Set y=0. f(x) = f(x)^ln(f(0)). apply ln to both sides. ln(f(x)) = ln(f(0))*ln(f(x)). Since ln(f(x)) is repeated on both sides, ln(f(0)) must be the multiplicative identity, or 1. ln(e) = 1. f(0) = e. Now set x=0. f(y) = f(0)^ln(f(y)) -> f(y) = e^ln(f(y)). From here we can easily see the function is e^k^x, as we have f(y) = e^(some value).
Sorry, but a=e^(ln(a)) holds for any number a > 0. The fact that you first take the logarithm of the number, and then substitute the result into the exponent will simply return you to the initial number. (sorry for my english)
@@shigen778 Nice observation. Yes, this is true. For any value greater than 0 this will just be an identity. If the value is less than 0, or negative, then f(y) must produce a positive value. Hence e^k^x still holds as a logical solution, as e^(negative power) will just be inversions of e, which are still > 0, and as you have stated, this equation is always true for any values greater than 0. Hence e^k^x should work as a solution.
@@ben69real Sorry, but you probably misunderstood me. I meant that just because f(y)=e^(ln(f(y))) , it doesn't follow that f(y)=e^k^y is the only solution. For example, you can try f(y)=1+y^2 and this function satisfies the condition. The functions e^x and ln(x) "destroy" when you apply them to each other. Moreover, in the notation f(y)=e^(ln(f(y))), the function is not specified explicitly, due to the fact that it occurs to the left and right of the "=" sign. If you simplify the expression on the right, you just get f(y)=f(y), which is obviously meaningless. The property x=e^(ln(x)) holds for all positive numbers, so you can't define a function that way.
@@shigen778 This is true, what I was trying to point out is that it is easy to see though that e^k^x is a solution just due to the fact that f(y) still equals e^some power, even if it does just cancel out to be an identity in that specific situation. That said, functions that don't employ values raised to even powers like x^1 or x^3 can easily be ruled out, as they'd yield negative values. Additionally, with even powered functions that would yield only positive values, we can apply the original statement to see that these functions clearly cannot work. Just from seeing that there's an ln in the original statement, something such as x^2 + 1 or anything else wouldn't work. Therefore we can see that the only one of those solutions that would actually work for all solutions is e^(something), or e^k^x as we're saying here. The thing that e^ln(f(y)) does is it just affirms the fact that e is involved, which we could pretty much already see from the ln in the in the first statement but shrug. It literally just has e^(some power), even if it does cancel out to f(y) = f(y). Alone it's pretty much useless as you've pointed out but with the original statement it is actually easy to see that that is the solution, just based off of a simple check.
The only continuous functions g (with domain all reals) which solve g(x+y) = g(x)g(y) are: g(x) = k^x (where k > 0) and g(x) = 0. x^k and kx, both with k 0 aren't solutions. k(ln(x)) has the wrong domain and isn't a solution even on the positive reals unless k = 0.
The determinant "le" is not part of the name. It's as if you were correcting someone in English "it's not "natural logarithm", it's "The natural logarithm ". You're over correcting something that's wasn't wrong in the first place
f(x + y) = f(x) ^ ln(f(y)) First, set y = 0 f(x) = f(x)^ln(f(0)) f(x) = 0, or ln(f(0)) = 1. Note that f(x) = 0 will result in ln(f(y)) = ln(0) = undefined everywhere, making the equation fall apart, so no good. So, ln(f(0)) = 1, or f(0) = e. Next, set y = 1. Then f(x+1) = f(x)^ln(f(1)). We'll note that ln(f(1)) is a constant, so we'll call it k. f(x+1) = f(x)^k f(x+2) = f(x+1 + 1) = f(x+1)^k = [f(x)^k]^k = f(x)^k^2 f(x+3) = f(x+2 + 1) = f(x+2)^k = [f(x)^k^2]^k = f(x)^k^3 Claim f(x+n) = f(x)^k^n Note that we already have base cases for n=1, 2 and 3. Suppose it's true for some value n = z. Then f(x+z+1) = f(x+z + 1) = f(x+z)^k = [f(x)^k^z]^k = f(x)^k^(z+1). So, at least for integer values, we have f(x + n) = f(x)^k^n. But we already know that f(0) = e, so we can write f(n) = e^k^n. This is looking promising, but we do still need to confirm it true for non-integer values of n. Consider x = 0, y = 1/q. Then we have f(1/q) = e^ln(f(1/q)). Denote ln(f(1/q)) as m, and we can find, with a similar proof by induction as above, that f(p/q) = e^m^p. But we also have a special case when p = q. Then f(1) = e^k. So, we have m^q = k, or m = k^1/q. But then, our formula just becomes f(p/q) = e^k^(p/q). So, it's true for all rational inputs. Finally, I'm just going to assume that the result is smooth, so if it's true for all rational inputs, it's true for all real inputs.
@@SyberMath Oh. I didn't end up showing that there were no limitations on k, and that we didn't find an extraneous solution. f(x) = e^k^x f(y) = e^k^y ln(f(y)) = k^y f(x+y) = e^k^(x+y) = e^[k^x * k^y] = [e^k^x]^(k^y) = f(x)^ln(f(y)) So, no limitations on k, and the solution holds.
