Solving sin(x)^sin(x)=2

Рет қаралды 396,017

Күн бұрын

We have two exponential equations with trigonometric functions (sin(x))^(sin(x))=2 and (sin(x))^(cos(x))=2. The tetration equation (sin(x))^sin(x)=2 requires us to use the Lambert W function, complex exponential, complex logarithm, and the quadratic formula. However, for (cos(x))^cos(x)=2, we can argue it has real solutions by intermediate value theorem but I do not know a way to solve it algebraically. I know we can use Newton's method to get an approximation for the solution but I wonder if we can somehow also use the Lambert W function to find a closed-form of the solution.
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0:00 I have a math conundrum
0:12 solving (sin(x))^sin(x)=2
7:40 why (sin(x))^cos(x)=2 has real solutions
10:16 can WolframAlpha solve (sin(x))^cos(x)=2?
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Пікірлер: 619

@blackpenredpen 2 жыл бұрын

What do you think about my "e-hoodie"?

@andy-kg5fb 2 жыл бұрын

I got till 2ln(2)×exp(2sinx)=2sinx. In sinx^(cosx)=2. Wolfram alpha can solve from there. I can't. Steps: First take ln on both sides. Then notice you have cos of an angle times something equal to something, so Pythagoras. Then exp both sides. Then simplify. Then Wolfram alpha. Edit: I messed up not 2ln(2)×exp(2sinx)=2sinx But exp(ln(2)²)×exp(sin²x)=exp(ln²(sinx))

@andy-kg5fb 2 жыл бұрын

Your "e-hoodie" is pretty cool.

@Shreyas_Jaiswal 2 жыл бұрын

e=2.7 1828 1828 45 90 45 235 360 ...

@andy-kg5fb 2 жыл бұрын

@@Shreyas_Jaiswal prove your claim.

@Shreyas_Jaiswal 2 жыл бұрын

@@andy-kg5fb From equation (i) and (ii) we conclude, LHS=RHS. Hence proved.

@sharpnova2 2 жыл бұрын

i found a perfectly marvelous closed analytical solution to this equation but the comment section is too small to contain it

@alham9656 2 жыл бұрын

so true

@ikocheratcr 2 жыл бұрын

360yr later and the line never gets old ;)

@dianeweiss4562 2 жыл бұрын

Give us the link to your video.

@maxwell1594 2 жыл бұрын

Are you rebirth(reincarnation) of Fermat 😊 ?!

@yunghollow1529 2 жыл бұрын

"The link, dude. You forgot the link" -Albert Einstein

@HungNguyen-rj3ek 2 жыл бұрын

It's actually okay to have an irrational number that cannot be explained in known functions and constants. To solve this equation, just make up a new function like Larmbert W for it.

@blackpenredpen 2 жыл бұрын

😆

@holyshit922 2 жыл бұрын

That could be an idea

@abhishekdevkota9538 2 жыл бұрын

@@blackpenredpen new Maths discovery kzfaq.info/get/bejne/bJl8nsqGqpibmmw.html

@gary.h.turner 2 жыл бұрын

We could call it the "BPRP" function: B(x) = (sin x)^(cos x)

@HungNguyen-rj3ek 2 жыл бұрын

@@gary.h.turner It should be: If f(x) = (sinx)^(cosx) Then B(x) = f^(-1)(x)

@gabequinn9796 2 жыл бұрын

How to make a homemade chain rule equation: 1. Preheat the oven to 400° 2. Gather all of your constants together, slowly mix with your variable(s) 3. Wrap this mix in parentheses, top off with a variable exponent or natural log 4. Bake for 45 min 5. Let stand for 5-10 min and serve Servings: 24 calculus students

@hellblazephoenix643 2 жыл бұрын

😂

@LIA-52 2 жыл бұрын

Looks like a recipe for the Chef language.

@chri-k Жыл бұрын

@@LIA-52 yes

@johndoe9659 2 жыл бұрын

Numerical approximations are, to a mathematician, what frozen ready meals are to a decent chef.

@jefflambricks 2 жыл бұрын

Is it a good thing or a bad thing?

@LiteralBacon 2 жыл бұрын

So you use them when you're feeling lazy and want something handy

@nope110 2 жыл бұрын

Used constantly but not enjoyed much? Chefs dont eat anywhere near as well as you'd think

@bowkenpachi7759 2 жыл бұрын

And this is why I hate it when it’s argued that .9 recurring is equal to 1

@oliverqueen5883 2 жыл бұрын

🤣

@wesleydeng71 2 жыл бұрын

Well, in the spirit of W Lambert function, let's define a Q function such that Q(x) is the solution of sin(t)^con(t) = x. Then obviously, the solution of the equation is Q(2).

@viharsarok 2 жыл бұрын

Agree. The Lambert W is a kind of cheating itself.

@DatBoi_TheGudBIAS 2 жыл бұрын

Give a numerical approximation of Q(2)

@aa01blue38 2 жыл бұрын

@@viharsarok well, so is sine and cosine

@pedrosso0 2 жыл бұрын

@@viharsarok Well, couldn't you say the same for ln? e^x=2, what is x? x=ln(2) where e^lnx := x Isn't that kind of cheating itself?

