Nice Trig Inequality

[@aaditolkar]

I don't understand most of this but your enthusiasm kept me super engaged! More teachers should be like you.

[@plaustrarius]

The turnaround @7:09 is epic!! Definitely cracked a huge smile when you said triangle identity i should have seen it earlier!! Thank you Dr. Peyam!

[@elbrohermanito3496]

Dr peyam! Thank you for teaching math in a very kind and friendly way!! Eres genial, muchas gracias!!

[@harrywotton7728]

Great video. An interesting thing is that mathematical induction shows up not only in integer-only(or rational-number-only) theories but also in theories of real number functions.

[@gregorio8827]

Draw a circle and choose two points on its circumference, such that they inscribe an angle of n*alpha. Divide the arc that inscribes this angle in n arcs of the same length, then, the cord of the big arc is sin(n*alpha) and the sum of n small cords is, n*sin(alpha) and because the shortest path is always the straight line, we get n*sin(alpha)>sin(n*alpha) as desired. There are some details but i think this proof can be extended to be valid to every real alpha.

[@thomasborgsmidt9801]

Interesting formula/inequality. It sort of explains why the higher harmonics to a tone makes the amplitude of the tone more "square" - and why the weight of the higher harmonics gets weighed with the e^(-1)*a*n. That is something every musician knows (or should know); but never explains WHY. Musicians can normally use simple a/b relations, but it took a Mozart to compose for the clarinet. Is that obscure to You - well it is bloody well meant to be!

[@dougr.2398]

Thomas Borgsmidt thé higher harmonics must be addd in the correct proportion to square up a pure tone (sine wave).

[@afrolichesmain777]

For current/future students, the triangle inequality something that is used over and over in higher level mathematics, get used to it. Great video as always!

[@High_Priest_Jonko]

Holder's Inequality is also one that every analyst should know, because it's going to be used constantly and assumed to be obvious

[@aruselizbaryan4042]

Bro I owe you my life. THANK YOU

[@dougr.2398]

This brings to mind a limit that might be proven using L’Hôpital’s rule or other means.... to show that in the limit as X --> 0 that [ N•sin(x) ] / sin (Nx) --> 1. and that N•sin (x) is always the larger of the two functions.

[@anacapillalozano2783]

I love how u explain maths ❤️ Thanks you!!!

[@sanjeevkumardhiman6783]

1:10 “jumping up and down” That was so funny. And can you show us the derivation of the formula for sin(a+b) [maybe with induction; coz I am loving it ]

[@drpeyam]

You can do it with complex numbers :)

[@High_Priest_Jonko]

It's not a proof but if you forget use Euler's Formula: e^i(theta) = cos(theta) + isin(theta). Replace theta with a + b. Then e^i(a+b) = (e^ia)(e^ib) = [cos(a) + isin(a)][cos(b) + isin(b)]. Expand this, then the real part will be cos(a+b) and the imaginary part will be sin(a+b)

[@drpeyam]

^ That’s actually the proof, I think!

[@Kdd160]

DR PEYAM and SYDNEY CARTON You are very intelligent!!; But in this case,we have to use cos(a+b) formula; and SPS we don't know that, what would we do??I think we must approach from an another way :)

[@Mclfarm2]

​@@Kdd160 I think you're right, it needs to be combined with other information. You need to have the properties of complex multiplication established beforehand. If you have established/assumed that multiplying complex numbers adds their angles on the Argand Diagram then combining this with what Sydney has commented should this should be a proof. I read about it on pages 73-76 of "Introductory Mathematics: Algebra and Analysis" by Geoff Smith in the Springer Undergraduate Mathematics series of textbooks.

[@ThePharphis]

This one isn't too bad! thanks for sharing. I think after drawing the picture it's also intuitive that the slope is greater or equal, initially for N*sin(x) and that it will not decrease as quickly (since they start with the same initial slope, I think, but one reaches a maximum sooner!)

