An Equation For Weierstrass Substitution

[@eurocouto]

This is a beautiful problem with a charming solution. I loved it! Thanks for this lovely gift!

[@AutoDisheep]

Ah this was the coolest way to discover the first equation you learn in trig

[@herbertsusmann986]

Nice! Who would have thought by substituting variables like this anything productive would result. Brilliant!

[@omograbi]

This is the pleasure of math, It doesn't stop from make us wonder.

[@baidonchandipo2804]

You know what! Because of how your brain keeps on improving everyday mathematically, trust me, you will keep on solving more hard olympiad math questions

[@BartBuzz]

Sometimes math just has to be fun. This was one of those times!

[@vladimir10]

Awesome development. But no uniqueness proof?

[@kirthiramaniyer4866]

You are my favorite teacher

[@ThAlEdison]

because I know the identities of the Weirstrauss substitution, I immediately saw that it was csc^x(t)-cot^x(t)=1

[@darickmendes969]

You never gonna lose people , you are great teacher haha

[@Pramit1156]

The spontaneous realization you had is the better way to go through , it will give rise to [cosec(2A)]^x - [cot(2A)]^x = 1 which is only possible if x = 2

[@Marcus-y1m]

I had no clue in the beginning, but even so I know in which level is this equation, thanks sir

[@franolich3]

A quick way to get the answer x=2... [1] ((1+a^2) / 2a)^x - ((1-a^2) / 2a)^x = 1 Let: p = (1+a^2) / 2a q = (1-a^2) / 2a Substituting p and q into [1]: [2] p^x - q^x = 1 Note: p + q = 1/a p - q = a Therefore: p^2 - q^2 = (p+q)(p-q) = (1/a).a = 1 So x=2 satisfies [2] and is therefore a solution to [1]. To prove uniqueness of this solution note that: 0 < a < 1 => 0 < a*2 < 1 => 0 < 1-a^2 < 1+a^2 => 0 < q < p -inf < ln(q) < ln(p) [3] So when x > 0: -inf < x.ln(q) < x.ln(p) -inf < ln(q^x) < ln(p^x) 0 < q^x < p^x [4] 0 < 1/p^x < 1/q^x [5] Also note: p - 1 = (1+a^2)/2a - 2a/2a = (1-a)^2/2a > 0 => p > 1 => ln(p) > 0 Given [3] and [4] then: ln(p).p^x > ln(q).q^x [6] Let: f(x) = p^x - q^x So [1] is equivalent to solving: [7] f(x) = 1 Case 1) x = 0: f(0) = 1 - 1 = 0 so x=0 not a solution. Case 2) x > 0: f'(x) = ln(p).p^x - ln(q).q^x > 0 because of [6] So f is strictly increasing and there can be at most one positive solution to [7] which is x=2. Case 3) x < 0: f(x) = p^-|x| - q^-|x| = 1/p^|x| - 1/q^|x| < 0 because of [5] So no negative solutions to [7]

[@secretsecret1713]

👍

[@xyz9250]

That’s how I got x=2, but your effort to prove that’s the only answer is great.

[@omm-o6i]

I didn't get the uniqueness of the solution

[@franolich3]

@user-fq7ft1tz9k The first part of the uniqueness proof establishes some inequalities. The second part uses these inequalities to show that p^x-q^x=1 has only one solution (equivalent to the original problem). I have added some more explanation to the proof but if this still does not make sense then let me know which part is giving you trouble.

[@sobolzeev]

To say the truth, the one familiar with an identity (1+A)² - (1-A)² = 4A will guess the solution x=2 quite fast. It is more important to explain why it is unique. This is where we really need the form (cosθ)ˣ + (sinθ)ˣ = 1 with 0

[@m.h.6470]

I don't know, where you got (2a/(1+a²))ˣ + ((1-a²)/(1+a²))ˣ = 1 from, but it is NOT the same as ((1 + a²)/(2a))ˣ - ((1 - a²)/(2a))ˣ = 1 For starters, your equation has a solution for a = 0, while the original equation does not.

[@sobolzeev]

@@m.h.6470 My sincere commiseration.

[@sobolzeev]

@@m.h.6470 First, I recommend you to study the original question and learn there that we solve the equation for an unknown x, while 00 as well. Hence we can multiply the equation by (2a/(1+a²))ˣ. I am sure you can do it, obtaining 1 - ((1-a²)/(1+a²))ˣ = (2a/(1+a²))ˣ Now you are left to add ((1-a²)/(1+a²))ˣ on both sides to the equation.