@Felipe-sw8wp 2 жыл бұрын

@@pedrosso0 all you guys contesting the elementary functions + lambert's, you have a point. However, all of those, including Lambert's W have a nice looking Taylor series. We could ask ourselves if that could also be the case for the Q function?

@69k_gold 2 жыл бұрын

My teacher always said "When you got a difficulty with algebra, use calculus to prove the sum" "When you got a difficulty with calculus, use algebra to prove the sum" "When both don't work, use trigonometry and get over it"

@Vortex-qb2se 2 жыл бұрын

Isnt trigonometry about triangles and Geometry shit 😭

@Thecurseofoctober 2 жыл бұрын

@@Vortex-qb2se the trigonometric values can be used to solve a lot of problems cause there are sooo many calculus formulas which revolve around trigonometry.

@Propane_Acccessories 2 жыл бұрын

@@Thecurseofoctober Trigonometric substitution wrecked me in calculus. Our professor didn't let us use a formula sheet, so we had to memorize all the substitutions and their derivations since there was not enough time to work them out manually. Luckily the class got a "do-over."

@TatharNuar 2 жыл бұрын

@@Propane_Acccessories For me, drawing out the triangle helped with trig substitution.

@yoav613 2 жыл бұрын

If you write in wolfram x^-sqrt(1-x^2)=2 you get the numeric sol x=0.4584 so x=arcsin(0.4584)=2.6653+2npi

@MichaelGrantPhD 2 жыл бұрын

Indeed, I think that once you concede that you can only solve it numerically, it's reasonably straightforward.

@yoav613 2 жыл бұрын

@@MichaelGrantPhD yes i think too that this can be solved only nemeicly,but i am not sure,anyway i guess that if wolfram gives only numeric sol,so this is the only way to solve it

@andrasfogarasi5014 2 жыл бұрын

@@benoitavril4806 I'm pretty sure it would violate Gödel's incompleteness theorem if WolframAlpha could give an analytical solution to all solvable problems.

@hOREP245 2 жыл бұрын

@@andrasfogarasi5014 Gödel's incompleteness theorem's have nothing to do with this.

@verifiedgentlemanbug 2 жыл бұрын

@@hOREP245 Gödel's incompleteness theorem have something to do with this

@goliathcleric 2 жыл бұрын

I haven't solved the second one (it's 5am and I just woke up) but my instinct is telling me the first step is going to be to convert both sin(x) and cos(x) to their complex equivalent using e. I think it'll be similar to cubic equations, where you have to journey through the complex world to find their real solution.

@Metalhammer1993 2 жыл бұрын

Good Idea, I thought it might legit be necessary to brute force it into a differential equation (utilizing cosine being the derivative of sine) to arrive at some insane formulation for sine that takes a week to prove that it actually IS sine and hope you can set that crazy MFer to the power of its derivative equal to two. Yeah I know the very idea is bonkers and can't work for several reasons I just don't see yer, but I'll try (and regret it) now xD

@fedem8229 2 жыл бұрын

I don't think that helps. I believe there's no nice answer and the best you can do is a numerical approximation

@Metalhammer1993 2 жыл бұрын

@@fedem8229 yeah most likely.

@williamcamp7665 2 жыл бұрын

@@Metalhammer1993 did you figure out how to solve it?

@scarmackd1498 2 жыл бұрын

Me in 10th grade like hmm yes I see 🤔

@Ploofles 2 жыл бұрын

Thank you for making these videos, there are very fun to watch, very educational and you explain difficult (at least for me) topics very well! You are blessing on this Earth! Thank you

@forgetittube5882 2 жыл бұрын

As any transcendental equation, it doesn’t have an algebraic closed form solution. Of course, as many have already observed, you could try to define/find an auxiliary transcendental function Z() (‘similar’ to the Lambert one), but you would only ‘shift the problem’ (e.g. finding a solution in terms of Z(k) would just convert the problem in terms of solving Z(K), by definition, transcendental). Anyway, looking the function f(x)=sin(x)^cos(x)-2, it is cyclical (2pi) and in the interval [0,pi] it is real, diverging near pi. So, there is for sure a solution to the equation f(x) == 0. Using Numerical Analysis, newton-rapson (Xn+1 approx. -f(Xn)/f’(Xn) ) behaves really badly (lol, diverges quite quickly… the derivative diverges near pi as well). Trying two different ‘recurring equations’ Xn+1 approx. arcsin (exp (log(2)cos(Xn))) And Xn+1 approx. arccos (log(2)/log(sin(Xn))) Using an initial value Xo=1, unfortunately, (interacting over the function with matlab) one gets X = 1.02197646023983-0.973667917229243 (lol, matlab… btw, f(X) = i * 0.2220446049e-15 which isn`t that bad, it is a solution, just that it is a complex one, outside the [0;pi] interval) So, as a last resort (to find a real value), I tried the most mundane of all approaches: simple bisecting, starting with two values Xa=2.5 and Xb=2.9, defining the next value Xc = (Xa+Xb)/2 and evaluating f(Xc) (if greater than zero, Xb=Xc, if less than zero, Xa=Xc, rinse repeat)… after a few interactions, finally, it results in X = 2.66535707927136 (f(X) = 1.332267e-15 (real)) So, at least two set of solutions for the problem X=1.02197646023983-i*0.973667917229243 + n*2pi X=2.66535707927136+n*2pi ======= Edit:: now an algebraic approach (lol, I think I found one) Being `creative`, in sin(x)^cos(x) = 2 replacing sin(x) by sqrt(1-cos(x)^2) and extracting the log of both sides, one gets Cos(x) * log ( sqrt ( 1 - cos(x)^ 2)) = log (2) which can be rewritten as cos(x) / 2 * log ( (1-cos(x) * (1+cos(x) ) = log(2) or just Log(1+cos(x)) + log(1-cos(x)) = 2*log(2)/cos(x) now, one can take the derivative in both sides (d/dx)… I know..”tricky” (formally dangerous) -sin(x)/(1+cos(x)) + sin(x)/(1-cos(x)) = 2*log(2)*sin(x)/cos(x)^2 which can be rewritten as 2*sin(x)*cos(x)/(1-cos(x)^2) == 2*cos(x)/sin(x) = 2*log(2)*sin(x)/cos(x)^2 that follows cos(x)^3 = log(2)*sin(x)^2 or just cos(x)^3 = log(2)*(1-cos(x)^2) so, we, finally, have cos(x)^3 +log(2)*cos(x)^2 - log(2) == 0 Lol, using wolframalfa to solve this last one x = 2pi * n +/- 0.789383 x = 2pi * n +/- ( 2.13703 +/- i * 0.759387 )

