Solving without my favorite tool.

[@roberttelarket4934]

Part 1 converting to sin function is really unique and ingenious!

[@diniaadil6154]

You could have found the minimum using your sin function. You can parametrize your function by letting theta in the first quadrant. Since pi/2 -arctan4 < arctan4 , the minimum is reached at theta = pi/2 , leading to sqrt(17/2) * 1/sqrt(17) = 1/sqrt(2)

[@TheAmazingOne]

Another way to find the max: square the expression: 8 - 15x + 8sqrt(x(1/2 - x)). Then observe that this can be rewritten as 17/2 + sqrt(u^2 - v^2) - u, where u = 15x + 1/2 and v = 17x - 1/2. Since u is positive, the square root term is never greater than u. And the maximum is achieved when v = 0.

[@JohnSmith-ux3xm]

Yes, do make a special relativity video, with your usual clarity. I look forward to it.

[@rubensramos6458]

Using the Lambert-Tsallis Wq function one gets, for the maximal value, f(x = 0.5Wq(1/16)) = sqrt(17/2), where Wq(x) is the Lambert-Tsallis Wq function and, in the present case, q = 2.

[@Vladimir_Pavlov]

Why move away when finding the minimum of the function from the resulting formula f(x)=f((sinθ )^2/2)= √(17/2)* sin(θ+atan4)? From the substitution x=(sinθ )^2/2 and the domain of the function definition 0≤x≤1/2, it follows that 0≤θ≤π/2. As θ increases from the value 0, sin(θ+atan4) grows until it becomes equal to 1 at θ=π/2 -atan4, which corresponds to the maximum of the function f(1/34)=√(17/2). With further growth of θ , sin(θ+atan4) monotonically decreases and reaches a minimum at θ=π/2. Therefore, the minimum of the function is at the boundary of the domain of definition: f(1/2)=√(17/2)*sin(π/2+atan4)= √(17/2)* cos(atan4) =√(17/2)* 1/√(1+4^2)=1/√2. Of course, it should be noted that f(0)=4/√ 2 > f(1/2).

[@michaelz2270]

These are olympiad people so they might do something like Cauchy-Schwartz on (1,4) and (sqrt(x), sqrt(1/2 - x)) to get the absolute value is bounded by sqrt(17/2). Then use that equality holds when sqrt(1/2 - x) = 4sqrt(x) which solves to x = 1/34. Minimum occurs at farthest angle which is x = 1/2 giving 1/sqrt(2).

[@amirb715]

the maximum could have been found very quickly by Cauchy Schwarz inequality: | given expr|^2

[@moshadj]

the fact that the domain limits x to 0 to 0.5 is what allows us to find theta such that sin^2(theta)/2 = x.

[@ElectricalMaths]

You said that we shouldn't worry about the absolute value of cosine and sine when taking the square roots because 0 ≤ x ≤ 1/2. But if you look at the substitution, with the domain restriction, sin^2(𝜃) is restricted between 0 and 1, which means that sin(𝜃) is between -1 and 1, meaning there is no restriction for 𝜃. However, removing the absolute value from both the sin and cos terms suggests that 𝜃 is in the first quadrant. I believe we do need the absolute values in this case.

[@janerphyll1996]

Can this be fixed by restricting the theta values from 0 to pi/2? The range is still 0

[@ElectricalMaths]

@@janerphyll1996 Yeah it would be fixed by such a restriction, which in turn will affect the combination of the sum of sine and cosine into a single trigonometric function. Anyway, I don’t think it will ultimately have an effect on evaluating the maximum of the function but it was just an observation

[@janerphyll1996]

@@ElectricalMaths I do not think it should affect the combination of the sum, the values are just restricted. There should be no change to the resulting trigonometric function but a smaller domain.

[@ElectricalMaths]

@@janerphyll1996 Yes, that was what I meant - the restriction will carry over to subsequent dependent calculations.

[@abrahammekonnen]

Cool problem. Thank you.

[@sahiltamang2352]

THANKS FOR NOKIA MATH IN THIS VIDEO!!!

