14:30 assuming, n is even, we have the two equations 1-λ=λ as shown in the video and the second equation becomes 1-λ=-λ , immediately leading to a contradiction, thus yielding no second solution.
This question screamed Lagrange multipliers and taking ln() to me. But first we need to justify that ln() is appropriate. The constraint tan(θ1)...tan(θn) = 1 means they're either all positive or there's an even number of negative factors in the product. Since tan() is an odd function, we can simply factor out the -1^(even power) and replace the tan() with the corresponding positive factor. So we can now assume that the tan(θi) is positive for all i and apply ln() without any concern. Since we're assuming that tan(θi) is positive for all i, this implies that θi belongs to (0,pi/2) for all i. Thus, sin(θ1)...sin(θn) is a product composed of all positive factors. Since ln() is a strictly increasing function, taking ln(sin(θ1)...sin(θn)) doesn't change the location of the maximum. From here, simply use the Lagrange multiplier formula applied to F(x) = ln(f(x)) and G(x) = ln(g(x)) ==> cos(θi)/sin(θi) = λ/[sin(θi)cos(θi)] for all i ==> cos^2(θi) = λ for all i. In other words, all the cos(θi) either equal each other or are negatives of each other. Similarly, cos^2(θi) = 1 - sin^2(θi) = λ ==> sin^2(θi) = 1 - λ for all i. So the sin(θi) also equal each other or are negatives of each other. And finally, since tan(θi) = sin(θi)/cos(θi), the tan(θi) also equal each other or are the negatives of each other. However, we've already established that the max will be attained if we fix θi in (0,pi/2) so we can further assume that the tan(θi)'s all equal each other. Plugging this back into our initial constraint: tan(θ1)...tan(θn) = [tan(θ)]^n = 1 ==> tan(θ) = 1 ==> θ = pi/4. Plugging this into our function: [sin(pi/4)]^n = 2^(-n/2)
Looks convincing to an amateur enjoying learning from you guy's skills. Can't cease the aggressions happening in reality so thankful an alternative is getting about what you've found in this 3 spacial and one temporal + multivariously spiritual NOW.
from Morocco thank you you are genious....i prefer the first method because the second is higher than my understanding...i have no idea about Lagrangian multiplier ...its an occasion and constraint for me for searching more about it...thank you..i shared your video in my facebook page
A spread of the various thetas randomly can be fed into the lateral part of the stochastic progress of a photon to make the hypoteneuses of a microscopic element add and give a random walk modified journey as a perturbation from straight path. These microscopic elements can be summed as an integral at their length approaching zero and give a probability distribution of paths through a very slender cone. Who needs a wave function collapsing when the improbable path of a photon with a randomised sidereal motion can be tweaked to look like an hamiltonian probability distribution.
can somebody please explain me the last line at th-cam.com/video/3VM_7wVGJZ8/w-d-xo.htmlsi=MA1pMDfKkfKuayz1&t=168 ? because setting n=1 and theta=30degrees gives you (1/2)squared = (1/2)times(root3/2), which is obviously false
You can also guess the answer by a simple argument. t1=...=tn=Pi/4 obviously satisfies the constraint with sin(t1)...sin(tn)=2^{-n/2}. Now suppose you have a variation, i.e. you increase t1,...,tk (say) to Pi/4+e1,...,Pi/4+ek while you decrease t_{k+1},...,t_{k+m} to Pi/4-e_{k+1},...,Pi/4-e_{k+m} s.t. the constraint remains satisfied. Since tan(t) is strictly increasing and convex on (0,Pi/2) while sin(t) is strictly increasing and concave on (0,Pi/2), sin(Pi/4+e1) ... sin(Pi/4+ek) * sin(Pi/4-e_{k+1}) ... sin(Pi/4-e_{k+m}) can only decrease when tan(Pi/4+e1) ... tan(Pi/4+ek) * tan(Pi/4-e_{k+1}) ... tan(Pi/4-e_{k+m}) remains unchanged. This is because tan(Pi/4+e1) ... tan(Pi/4+ek) will always increase more relative to tan(Pi/4-e_{k+1}) ... tan(Pi/4-e_{k+m}) than sin(Pi/4+e1) ... sin(Pi/4+ek) will relative to sin(Pi/4-e_{k+1}) ... sin(Pi/4-e_{k+m}).
I would log it before doing Lagrange
6:11 in that case the tan product would be -1, not 1. So it won't work
Then instead take θ_1 = 5π/4. Then both sine and cosine become negative, so the problem with tangent vanishes
14:30 assuming, n is even, we have the two equations 1-λ=λ as shown in the video and the second equation becomes 1-λ=-λ , immediately leading to a contradiction, thus yielding no second solution.
