1:50: To solve for 0^0, you need to apply L'Hopital's rule, if possible. When y=x, x^y = x^x lim x->0 (x^x) = 0^0 = indeterminate. Take the derivative of both parts (base & exponent) and try again. lim x->0 (1^1) = 1 Therefore, the limit exists at (0, 0).
Off topic, but I tried seeing if integrating (e^sinx - e^-sinx)/2 was possible. Couldn’t find a way, so I decided to plug the function into desmos, and weird enough, it looks exactly like a sin function. In fact, it looks EXACTLY similar to 1.175 sin x. There are very VERY small differences between f(x) = (e^sinx - e^-sinx)/2 and f(x) = 1.175 sin x. Each function had the same maxima and minima, and also the same x-intercepts. ALL I WANT TO KNOW, is how and why does the number 1.175 relate to (e^sinx - e^-sinx)/2
This is sinh(sin(x)). The number 1.175 comes from when you look at the peak of the sine function, which is when x=pi/2 and thus sin(x)=1, the function you're looking at becomes (e - 1/e)/2 = 1.175. If you look at the power series of sinh(y) you'll get y + y^3 / 6 as the leading term and since y is only ever between -1 and 1 in this case, that will be a relatively good approximation as well.
I was thinking about this in my head all week. For no reason. I got the trivial cases y=x, (4,2) and (2,4) , the asymptotes and no negatives but needed a guide to take me home. Thank you.
I have actually found that there *are* negative real numbers that satisfy the equation, they just produce complex numbers in the process, but x and y are still real numbers (for example, x=2 produces not only y=2 and y=4, but also y≈-0.76666469596212…) The main problem is that there is not a continuum of these numbers, hence they would only be disconnected points and not a curve (even though as x or y grows, the points get closer and closer)
@@mytriumph can you explain what method you used to get that? I know that blackpenredpen used lambert w function to get -0.76666469596212..., but are there other ways?
I was hoping for some great content from this one, after seeing part one - and you delivered! I had to stop and rewind a few times to make sure I fully understood what you had done, but I got it in the end. Thank you.
It's worth substituting t=1/u. You'll find that when the parameter is greater than 1, the x and y values are the y and x values for 1/t. So you don't need to work out that region even if you're interested in where the curve is for values of the parameter.
I see that you've only managed to consider the first quadrant, but that leaves you short of some solutions to x^y = y^x. For example, the function f(x, y) = x^y - y^x will have the solutions to x^y = y^x whenever f(x, y) = 0. But if y is an even positive integer, then f(x, y) is large and negative when x is large and negative; and f(x, y) is large and positive when x is large and positive. Given that y is an even positive integer, f(x, y) is clearly continuous, and therefore the intermediate value theorem provides that there will be an odd number of solutions to f(x, y) = 0. At 19:30 I can see two solutions for y=2, but where's the third one? Likewise for y=4, 6, 8 ... Are those third values for x inevitably negative? Are they all close to -0.8? Is there a generating function for them?
Two quick comments: 1. Substitution with t=1/s algebraically shows the symmetry of x and y 2. Sometimes when L'Hôpital's rule is applied, the fraction is just the definition of derivative at a point
blackpenredpen and for the first part you can substitute t=1+z where z approaches 0, and the resulting expression for x is the literal limit definition of e (well with a function limit rather than the usual series limit, but whatever)
very nice.... much easy if you root both sides by (x*y) it makes LHS and RHS separable power(X,1/X) = power(Y,1/Y) =t...........1 Also, there are only 2 real solutions here. If Y > e then the other solution is X < e If Y < e then the other solution is X > e and at Y =e both roots are e only You can use the binomial theorem to expand (X-Y)(X-X2)*F(X,Y), X2 is real where F(X,Y) has only complex solutions
If x^y=y^x, it follows that e^[y*ln(x)]=e^[x*ln(y)], and y*ln(x)=x*ln(y), and ln(x)/x=ln(y)/y. If you consider the function f(t)=ln(t)/t, it follows that f'(t)=[1-ln(t)]/t^2, therefore f(t) will be increasing when t is less than e, and decreasing afterward. So each pair of x and y will have one greater than e, and the other less than e!
At 12:18, couldn't you just use the fact that y = tx to solve everything more quickly? lim(t->1) tx = 1 * lim(t->1) x And since we already know that limit is e, then y must also be equal to e.
Some interesting insight on the 0^0 "debate"... Using an arbitrarily-small number b to represent "almost 0", beginning with a value of 1 and graphing y=x^b, and y=b^x, we get the lines y=x and y=1, which disagree in what to return at x=0 to begin with. As we decrease b, they both approach different lines, but their y-intercepts both remain firmly at 0 and 1 respectively. Funnily enough, y=x^b approaches y=1, while y=b^x approaches y=0 (they seem to have swapped places, is the joke), and y=x^b conforms to y=1 much more quickly than y=b^x conforms to y=0 (at b = 2^(-10), 0.1^b ~=0.998, and b^0.1 = 0.5, as an example of how quickly the two lines conform) And, Desmos says that y=0^x has two values for x=0, both (0, 0) and (0, 1). (Though, y=x^0 only gives (0, 1)) However, I think 0^0=0^0 simply due to the fact that it looks strikingly like an identity. Anything, even when undefined, _should_ be equal to itself, right?
