Solving the Functional Equation f(x+y)=f(x)f(y)
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- Опубліковано 10 сер 2021
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Me: sees the thumbnail , my subconscious mind: a^x!
Me too
Once you reach f(x) = m e^(c x), you can use f(0) = 1 to see immediately that m = 1 is the only (non-zero) choice. :)
That's right!
This style of problems is my favorite, not insanely complex but just tricky enough to be intriguing. Great job!
Thank you very much! 💖
yes
Directly differentiate both sides taking y as constant you will get f'(x+y)=f'(x)f(y) Now put x=0 and y=x you will get f'(x)=f'(0)f(x) now proceed with the video
Nice!
nice 👍😊
YES 👍
@@SyberMath I don't like these because it's NEVER clear..is y a dummy variable like x or is it another function of x? You didn't say.
@@SyberMath Why in hell would anyone think kf doing it this way..come on no one would ever thonknof all this convoluted nonsense..surely there's another way that's more intuitive and clear..and smarter?
It's like re-watching your favorite movie: I know where it's going but it's still fun to watch it get there. Great video!
A nice way to put it! Thank you! 🤩
I enjoy this problem because it can even be offered to strong algebra students, who can be challenged to think deeply about the operations.
Set y = 0... f(x+0) = f(x) • f(0). So either f(x) = 0 or you can divide by f(x) to get f(0) = 1.
Set y = 1... f(x+1) = f(x) • f(1). So f has a constant factor of change, whatever value f(1) is.
A function with a constant factor of change - that's what exponential functions are all about!
So f(x) = 1 • f(1)^x.
(One can rewrite the value f(1) other ways, say, b or e^k.)
well done Syber, thanks for sharing
You're welcome! Thank you!
f(x + y) = f(x) * f(y)
First, by exchanging x by y, we get exactly the same expression. This allows us to conclude that the expressions for f(x) and f(y) have exactly the same form. Also, making x = 0, it follows that f(y) = f(0) * f(y), that is, f(0) = 1.
Differentiating both sides with respect to x (using the product rule on the right side of the equality):
df(x+y)/dx = f(x) * df(y)/dx + f(y) * df(x)/dx
Using the starting equation:
df(x+y)/dx = [ f(x + y)/f(y)] * df(y)/dx + [f(x + y)/f(x)] * df(x)/dx
[1/f(x + y)] * df(x+y)/dx = [1/f(y)] * df(y)/dx + [1/f(x)] * df(x)/dx
Since f(y) is not a function of x, df(y)/dx = 0. Therefore:
[1/f(x + y)] * df(x+y)/dx = [1/f(x)] * df(x)/dx
Since x and y are independent, the only way for equality to hold is if both sides of the equation are a constant (k). So:
[1/f(x)] * df(x)/dx = k
df(x)/f(x) = k * dx
Integrating both sides, we get:
f(x) = A e^kx
Since f(0) = 1, we have A = 1. Therefore, f(x) = e^kx.
Thankyou so much I had been searching for same time for about 30 min.😊
Wow how a functional equation ends up with the definition of derivative 😍
yes 🥰 he used the definition of derivative
Thank you I actually was trying to find how to solve such problems.
The highlight is f(0)=1, which is used to express f’(0). SO COOL👍
I never knew about these types of problems. Functional equations are my new favourite thing. You could easily see that e^Cx is a solution for f(x), but doing it rigorously revealed that it is the only solution (apart from the trivial case).
Thanks a lot syber so so much syber OMG i can't believe. How much u helped me!!!!!!
Good to hear! You’re welcome!
f(t) = C^t ... And C becomes the definition ... Thank you I now understand how you are factoring this.
nice video AND approach, I always learn something watching you! Another way: y=x both side. Then, u=2x only left side. Derive both sides using leibnitz notation and Chain rule. Solve ODE. Greetings from Argentina!
💖
I love it, you survived me 😁
Wow that's a really cool solution :0
I've only known this problem with the given that is continuous, not differentiable. It's a much longer solution if we don't assume differentiability.
@@carlgauss1702 I mean you can prove it's always positive even quicker. But what does that matter? It's one step in the process, yes, but there's a lot more to do to prove f=exp
@@carlgauss1702 that does not imply that f=exp. Unless you define exp in that way of course. When you define it as its power series (xⁿ/n!), you have to do a lot of steps. I know this because this was an actual question on my real analysis exam last semester.
