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.
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. EDIT: Oops, correction: it's not enough to remark that 4.7694 is less than 5 and that f(x) is continuous and differentiable. Obviously 4.7694 cannot be a maximum, but f"(x) could be equal to 0 in which case we're dealing with an _inflection_ point, which would mean that it's not a minimum either. Therefore, it's necessary to show that f"(x) is positive at this point.
Can you do a video about the inverses of absolute values? Much appreciated.
15 วันที่ผ่านมา
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.
I have a question at 5:59. How can you get -sqrt(8) when multiplying a square with an exponent (which is never negative)? Would it be easier to get rid of the minus in the ± before applying the Lambert W function if you said that multiplying a square and an exponent would never give you a negative result?
@@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.
4:02 Is it only me, but the Lambert W function seems too excessive to me. I mean, I can define a function called Lambert Ğ, which is defined as Ğ(x(lnx)^2) = x, and the minimum x value would be Ğ(ln8) in this question. I hope you understood my question, can we define infinite amounts of functions that we can use to solve irregular equations?
We already have the Lambert W function, so why would we also need your Lambert Ğ function? Moreover, (every branch of) the Lambert W function is well-defined over the complex plane; would that also be true for your Lambert Ğ function? (Note that z*e^z is a function for complex-valued z , but z*(ln(z))² isn't a function because ln(z) isn't uniquely defined for complex-valued z . For example, what values should Ğ(-1*(ln(-1))²) , Ğ(π²) and Ğ(e^[i*3π] * (i*3π)²) = Ğ(9π²) produce?)
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
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)?
Hey bprp i know there might be no chance you see this, but i love your content. I think you are one of the reasons i love calculus so much. I really like three way you explain everything, I watched through the whole 100 polynomial factorization to learn factoring. I recently decided that i will be ending it all soon but, i just wanted to let you know that you are very important to me and that you are one of the greatest calculus and math teachers in the world
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.
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.
Sorry, but your step from y² = ln(8) * e^(ln(8)/y) to y² * e^(-1/y) = W(ln(8)) doesn't make sense. It seems you're also performing multiple operations in just one step, which may have led to errors of confusion. The proper way to apply the Lambert W function, is when you have an equation of the form M * e^M = C (where M and C can be complicated expressions). The step to take is then to say that M can be found by applying the Lambert W function on the expresson of the righthandside: M = Wₖ(C) I write Wₖ , because the Lambert W function has multiple branches, and it depends on the conditions which branch you want (usually W₀ and/or W₋₁ ; for example, if C is real and positive, and you want M to be real, then you take W₀).
@johangonzalez4894 unfortunately Wolfram Alpha and it didn't give steps. But the values are correct to 6 decimals. A nicer formulation is: try to minimize x*y. There are only two possible pairs of x and y anyway
So what if we take logx of both sides x>1 so no provlem then we get y=logx(8) then just say y=f(x) and we want min of f(x) +x so just take the derivative and then equal to zero would it work?
question: when we had x ln(x)^2 = ln(8), could we not have divided by ln(x) to get x ln(x) = ln(8)-ln(x)? Or is x = W [ln(8) - ln(x)] a train wreck to calculate?
That doesn't work for two reasons: 1- Dividing by ln(x) gives xln(x) = ln(8)/ln(x) ≠ ln(8) - ln(x). You're confusing the identity ln(a/b) = ln(a) - ln(b) for a, b positive real numbers. 2- Writing x = W[ln(8) - ln(x)] doesn't lead to anything because you haven't isolated x.
@@johangonzalez4894 I'm not the author of the comment to the video, but you're right, I've also confused the identity ln(a/b) = ln(a) - ln(b) for a, b positive real numbers. But the problem goes beyond this, writing x = W[ln(8) - ln(x)] doesn't lead to anything since x isn't isolated.
@diegocabrales I just noticed that too. I shouldn't try to do math at 1:00 AM. I'm also not super familiar with the W function. Like I see that for x ln x = p, then x = W (p), but what the heck does that mean? Does it look like y ln y = x so that y = W(x), because if it is, then there are multiple values of y for some x values (like -0.5
"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.
I am very confused. Isn't the goal to minimise x+y? In which case the sum of the answers given is higher than 5 (2+3=5). I hope someone can explain this to me.
I took maths in my bsc just because i didnt focus on it during my highschool but still i wanted to learn maths I still want to learn maths but i understand nothing from my textbook, i do self study 😢 My college is distance based so there are no classes, all i have to do is study maths on my own and mostly i understand nothing I have tried getting TH-cam's help but there are no videos related to the content in my universities' textbooks
"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.