@sithlordbinks 2 жыл бұрын

Wow, nice solution

@lih3391 2 жыл бұрын

I think the solution found with bisecting is the only correct real number solution looking at the graph on desmos. I can't verify the complex ones, but I don't think just taking the derivative on both sides works without integrating it back. For example, x^2=5 if you take the derivative on both sides, 2x=0 and I think you see the problem here

@saimohnishmuralidharan5440 2 жыл бұрын

Great Effort. Use the Cubic Equation to not make it as an approximation.

@forgetittube5882 2 жыл бұрын

@@lih3391 I know, that`s why I wrote `formally dangerous`… and it`s tricky, because, if you integrate it back you have two problems, obtaining the correct constant integral and solving the `resulting equation` (back to square one, it`s a transcendental equation again (I did the effort, no go…))

@forgetittube5882 2 жыл бұрын

@@saimohnishmuralidharan5440 with wolfram alpha you can obtain the complete, exact, solution, the expression is just painful long/complex (no point in even trying to copy and paste it here…)

@hassanalihusseini1717 2 жыл бұрын

That were two similiar equation with a surprising different solution. Thank you for that!

@kashgarinn 2 жыл бұрын

You already know x has to be between pi/2 and pi, and you already know you’re looking for a negative power, which means it becomes about a ratio between sinx and cosx that must equal 2, i.e. Sinx/cosx = 2 where pi/2 < x < pi. I don’t know whether replacing cosx with a trig equality, or doing ln(sinx)-ln(cosx)=ln(2) would lead to an answer through switching to euler representation.

@petrie911 2 жыл бұрын

You can reduce the complex analysis in the first part considerably by solving (cos x)^(cos x) instead, then using cos(ix) = cosh(x) to end up with a purely real equation. Then use sin x = cos(pi/2 - x) to get the solution to the original equation.

@vanderavongola 2 жыл бұрын

hi @blackpenredpen! This is such a great video! I solved this equation almost the same way as you did but I had a different approach starting from 4:40. I expressed e^ix as cos x + i sin x and grouped real and imaginary terms. What I got was x = sin^(-1) {exp[W(ln2) \pm \sqrt(exp[2*W(ln 2)]-1)} which is a real solution if exp[2*W(ln 2)]-1. Is this consistent with your solution? Hope you see this!

@paulkolodner2445 2 жыл бұрын

The first equation is easy to solve using a calculator: with y=sinx, we have y = exp(ln2/y). Guess a value of y for the RHS, obtain a new value using this equation. Iterating leads to y=1.5596105... Figuring out the complex value of x is your problem. The second equation requires more button pushing because you have to compute y = exp(ln2/(y^2 - 2)). Iterating leads to y = 0.4280110...

@myuu22 2 жыл бұрын

Wolfram Alpha might have timed out, but Desmos did not. I just graphed y=(sin x)^(cos x) and y=2 and found where the two graphs intersected. The intersection points are, to three significant figures, 2.665+2πn

@orangenostril 2 жыл бұрын

Desmos is god tier

@JemEklery 2 жыл бұрын

Desmos "brute-forces" such equations and calculates y for every x on screen. We are looking for a clean solution

@sergiokorochinsky49 2 жыл бұрын

WolframAlpha does not time out, he just doesn't know how to use it. Try writing the equation using Mathematica Language, that is, Sin[x]^Cos[x]=2, and you will have much better information than that given by Desmos. (Never mind using the correct Mathematica command: Solve[Sin[x]^Cos[x]==2,x])

@jbrady1725 7 ай бұрын

You're right. Wolfram Alpha actually gives a result with that input, in exact form.

