6842]

Your formatting: bad Bye