Solving sin(x)^sin(x)=2
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- Опубликовано: 7 июл 2024
- 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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What do you think about my "e-hoodie"?
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))
Your "e-hoodie" is pretty cool.
e=2.7 1828 1828 45 90 45 235 360 ...
@@Shreyas_Jaiswal prove your claim.
@@andy-kg5fb From equation (i) and (ii) we conclude, LHS=RHS. Hence proved.
i found a perfectly marvelous closed analytical solution to this equation but the comment section is too small to contain it
so true
360yr later and the line never gets old ;)
Give us the link to your video.
Are you rebirth(reincarnation) of Fermat 😊 ?!
"The link, dude. You forgot the link" -Albert Einstein
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.
😆
That could be an idea
@@blackpenredpen new Maths discovery ruclips.net/video/63LkgUE27j8/видео.html
We could call it the "BPRP" function: B(x) = (sin x)^(cos x)
@@gary.h.turner It should be:
If f(x) = (sinx)^(cosx)
Then B(x) = f^(-1)(x)
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
😂
Looks like a recipe for the Chef language.
@@LIA-52 yes
Numerical approximations are, to a mathematician, what frozen ready meals are to a decent chef.
Is it a good thing or a bad thing?
So you use them when you're feeling lazy and want something handy
Used constantly but not enjoyed much? Chefs dont eat anywhere near as well as you'd think
And this is why I hate it when it’s argued that .9 recurring is equal to 1
🤣
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).
Agree. The Lambert W is a kind of cheating itself.
Give a numerical approximation of Q(2)
@@viharsarok well, so is sine and cosine
@@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?
@@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?
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"
Isnt trigonometry about triangles and Geometry shit 😭
@@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.
@@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."
@@Propane_Acccessories For me, drawing out the triangle helped with trig substitution.
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
Indeed, I think that once you concede that you can only solve it numerically, it's reasonably straightforward.
@@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
@@benoitavril4806 I'm pretty sure it would violate Gödel's incompleteness theorem if WolframAlpha could give an analytical solution to all solvable problems.
@@andrasfogarasi5014 Gödel's incompleteness theorem's have nothing to do with this.
@@hOREP245 Gödel's incompleteness theorem have something to do with this
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.
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
I don't think that helps. I believe there's no nice answer and the best you can do is a numerical approximation
@@fedem8229 yeah most likely.
@@Metalhammer1993 did you figure out how to solve it?
Me in 10th grade like hmm yes I see 🤔
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 )
Wow, nice solution
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
Great Effort. Use the Cubic Equation to not make it as an approximation.
@@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…))
@@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…)
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
That were two similiar equation with a surprising different solution. Thank you for that!
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.
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!
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.
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...
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...
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)
The approx is 2.6653
Espectacular. Impresionante como resuelves en el campo complejo. Yo intenté resolver la primera ecuación y lo hice con W de Lambert como tú. Mi admiración completa. Eres un maestro. Muchas gracias. Disfruto viendo tus videos.
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
Desmos is god tier
Desmos "brute-forces" such equations and calculates y for every x on screen. We are looking for a clean solution
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])
You're right. Wolfram Alpha actually gives a result with that input, in exact form.
You seemed so devastated when you couldn’t solve the second one. I felt your pain there. Great vid :)
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 :)
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) ]
I think x values are not equal anymore once you integrate
@@chikenwingsteve if two things are equal (eg. a=b) then applying the same operation on both sides gives an equivalent result.
@@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.
Brilliant Yung Man, i like your videos.. Thanks for enlightening this yung hollow's mind.
I love how you juggle the markers so easily.
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)))))
new Maths discovery ruclips.net/video/63LkgUE27j8/видео.html
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.
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
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.
@@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?
@@chenghowkoh2178 If you can remove all the constants then it will hold
@@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.
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
This is the time for the mathematics world to stand up and generalize the Lambert function wider than ever
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.
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))
are you sure your value isn’t off due to floating point imprecision?
This is the fastest I havent understood anything in a video in a long time
Would a maclaurin/ taylor series be useful for the sin(x)^cos(x) equation
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?
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.
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?
Its wrong
I get 152.714≈
@@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
Put ur x value on the calculator it doesn't work
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
So cool
Way out of my depth but I'm excited to see the solutions people find
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.
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
Good work my friend
I admire what u're doing
Little fact: e^(W(x)) = x/W(x). So one more step of simplification.
Isn't it e^((W(x)) = W(x)*e^(W(x))/W(x) = x/W(x) ?