@andrewshaw6921 2 жыл бұрын

You seemed so devastated when you couldn’t solve the second one. I felt your pain there. Great vid :)

@PeterBarnes2 2 жыл бұрын

Tried out the sin^cos a bit. Following from your next step, I substituted sin(x) for u, giving +/- sqrt(1-u^2) * lnu = c Where c is ln(2). I saw the lnu and thought that could be useful if I was integrating, so I integrated. It isn't pretty, but you can integrate. But that's about it. I have no idea how to use the integrated... thing, now. I'll put it here in case anyone has any ideas: Where c_1 = ln(2), c_1*u + c_2 = +/- [ arcsin(u)/2u - sqrt(1-u^2)/2 + ln(1+sqrt(1-u^2) / u) ]

@chikenwingsteve Жыл бұрын

I think x values are not equal anymore once you integrate

@gamerpedia1535 Жыл бұрын

@@chikenwingsteve if two things are equal (eg. a=b) then applying the same operation on both sides gives an equivalent result.

@chikenwingsteve Жыл бұрын

@@gamerpedia1535 alright, let me test your logic : Let's say the parabola x^2 -1 = x +5 I integrate both sides (even tho there are no dx) and get : (1/3)x^3 - 1x should equal (1/2)x^2 + 5x Which means x should equal to either : - 0 - about -3.56 - about 5.06 Now, let's test all of these solutions in our original equation : x^2 -1 = x+ 5 Let's try 0 first : It would mean that -1 is equal to 5, which is absurd, so get that solution out. Let's try -3.56 : It would mean that 11.6736 is ruffly equal to 1.44, which is absurd. Let's try the very last solution... : 24.6036 should ruffly be equal to 10.06 Would you look at that... None of the solutions are possible. Integrating changed the entire equality. Therefore, you cannot just randomly integrate both sides and hope that the x's are still equal. In order to integrate, you must have a differencial equation. In other terms, you must have a dx, or a dy, or whatever variable that you need to integrate. Tell me if you see any mistakes in my reasonning. Thanks.

@yunghollow1529 2 жыл бұрын

Brilliant Yung Man, i like your videos.. Thanks for enlightening this yung hollow's mind.

@wiseSYW 2 жыл бұрын

you'll need to define a new function like lambert W, I guess using google, putting in (sin x) ^ (cos x) and moving the mouse at the graph, you'll get x = 2.666... when y is about 2.0003...

@gregwochlik9233 2 жыл бұрын

I used your suggested x-point, and got "Solution found at x = 2.66535707927136 (c = 2.0)" from my Python script, which uses the secant method (similar to Netwon's)

@DatBoi_TheGudBIAS 2 жыл бұрын

The approx is 2.6653

@jschnei3 2 жыл бұрын

7:21 I'm tempted to keep massaging the algebra here. Factor e^(2W(ln2)) out of the radicand. Since it's square, this factor emerges from the radical as e^(W(ln2)). It then factors out of the argument of the log, causing the log to split. It then cancels with the log. The result is gorgeous: π/2 − i (W(ln2) + ln(1 ± √(1 − e^(−2W(ln2)))))

@abhishekdevkota9538 2 жыл бұрын

new Maths discovery kzfaq.info/get/bejne/bJl8nsqGqpibmmw.html

@ubern3rd 2 жыл бұрын

I was watching this at work and decided to try something on the second one. Instead of (sin(x))^cos(x) = 2, why not do (sin(x))^cos(x) = e? Since 2 is just a real number, we can use another real number to understand what it's doing, also I'm not very familiarized with the Labert W function with the exception that it gives you the fish back. I get stuck, you'll see where, but I wanted to lay out my thought process to see if anyone had any thoughts on it. Start: (sin(x))^cos(x) = e Take the cos(x) root of both sides: sin(x) = e^(1/(cos(x))) Divide by cos(x) on both sides: sin(x)/cos(x) = (1/cos(x))*(e^(1/(cos(x)))) We have a tangent function on the left with a n*e^n expression on the right! (I knew that based on the wolfram graph that this looked kinda like a tan function). Simplify: tan(x) = (1/cos(x))*(e^(1/(cos(x)))) Lambert W Function: W(tan(x)) = 1/cos(x) I don't know how to simplify W(tan(x)) further, so I guess that's where I stop. If we can find this out, then theoretically, we can do it for (sin(x))^cos(x) = 2, just convert 2 to e^n and go from there? Just a thought and any thoughts on this would be cool. I'd love to see bprp solve for this :)

@TatharNuar 2 жыл бұрын

I love how you juggle the markers so easily.

@chenghowkoh2178 2 жыл бұрын

I tried solving it be letting y=sin x, then manipulating it to differentiate both sides. Where we have, ln y/ln 2 = +- 1/sqrt(1-y^2). If we input this pre differentiation, we will get numerical solution for y = 0.458437... when considering the negative side of the equation. However, after differentiation, you will get a cubic equation but my y values are all complex at least considering the negative one, while the positive one will yield a real value which still differs from the y= 0.45... solution as above

@Fematika 2 жыл бұрын

Just because two sides are equal at one point, doesn't mean their derivatives are. For instance, x^2 + 1 = x^3 has a solution, and so does 2x=3x^2, but they are difference values of x.

@chenghowkoh2178 2 жыл бұрын

@@Fematika yep, I just realised that! However can you explain when equality holds after differentiation and when doesn’t it, because I seem to remember when I was still in school that it also could be used to solve?