@@carlgauss1702 what? Is there anything wrong? Would you like to say something? Because I know for a fact that I'm correct
@@carlgauss1702 you can't just back out of an argument by commenting smh when you're starting to lose. Extremely childish imo
Thank you for the kind words, Skylar! 💖
We can let e^c=b for some b>0 and then e^cx=(e^c)^x=b^x
Which we can combine with the solution f(x)=0 to say the solution is
f(x)=b^x for b greater than or equal to 0
Do we really need a b value that is greater than or equal to 0? A negative b value will satisfy it, too. Thus, no need to limit b as positive or 0.
@@fatihsinanesen b has to be positive since e^c is always positive
@@Happy_Abe We don't need any e's in the solution. Just a constant is enough.
@@fatihsinanesen I’m going off the video and in it he uses e^c which must be positive
Furthermore if we have f(x)=a^x where a
@@Happy_Abe You're right. x=1/2 was a good counterexample. Thank you. 👍🏻
I love this problem so much, I have not studied functional equations yet, so after seeing problem I was like "The answer must be a^x" but the hell can I prove that. And at each step I was like "wow, that's genius" and towards the end I caught on. The solution was pretty interesting too👏
Good to hear!
Well a in a^x is just any constant,
How about you put e^C as 'a' see both are same, it's just we in our functional equation directly :)
Hello, @SyberMath I deduced that f(x)=0 for all real x. This, being a constant function, is differentiable for all real x.
Hi! Is this the only one?
@@SyberMath Yes. f(x) can be a constant function. It can be the constant function given by f(x)=0 for all x, or it can be the constant function given by f(x)=1 for all x. If f is assumed to not be constant, then, f(x)=e^(cx) as shown in the video.
I think it can get weird of f doesn't have to be continuous. Like I think you have to choose f(E_k) for a basis of the reals over the rationals or something, and the rest is uniquely determined. It's been a while since I looked at that though.
Thank you for the quality of your videos. What's the name of software you are using?
Notability
Thanks!
In fact the solution of this functional equation is obvious: f(x) = exp(x), what can be easily verified by substitution exp(x) to the equation. But the result may be achieved by more strong approach.
1) f(x) never can be negative, what follows from partial case y = x, so f(2x) = f(x)^2 > or = 0
2) f(x) is not 0 for either value of argument, except of the trivial case f(x)=0 for all x.
This follows from modified equation
f(z) = f(x) • f(z - x)
So if f(x) = 0 for some x, then f(z)=0 for all z
3) From anothe partial case y=0 we can find f(0):
f(x) = f(x) • f(0)
so f(0) = 1 except the trivial case
4) let's logarithm both sides of equation. We may do this because f(x) > 0
ln(f(x+y) = ln(f(x)) + ln(f(y))
Denoting g(x) = ln(f(x)) we'll get
g(x+y) = g(x) + g(y)
and the only solution of this equation is a linear function f(x)=x
classical properties of exponential function as for logarithm (reciprocal function of exponential) f(xy) = f(x) + f(y)
That's only one way of the implication. "if f = exp(ax), then f(x+y)=f(x)f(y)"
We essentially want to prove the converse: "if f(x+y)=f(x)f(y), then f=exp(ax)"
@@skylardeslypere9909 One definition of exponential function is the function which is the inverse of ln (logarithm)
ln (exp a exp b) = ln exp a + ln exp b = a + b.
then exp(ln((exp a exp b)) = exp(a+b)
exp(a).exp(b) = exp(a+b)
4:57 Do you c what I c?
That's an interesting way to proceed! In fact you don't need to know that f is differentiable to get to this result. Any continuous solution f will be of the form a^x (or constant 0). As for non continuous functions... it will depend on the axioms of mathematics that you chose to use. There are universes of maths where you can remove the hypothesis "f is differentiable" and the solutions will remain exactly the same!
That's right!
f(x) = 1 for all x, also solves this equation. So there are two trivial solutions.
f(x) = e^cx
covers that case with c = 0
f(x) = e^(0 x) = e^0 = 1
Thanx sybermath!
I have two questions:
1. why did you replace y with 0 at the beginning?
