A Functional Equation from Taiwan's International Mathematics Olympiad Training.

In this video, I solved a functional equation from Taiwan. The main Idea was to reconfigure the function into a Cauchy functional equation. The rest was easy to see.
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[@glorrin]

I believe there are infinitely more solutions. But they are easy to represent f(x) = x + c with c in R when k = 1 then k-1 is 0 and there is no restrictions on c so any c should do the trick.

[@Dshado]

That's what I thought as well. Given k=1, c is any real number

[@_HAKSOZ]

Yea I agree

[@EnzoMariano]

It is correct

[@JourneyThroughMath]

I agree as well

[@xavigb5991]

Then, where is the error? 🙂

[@jirisykora9926]

Oh my god, that's why i love math! An amazing teacher you are! Please, continue your excellent work!

[@scottparkins1634]

If you assume f is differentiable then there is a quick proof that f must be linear in x. Proof(ish). Since f(f(x+y)) = f(x) + f(y), this implies f(x+h) + f(0) = f(x) + f(h) for every x, h (to see this, note that x+h = (x+h) + 0 to express the initial equation in two different ways, and equate them). Rearranging this gives f(x+h) - f(x) = f(h) - f(0). Therefore taking first principles definition of the derivative, dividing the last equation by h and taking h to zero gives f’(x) = f’(0) for every x. So the derivative of f must be constant everywhere!

[@user-nw4sv4ki3g]

never stop teaching !!!

[@dannieee333]

such an amazing teacher😭❤

[@TheIndianPrince_]

THANK U SIR!!.. U are an amazing teacher

[@Ron_DeForest]

Ok. Can you do a video on what a Cauchy Functional Equation is and how it’s derived? That be handy.

[@ibrahimmassy2753]

x+c is also a solution where c is real

[@PrimeNewtons]

Yes. Now that I think about it

[@JourneyThroughMath]

I actually took notes for this video

[@ShaunakDesaiPiano]

3:51 meme material

[@harshplayz31882]

Got confused after 5 min Then again watched it very carefully😅

[@motivationvinod4893]

3:46 f(x+y) = f(x + y) +2f(0), Does it can be possible sir ?

[@9adam4]

I solved it myself and jumped to the end. Why isn't any x + c a solution?

[@brenobelloc8617]

Thanks you so much Prime Newtons. I enjoy deeplly all your problems. In the case of funcional equations, im having troubles to understand the deep of this at all, but i will try again in home to study it. Please if you can and think thats is good idea, do more problems of this family.

[@Dshado]

Hi, I wrote an explanation in the comment by donwald3436, which could help you understand the intrecacies a little bit more

[@brenobelloc8617]

@@Dshado hi fellow. I really apreciate your comment and your explanation. I will check and practice it for sure. 🤗

[@Dshado]

@@brenobelloc8617 hi, if you get stuck just messege here, i might be able to help :D

[@mircoceccarelli6689]

👍👍👍

[@sobolzeev]

Remark: f need not be continuous everywhere. It can be discontinuous on a set of measure zero (Lebesgue function) or on a meagre set (Bair function). It is a general question: what additional property should a function have, so that the only solution to the Cauchy equation is a linear function? One cannot drop continuity entirely, as there are counterexamples.

[@admkaan504]

cool necklace

[@collapsingwavefunction_.3356]

Who else got lost at step 3? 😅🤦🏽

[@oliverkoller5468]

x + y just represents some input into the double composition. So, if you replace all the y terms in line 2 with x + y, you get line 3

[@rayyt5566]

I think step 3 means that he’s drawing a conclusion: since f(f(x)) = f(x) + f(0) and f(f(y)) = f(y) + f(0) ➡️ f(f(whatever)) = f(whatever) + f(0)

[@collapsingwavefunction_.3356]

@@rayyt5566 oooooooo

[@collapsingwavefunction_.3356]

​@@oliverkoller5468 gotcha thx

[@rayyt5566]

@@collapsingwavefunction_.3356 you’re welcome : )

[@Zzz12367]

You could assume it’s a differentiable function and differentiate it wrt to x keeping y as a constant, then keep y=0 and solve, you can get the values of constant later on by comparison

[@Dshado]

You can't assume it is differentiable from it being continuous.

[@skwbusaidi]

f(f(y))= f(y)+ C , c=f(0) Replace f(y) with x f(x) = x+ c f(f(x+y) = f(x)+f(y) f(x+y+c) = x+c + y+c x+y+c+c = x+c+y+c which is always true f(x) = x+c

[@ben_adel3437]

Couldn't we just said that f(x) is gonna be a function like ax+b after we found f(f(x))=f(x)+f(0) which would mean that f(0) is a constant and then just found the answer normally

[@chaosredefined3834]

How do you know that f(x) will be of the form ax+b?

[@ben_adel3437]

@@chaosredefined3834 well f(0) is constant and if f(x+y)=f(x)+C then just like in the video you can find that its in that form and its also easier because its skips a lot of steps

[@chaosredefined3834]

@@ben_adel3437 If f(x+y) = f(x) + C, then f(y) = f(0) + C when x = 0, so f(y) is a constant.

[@ben_adel3437]

@@chaosredefined3834 well yeah you made both sides constants which wouldn't give you info about the function

[@chaosredefined3834]

@@ben_adel3437 You said that we get f(x+y) = f(x) + C. For this to be true, it needs to be true for all values of x. So, it needs to be true for x=0.

[@donwald3436]

How do you justify just setting y to 0 then saying that is equivalent to the original function?

[@Dshado]

Its a functional equation so its true for ANY value of y, so he substituted a value of 0 instead, since it needs to be true for any and all values

[@donwald3436]

@@Dshado Okay so I set y to 0 and x to 0 so f(f(0)) = f(0) + f(0) and somehow that is just as general as the original function?

[@Dshado]

@@donwald3436 its not as general, but it still is true. Giving the x and y values, does not make it NOT true

[@donwald3436]

@@Dshado That's my point, the problem said find all solutions.

[@Dshado]

​@@donwald3436 Ok, I'm picking up what you're putting down now. The problem sais find all solutions for f not x,y. x,y are just random variables. So let's say we have f(x)+f(y)=5, and we are looking for f. so we can substitute x=y=0 into the same equation, without changing f. Since f is true for ANY and ALL values of x,y. 2f(0)=5, f(0)=2.5, for any value of x or y. We found something about the FUNCTION without knowledge of variables. Knowing f(0), we can get f(x) using f(x)+f(0)=5, f(x)=2.5 Since f is the same, f(y)=2.5 as well, making the above FUNCTIONAL equation true for all x and y. Using the functional equation (with answer) from the video: f(f(x+y))=f(x)+f(y). using f(x)=x+c (f(x)=kx+c ,k=1) f((x+y)+c)=x+c+y+c ((x+y)+c)+c=(x+y)+2c x+y+2c=x+y+2c which means this solution is correct. What the values of the variables are doesn't matter in a functional equation, because the given equation is true for ANY and ALL x,y pairs. I hope this made it clearer

[@guest2649]

i smart f(x) = x

[@surojitray2639]

Pl.try to write a little bold so that it can be followed clearly I am from india.