A Functional Equation f(x+y)=f(x)+f(y)-xy

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[guisav]

Great manipulation! Two comments: keep in mind that g(x) = ax only if you either assume x is rational, or do some other continuity assumption. Otherwise, there are other (pretty wild) solution.
Anyway, here is my way to solve the problem (roughly speaking):
1) Assume y=0 and you will see that f(0)=0
2) Assume y=-x and (together with f(0)=0) will show that f(x)+f(-x)+x^2=0
3) Now we know that each function is the sum of an odd and an even functions. I called them f(x)=o(x)+e(x). Replace them in 2): o(x)+e(x)+o(-x)+e(-x)+x^2=0
4) Since e(-x)=e(x) and o(-x)=-o(x), o(x)+e(x)-o(x)+e(x)+x^2=0, 2e(x)+x^2=0 so e(x)=-x^2/2, or f(x)=o(x)-x^2/2
5) Replace in the original equation, and you will see everything disappear leaving o(x+y)=o(x)+o(y)
6) It was already discussed that, given some assumptions, this mean o(x)=ax, so f(x)=ax-x^2/2

[f-th]

Proving g(x) = ax is a stndard problem you migt get in real analysis.
Let g(1) = a. Then for natural n, g(n) = g(1+1+...+1) (n copies of 1) = g(1)+g(1)+...+g(1) (by additivity) = ng(1) = an.
For negatives integers, we already showed g(-x) = -g(x). So for any integer n, g(n) = an.
For rationals p/q, ap = g(p) = g(q*p/q) = q*g(p/q) by additivity, so ap = q*g(p/q), so g(p/q) = ap/q.
So for rational x, g(x) = ax. The equality extends to all reals via continuity.

[verystrongseju]

we learned this one in high school class
its actually really easy if you use some skills and memorize the equation.
you can solve it in like 10sec
If f(x+y)=f(x)+f(y)+ax+b,
f(x)=1/2ax^2+cx-b
and you can put whatevers offered in the question.
for example, f(1)=0
something like that.
so in this question, the equation is
f(x+y)=f(x)+f(y)-x^2
so you can easily know that f(x)=-1/2x^2+ax
according to the formula
we call this the '함수방정식'
its a skill to solve questions faster.

ps. sorry if you couldnt understand what i was saying. Poor english skills ㅠㅠ
if you have anything to ask, please feel free to ask! thanks

[justinnitoi]

At 5:58 you can easily prove that g(x) is linear by setting y = 1 to get: g(x+1) - g(x) = g(1). g(1) is constant and the slope of g(x)

[engjayah]

Another solution as follows:
set y = 0; then we get f(0) = 0
set y = 1; then we get f(x+1) = f(x) + f(1) - x; and rearranging terms
f(x+1) - f(x) = f(1) - x; now this is a sequence/array in x;
keep substituting x = 1, and then obtain the summation from which we arrive at
f(x) - f(1) = (x-1)f(1) - x(x-1)/2; by simplifying
f(x) = (f(1)+0.5)x - 0.5x^2 = ax - 0.5x^2; where a = f(1) + 0.5 = constant;

[Endymion]

Let y=0, then f(0)=0. Let y=-x, then f(0)=0=f(x)+f(-x)+x^2. Then fe(x) (even part of f) is x^2/2. For the odd part, consider 1/2(f(x+y) - f(-x-y)) =fo(x+y)=fo(x) +fo(y), which is a Cauchy functional equation, then fo(x) =cx. Then f(x) =fe(x) +fo(x) =cx - x^2 /2, under assumption f is continous

[trnfncb]

Since there is a nonlinear interference term, the easiest is to try f(x)=ax^2+bx, which, upon substitution, leads immediately to a=-1/2 and b=any.

[danilonascimentorj]

you can set y=h and rearrange the terms such as f(x+h)-f(x) = f(h)-hx. Then dividing all by h and applying the limit when h->0. We have f'(x) = a-x. Integrating both sides f(x) = ax-x^2/2+c, we know that f(0)=0. Then f(x) = ax-x^2/2.

[agni]

Man missed so many of your vids lately, but glad I didn't miss this...one of my most favourite problems. Btw You could've totally done this by finding f(0) then proving f'(x) is linear and finally integration to get a quadratic as f(x)

[thierryvieville]

Yes thanks to SyberMath clever manipulation, with g(x) = f(x) + x^2/2 the functional equation becomes the Cauchy
g(x+y) = g(x) + g(y)
additive function equation, which is of the form g(x) = u x as soon as continuous at x = 0, or if g(x) is only taken on fractional numbers, otherwise there also are discontinuous solutions, as alluded in the video, that have ``not only “regular” properties but some surprising or even intriguing ones´´ (see Discontinuous additive functions: Regular behavior vs. pathological features Claudio Bernardi, the paper is avialbl;e on line)

Very interesting. Thanks a lot !