2. The solutions could also be a^cx and not only e^cx, so why did we get only e^cx?
a^cx=e^cx*lna, donc c’est la même chose!
The base does not matter. They are equivalent
Nice way to solve this.
Nice problem
I love using calculus on functional equations
Great!
in the end of the video, why can you conclude that f(x) is an exponential function? while at the end you only get the conclusion that f(x)=e^(cx)?
What kind of softwere do you using during solving problems.
I am asking the writing technique on the black board.
Apple Pencil on iPad using Notability as my blackboard
@@SyberMath Danke Schoen !
Very helpful
Glad you think so!
We don't need e, indeed. f(x)=c^x is the solution for any arbitrary constant c>0. Let's check: f(x+y)=f(x)f(y) -> c^(x+y)=(c^x)(c^y) -> c^(x+y)=c^(x+y)
c >= 0 and you will wrap in the other solution f(x) = 0 as well.
السلام عليكم هل من ممكن شرخ امثله عن تطبيقات حول المشتقات وجزيل الشكر لكم
After seeing the live chat, I realize that I messed up and did not follow the plan. Today should be a radical equation so we'll have to switch today and tomorrow and we'll do the radical equation tomorrow! Sorry for the confusion
EDIT: The post has been edited to reflect the changes!
ua-cam.com/users/SyberMathcommunity
It's no big deal, you have great videos! Greetings from Russia))
@@predatorymink3400 Thank you! 😊
@@SyberMath Hello sir, great video sir, but i want to ask, in the end of the video why you can conclude that f(x) is an exponential function? while you get in tge end is f(x)=e^(cx)
Very nice
log both sides, linear cauchy functional equation. which implies that f is exponential or 0
grate job master 🔥🔥🔥💪☕⚓👑
Thank you!!! 😊💖
In general, solution is f(x)=a^cx, where a>0.
But a^x can also be a solution where a >0???
wow nice !
Thank you! 🤩
Why didn’t you take f(x)=1 in addition to taking f(x)=0 when accounting for constant functions ?
Is there other function f other than any exponential (not necessarily base e)
f(x)=0
Nice proof! Maybe not as rigorously, but from f(x+a+y-a)=f(x+y) one sees immediately that f(a)=f(-a)^-1. Only f(x)=u^x, for any u, satisfies this.
This is not true. Let f be ANY function defined on [0,infinity) with f(x)>0 for all x in [0,infinity) and f(0)=1. Now extend f to (-infinity,0) by setting f(-x)=f(x)^-1. Then this function has your stated property that f(a)=f(-a)^-1.
@@willnewman9783 Right. Thanks.
I'm a little confused because
f(x) = k^(cx)
gives us
f(x+y) = k^[c(x+y)] = k^(cx+cy) = k^(cx)•k^(cy) = f(x)•f(y)
Hence, k only needs to be greater than 0 and not the specific value k=e
Very nice problem
Thanks
I was imagining that I need to get solutions without using calculus knowledge:-). So
1, obtain f(0)=1 by using the original equation enforcing x=y=0;
2, obtain log f(0)=0 by applying log to both sides of result in step 1,
3. Assuming g(x) = log f(x) then we know g(x+y)=g(x)+g(y); and g(0)=0;
4. Now apparently g(x)=cx can be a solution. Or f(x)=a^cx, with a>0.
Putting x=y=０in the eq. yields f(0)=f(0)², so f(0)=0 or 1. f(0)=0 is possible.
f(x)=0 is one of solutions. Assuming that f is a real-valued function, we need to show f(x)>0
for all x if logarithm of f is required.
How do you solve it without the assumption that f is differentiable?
good bro
Great, exciting solution. Suggestion problem: find all complex x, y with non-zero imaginary part such that x^y is real. I have tried by myself and found only one solution i^i. Is it only solution?
i^ki would also work, I think
@@SyberMath , yeah, it also works. And what about x, y with also non-zero real part? I was aimed to find this type of solutions, to be honest. In my solution, I found a formula, that generate infinite amount of solutions, but it didn't work and I don't know why. If you interested, I can share my solution
So what if f needn't be differentiable? I remember I read somewhere that it's not possible to give any more solutions concretely, but using the axiom of choice, it can be proven that non- continuous solutions do exist. Crazy shit... :D
F(x)=1 is a solution.