@@Gniaum yeah that was just a slip of typography
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)
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]
That's what I did in my head
For y in [0,1], y^y is not less than or equal to y lol
@@TrinoElrich oh ya
That's true
@@TrinoElrich Oops, you're right. It is true that it's less than or equal to 1 though
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?
One thing that's worth remembering is that sin^-1 x for all real x>1 has real part pi/2 plus 2npi.
I liked it. But how about we use log2 base 2 instead of Ln?
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)
What about trying to put cos(x) as sin(pi/2 - x) or sin(pi/2 + x)
I loved your hoodie with the Neper Constant.🌺
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.
I love your channel too much♥️♥️, but I want to know your strategy in thinking to solve mathematical problems.
Someone give this man a bigger board! ! !
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
In the meanwhile best parenthesis closing ever :D
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 !
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!
x=arccos(-1/n) where n is solution of the equation 1-1/n^2=1/4^n.
Numerical Method seems to be the only way.
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)_/¯
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.
Bc the limit of 1/x is 0 as x goes to inf. So we are like adding nothing.
Ok, I haven’t learned about limits yet so that helps.
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
Wow, Great explanations Dear
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
That hoodie is awesome! Where can I get it!?
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 🤣
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.
You forgot to add the multiples of 2pi*i as complex log is a multifunction :)
I noticed that too, but he made up for it by noticing that sin is periodic. This gives back the same result
@@bbqandchill8631 those are real integer multiples of 2pi, complex log is imaginary integer multiples of 2pi
@@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
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.
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
{1-y²}^(y/2)=2
Solving for y and putting cos x=y.
Nice to learn something new
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
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!
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
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
You can approximate using Newton's method. x(n+1)=x(n)-f(x(n))/f'(x(n))
π/2
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))
Squaring makes it sinx^(2sinx)=4
@@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)
@@pk1proMore importantly, the mistake you made here was referring to the first equation, when I explicitly told I was talking about the second equation
@@matniet43 ohh yeah makes sense
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.
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.
does it help if you use a different identity for cosine, like cos(x) = sin(pi/2 - x)?
You cant use many other trig identities then.
Nope
Three ways to get infinite answers:
-Lambert W branches
-Sine period
-Polar form of i has infinite angles
You could try something with newtons method or with a taylor series
I'm confused around the 3:10 mark, what happens to -e^(-ix) when he multiplies through?
Now I feel like I must solve it.
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?
The blackpenredpen I function (I for I dunno). ;)
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.
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
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.
Me searching for this in English and not even understand it with my native language 😂
One solution of (sin(x))^(cos(x))=2 can be given by
y*ln(1-y^2) = 2ln(2) with y=cos(x)
The first equation can be solved numerically (just like the lambert W function), and the rest should foilow.
For the eq. Sinx^(cosx) = 2 , wolfram alpha actually gave an solution of x=2.66536
Can you please help me, what value does the function cos(cos(cos(cos(...(cos(x)))))...) (infinitely many cos) approach in a nice form? I found out accidentaly on the calculator, that this value is almost the ssrt of 0,8, but I am not sure about that. Thank you! :)
It's called the Dottie Number, and skimming wikipedia it seems that it can be represented as an infinite series, but that's it.
@@simples6475 Oh, I see it now. I have never heard of this constant before. :)
Solution of (Sinx)^Cosx=2 is approximately 163*pi/192
Have u tried with rest theory? Maybe using e^(e^ix) = 2 u can solve it
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.
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.
I tried it on wolfram alpha and it worked for me. It gave me a pretty long equation though and I don’t really understand it.
Please check this out guys-
Sinx^cosx = 2
cosxlnsinx = ln2
Cotxcosx - lnsinx x sinx = 0
( Take derivative)
cotxcosx = sinx lnsinx
cos²x = sin²x lnsinx
1 = sin²x(1 + lnsinx)
1 = t²(1+lnt)
1 = t² + t²lnt
2t + t²/t + 2tlnt = 0
3t = -2tlnt
lnt = -3/2
t = 0.22
hence sinx = 0.22
X 0.22 ??
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?
Did you mean y^2 + z^2 = 1?
@@ididagood4335 Indeed. I suppose that was a reflexive x^2
Certainly looks potentially promising!
Hey bprp, I sure ain't mathematically smart enough to get the answer, but I think I can do it computationally. My tip is, just use some sort of gradient descent and define the loss function as (sinx^cosx - 2)². Then find the gradient and iterate through the process. I know its not what you have in mind as a mathematician, but I guess it's an alternative
Wow this is clever 🤔
Seems to me like basic newton iteration