[@m.h.6470]

@@sobolzeev I agree with your calculation, but your comment clearly stated, that the original equation can be rewritten to your equation. You don't mention 0 < a < 1 AT ALL. And without that distinction your comment is simply wrong. With the inclusion of 0 < a < 1, it is correct, but it needs to be made clear, that any result of this new equation is only valid inside these boundaries, while the original equation is NOT limited by these boundaries. The original equation only has the limitation of a ≠ 0, just based on the terms themselves.

[@sobolzeev]

@@m.h.6470 Please accept my even deeper commiseration. You did not observe ((1-a²)/(2a))ˣ. You cannot raise a non-positive base into a real power. Thus, the bounds 0

[@prabhatrexkira398]

Can we do it without Trigonometry?

[@m.h.6470]

You can prove that x = 2 is a solution without trig, but proving, that it is the only solution is tricky - if not impossible - without it.

[@kushagrasharma5983]

Sir could you please teach how to solve cubic equations without hit and Trial method?

[@Alians0108]

There's an extremely long formula if you are super commited. Then whatever that factor is, will yield a quadratic * (x-ă)

[@chintu4398]

@@Alians0108 can you pls tell what the formula is...or where can I find it?any website??

[@GURPARASSINGH-sg8sv]

Lagrange resolvent​@@chintu4398

[@ThAlEdison]

Given a cubic ax^3+bx^2+cx+d=0 x^3+(b/a)x^2+(c/a)x+d/a=0 let x=y-(b/3a) (y-(b/3a))^3+(b/a)(y-(b/3a))^2+(c/a)(y-(b/3a))+d/a=0 y^3-(b/a)y^2+(b/a)^2(y/3)-(b/a)^3(1/27)+(b/a)y^2-(b/a)^2(2y/3)+(b/a)^3(1/9)+(c/a)y-(bc/a^2)(1/3)+(d/a)=0 y^3-(b^2-3ac)y/3a^2+(2b^3-9bca+27da^2)/27a^3=0 rearrange to y^3=(b^2-3ac)y/3a^2-(2b^3-9bca+27da^2)/27a^3 let y=u+v (u+v)^3=u^3+3u^2v+3uv^2+v^3 = 3uv(u+v)+(u^3+v^3) =(3uv)y+(u^3+v^3) =(b^2-3ac)y/3-(2b^3-9bca+27da^2)/27a^3 3uv=(b^2-3ac)/3a^2 uv=(b^2-3ac)/9a^2 (uv)^3=(b^2-3ac)^3/729a^6 u^3+v^3=-(2b^3-9bca+27da^2)/27a^3 u^6+(2b^3-9bca+27da^2)u^3/27a^3+(uv)^3=0 u^6+(2b^3-9bca+27da^2)u^3/27a^3+(b^2-3ac)^3/729a^6=0 u^3=(-(2b^3-9bca+27da^2)+sqrt((2b^3-9bca+27da^2)^2-4(b^2-3ac)^3))/54a^3 u=cbrt(-8b^2+36bca-108da^2+sqrt((8b^3-36bca+108da^2)^2-64(b^2-3ac)^3))/6a v^3=(-(2b^3-9bca+27da^2)-sqrt((2b^3-9bca+27da^2)^2-4(b^2-3ac)^3))/54a^3 v=cbrt(-8b^2+36bca-108da^2-sqrt((8b^3-36bca+108da^2)^2-64(b^2-3ac)^3))/6a y=u+v=(cbrt(-8b^2+36bca-108da^2+sqrt((8b^3-36bca+108da^2)^2-64(b^2-3ac)^3))+cbrt(-8b^2+36bca-108da^2-sqrt((8b^3-36bca+108da^2)^2-64(b^2-3ac)^3)))/6a x=y-b/3a=(cbrt(-8b^2+36bca-108da^2+sqrt((8b^3-36bca+108da^2)^2-64(b^2-3ac)^3))+cbrt(-8b^2+36bca-108da^2-sqrt((8b^3-36bca+108da^2)^2-64(b^2-3ac)^3))-2b)/6a x=(cbrt(-8b^2+36bca-108da^2+sqrt((8b^3-36bca+108da^2)^2-64(b^2-3ac)^3))+cbrt(-8b^2+36bca-108da^2-sqrt((8b^3-36bca+108da^2)^2-64(b^2-3ac)^3))-2b)/6a

[@HighKingTurgon]

I am enjoying the Scriptures at the ends of videos, Mr Newtons. Could you follow the 1/tan2a thread in a separate video? I was hoping you'd come back to that. I mean, I COULD do it myself, but I love walking through problems with your guidance.

