That was a much better solution than mine haha, I decided to use lagrange multipliers. Constraint Function: g(x,y) = x^y = 8 Optimizing Function: f(x,y) = x + y grad(g(x,y)) = grad(f(x,y)) = 1 = yx^(y-1) * lambda 1 = (x^y)(lnx) * lambda yx^(y-1) = (x^y)(lnx) xlnx = y then plugged into the constraint x^(xlnx) = 8 x(lnx)^2 = ln8 which is what you had and I solved it the same from there using lambert W
Your argument for dismissing the negative value at 7:20 is wrong. Yes, W(z) can only be real when z is real and z ≥ -1/e . But the lefthandside is a function in x , and that function (and hence the value of W(z)) doesn't need to be real in order for x to be real. The fact of the matter is that even if -½√(ln(8)) was greater than -1/e , this negative value must have been dismissed earlier, namely at 6:30 when you've completed taking the square root: ln(x) * e^[½*ln(x)] = ±√(ln(8)) Since x is real and x > 1 , the lefthandside is positive, and hence the negative value of the righthandside can be dismissed. In other words, the dismissal has nothing to do with properties of the Lambert W function.
Alternatively, we can consider that x=8^(1/y) and find the minimum of y + 8^(1/y). If f(y)=y + 8^(1/y)=y + e^(ln8/y), then f'(y)=1 + e^(ln8/y)*-ln(8)/y^2=1 - ln(8)*8^(1/y)/y^2. If f'(y)=0, then ln(8)*e^(ln8/y)/y^2=1, therefore y^2=ln(8)*e^(ln8/y), therefore y^2*e^(-1/y)=W(ln8), therefore y*e^(-1/2*y)=[W(ln8)]^1/2, therefore (2*y)*e^(-1/2*y)=2*[W(ln8)]^1/2, therefore (1/2*y)*e^(1/2*y)=W(1/2*y)=(1/2)*W(ln8)^(-1/2), therefore y=(1/2)*W^(-1)[(1/2)*W(ln8)^(-1/2)]^(-1/2). I think.
How does this argument get us a minimum? We just used critical points, it could’ve also been a maximum, we would need to say a bit more to conclude with minimum
Yeah I feel like it needs some sort of argument for why it’s the minimum since it’s not trivial at a glance, but maybe I’m just missing something obvious
@@Happy_Abe 4.7694 is less than 5 , so it has to be the minimum (since f(x) = x + ln(8)/ln(x) is continuous and differentiable for x > 1 ). But yes, for the sake of completeness and correctness, he should have derived the second derivative f"(x) and demonstrated that its value is positive when x has the calculated value from f'(x) = 0 . It's a nice exercise for the viewer, though.
@@Happy_Abe Not x = 5 , but x+y = 5 ; which is the example in the thumbnail.
9 годин тому
Not to confuse. x^y = 8. With the same x and y, y^x would be less than 8, so would have to increase one or both of them and we would not get a minimum for x + y, since x^y must reach 8 from the statement of the problem. So y is xlnx.
Shouldn't the minimum value m of x+y occur when x+y=m is tangent to x^y = 8, which would be when y' = -1. Then using implicit differentiation we solve and find y = x lnx and therefore we just need to solve x^x = 8 using the super-root (or Lambert W if we want a longer-lookong answer)?
"it's a little less than 5" is a true answer but not a good answer in class. I don't have a feel for the W function in my head, so I could not have approximated it more closely other than by trial and error.
9 годин тому
The problem of course is to compute the Lambert W function. Since one possibly could say that I am at least indirectly involved in the development of that I didn't mention it.
@@vata7_ You wrote "W(xe^x) = x" , but that's not necessarily true. It's rather W(xe^x) = c ==> ce^c = xe^x but c isn't necessarily equal to x (because different values of x can lead to the same value of xe^x ). That's why the Lambert W function has multiple _branches_ Wₖ(x) , with index k being a (positive or non-positive) integer.
"If you put x to br closer and closer to 1 then the result will be bigger abd bigger" Not always true, if x = 8 and y = 1 then x + y = 8 + 1 = 9 but if x = 2 and y = 3 then x + y = 2 + 3 = 5 and 5 < 9.