@user-dh8oi2mk4f 2 жыл бұрын

@@chenghowkoh2178 If you can remove all the constants then it will hold

@Fematika 2 жыл бұрын

@@chenghowkoh2178 If the equality is always true on some open interval (a,b) with a < b, then so are all of the derivatives. This is how you derived Taylor series, because you want the functions to be equal on an entire interval, not just at one point. Basically, if in some region they are equal, then their derivatives are equal on that region, but not just at one point.

@rossjennings4755 2 жыл бұрын

A tangent half-angle substitution (always a good trick) puts (sin x)^(cos x) = 2 into the juicy-looking form (2/(1+t^2) - 1)*ln(2t/(1+t^2)) = ln 2, where t = tan(x/2). Unfortunately I don't think any more progress can be made from there -- even though you can get a term like A ln A, where A = 2/(1+t^2), by expanding the left side, there's a bunch of other terms too that throw a wrench in things, so you still can't use the Lambert W function. Maybe there's some other sneaky trick that can make it work, but I doubt it.

@csehszlovakze 10 ай бұрын

4:25 I think if you don't bring the 2 in the front you'll have a sqrt(a^2-b^2) inside, which would simplify a lot of things

@youkaihenge5892 2 жыл бұрын

Since we know that cos(x) and sin(x) are complex exponentials couldn't we write these as a Complex Fourier Series raised to another Complex Fourier Series? And just find the coefficients such that when they are plugged into the exponential Fourier that it produces 2?

@darzkzthelegend9667 2 жыл бұрын

Would a maclaurin/ taylor series be useful for the sin(x)^cos(x) equation

@ankurage 2 жыл бұрын

This is the time for the mathematics world to stand up and generalize the Lambert function wider than ever

@jackossie 2 жыл бұрын

Have you tried substituting cos(x) by sin(x+pi/2) in the 2nd problem and then solve it in the same way as you did in the first problem?

@alexk5990 2 жыл бұрын

This is the fastest I havent understood anything in a video in a long time

@billcad15 2 жыл бұрын

The solution to the second problem is close to 2.6653571 radians, which I calculated by iteration. Can you show how to prove whether the answer is going to be rational or irrational? BTW, your videos are fantastic.

@tamasburik9971 2 жыл бұрын

So cool Way out of my depth but I'm excited to see the solutions people find

@factsheet4930 2 жыл бұрын

If you have a good enough computer, and you work out the period of the function, you can then work out pretty easily (given you have high enough precision) an approximation to the value of x. The function happens to be continuous on [2, 3], so just use the intermediate value theorem to narrow down like a billion times. I got x ~ 2.6653570792713603... Code: from math import sin, cos def func(x): return (sin(x) ** (cos(x))) - 2 def intermid(small, big): mid = (small + big) / 2 for i in range (1000): if func(mid) < 0: small = mid else: big = mid mid = (small + big) / 2 return mid print(intermid(2, 3))

@rsv9999 7 ай бұрын

are you sure your value isn’t off due to floating point imprecision?

@120Luis 2 жыл бұрын

I was about to get mad that you didn't consider the log branches, until you did lmao That was a pretty clever way to bring that +2nπ without going into the complex log definition

@smartube4828 2 жыл бұрын

I liked it. But how about we use log2 base 2 instead of Ln?

@chessematics 2 жыл бұрын

Little fact: e^(W(x)) = x/W(x). So one more step of simplification.

@Gniaum 2 жыл бұрын

Isn't it e^((W(x)) = W(x)*e^(W(x))/W(x) = x/W(x) ?

@chessematics 2 жыл бұрын

@@Gniaum yeah that was just a slip of typography

@nicholasng5227 2 жыл бұрын

My suggestion is to square both sides of the equation and get sin ²x ^cos x=4 (1-cos²x)^cos x=4 Let u=cos x then (1-u²)^u=4 By using intermediate value theorem, we can show that u is in between -0.88 and -0.89(Yes I did a lot of trial and error) Then use Newton Raphson Method, we can get a great approximation of u(u≈ -0.8894), then u=cos x, x≈ 2.6668 Perhaps if we want to find the exact value, maybe we should introduce a new function like a Lambert W function?

@awkwardhamster8541 Жыл бұрын

Its wrong

@awkwardhamster8541 Жыл бұрын

I get 152.714≈

@nicholasng5227 Жыл бұрын

@@awkwardhamster8541 Well, maybe try use False Position Method? I learnt this method now and is indeed easier than Newton-Raphson since False Position does not required to find the derivative of f(x). Probably there is an error in your derivative of f(x), I guess? I tried on my own, answer is x=2.6654

@awkwardhamster8541 Жыл бұрын

Put ur x value on the calculator it doesn't work

@awkwardhamster8541 Жыл бұрын

Anyways here's how I solved it ... I literally just estimated that sinx would be anywhere in between 1/2 and cosx will be something like -1 ... So in four quadrants the second quadrant will be where x lies on ...so by trial and error I get that someting in between 152.5-153 gives a real good approximation for this equation . So ye

@holdenmacock8526 2 жыл бұрын

Hi, could you make a video about adding 1/x to different functions? More specifically, why is the new function is asymptotic to the original? I know that when you add functions together they combine characteristics, but I do not know how to prove this mathematically for any example other than x^2 + 1/x.