[@barryzeeberg3672]

Groucho Marx came up with the "why-a-duck" substitution.

[@balubaluhehe2002]

At 4:27, you could have used 1/tan(2A) because it would have still led to cos(2A)/sin(2A), so it wouldn't have changed anything

[@SiladityaSen1993]

One question though: the question says 0

[@prabhatrexkira398]

U r a Magician ❤

[@laman8914]

We are not mathematicians, but our thinking tells us that this is just one solution for a particular situation, which is within the trigonometry dimension. But there may be other dimensions outside of this one. Is it proper to mention the limitations of this solution?

[@ethandasilva8243]

Excellent!

[@belindedireds]

Great development, but how can i be sure that this is the only solution?

[@jameyatesmauriat6116]

Which book contains these hard concepts?

[@hammadsirhindi1320]

The condition 0

[@zintows]

I love your videos! You are a very charming person

[@PrimeNewtons]

You are so kind

[@dhruvm.s.1103]

Try jee advanced questions.....

[@nasrullahhusnan2289]

u=(1+a²)/2a =½[(1/a)+a] v=(1-a²)/2a =½[(1/a)-a] u+v=1/a, u-v=a and (u^x)-(v^x)=1 u²-v²=1 Comparing (u^x)-(v^x)=1 to u²-v²=1 it is clear that x=2

[@ahmedkafi7524]

Thank you.

[@dirklutz2818]

Amazing!

[@SidneiMV]

wow! how awesome!

[@Harrykesh630]

a = tan(x) ??

[@nothingbutmathproofs7150]

I would normally say that this was sweet, but to use your terminally I'll say that this was smooth. Is there even another way to solve this?

[@PrimeNewtons]

Yes. Someone posted a solution in the comments.

[@GURPARASSINGH-sg8sv]

Sir when you were avoiding to put the value of tan 2a My soul was speaking please please please no no!!!!

[@KazACWizard]

i just retwrote each term as cosec(a)^x - cot(a)^x=1 then rearranged to get the end result.

[@robot8324]

Thanx❤❤❤❤❤

[@comdo777]

asnwer=1ax

[@comdo777]

cos + tan what =1 but asnwer=2x

[@AssetOspanov]

x=2

[@m.h.6470]

Solution: (no trig) with 0 < a < 1: ((1 + a²)/(2a))^x - ((1 - a²)/(2a))^x = 1 assuming a = 0.1: ((1 + (0.1)²)/(2(0.1)))^x - ((1 - (0.1)²)/(2(0.1)))^x = 1 ((1 + 0.01)/0.2)^x - ((1 - 0.01)/0.2)^x = 1 (1.01/0.2)^x - (0.99/0.2)^x = 1 (10.1/2)^x - (9.9/2)^x = 1 (5.05)^x - (4.95)^x = 1 With a keen eye, and knowing about the difference of two squares, you can see, that (5.05)² - (4.95)² = 1 because: (5.05 + 4.95)(5.05 - 4.95) = 1 10 * 0.1 = 1 1 = 1 assuming a = 0.9: ((1 + (0.9)²)/(2(0.9)))^x - ((1 - (0.9)²)/(2(0.9)))^x = 1 ((1 + 0.81)/1.8)^x - ((1 - 0.81)/1.8)^x = 1 (1.81/1.8)^x - (0.19/1.8)^x = 1 (18.1/18)^x - (1.9/18)^x = 1 It is a little bit more difficult to see, but: (18.1/18)² - (1.9/18)² = 1 (18.1/18 + 1.9/18)(18.1/18 - 1.9/18) = 1 (20/18)(16.2/18) = 1 324/324 = 1 1 = 1 so in general: ((1 + a²)/(2a))² - ((1 - a²)/(2a))² = 1 ((1 + a²)/(2a) + (1 - a²)/(2a))((1 + a²)/(2a) - (1 - a²)/(2a)) = 1 ((1 + a² + 1 - a²)/(2a))((1 + a² - 1 + a²)/(2a)) = 1 (2/(2a))((2a²)/(2a)) = 1 (1/a)(a) = 1 a/a = 1 1 = 1 so with x = 2, a only has to be a ≠ 0

[@badralshammari8004]

The light is only from god brother