I just have a bad reaction to the W function. It's just not a very satisfying solution to a problem. If you're going to iteratively compute a numeric answer, just start from the original problem. Yeah, that would be ugly for this particular problem, but still.
Yes, but for a non-negative argument (as is the case here) you just get one value, found by applying W₀. You would get a 2nd real value for an argument strictly between -1/e and 0, found by applying W₋₁.
15 วันที่ผ่านมา
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.
For any constant c , tan(c + i*∞) = i tan(c - i*∞) = -i Verify this by using tan(z) = sin(z)/cos(z) sin(z) = -i(e^(iz) - e^(-iz))/2 cos(z) = (e^(iz) + e^(-iz))/2 ==> atan(i) = c + i*∞ atan(-i) = c - i*∞
How to solve a^x+bx+c=0 (NOT quadratic): th-cam.com/video/rxVK5cWLRKQ/w-d-xo.html
Master, could you teach us how to solve x^x^x=a?
that's not the minimum.
4^1.5= 4^1 *4^(1/2)=8 where x=4 and y=1.5, then x+y=6.5
THE FISHH!!!!
THE EYEBROWS!1!!
Strikes again!
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.
I'm watching from Turkey. You explain it very well
I have a question, does the fish have to be mischievous?
It has to be equally mischievous in both places.
Welcome back, BPRP! Happy New Year! 🎉
4:27 oh no! the mischievous fish!
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.
EDIT: Oops, correction: it's not enough to remark that 4.7694 is less than 5 and that f(x) is continuous and differentiable. Obviously 4.7694 cannot be a maximum, but f"(x) could be equal to 0 in which case we're dealing with an _inflection_ point, which would mean that it's not a minimum either. Therefore, it's necessary to show that f"(x) is positive at this point.
@@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.
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
That's so simple bruhh
Can you do a video about the inverses of absolute values? Much appreciated.
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.
I have a question at 5:59. How can you get -sqrt(8) when multiplying a square with an exponent (which is never negative)? Would it be easier to get rid of the minus in the ± before applying the Lambert W function if you said that multiplying a square and an exponent would never give you a negative result?
Yes, you are correct.
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.
4:02 Is it only me, but the Lambert W function seems too excessive to me. I mean, I can define a function called Lambert Ğ, which is defined as Ğ(x(lnx)^2) = x, and the minimum x value would be Ğ(ln8) in this question. I hope you understood my question, can we define infinite amounts of functions that we can use to solve irregular equations?
We already have the Lambert W function, so why would we also need your Lambert Ğ function?
Moreover, (every branch of) the Lambert W function is well-defined over the complex plane; would that also be true for your Lambert Ğ function?
(Note that z*e^z is a function for complex-valued z , but z*(ln(z))² isn't a function because ln(z) isn't uniquely defined for complex-valued z . For example, what values should Ğ(-1*(ln(-1))²) , Ğ(π²) and Ğ(e^[i*3π] * (i*3π)²) = Ğ(9π²) produce?)
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
optimization issues like these tend to have e as answear, so i guess e+ln(8)
can't we write y in terms of x in the first equation and substitute x in x+y and differentiate to get possible values of the function
New year, new board (?)
Can we do a 100 trignometry problem would like to watch that too
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)?
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)
Is this solvable by Lagrange multipliers?
Hey bprp i know there might be no chance you see this, but i love your content. I think you are one of the reasons i love calculus so much. I really like three way you explain everything, I watched through the whole 100 polynomial factorization to learn factoring. I recently decided that i will be ending it all soon but, i just wanted to let you know that you are very important to me and that you are one of the greatest calculus and math teachers in the world
What a nice message and thank you so much for telling me! Wishing you all the best!
Same result I have. Nice problem
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.
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.
Sorry, but your step from
y² = ln(8) * e^(ln(8)/y)
to
y² * e^(-1/y) = W(ln(8))
doesn't make sense. It seems you're also performing multiple operations in just one step, which may have led to errors of confusion.
The proper way to apply the Lambert W function, is when you have an equation of the form
M * e^M = C
(where M and C can be complicated expressions).