[@bvwalker1]

Can someone explain to why in detail he is able to set the the coefficients to each side of the trig function equal to each other? He mentioned something about linearly independent?

[@ohjahohfrick9837]

You cannot turn cosine into sine or vice versa by multiplying by and/or adding some constants. That is the definition of being linearly independent.

[@manucitomx]

I like problems like this. Thank you, professor.

[@charly6052]

Hi i have a hard exercice

[@General12th]

Hi Dr.! Very cool techniques!

[@goodplacetostop2973]

12:41

[@GreenMeansGOF]

I was expecting that the minimum was negative of the maximum but it would seem that there are restrictions on the trig functions.

[@DavidSavinainen]

Remember that theta may only take values between 0 and π/2, so that x remains between 0 and 1/2 as required by the original function. So the sine cannot acquire its typical minimum at negative the amplitude, because the argument is not allowed to reach the required value.

[@edwardlulofs444]

Very clever. Thanks. And ... the moral is ... not using calculus is like building a house with only one arm.

[@tordjarv3802]

It would be very interesting to get a mathematicians view of special relativity so I’m all for you making a video about it

[@BrianGriffin83]

It's much easier to calculate A by squaring Asinα and Acosα and add them up.

[@Szynkaa]

Cauchy-Schwartz inequality does the job here in 10 seconds.

[@GreenMeansGOF]

Could you explain?

[@Szynkaa]

@@GreenMeansGOF (a_1*b1+a_2*b_2)

[@GreenMeansGOF]

@@Szynkaa Ok, that makes sense. What about for the minimum?

[@Szynkaa]

@@GreenMeansGOF don't need any special tools to find minimum, i did it very similiarly to Michael

[@theamazingworldofgusball1852]

I would love to see some physics videos!!

[@ronyadin3958]

I have a question regarding the min. If for the first part we used the substitution x=sin^2(theta)/2 For every theta we will get a valid x (from 0 to 1/2) Also we got a closed formula of f(x)=sqrt(17/2) sin(theta + arctan4) Then wouldn't pluging theta = pi - arctan4 result in a min of f(x)=0?

[@carstenmeyer7786]

The closed form is only valid as long as *sin(𝜃), cos(𝜃) ≥ 0,* as the absolute values were omitted at 2:18ff. Or equivalently: *𝜃 ∈ [0; 𝜋/2] + 2𝜋k, k ∈ ℤ* The value *𝜃 = 𝜋 - atan(4)* does not lie inside any of those allowed intervals, so we may not use it!

[@ronyadin3958]

@@carstenmeyer7786 Ohh makes much more sense now. Thank you for clerafication

[@MrFtriana]

Hell yeah! Please, do the special relativity series. It will be awesome!

[@HershO.]

+1. Hopefully he does do this

[@MrFtriana]

@@HershO. it's a must. There is a lot of Lie algebras in the formulation of the relativity and also a bit of topology of metric spaces, so could be awesome see how Michael tackle that parte of physics from the point of view of a mathematician. There is a lot to talk about it.

[@HershO.]

@@MrFtriana It sure is a lot of content, so even from an economical POV, this is a must. It will massively help him reach the pi-100k sub goal and would also attract a lot of new audiences. I really hope Michael considers this.

[@rogercastellanosfernandez3551]

I have a question: if the domain would not be restricted, could we do the initial substitution x= (sin^2(theta)) / 2 as well?

[@MessedUpSystem]

Problem: you must do it without calculus! Michael: ok but can I get inspired by calculus?

[@matheusjahnke8643]

You need calculus to know how to solve it without using calculus. And this is not the first time I saw something like that.

[@takyc7883]

who even hinks of that substitution - so smart

[@jonasl7323]

Can someone nicely proof that this value for x_0 can be written as 1/34 (the value you find using calculus)?

[@jopicocco]

Prof penn, i have a question: Determine How many Numbers nm (not multiplication but juxtaposition of two natural numbers ), with n,m natural number with the following property nm=mn Example If We consider 13 and 1313 We have 131313. There is something about permutations and cardinality of perm. Set.