@JARG-Random_Guy The Lambert W function has multiple "solutions" _everywhere_ , not just between -1/e an 0 . However, W(z) has two _real_ outcomes for real variable z between -1/e and 0 , namely W₀(z) and W₋₁(z) (which are respectively the 0'th branch and the (-1)'th branch of the Lambert W function; and both are then negative reals). But in this problem, the Lambert W function is applied on the value of ½√(ln(8)) , not on x ; so the domain of x isn't relevant to the behavior of the Lambert W function. In other words: there is no "x" on which the Lambert W function is applied.
The remark that y = xlnx is for the case of minimum or maximum depending on the value. Since it here is less than 2e it is a minimum. Of course one can get an arbitrarily large value without restrictions. But the sum 4.769378247 gives x = 2.4926758599...because it is x + xlnx.
How to solve a^x+bx+c=0 (NOT quadratic): ua-cam.com/video/rxVK5cWLRKQ/v-deo.html
Master, could you teach us how to solve x^x^x=a?
THE FISHH!!!!
THE EYEBROWS!1!!
Strikes again!
Welcome back, BPRP! Happy New Year! 🎉
4:27 oh no! the mischievous fish!
That was a much better solution than mine haha, I decided to use lagrange multipliers.
Constraint Function:
g(x,y) = x^y = 8
Optimizing Function:
f(x,y) = x + y
grad(g(x,y)) =
grad(f(x,y)) =
1 = yx^(y-1) * lambda
1 = (x^y)(lnx) * lambda
yx^(y-1) = (x^y)(lnx)
xlnx = y
then plugged into the constraint
x^(xlnx) = 8
x(lnx)^2 = ln8
which is what you had and I solved it the same from there using lambert W
Your argument for dismissing the negative value at 7:20 is wrong. Yes, W(z) can only be real when z is real and z ≥ -1/e . But the lefthandside is a function in x , and that function (and hence the value of W(z)) doesn't need to be real in order for x to be real.
The fact of the matter is that even if -½√(ln(8)) was greater than -1/e , this negative value must have been dismissed earlier, namely at 6:30 when you've completed taking the square root:
ln(x) * e^[½*ln(x)] = ±√(ln(8))
Since x is real and x > 1 , the lefthandside is positive, and hence the negative value of the righthandside can be dismissed.
In other words, the dismissal has nothing to do with properties of the Lambert W function.
Happy new year blackpenredpen
2:49 But Why + x, how did he come up with that?
because f(x)=x+y, and y=ln8(lnx)^-1
Because our goal was to calculate x + y and he lets that f(x) = X + y and he replaces y by the value he gets before if you look at the begin
@ Makes sense
I have a question, does the fish have to be mischievous?
It has to be equally mischievous in both places.
making it y=x (so that x^x = 8) actually did not give the minimum value
No, apparently the minimum occurs when
x = y²/ln(8)
New year, new board (?)
Alternatively, we can consider that x=8^(1/y) and find the minimum of y + 8^(1/y).
If f(y)=y + 8^(1/y)=y + e^(ln8/y), then f'(y)=1 + e^(ln8/y)*-ln(8)/y^2=1 - ln(8)*8^(1/y)/y^2.
If f'(y)=0, then ln(8)*e^(ln8/y)/y^2=1,
therefore y^2=ln(8)*e^(ln8/y),
therefore y^2*e^(-1/y)=W(ln8),
therefore y*e^(-1/2*y)=[W(ln8)]^1/2,
therefore (2*y)*e^(-1/2*y)=2*[W(ln8)]^1/2,
therefore (1/2*y)*e^(1/2*y)=W(1/2*y)=(1/2)*W(ln8)^(-1/2),
therefore y=(1/2)*W^(-1)[(1/2)*W(ln8)^(-1/2)]^(-1/2).
I think.
How does this argument get us a minimum?
We just used critical points, it could’ve also been a maximum, we would need to say a bit more to conclude with minimum
Yeah I feel like it needs some sort of argument for why it’s the minimum since it’s not trivial at a glance, but maybe I’m just missing something obvious
The function is concave and thus there only is a global minimum
@@Happy_Abe 4.7694 is less than 5 , so it has to be the minimum (since f(x) = x + ln(8)/ln(x) is continuous and differentiable for x > 1 ).
But yes, for the sake of completeness and correctness, he should have derived the second derivative f"(x) and demonstrated that its value is positive when x has the calculated value from f'(x) = 0 . It's a nice exercise for the viewer, though.
@@yurenchuagreed but you’d have to calculate f(5) and I don’t remember that in the video
@@Happy_Abe Not x = 5 , but x+y = 5 ; which is the example in the thumbnail.