The step to take is then to say that M can be found by applying the Lambert W function on the expresson of the righthandside:
M = Wₖ(C)
I write Wₖ , because the Lambert W function has multiple branches, and it depends on the conditions which branch you want (usually W₀ and/or W₋₁ ; for example, if C is real and positive, and you want M to be real, then you take W₀).
maximum value of f(x) = (a2-b2)/root(a2secsquaretheta + b2cosecsquaretheta)
solve this
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.
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)
How did you find it?
@johangonzalez4894 unfortunately Wolfram Alpha and it didn't give steps. But the values are correct to 6 decimals. A nicer formulation is: try to minimize x*y. There are only two possible pairs of x and y anyway
So what if we take logx of both sides x>1 so no provlem then we get y=logx(8) then just say y=f(x) and we want min of f(x) +x so just take the derivative and then equal to zero would it work?
That is practically the same as what he did.
I didnt watch the video but thank you
question: when we had x ln(x)^2 = ln(8), could we not have divided by ln(x) to get x ln(x) = ln(8)-ln(x)? Or is x = W [ln(8) - ln(x)] a train wreck to calculate?
That doesn't work for two reasons:
1- Dividing by ln(x) gives xln(x) = ln(8)/ln(x) ≠ ln(8) - ln(x). You're confusing the identity ln(a/b) = ln(a) - ln(b) for a, b positive real numbers.
2- Writing x = W[ln(8) - ln(x)] doesn't lead to anything because you haven't isolated x.
You made a mistake. ln(a)/ln(b) ≠ ln(a)-ln(b).
@@johangonzalez4894 I'm not the author of the comment to the video, but you're right, I've also confused the identity ln(a/b) = ln(a) - ln(b) for a, b positive real numbers. But the problem goes beyond this, writing x = W[ln(8) - ln(x)] doesn't lead to anything since x isn't isolated.
@diegocabrales I know. I was responding to the OP.
@diegocabrales I just noticed that too. I shouldn't try to do math at 1:00 AM. I'm also not super familiar with the W function. Like I see that for x ln x = p, then x = W (p), but what the heck does that mean? Does it look like y ln y = x so that y = W(x), because if it is, then there are multiple values of y for some x values (like -0.5
"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.
I am very confused. Isn't the goal to minimise x+y? In which case the sum of the answers given is higher than 5 (2+3=5). I hope someone can explain this to me.
The minimum in the end is ~4.7694.
Hey, bprp, here's a fun limit I want you to attempt to solve: lim x->1 x^x/x^∞
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
There are series expansions for it, but they are not nice. However, given enough time, paper and pencils one can do it.
@@dlevi67 Heh. Might make a good video for someone to do....
Evaluate the following expression modulo :
1 + 2^2 + 3^3 + 4^4 + \dots + 2025^{2025} \equiv x \ (\text{mod} \ 7).
I took maths in my bsc just because i didnt focus on it during my highschool but still i wanted to learn maths
I still want to learn maths but i understand nothing from my textbook, i do self study 😢
My college is distance based so there are no classes, all i have to do is study maths on my own and mostly i understand nothing
I have tried getting TH-cam's help but there are no videos related to the content in my universities' textbooks
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).
"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
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.
I just have a bad reaction to the W function. It's just not a very satisfying solution to a problem. If you're going to iteratively compute a numeric answer, just start from the original problem. Yeah, that would be ugly for this particular problem, but still.
Helpful & Give me more math experiences...
New year new marker??
I mean, if x and y don't have to be rational... or whole... or positive...
W video with the W function
Why are my comments in this comment section being deleted? 😠
Lambert function does not give one solution. It gives 2 in some cases. Can give even more in some very specific cases
Yes, but for a non-negative argument (as is the case here) you just get one value, found by applying W₀. You would get a 2nd real value for an argument strictly between -1/e and 0, found by applying W₋₁.
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.
lagrange multiplier in play
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)
Nice!
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
WHAT THE FISH?!
4.7694
Integrate (x^x)(x^(x^x))(1/x+ln(x)+(ln(x))^2)
These type of problems are always fishy.
can you do the arctangent of i
For any constant c ,
tan(c + i*∞) = i
tan(c - i*∞) = -i
Verify this by using
tan(z) = sin(z)/cos(z)
sin(z) = -i(e^(iz) - e^(-iz))/2
cos(z) = (e^(iz) + e^(-iz))/2
==>
atan(i) = c + i*∞
atan(-i) = c - i*∞
Unrelated: th-cam.com/video/uqwC41RDPyg/w-d-xo.html
hi guy
nice222
“Fish” variable in big 2025 💀
the fishe is everyone's favourite anime character