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.
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
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.
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)
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.
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?
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
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.
@@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.
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.
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
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)
@@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
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)
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.
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.
@@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
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 UA-camrs with stupid arguments.
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!
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
Translation op👌🏻
@SNEHANSH MAHARAJ Unfortunately, this is a trade secret that cannot be shared
@SNEHANSH MAHARAJ I have written a truly marvelous explanation of that trick, which this margin is too narrow to contain
bruh. invaded by indians?
Perfect translation
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.
Yes bro I have also applied a differential way.
In the equation we see that if x and y are inter changed then also the equation is still same
So we can partially difference wrt x and put x=0 then we get f'(x)=f'(0)f(x)-2sinx
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
@@rudradevaroy1233 bro I did the same and arrive at the same ans
But how did you find f'(0) to substitute for the constant?
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.
2:46 love the look and pause after the pun. Classic. 😂
Homework is proven by setting x=y=pi/2
Very interesting problem! Thanks for the upload
set x=0 for the fourth equation
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)
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.
🤡
Kuch bhi
😂 lul
You nailed it !!
Lmfao 😂
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).
No?? Cus that leads to -1=-2 which is obviously a contradiction
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?
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
WHAT ABOUT THE F(-X) DO YOU KNOW THE VALUE WE HAVE TO GET IT ONLY FOR F(X)
I haven't had a lot of experience with functional equation problems. I found this video very helpful.
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.
Another method is to expand f(x+y+z) in two ways.
Great Video!
10:11 এটি থামার জন্য ভালো জায়গা 😁😁
You summon us Indians by making such videos
🤓
Great explanation👍
Sound is more and more quiet for every clip
The sound is good enough
Brilliantly framed question. Thanks for this Question.
Nice to see one of these where the answer isn't f(x)=1 or 0
MY ANSWER IS ALSO COMING 1 AND HENCE IT IS REAL NUMBER SO IT CAN BE
I think we must rewrite π as π/2+π/2 then plug it in f(x+y): f(π/2+π/2)
Lmao!, looks like India has discovered Michael Penn. Now that's indeed a Good place to stop!
This was neat!
Well you proved that cos-sin and cos+sin are the only two possible solutions, you didn't prove these are actual solutions
True. It's probably homework. XD
"That Is A Good Place To Stop".
0=y=1 so 0=1
QED, I successfully managed to break maths!
At long last!
I solved this without substituting any number. Was harder than it was supposed to, but also more satisfying.
Liking the intro music!
fascinating.
At 5:59 plug x=0 in the last equation to get the homework
Nice!
@@Sush ARE YOU MAD HOW IT CAN BE IF WE HAVE TO PUT TWO VALUES OF X AND Y AND THAT WILL PIE /2
@@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.
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.
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
I think that you should check your answer
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
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)
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!
@@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
@@anshumanagrawal346 Yeah I don't think so, I think technically he would've had to verify it as well.
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)
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.
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)?
Wonderful 💖💖💖
Will use of calculus make the problem easier to solve ?
No way did I just solve a functional equation on my own with ABSOLUTELY NO HINTS ????????!!!!!!
For the first calculation, I think you write a mistake. y=0 and not y=1.
That’s can be thought of a degenerate case of a very thin zero with infinite eccentricity 😅
Like it !
There is a countable and compact set in same time?
this question was best ... I like more solutions
Hi.
From 🇮🇳
Put zero on x+pi equation.
Good one
So in the end, you solved the homework problem for us! 🎉
With y=-x i found f(x)=√cos(2x), it's possible?
I notice that it's R→R so it's not a valido solution
@@matteoanoffo1447 why is it not valid? because cos(2x) can be < 0?
@@olau5478 yes because the Dominio Is R and √cos(2x) doesn't exist for Every number
f(x)+f(-x)+2sin² x=1
How can you get that function, when you don't know the value of f(-x)
@@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)
Hw solution x=y=pi/2
317th viewer from India
And very old subscriber 😁
isn't this a solution?
2cos(x+y)=2cosxcosy-2sinxsiny
y=1 (zero)
And that's a good place to stop.
Great vid
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)?
So, I did this problem and my doubt was the coefficient of cosx and sinx.
1:06 you wrote 1 and said 0
I found the solution by guessing but this solution is great!!
This is an easy function equation problem, i can solve it just by doing some classic substitution
ok
@@olau5478 SO REPLY
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.
The homework can be done by setting x = y
Or putting x=y=π/2
nice video, but mainly though, nice camera heh
y is 0，cool
This
Is
Impossible!
Nice intro
Hii 🙏🏻🙏🏻🙏🏻
Nice thumbnail .
come on fellow Indians... let's invade here
Kuch dimaag se paidal ho kya
Har jagah bachabazi Achi nhi lgti
@@thedarkknight1865 dude :/
@@thedarkknight1865 ok uncle, jao aap apne bachon ko sambhalo..... ye sab youtube pe majak aapki bas ki nhi
@@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
@@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
Vocal fry at its best. Videos may be as excellent without this US vocal habit.
IndianJEE aspirants!! Assemble!!
Lol
Assembled! Ready for durther command.
I can’t believe I solved a functional equation
LIKE A VIKRAM BATRA RIP 😭😭😭!!! 😁😁👍👍
I can understand but it is not really relevant in this math channel
@@Miyamoto_345 true
What with the flag?? It's a damn math problem.
It's a marketing strategy i think most of his viewers are indian and we watch anything that mentions indian in a video
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 UA-camrs with stupid arguments.
@@ilickcatnip true
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!
sound quality,bad.video quality,bad.
bye.
Your formatting: bad
Bye