@blackpenredpen 2 жыл бұрын

Bc the limit of 1/x is 0 as x goes to inf. So we are like adding nothing.

@holdenmacock8526 2 жыл бұрын

Ok, I haven’t learned about limits yet so that helps.

@dennisren5786 2 жыл бұрын

if im undertanding you right, then you can combine the fractions to show that there is always an asymptote at x=0 ex: sin(x)+1/x = sin(x)/1 + 1/x = xsin(x)/x + 1/x = (xsin(x)+1)/x

@abdoa2477 2 жыл бұрын

Good work my friend I admire what u're doing

@abdallahgamal6250 2 жыл бұрын

I love your channel too much♥️♥️, but I want to know your strategy in thinking to solve mathematical problems.

@user-pr6ed3ri2k Жыл бұрын

if you already know the value of the real square superroot of 2 (x^^2 = 2 or x^x =2), the sin one is obvious but tbh not many people do tetrational root stuffs

@srideviganesh441 2 жыл бұрын

What about trying to put cos(x) as sin(pi/2 - x) or sin(pi/2 + x)

@lachouetteaveugle4893 2 жыл бұрын

the final solution of the first equation look like the arcos and the arcsin fonction. I didn't rly understood what's the W fonction you used. But the othere part are really good thx for this one !

@crane8035 2 жыл бұрын

actually if you are good with x being as an inverse function of cosine it is possible we can write the sinx^cosx as ( if we let ln(sinx)=a and ln(2)=b) cosx*(e^a)=e^b cosx=e^(b-a) cosx=e^(ln2-ln(sinx)) ln(cosx)= ln(2)-ln(sinx) and after a tedious bout of calculation x=cos^-1((0.5(1+i*(15^0.5)))^0.5)

@JLConawayII 2 жыл бұрын

How you solve this is you make a graph and find the intercept (or make a guess, it's clearly in the second quadrant somewhere), then make a series of increasingly accurate approximations using Newton-Raphson to get as close to the actual value as you need. I get x=2.6653570792714 after a few iterations. If you're hoping for some closed-form solution, you're going to be waiting a long time.

@akhilrao2015 2 жыл бұрын

sin(x)^(cos(x)) blows up to infinity at x =pi; ie 0^-1 hence the equation has real roots. I do not know exactly how to solve the equation probably talyor series, but you can approximate it by just guessing values!

@shreejipatel2084 2 жыл бұрын

Someone give this man a bigger board! ! !

@Jrcoaca 2 жыл бұрын

Best I could do is Let y = cos(x) y^2 = 1 - 4^(1/y) Using logarithms. If you numerically solve for y and take arccos you will get x, then if you plug that x in, you will get 2. Can’t find out how to analytically solve it though.

@iabervon Жыл бұрын

One thing that's worth remembering is that sin^-1 x for all real x>1 has real part pi/2 plus 2npi.

@MrALFA1 2 жыл бұрын

I loved your hoodie with the Neper Constant.🌺

@arimermelstein9167 2 жыл бұрын

It’s a pain the butt, but you could use Newton’s method or bisection to find an approximation to the real solution. I’m not sure how to do it analytically either.

@georget8008 2 жыл бұрын

An approach to find a solution. 1. We prove that the function f(x)=(sinx)^cosx-2 is increasing in the domain [π/2,π]. 2. We observe that f(π/2)=-1 and f(2*π/3)=2^(sqrt(3)/2) -2>0 3. We apply the binary search algorithm to find the solution in the (π/2, 2*π/3)

@skylardeslypere9909 2 жыл бұрын

Proving sinx^sinx = 2 has no real solution (without finding the complex solution) can be done as follows: Suppose sinx^sinx = 2 and let y=sin(x). Suppose y ≥ 0 (because exponents aren't really well defined for negative bases). Because y ≤ 1, we have y^y ≤ 1 which means y^y = 2 can never be satisfied for y in [0,1]

@anshumanagrawal346 2 жыл бұрын

That's what I did in my head

@TrinoElrich 2 жыл бұрын

For y in [0,1], y^y is not less than or equal to y lol

@anshumanagrawal346 2 жыл бұрын

@@TrinoElrich oh ya

@anshumanagrawal346 2 жыл бұрын

That's true

@skylardeslypere9909 2 жыл бұрын

@@TrinoElrich Oops, you're right. It is true that it's less than or equal to 1 though

@Ethiomath16 2 жыл бұрын

Wow, Great explanations Dear

@bol9332 7 ай бұрын

I am wondering if the result of sinx to the cosx power is a periodical number. After some experimenting (not good experimenting tho), I found something interesting. The approximate result is 2.665356. But if you add the part 65356 as many times as you want to it, you get closer and closer to 2. Example: sinx to the cosx power ≈ 2 for x = 2.665356653566535665356. The exact result is about 1.999998228 = x

@LunizIsGlacey 2 жыл бұрын

The second (just looking at intersection of x^2+y^2=1 and x^y=2, this wasn't my idea but someone else's) has a solution close to around x=2.665 radians. It's exact value, ¯\_(ツ)_/¯

@Goku_is_my_idol 2 жыл бұрын

Nice to learn something new

@AbouTaim-Lille 7 ай бұрын

The main problem is just solving U^u = C and it is pretty easy if we have a tool called Lambert W function then the rest is just replacing u with sin X and taking the arcsin.