Note: As an alternative derivation, we may also have 1st looked for regular solutions, by differentiation, as Random Jin suggest, because :
f(x+y) = f(x) + f(y) - x y
yields after differentiation with respect to y and then considering y = 0
f'(x) = f'(0) - x
thus
f(x) = - x^2/2 + u x + v
for some constant u, while v = 0 because f(0) = 0 and we verify such function verifies the functional equation, we may write without loss of generality f(x) = g(x) - x^2/2 + u x and look for general g(x) on a simplified equation.


Note: We also would have found the same result by defining a recurrent equation considering y = 1 and use prolongation to real function, or by intuiting that because of the ``-x y´´ then a solution might be of the form f(x) = g(x) + t x^2 + u x + v for some t, u and v and identify, as partailly proposed by Quentin Chin.

Note: To obtain that f(0) = 0 you just have to set x = y = 0 in the equation and solve.

[georget]

Set x=y=0 and solve for f(0)
Set x<>0 and y=0 and find f(x).
Would that be OK?

[AliBarisa]

These problems are so nice to look at when someone else is working on them. But very difficult to work out by oneself.

[Rbmukthegreat]

Today marks my 4 month subscription! Also wonderful solution!

[iquisickis]

From g(x+y) = g(x) + g(y) I deduced that g must be a commutative operation with the addition so it must be g(x) = ax. Did I jump to a conclusion?

[별의별-hb]

I thought about it. Why can't I find the coefficient of term x?
After watching the video, I found out that it was unknown.



And my solution was to move f(x) to the left, divide y on both sides, and then take the limit of y to zero and solve it. It is using differential. And we found f(x) by integrating.

[dumplet]

Cauchy + continuous -> linear function

[angelmendez-rivera]

Challenge: Find every f on R such that f(x + y) = f(x) + f(y) – x·y everywhere and f is continuous everywhere.

Let y = 0. Hence f(x + 0) = f(x) = f(x) + f(0) – x·0 = f(x) + f(0) for every x in R. Thus f(0) = 0. Let f = g + h. Thus f(x + y) = g(x + y) + h(x + y) and f(x) + f(y) – x·y = g(x) + g(y) + h(x) + h(y) – x·y. Hence g(x + y) – g(x) – g(y) + h(x + y) = h(x) + h(y) – x·y. This is equivalent to g(x + y) – g(x) – g(y) = 0 and h(x + y) = h(x) + h(y) – x·y. Notice that g(x + y) – g(x) – g(y) = 0 is equivalent to g(x + y) = g(x) + g(y), Cauchy's functional equation, which implies that if g is continuous, then g(x) = A·x for some A in R everywhere. Hence f(x) = A·x + h(x), and h(x + y) = h(x) + h(y) – x·y, being an example of a solution to the equation. The other solutions f are found by adding g to h. Hence, we are still solving the original equation, but now we only need a particular solution.

h(x + y) = h(x) + h(y) – x·y looks very similar to (h0)(x + y) = (h0)(x) + (h0)(y) + 2·x·y, which we know for a fact from the binomial theorem is true if (h0)(x) = x^2 everywhere. In fact, (h0)(x + y) = (h0)(x) + (h0)(y) + 2·x·y implies –(h0)(x + y)/2 = –(h0)(x)/2 – (h0)(y)/2 – x·y. So we have h(x) = –x^2/2. Hence f(x) = A·x – x^2/2 everywhere for some A in R.

[andreadevescovi]

Sorry, but i belive that a=0 always:
f(x)=-(x^2)/2
In fact: f(0)=0
And if x=y: f(2x)=2f(x)-x^2
and if y=-x: f(0)=f(x)+f(-X) +x^2=0. f(x)=-f(-X)-x^2
Add theese: 3f(x)=f(2x)-f(-x). So: . So: 2f(x)=-x^2 and f(x)=-(x^2)/2

[jayantaboral]

I think the more general solution would be g(x) = ax^(2m+1), where in m=0 or any positive integer. That would give f(x) = ax^(2m+1)- x^2/2

[mohammedal-haddad]

That was a neat twist. I enjoyed this one very much.