This question screamed Lagrange multipliers and taking ln() to me. But first we need to justify that ln() is appropriate. The constraint tan(θ1)...tan(θn) = 1 means they're either all positive or there's an even number of negative factors in the product. Since tan() is an odd function, we can simply factor out the -1^(even power) and replace the tan() with the corresponding positive factor. So we can now assume that the tan(θi) is positive for all i and apply ln() without any concern. Since we're assuming that tan(θi) is positive for all i, this implies that θi belongs to (0,pi/2) for all i. Thus, sin(θ1)...sin(θn) is a product composed of all positive factors. Since ln() is a strictly increasing function, taking ln(sin(θ1)...sin(θn)) doesn't change the location of the maximum.
From here, simply use the Lagrange multiplier formula applied to F(x) = ln(f(x)) and G(x) = ln(g(x)) ==> cos(θi)/sin(θi) = λ/[sin(θi)cos(θi)] for all i ==> cos^2(θi) = λ for all i. In other words, all the cos(θi) either equal each other or are negatives of each other. Similarly, cos^2(θi) = 1 - sin^2(θi) = λ ==> sin^2(θi) = 1 - λ for all i. So the sin(θi) also equal each other or are negatives of each other. And finally, since tan(θi) = sin(θi)/cos(θi), the tan(θi) also equal each other or are the negatives of each other. However, we've already established that the max will be attained if we fix θi in (0,pi/2) so we can further assume that the tan(θi)'s all equal each other.
Plugging this back into our initial constraint: tan(θ1)...tan(θn) = [tan(θ)]^n = 1 ==> tan(θ) = 1 ==> θ = pi/4.
Plugging this into our function: [sin(pi/4)]^n = 2^(-n/2)
Looks convincing to an amateur enjoying learning from you guy's skills. Can't cease the aggressions happening in reality so thankful an alternative is getting about what you've found in this 3 spacial and one temporal + multivariously spiritual NOW.
Nice solutions!
from Morocco thank you you are genious....i prefer the first method because the second is higher than my understanding...i have no idea about Lagrangian multiplier ...its an occasion and constraint for me for searching more about it...thank you..i shared your video in my facebook page
Would proof by induction work also? I can see that getting messy.
R u dumb, this isn't a proof
A spread of the various thetas randomly can be fed into the lateral part of the stochastic progress of a photon to make the hypoteneuses of a microscopic element add and give a random walk modified journey as a perturbation from straight path. These microscopic elements can be summed as an integral at their length approaching zero and give a probability distribution of paths through a very slender cone. Who needs a wave function collapsing when the improbable path of a photon with a randomised sidereal motion can be tweaked to look like an hamiltonian probability distribution.
can somebody please explain me the last line at th-cam.com/video/3VM_7wVGJZ8/w-d-xo.htmlsi=MA1pMDfKkfKuayz1&t=168 ? because setting n=1 and theta=30degrees gives you (1/2)squared = (1/2)times(root3/2), which is obviously false
The condition for angles is that product of all tangents must be 1. This exclude your choice of the thetas.
good day
Shii
The second method can be regarded as an optimization approach using Lagrangian multipliers.
As he said in the beginning of the video
You can also guess the answer by a simple argument.
t1=...=tn=Pi/4 obviously satisfies the constraint with sin(t1)...sin(tn)=2^{-n/2}. Now suppose you have a variation, i.e. you increase t1,...,tk (say) to Pi/4+e1,...,Pi/4+ek while you decrease t_{k+1},...,t_{k+m} to Pi/4-e_{k+1},...,Pi/4-e_{k+m} s.t. the constraint remains satisfied. Since tan(t) is strictly increasing and convex on (0,Pi/2) while sin(t) is strictly increasing and concave on (0,Pi/2), sin(Pi/4+e1) ... sin(Pi/4+ek) * sin(Pi/4-e_{k+1}) ... sin(Pi/4-e_{k+m}) can only decrease when tan(Pi/4+e1) ... tan(Pi/4+ek) * tan(Pi/4-e_{k+1}) ... tan(Pi/4-e_{k+m}) remains unchanged.
This is because tan(Pi/4+e1) ... tan(Pi/4+ek) will always increase more relative to tan(Pi/4-e_{k+1}) ... tan(Pi/4-e_{k+m}) than sin(Pi/4+e1) ... sin(Pi/4+ek) will relative to sin(Pi/4-e_{k+1}) ... sin(Pi/4-e_{k+m}).