[@beniborukhov9436]

I'm getting a minimum of 0 using the trigonometric substitution. Since f(x) is defined as root 17/2 times sin(theta + arctan(4)), and sin can reach 0, f(x) is 0 when the sin portion is 0. Sin is 0 at pi so theta will be pi - arctan(4) or about 1.816. If we put that value back into our definition of x as sin squared theta over 2, we get an x value of about 0.471 which is within the range [0, 1/2]. Am I missing something?

[@carstenmeyer7786]

The transformed version of *f(x)* is only valid for *sin(𝜃), cos(𝜃) ≥ 0,* as the absolute values were omitted at 2:18ff *=> 𝜃 ∈ [0; 𝜋/2] + 2𝜋k, k ∈ ℤ* The value *𝜃 = 𝜋 - atan(4)* does not lie inside any of those allowed intervals, so we may not use it!

[@nestorabad7743]

The problem is that this sin cannot reach 0, because 0≤θ≤π/2 and then sin(θ+arctan(4)) reaches its minimum when θ+arctan(4) = π/2+arctan(4) (a picture clarifies a lot the situation). This minimum is sin(π/2+arctan(4)) = cos(arctan(4)) = 1/√17. In fact this is the way I found the minimum of f: √(17/2)·1/√17 = 1/√2.

[@gibbogle]

Did you show that x at the max value of sqrt(17/2) is in the range (0,0.5)? If so I missed it.

[@jellyfrancis]

Well the general relativity is needed ❣️💝

[@loonycooney22]

The rms am inequality should pull out the min easily enough I think.

[@synaestheziac]

Relativity please!

[@anon6514]

I bottled out and used calculus to solve it. >< Always forgetting about trig substitutions!

[@Alluminum-z6n]

I want you to tell us about relativity, mr. Penn

[@wolfwerewolf4754]

The last bit with the minimum value seams to be incorrect. Since showing that f^2 is a sum of two functions do not tells us that the minimum of f^2 will be in a point of "joint" minimum. Let f^2(x) be equal to a(x)-b(x), where b(x) is always increasing while a(x) has a minimun value of a(x0)=0. If we continue increasing the x value after the point x0, a(x) will start increasing as well, but b(x) may outweigh this growth and overall value of f^2(x) will decrease, even though a(x) adds a positive value to the sum.

[@colonelburak2906]

But note that the domain for this square root function is [0,1/2], and it is equal to 0 at both endpoints and positive for all interior points. Hence x can not be larger than 1/2. Since we want x to be as large as possible for the 8-15x term, we have to choose x = 1/2 in order to minimize the entire expression.

[@wolfwerewolf4754]

Yes, after mentioning the end of the domain the solution does indeed become valid, you are right. But this still should be mentioned while the author skips it

[@speedsterh]

I too found that this part was a bit sloppy and needed further explanations

[@mke344]

How to do it with calculus

[@PeperazziTube]

The answer is correct, but what is not proven is that x0 is in the domain [0,1/2] of the function

[@TechToppers]

Just plug x = 1/2

[@Hiltok]

x0 (for maximum) = 1/2 (sin(pi/2 - arctan4))^2 = 1/2 (cos(arctan4))^2 = 1/2 * 1/17 = 1/34 which is certainly in [0,1/2].

[@gibbogle]

@@Hiltok I think the OP's point was that this should have been included for completeness.

[@mehmetozdemir8115]

How can you guarentee that theta+Alpha equals pi/2

[@HershO.]

Curious as to whether there's a closed form for arctan(4).

[@sttlok]

Well... There's not.

[@HershO.]

@@sttlok interesting

[@Nikolas_Davis]

You could have simplified the position of the maximum quite a bit. It's not hard to show that x0 = (1/2) sin^2 (π/2 - arctan4) = (1/2) cos^2(arctan4) = (1/2) (1/17) = 1/34

[@t567698]

I want the video about general relativity.😎

[@yasharthyash17]

Pretty esy questions without calculus, it will be sexy and esy going

[@charly6052]

Hi sir , i want to Ask you , how to proof that f IS dérivable n Time

[@leofabregues5824]

mdr

[@charly6052]

@@leofabregues5824 how?