@LuigiElettrico 2 жыл бұрын

In the meanwhile best parenthesis closing ever :D

@narfee7529 2 жыл бұрын

That hoodie is awesome! Where can I get it!?

@sylowlover 2 жыл бұрын

You forgot to add the multiples of 2pi*i as complex log is a multifunction :)

@bbqandchill8631 2 жыл бұрын

I noticed that too, but he made up for it by noticing that sin is periodic. This gives back the same result

@sylowlover 2 жыл бұрын

@@bbqandchill8631 those are real integer multiples of 2pi, complex log is imaginary integer multiples of 2pi

@bbqandchill8631 2 жыл бұрын

@@sylowlover but the whole function gets multiplied by i, meaning if you write out the imaginary multiples of 2pi, you get the real multiples if you take it out if the brackets

@rikschaaf 2 жыл бұрын

You could try something with newtons method or with a taylor series

@coreyclemons7573 2 жыл бұрын

Late to the party. I tried my own approach for p2, I haven’t seen anybody try it this way. I got stuck near the end but maybe somebody can piggyback off of me. I split the sin and cos bits by their half angle formulas so that the whole system was in terms of cos, no sin involved. From there I could do a lot of simplifying and conjugate multiplying. The new form of the equation that I ended up with was: -sin(2x)/2 • ln(csc(x)) • e^ln(csc(x)) = ln(2) Left side of the equation lines up with the log of the old expression in desmos so I must have done something right. Plugging this into wolfram doesn’t do me any good. If anybody can take this further, please give it a try.

@jeremymwilliams 2 жыл бұрын

Numerical Method seems to be the only way.

@FrostBurn69Thingy Жыл бұрын

Me searching for this in English and not even understand it with my native language 😂

@kevinibarravera9265 2 жыл бұрын

x=arccos(-1/n) where n is solution of the equation 1-1/n^2=1/4^n.

@yinsdemise 4 ай бұрын

Hey, I thought of giving the (sinx)^(cosx)=2 a try and I may have found something (got stuck). Here is what I tried to use to solve it: sin(x) = 2*tan(x/2) / (1+(tan(x/2))^2) cos(x) = (1-(tan(x/2))^2)/(1+(tan(x/2))^2) I then tried to make a u-sub by setting (1+(tan(x/2))^2) = u. The formulas above (not replacing sin and cos) would look like: sin(x) = (2* sqrt(u-1))/u cos(x) = (2-u)/u Where: sqrt = square root (tan(x/2))^2 = u-1 => tan(x/2) = +/- sqrt(u-1) (took the positive part) (sinx)^cos(x) = 2 => [(2*sqrt(u-1))/u]^[(2-u)/u] = 2 and this is where I got stock 🤣

@stlemur 2 жыл бұрын

does it help if you use a different identity for cosine, like cos(x) = sin(pi/2 - x)?

@rikthecuber 2 жыл бұрын

You cant use many other trig identities then.

@anshumanagrawal346 2 жыл бұрын

Nope

@ayaan5540 2 жыл бұрын

Weird... when I input the [(sin x)^(cos x)] = 2 function into Wolfram Alpha I am getting an answer as well as an equation for x, which is quite long and involves tan and exp and log, but is kinda similar to the solution you arrived at for the other equation. I also verified the value from Desmos (x ≈ 2.66536). Still not sure how to arrive at the given equation though, but maybe you can if you tried working backwards. I can send it if you want.

@AndrewJohnson-ur3lw Жыл бұрын

For the second version sin(x)^cos(x) could you use sin(x)^sin(x- 3/2 Pi) as the starter then follow the same steps of sin^sin

@yolanenarkice4844 2 жыл бұрын

why s he holding a pokeball?

@Abhay0505 Ай бұрын

It's his microphone

@TheInvisiblePickaxe 2 жыл бұрын

I'm confused around the 3:10 mark, what happens to -e^(-ix) when he multiplies through?

@kasuha 2 жыл бұрын

I can't put my finger on it (my math is rusty) but I have feeling it could help to use substitution like y=x+pi/4. Or maybe not, I don't know.

@JakeFace0 2 жыл бұрын

Is there a reason why we didn't say " ln(i) = (π/2)i + (2π n)i "? Other than the fact that we were going to add 2pi*n later anyway?

@kavyapatel3936 2 жыл бұрын

I am a 11 sci student of India and in functions chapter we learnt that putting and power to a function doesn't affect its range, so I knew it would be wrong to put sinx in power and expect 2 but was amused by the answer as it came out a value close to it's range just outside it

@matniet43 2 жыл бұрын

For the second one I think you could square both sides and rewrite sin²x as 1-cos²x, so that there aren't any radicals in the exponent and just a variable (cos(x))

@pk1pro 10 ай бұрын

Squaring makes it sinx^(2sinx)=4

@matniet43 10 ай бұрын

@@pk1pro That's right! I didn't notice that at first. Maybe before squaring you could take the ln if both sides so that the exponent becomes a factor and we have really just one variable sin(x)

@matniet43 7 ай бұрын

@@pk1proMore importantly, the mistake you made here was referring to the first equation, when I explicitly told I was talking about the second equation

@pk1pro 7 ай бұрын

@@matniet43 ohh yeah makes sense

@XJWill1 2 жыл бұрын

I tried a Weierstrass substitution, t = tan(x/2) , sin(x) = 2*t/(1+t^2) , cos(x) = (1-t^2)/(1+t^2) but that did not help. I could not find a closed form solution for t, and neither could Wolfram.