Not to confuse. x^y = 8. With the same x and y, y^x would be less than 8, so would have to increase one or both of them and we would not get a minimum for x + y, since x^y must reach 8 from the statement of the problem. So y is xlnx.
Shouldn't the minimum value m of x+y occur when x+y=m is tangent to x^y = 8, which would be when y' = -1. Then using implicit differentiation we solve and find y = x lnx and therefore we just need to solve x^x = 8 using the super-root (or Lambert W if we want a longer-lookong answer)?
"it's a little less than 5" is a true answer but not a good answer in class.
I don't have a feel for the W function in my head, so I could not have approximated it more closely other than by trial and error.
The problem of course is to compute the Lambert W function. Since one possibly could say that I am at least indirectly involved in the development of that I didn't mention it.
can someone explain what W( ) do and what's its purpose?
W(x) is the solution of W e^W = x
W(xe^x)=x
W(x)e^W(x)=x
It's to solve these kind of problems!
inverse function to f(x) = x*e^x.
W(f(x)) = x
@@vata7_ You wrote "W(xe^x) = x" , but that's not necessarily true. It's rather
W(xe^x) = c ==> ce^c = xe^x
but c isn't necessarily equal to x (because different values of x can lead to the same value of xe^x ).
That's why the Lambert W function has multiple _branches_ Wₖ(x) , with index k being a (positive or non-positive) integer.
Same result I have. Nice problem
Other than using WA, how can you derive the value of the Lambert W function? Is there a means to do it with a lot of paper and pencils?
Man you could use newton method with Euler method do do a circular way tô aproximate in paper
@@eduardomalacarne9024 Might make a good video for someone to do....
There are series expansions for it, but they are not nice. However, given enough time, paper and pencils one can do it.
Why do the solutions have to be real numbers? He never said what set we are looking for solutions in, could be mod(5) or something.
W video with the W function
I thought the question was: x^y=8; x+y=5 what is the minimum value of y? (solution: x≈3.22333, y≈1.77667)
Hey, bprp, here's a fun limit I want you to attempt to solve: lim x->1 x^x/x^∞
WHAT THE FISH?!
"If you put x to br closer and closer to 1 then the result will be bigger abd bigger"
Not always true, if x = 8 and y = 1 then x + y = 8 + 1 = 9 but if x = 2 and y = 3 then x + y = 2 + 3 = 5 and 5 < 9.
with x already being close to 1
Why is everything productlog and why am I watching this at 3am lol. Are there other cool functions nobody has heard of?
Ackermann
@@yurenchu Bessel (just because the name begins with B)
I got pretty close by just quessing it would be close to e+3*ln(2) which is just 0.028 or so diffrent(4,7977).
New year new marker??
Helpful & Give me more math experiences...
Why are my comments in this comment section being deleted? 😠
Nice!
What if x is less than 1?
@@ziplock007 If 0 < x < 1 , then y can be as negative as you want, and hence the sum x+y can be as negative as you want; so there is no minimum.
Between 0 and -1/e, lambert W func. Has multiple solutions. You can see it yourself, type in Desmos x=ye^y
@JARG-Random_Guy The Lambert W function has multiple "solutions" _everywhere_ , not just between -1/e an 0 . However, W(z) has two _real_ outcomes for real variable z between -1/e and 0 , namely W₀(z) and W₋₁(z) (which are respectively the 0'th branch and the (-1)'th branch of the Lambert W function; and both are then negative reals).
But in this problem, the Lambert W function is applied on the value of ½√(ln(8)) , not on x ; so the domain of x isn't relevant to the behavior of the Lambert W function. In other words: there is no "x" on which the Lambert W function is applied.
lagrange multiplier in play
I mean, if x and y don't have to be rational... or whole... or positive...
4.7694
The remark that y = xlnx is for the case of minimum or maximum depending on the value. Since it here is less than 2e it is a minimum. Of course one can get an arbitrarily large value without restrictions. But the sum 4.769378247 gives x = 2.4926758599...because it is x + xlnx.
hi guy
Why go through the complex solution while we have simple one!!
Complex analysis? I'm sorry, I only do simple analysis 😈
find the "min"
@@dolphin-314 Big maths wants you to think there is no min in a subset of C, open your eyes sheeple
These type of problems arebalways fishy.
nice222