[@leofabregues5824]

@@charly6052 by induction, but you can do it simply with the function, it just depends how "complex" is constructed the function

[@charly6052]

@@leofabregues5824 give me your social to send you a nice exercice

[@siwon9924]

@@leofabregues5824 il pense qu'on a pas capté qu'il était français je crois bien

[@ccg8803]

Me, doing a calculus exam

[@MyOneFiftiethOfADollar]

Dude, take the extra 30 seconds to double check the thumbnail which omitted the constant 4. Thx for problem and it looks like Cauchy Schwarz inequality may get the job done in lieu of the trig substitution you did.

[@sttlok]

Plus the thumbnail is just a random equation for some reason.

[@carstenmeyer7786]

I'd agree with _Cauchy-Schwarz (CS)_ if the assignment was "Find an upper bound for *f* ". To find a min/max, we need to show that *f* actually takes on these values somewhere. For the max it is possible (CS turns into equality iff both vectors are linearly dependend), but I don't see how to do that for the min.

[@MyOneFiftiethOfADollar]

College algebra tells us the domain of the function in [0,1/2]. If we can agree to call sum of two square root functions continuous without mentioning calculus, then mins and maxes will happen at 0, 1/2 or somewhere in between.

[@carstenmeyer7786]

@@MyOneFiftiethOfADollar I agree with your statement above - we know the min lies _somewhere_ in the domain. But the assignment was to actually find the min without advanced calculus, and honestly I still don't see how CS will help with that. Finding the max is easy, as CS turns into equality iff *(√x; √(1/2 - x) ) = t * (1; 4) => √(1/2 - x) = 4 * √x*

[@MyOneFiftiethOfADollar]

@@carstenmeyer7786 I don't see how to use CS to prove the min. CS equality condition tells us how to find the x coordinate of the max. If we could show function is decreasing, without Calculus, from that x coordinate to 1/2, then min clearly occurs at 1/2. CS does show sqrt(17/2) is an absolute max on the domain [0, 1/2] without having to consider the equality condition. Also I did not notice, at the beginning, if the video stated the domain is [0, 1/2] which is alway a good idea. Just did video with instructions to find the max without calculus OR Trig :)

[@physjim]

you are a bit sloppy in places. Like hiding the linear independence under the carpet, ignoring the absolute value of the trig functions, writing f(x)= a sum of trig functions of theta and some other minor things. You tend to speed up when interesting issues arise and ignore them by hand-waving arguments. This video looks like you saw the solution somewhere and you're just repeating it here.

[@MyOneFiftiethOfADollar]

This is an excellent opportunity for you! Produce a video on your own exemplary math channel that improves upon your “perceived flaws” in Dr Penn’s video. I am certain your charm, presence, gravitas, and latent math acumen would attract many viewers and subscribers.

[@carstenmeyer7786]

1:58 You may restrict *𝜃 ∈ [0; 𝜋/2] =: D,* as that is all you need to reach every *x ∈ [0; 1/2].* After the restriction you don't need to worry about signs anymore, because both *sin(𝜃), cos(𝜃) ≥ 0* for any *𝜃 ∈ D:* *f( 1/2 * sin^2(𝜃) ) = √(17/2) * sin( 𝜃 + atan(4) ) =: g(𝜃) // 0 < atan(4) < 𝜋/2 < atan(4) + 𝜋/2 < 𝜋* As *g* is a sine shifted by *atan(4)* to the left, it is increasing for *0 ≤ 𝜃 < 𝜋/2 - atan(4)* and decreasing for *𝜋/2 - atan(4) < 𝜃 ≤ 𝜋/2:* - one local maximum: *g( 𝜋/2 - atan(4) ) = f(1/34) = √(17/2)* - two local minima: *g(0) = f(0) = 1, g(𝜋/2) = f(1/2) = 1/√2* The local maximum is also the global maximum, while the global minimum is *f(1/2) = 1/√2*