@herbie_the_hillbillie_goat 2 жыл бұрын

Now I feel like I must solve it.

@rongjunhuang258 2 жыл бұрын

It does have the exact solve wolframalpha just need more time to compute it -2+exp((log(tan(x/2)/(tan^2(x/2)+1)) (1-tan^2(x/2)))/(tan^2(x/2)+1)+(log(2) (1-tan^2(x/2)))/(tan^2(x/2)+1))+2 \[Pi] n

@JayTemple 2 жыл бұрын

If you raised both sides to the cox x power, you'd have (sin x) ^ (cos^2 x) = 2 ^ cos x, and you could replace the exponent on the left with 1 - sin^2 x. I don't know if that's helpful.

@terezakot2921 2 жыл бұрын

Three ways to get infinite answers: -Lambert W branches -Sine period -Polar form of i has infinite angles

@ZMax36 2 жыл бұрын

Have u tried with rest theory? Maybe using e^(e^ix) = 2 u can solve it

@metorasay 2 ай бұрын

for the second i put in desmos sin(x)^{-[1-SQRT(sin^2(x))]} because only the negative posibility is right (as you said so yourself) and it was equal to 1 when x is 2 i think thats the answer

@Nine-2545 Ай бұрын

in the sin(x)^cos(x) = 2, we know that π/2 < x < π ,so we can say x = π/2 + y. After that, we will have sin(x)^cos(x) = sin(x)^(-sin(y)) =2

@lumina_ 13 сағат бұрын

that first equation was so cool

@TheQEDRoom 2 жыл бұрын

the answer for the second is about 2.665357079 radians so we need to solve a=(2^a)/(sqrt(4^a-1)), raise 0.5 to a, then get the arcsine, and subtract from pi. but i'm not sure how to solve this without using numerical solution

@BeginWithDoubt 2 жыл бұрын

What if you say y=sin x and z=cos x and reinterpret the problem as the intersection of y^z=2 and x^2 + z^2 = 1?

@ididagood4335 2 жыл бұрын

Did you mean y^2 + z^2 = 1?

@BeginWithDoubt 2 жыл бұрын

@@ididagood4335 Indeed. I suppose that was a reflexive x^2

@LunizIsGlacey 2 жыл бұрын

Certainly looks potentially promising!

@flavioerrico8965 2 жыл бұрын

Thank you for all your work, if you can could you help my with this problem. Thank you so mutch How many solution The equation in the real variable x has? 10^x = x^3 - x

@rogerkearns8094 2 жыл бұрын

The blackpenredpen I function (I for I dunno). ;)

@christopherthomas6124 Жыл бұрын

I was scrolling through math stackexchange and saw someone was able to write out the real solution to this (sin x)^(cos x) = 2! Can you update this video and do this solution for us!?

@ssifr3331 2 жыл бұрын

So I tried wolfram alpha with input y=(sin(x))^(cos(x)) and it gives a nice graph. Replacing the y with 2 it gives the solution of 2.665...(+/-) 2pi.

@mryip06 2 жыл бұрын

Maybe u can invent a bew function to solve the 2nd equation. Without Lambert W function, the 1st one cannot be solved, right?

@blank0s162 2 жыл бұрын

let y = cosx we know sinx > 0, so we can assure that √(1-y²) = sinx. with this we have (1-y²)^(y/2) = 2 (1-y²)^y = 4 yln(1-y²) = ln(4) this is something wolfram alpha can help us with, though it doesn't give an exact solution. But the numeric approximation of -0.8887 works well enough. cosx = y ≈ -0.8887 x ≈ acos(-0.8887) ≈≈ 2.6653 radians.

@yumirai4 2 жыл бұрын

Upon messing around with Taylor series i managed to "reduce" this equation to t^3 + t^5/2 + t^7/3 + ... = -2ln2 for t = cos(x). Now we only gotta solve an infinite polynomial, EZ!

@biggybrolunch3809 2 жыл бұрын

Would it be helpful to use u=sinx and u'=cosx? Wolfram does give a solution for u(x) in this case, which does include the W function.

@Tzizenorec 2 жыл бұрын

Nah, if you just give Wolfram "u^u'=2" without specifying that u=sin(x), then Wolfram proceeds to tell you that u is something else that is definitely _not_ sin(x).

@piraptor4963 2 жыл бұрын

I think this somehow works you will turn your equation into differential equation problem

@biggybrolunch3809 2 жыл бұрын

u=sinx would actually be a trivial solution, because we defined it that way in the first place. What I'm asking is whether the differential form gives a solving advantage, where sinx is equal to a general solution u to u^u'=2. Looking at that solution, though, it seems like it would not be helpful, because 1) we would still have functions of x on both sides, and 2) the initial condition u(0)=0 appears to be instead an indeterminate hole requiring limits. Therefore, my bad.