Task: find all continuous f : R -> {x in R : 0 < x} such that for all x, y in R, f(x + y) = exp(ln(f(x))•ln(f(y))). Notice that the above is equivalent to, for all x, y in R, ln(f(x + y)) = ln(f(x))•ln(f(y)). Let g = ln°f, hence for all x, y in R, g(x + y) = g(x)•g(y). Therefore, for some k in R, and for all x in R, g(x) = exp(k•x) or g(x) = 0. Since g = ln°f, f = exp°g, hence for all x in R, f(x) = exp(exp(k•x)) or f(x) = 1.
The given functional equation implies the condition f(0)=e. Without loss of generality, g(x) should be equal to k^(a*x), where a is a constant. I don’t think we have enough information to determine a, say a=1. Is it correct?
Zakaj bi ugibali? g(z)=g(x)g(z-x) difernciramo po x: g'(x)g(y)-g(x)g'(y)=0 dobimo: g'(x)/g(x)=g'(y)/g(y)=konst. in [ln(g(x))]'=konst. ter ln(g(x))=konst*x+konst_1 in g(x)=A*exp(konst*x) A^2=A : trivialna rešitev je A=0, sicer pa A=1. Takšno enačbo dobite npr. v statistični mehaniki pri izpeljavi ravnovesne porazdelitve v faznem prostoru, kjer je g ≡ ρ verjetnostna gostota, x,y ≡ E_1,E_2 sta energiji podsistemov v termičnem ravnovesju ter konst ≡ - β = 1/kT je inverzna temperetarura Rešitev ρ=A*exp(-β*E) pa je Bolzmannova utež. A je normalizacijska konstanta.
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Set y=0. f(x) = f(x)^ln(f(0)). apply ln to both sides. ln(f(x)) = ln(f(0))*ln(f(x)). Since ln(f(x)) is repeated on both sides, ln(f(0)) must be the multiplicative identity, or 1. ln(e) = 1. f(0) = e. Now set x=0. f(y) = f(0)^ln(f(y)) -> f(y) = e^ln(f(y)). From here we can easily see the function is e^k^x, as we have f(y) = e^(some value).
Sorry, but a=e^(ln(a)) holds for any number a > 0. The fact that you first take the logarithm of the number, and then substitute the result into the exponent will simply return you to the initial number. (sorry for my english)
@@shigen778 Nice observation. Yes, this is true. For any value greater than 0 this will just be an identity. If the value is less than 0, or negative, then f(y) must produce a positive value. Hence e^k^x still holds as a logical solution, as e^(negative power) will just be inversions of e, which are still > 0, and as you have stated, this equation is always true for any values greater than 0. Hence e^k^x should work as a solution.
@@ben69real Sorry, but you probably misunderstood me. I meant that just because f(y)=e^(ln(f(y))) , it doesn't follow that f(y)=e^k^y is the only solution. For example, you can try f(y)=1+y^2 and this function satisfies the condition. The functions e^x and ln(x) "destroy" when you apply them to each other. Moreover, in the notation f(y)=e^(ln(f(y))), the function is not specified explicitly, due to the fact that it occurs to the left and right of the "=" sign. If you simplify the expression on the right, you just get f(y)=f(y), which is obviously meaningless. The property x=e^(ln(x)) holds for all positive numbers, so you can't define a function that way.
@@shigen778 This is true, what I was trying to point out is that it is easy to see though that e^k^x is a solution just due to the fact that f(y) still equals e^some power, even if it does just cancel out to be an identity in that specific situation. That said, functions that don't employ values raised to even powers like x^1 or x^3 can easily be ruled out, as they'd yield negative values. Additionally, with even powered functions that would yield only positive values, we can apply the original statement to see that these functions clearly cannot work. Just from seeing that there's an ln in the original statement, something such as x^2 + 1 or anything else wouldn't work. Therefore we can see that the only one of those solutions that would actually work for all solutions is e^(something), or e^k^x as we're saying here. The thing that e^ln(f(y)) does is it just affirms the fact that e is involved, which we could pretty much already see from the ln in the in the first statement but shrug. It literally just has e^(some power), even if it does cancel out to f(y) = f(y). Alone it's pretty much useless as you've pointed out but with the original statement it is actually easy to see that that is the solution, just based off of a simple check.
Just amazing 😍
Incredible!
The only continuous functions g (with domain all reals) which solve g(x+y) = g(x)g(y)
are: g(x) = k^x (where k > 0) and g(x) = 0.
x^k and kx, both with k 0 aren't solutions. k(ln(x)) has the wrong domain and isn't a
solution even on the positive reals unless k = 0.
In 3:00
To correct , in french it's not
" Logarithme naturel "
ln is called in French :
"Logarithme népérien"
It's latin.
Tu peux dire les deux
The determinant "le" is not part of the name. It's as if you were correcting someone in English "it's not "natural logarithm", it's "The natural logarithm ". You're over correcting something that's wasn't wrong in the first place
@@alexandreocadiz9967 Yes , you are right
@@ulysse6825 Oui je sais mais je parle de l'origine de la notation ln , je crois que c'est logarithme népérien
Also f(x) = 1, which is not included in the family of functions f(x) = exp(k*exp(x)), since when k = 0, it's not defined at x = 0.