F(x)=f(y) =1=f(x) f(y)= f (x+y)
yess i solved it
There is another solution you did not talk about. Which is f(x) = const, not just f(x) = 0, referring to m /= 0 and c == 0
A constant different than 0 or 1 is not a solution since CxC not equal to C
Actually, no kidding but, this is how you could potentially derive the Boltzmann probability density function from probability laws.
Wow!
f(x) = e^x f(y) = e^y
Any constant to the x power will do
This is covered by the values of C, if for example you pick C = ln2, the function e^(cx) becomes 2^x.
m = (+/-) exp(k) and thus it can't be equal to 0
🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥
Ln(f(x+y))= ln(f(x))+ ln (f(y)) et f est continue ,donc ln•f est linéaire.
ln(f(x))= ax, et donc f(x)= e^ax
Oui, main ce n'est pas necessaire à employer ln du tout. (Regardez mon autre commentaire.)
x=y=0 gives f(0)=f(0)². Thus f(0)=0 or 1. If f(0)=0, then f(x)=f(0+x)=f(0)f(x)=0.
If f(0)≠0, then f(0)=1. f(x)=f(x/2+x/2)=[f(x/2)]²≧0. Can we exclude the case f(x)=0?
For sybermath why the base of exponent must be e???? for every positive real number a^(x+y)= a^x×a^y 2^(x+y)=2^x×2^y (1/3)^(x+y)=(1/3)^x×(1/3)^y and so on why they dont came out in your solution???
a^x = e^(ln(a)x), so they come out for the good constant c. Actually the base is not e but e^c in the final solutions
Yes
For example, 2^x is the same thing as 1•e^(ln(2)•x). So all positive real number already came out in his solution m•e^(cx)
e for everything! 😁
@@SyberMath oh lol
f'(x)=f'(0)*f(x) then f'(0) does not exits?
why not?
@@SyberMath Сорнян показалось f'(0)=f'(0)*f(0)
How about log functions ?
No
Woah thats one of those i solved.
Nice!
f is an exponential function
so, the only solution is f(x)=e^(cx) ?
K^x behaves this way.
f(x)=antilog(x)
f(x) = a^(b x)
Why is that not a solution?
f(x+y) = a^(b (x+y)) = a^(b x + b y) = a^(b x) * a^(b y) = f(x) * f(y)
No? Your solution seems to unnecessarily force a = e
Your solution had: ln( |f| ) = c x + k
Divide both sides by ln(a), where a and hence ln(a) are constants
ln( |f| ) / ln(a) = (c/ln(a)) x + (k / ln(a))
log_a( |f| ) = b x + d ---- replacing constants (c/ln(a)) = b and (k / ln(a)) = d
Now simplifying the same way as you did before:
f(x) = a^(b x)
f(x) = a^(b x) = (e^(ln(a))^(b x) = e^(ln(a) b x) = e^(c x)
So those two are equivalent. Yours might be better since it combines both the constants into one term, but as a potential family of functions, it just seems more obvious to me to say,
f(x) = a^(b x)
Actually, a better answer is:
f(x) = a^x
Now that covers all the functions in one constant and eliminates the arbitrary constant e in your solution.
Actually your solution is:
f(x) = 0 *OR* f(x) = e^(c x)
*while* f(x) = a^x where a >= 0 covers all cases, since a can take the value 0, giving f(x) = 0 and a can take the value 1, giving f(x) = 1
*while* in your case c would have to take on the value of minus infinity to cover the case for f(x) = 0, so one can actually argue that the solution f(x) = a^x is the better one, since you don't need to special case the f(x) = 0.
What if m=-1?
does that make a difference?
the first comes to my mind is e
But other basses also work
@@aashsyed1277 of course, but in general it's an exponential function
e for everything! 😁
It's exp(x) function
👍👍👍👍👍👍👍👍👍👍👍
Sogh..isnt it another way to do this? A way that is more intuitive. .you dont even know if y is a function of x or just a dummy variable..that matters..yea you can assume its one and then the other case but you shouldn't have to do that..
We're going to be solving to f of x. Too easy, if f of x+y equal f of x times f of y, f of x equal f of x+y over f of y. 😂 And this is the answer LOL