[@dohyun031]

Thank you for your great video! Is it possible to use the mathematical induction as follows? (n+1)|sinx| = n|sinx| + |sinx| >= |sin(nx)| + |sinx| >= |sin(nx)||cosx|+|sinx||cos(nx)| >= |sin(n+1)x| (I omitted the triangular inequality step)

[@High_Priest_Jonko]

We have |sin[(n+1)x]|

[@Borja_Moreno]

Oh yeah alright but isnt that obvious? I mean, sin(x) is a function that always go from -1 to 1 and if you multiply that function to a natural number always gives you a number that is larger in absolute value than any number between 0 and 1. I dont know where came from that surprise

[@عمرانآلعمران-و7خ]

Hi you know that the function sin x is uniformly continuous on the real line Is it true for sin(Nx)? if so ,can we prove it by using the above inequality in the blackboard I appreciate your work ,thank you a bunch!

[@drpeyam]

Interesting question! I don’t think it follows from this, but you could just apply the mean value theorem to sin(nx) to get uniform continuity

[@clorulo]

Hi Dr. Peyam, I have a challenge for you! I have found an infinite sum that Wolfram alpha can evaluate but cannot prove. The series is Σ from 1 to infinity of n!/(n^n). The sum is equal to the integral from 0 to infinity of e^t/(e^t-t)² dt.

[@tygodankers6526]

it does check out!

[@rudolfgyorkei9558]

The best on KZbin

[@ramk4004]

Great explanation. Thank you.

[@thomasborgsmidt9801]

How does this apply to conplex x'es?

[@gurindersinghkiom1]

Very nicely explained....!!!!!

[@giannismaris13]

Verry useful for Dirichlet kernel stuffs:)

[@carlosmonteiro4263]

Great problem!

[@haoli9220]

is is possible to show that they meet each other with tangents of gradient = n and -n at some points

[@haoli9220]

At every x intercept of n|sin x|

[@Sahanie]

I just realized you can simplify your name to be πm.

[@neilgerace355]

| sin nx | must be

[@grawe687]

Yeah, the easiest way to prove this is to make that observation and use the fact that |sin(x)| is also less or equal to 1 and to this point the proof is trivial (you're left with n>=1 which is true for all natural number). The fact is that he's trying to explain how to prove something using induction and he's using this exercise as an example

[@johannesh7610]

you still have to prove it for the range near zero where |sin(x)| < 1/n

[@grawe687]

@@johannesh7610 sure, just Taylor expand to the third order and after some simplifications you'll be left with an inequality which holds for all n

[@johannesh7610]

@@grawe687 That sounds like not really a proof. You'd have to estimate the rest of the Taylor series either by this integral term I never found compelling or by estimating the series of the absolute values. Therefore I'd hesitate to approximate a Taylor series in a proof. But maybe you can actually argue with a value of |sin(x)| in sin^-1([-1/n,1/n]) divided into its regions of monotony

[@grawe687]

@@johannesh7610 yeah that's not the way to really prove it, the method you proposed is more adequate

[@algobrax]

Thank you!!

[@sergioverde1322]

If you have 1/n instead of n, the inequality swaps over

[@yyc3491]

I like the figure proof more😀

[@SaojininiranjalaNethu-xi8mw]

❤️🔥👍🏻

[@easymathematik]

Here was some bullsh... :D Big sorry. :)

[@drpeyam]

In fact your identity with cosine is wrong, try x = pi/2 and n = 2. There’s a reason why I chose sin

[@drpeyam]

It’s not as trivial as you think

[@easymathematik]

@@drpeyam Sorry, I was too fast. :D Forget everything, what I said. :P

[@drpeyam]

It’s ok lol

[@sparsetable]

Am I the only one who thinks that Peyam lives in a mansion?

[@drpeyam]

Mansion? 😂😂😂 I live in a studio apartment

[@sparsetable]

@@drpeyam nice studio apartment

[@tylershepard4269]

Did I hear someone say Chebyshev?

[@thedoublehelix5661]

Isn't this one obvious?

[@matheolentz4344]

obvious doesn't mean true

[@easymathematik]

@@matheolentz4344 Obvious doesn´t mean true. This is correct, but since sin(x) is bounded by -1 and +1 by definition (!), this result is trivial. It is like saying 1

[@drpeyam]

@Easy Mathematics No it’s not...

[@thedoublehelix5661]

@@easymathematik yeah that's not correct reasoning

[@easymathematik]

@@drpeyam Can I remind you, that you used the fact (in your words) cosine of junk is less or equal to 1. in your proof?