The family of solutions is not exp(k•exp(x)).
@@angelmendez-rivera351
Yes, you're right. It was a careless error. How about exp(exp(x*ln(k)))?
@@esteger1 Yes, that works
@angelmendez-rivera351
I owe you a beer. If you're ever in the neighborhood of Foothill College in Los Altos CA, look me up.
Just for the linguistics, ln in french does not stand for "natural" but for "logarithme népérien" hence ln
de John Neper l'inventeur du logarithme
k>0 in the definition of exponential function
f(x + y) = f(x) ^ ln(f(y))
First, set y = 0
f(x) = f(x)^ln(f(0))
f(x) = 0, or ln(f(0)) = 1. Note that f(x) = 0 will result in ln(f(y)) = ln(0) = undefined everywhere, making the equation fall apart, so no good. So, ln(f(0)) = 1, or f(0) = e.
Next, set y = 1. Then
f(x+1) = f(x)^ln(f(1)). We'll note that ln(f(1)) is a constant, so we'll call it k.
f(x+1) = f(x)^k
f(x+2) = f(x+1 + 1) = f(x+1)^k = [f(x)^k]^k = f(x)^k^2
f(x+3) = f(x+2 + 1) = f(x+2)^k = [f(x)^k^2]^k = f(x)^k^3
Claim
f(x+n) = f(x)^k^n
Note that we already have base cases for n=1, 2 and 3. Suppose it's true for some value n = z. Then f(x+z+1) = f(x+z + 1) = f(x+z)^k = [f(x)^k^z]^k = f(x)^k^(z+1).
So, at least for integer values, we have f(x + n) = f(x)^k^n. But we already know that f(0) = e, so we can write f(n) = e^k^n. This is looking promising, but we do still need to confirm it true for non-integer values of n.
Consider x = 0, y = 1/q. Then we have f(1/q) = e^ln(f(1/q)). Denote ln(f(1/q)) as m, and we can find, with a similar proof by induction as above, that f(p/q) = e^m^p. But we also have a special case when p = q. Then f(1) = e^k. So, we have m^q = k, or m = k^1/q. But then, our formula just becomes f(p/q) = e^k^(p/q). So, it's true for all rational inputs.
Finally, I'm just going to assume that the result is smooth, so if it's true for all rational inputs, it's true for all real inputs.
Wow! nice
@@SyberMath Oh. I didn't end up showing that there were no limitations on k, and that we didn't find an extraneous solution.
f(x) = e^k^x
f(y) = e^k^y
ln(f(y)) = k^y
f(x+y) = e^k^(x+y) = e^[k^x * k^y] = [e^k^x]^(k^y) = f(x)^ln(f(y))
So, no limitations on k, and the solution holds.
@@SyberMathey, check out this problem. x^2+(x^2/(x+1)^2)=3
There are four solutions can you do a video on this problem
Try this problem
x^2+ x^2/(x+1)^2=3
Hint: there are 4 solutions
Nice.
Thanks!
Task: find all continuous f : R -> {x in R : 0 < x} such that for all x, y in R, f(x + y) = exp(ln(f(x))•ln(f(y))).
Notice that the above is equivalent to, for all x, y in R, ln(f(x + y)) = ln(f(x))•ln(f(y)). Let g = ln°f, hence for all x, y in R, g(x + y) = g(x)•g(y). Therefore, for some k in R, and for all x in R, g(x) = exp(k•x) or g(x) = 0. Since g = ln°f, f = exp°g, hence for all x in R, f(x) = exp(exp(k•x)) or f(x) = 1.
The given functional equation implies the condition f(0)=e. Without loss of generality, g(x) should be equal to k^(a*x), where a is a constant. I don’t think we have enough information to determine a, say a=1. Is it correct?
correct
Since k^(a*x) = (k^a)^x, and there were no restrictions put on k in the video, I think it still works generally because k^a is also unrestricted
Noice
f(x+x)=f(x)^ln(f(x))=f(x) and so on f(nx)=f(x)=f(0) so f(x)=constant
Zakaj bi ugibali?
g(z)=g(x)g(z-x)
difernciramo po x: g'(x)g(y)-g(x)g'(y)=0
dobimo: g'(x)/g(x)=g'(y)/g(y)=konst. in [ln(g(x))]'=konst. ter ln(g(x))=konst*x+konst_1 in g(x)=A*exp(konst*x)
A^2=A : trivialna rešitev je A=0, sicer pa A=1.
Takšno enačbo dobite npr. v statistični mehaniki pri izpeljavi ravnovesne porazdelitve v faznem prostoru, kjer je g ≡ ρ verjetnostna gostota, x,y ≡ E_1,E_2 sta energiji podsistemov v termičnem ravnovesju ter konst ≡ - β = 1/kT je inverzna temperetarura
Rešitev ρ=A*exp(-β*E) pa je Bolzmannova utež. A je normalizacijska konstanta.
Easy f(x)=1