You should do case (2) and (3) first and calculate just x. For y, instead of doing complicated calculations, you can just use "y = t x" and get the y values right away. Case (1) is by symmetry with case (3). Also, this is not all the solutions in real space. y^x = x^y ==> y^(1/y) = x ^(1/x) The advantage of the latter form is that it is symmetric and has variable separation between x and y. Now you can go to desmos and plot "y = x^(1/x)". Drawing lines parallel to the X-axis (y = c), that intersect this curve in more than 1 location yield (x, y) pairs that are solutions to the original equation. Here the two intersection points with "y = c" would be "(x1, c)" and "(x2, c)". The (x, y) pairs that are solutions to original equation correspond to the x values of the intersection points (x1, x2). Note: The curve "y = x^(1/x)" rises from (0, 0) through (1, 1) to (e, e), like an S-curve, then peaks at (e, e) and finally monotonically decreasing for higher values of x, with the y value being asymptotic to 1 (going towards (infinite, 1)). All positive (x, y) solution pairs will lie on this curve in the range ((1, 1), (infinite, 1)) with (e, e) being the only unpaired solution in this range. There are discrete valued solutions for negative x values that are real. In the range x = [-e, -infinite], if y is real and positive, it monotonically increases in the range from about [~0.692, 1]. All of these values pair up with exactly one positive value from the range [(~0.755, ~0.692), (1, 1)]. The two x values (one positive and one negative) are all solution pairs for the original equation. In the x value range [-1, -e], the graph is again decreasing and some positive real y values are created in range [-1, ~0.692], which pairs up with exactly one positive x value solution in the range [(0, 0), (1, 1)], which is the rising part of the S-curve. These can also potentially pair up with one of the y values generated in the x value range [-e, -infinite], creating 3 x values with the same y value, meaning (x, y) solutions to original equation of the form (x1, x2), (x1, x3), (x2, x3). There are some positive solutions in the x value range [0, -1], that have real y values between 1 and e. These will pair up with two positive solutions creating 3 solutions to the original equation similar to the three point case above. There are many negative y value solutions corresponding to all odd negative integers less than -2 of the form "(-n, - n^(1/n))", where n is an odd positive integer greater than 2. The y-values monotonically increase from ~ -0.693 to -1. All of these potentially pair up with negative x-values in the range [-1, -2], yielding more solutions. Finally, there are solutions in the complex plane, where y takes on the same complex value for two or more negative values of x. One very interesting solution of original equation with negative integers and complex y values is, e.g. (x, y) --> (-2, -4) => (-2)^(-4) == (-4)^(-2) == 1/16
You don't need to find that t=1 point is useful just by chance. You know it is asynptotic to x=1 and y=1and is symmetric about x=y. Plug in x=y, so t^(1/(t-1))=t^(t/(t-1)), which gets you t=1. Then you plug in 1 and find your x or y, whichever is easier, and you have a point perpendicular to x=y at that point.
I have a pretty heird question about that thing of "0^0 = 0^0", it's "ok" to say that "1/0 = 1/0"? I know that it may not exist something like "it's ok" sort of thing here, but it made me think a lot
I think I'm missing something basic. If x is plotted on the x axis, and y is plotted on the Y axis, then you get the diagonal curve y=x. Are you plotting t on the z axis?
I fully understand what you're doing, but I can't for the life of me grasp how you come up with the various strategies, but I guess it comes with practice ;-)
I'm not super strong with parametric functions, but I don't think he's actually assuming anything by saying that. He was initially inspired to set up that equation by the case of 2^4 = 4^2, for which t happens to be an integer, but he's evaluating the function for all values of t, including non-integer ones. So all it's really saying is that for each pair of values x and y, there exists some number t by which x can be multiplied so that it equals y, and that's true of any pair of real numbers
Actually, he derived the equations with t by plugging y=tx into the original equation and solving stuff. Since we’re not going to be working with 0s or any such with this equation, you can be confident that this is always meaningful for any numbers you could plug into the equation, so solutions won’t disappear this way.
hey we know how to flip some curve around the plane but do you think about rotating it with some degree use parametric equations and try to plane the rotaed graph
But still there are a set of solutions missing right? For example, 2 has 3 solutions: [-0.766665, 2, 4] I wonder what would the graph look like if we even include those ones!
Ach, I've been looking into all possible values that satisfy x^y = y^x if you allow complex x and y... this video was good revision but... please... RPBP... are you going to consider the (admittedly multivalued) case of (a+bi)^(c+di) = (c+di)^(a+bi)?
I hope you make a video to answer this question, if limit((1+1/n)^n,n->infinity) equal to e, then limit(product(1+1/(n/2+k),k from 0 to n),n->infinity) is equal to?
just tried that and it doesn't work lol. The solution for t is 1 in that case. It's a limit situation wherein 1 to the infinity is e. Not sure how to prove that
You can use the classic (1+1/t)^t, which in the limit as t->inf is equal to e. I know that BPRP has a video showing this, and that this form (and it's relative) are the definition of 'e'.
You r doing a great job sir. But i have a small question, in here for the same x we r getting two values of y, but we know functions dont do that. So i am a little confused. Please explain this little ambiguity!!
It's cause you are getting the value of e^(f(t)) while "t" goes to 0 and f(t) being the equations in the second blue point, that is because using some algebra you can move the limit from the base to the exponent, so you are essencially getting a limit but this time the limit is the exponent (sorry about my english)
Gergő Dénes Let z = x^y Then z = 1 for y=0 and x is not zero but approaches zero. Also z = 0 for x = 0 and y is positive, approaching zero. So far, we have two line sgments approaching (0,0) leading us to expect two different values. Now try other values of x and y both approaching zero, for example x = y = -0.000001 You will quickly find that z has many values as x,y approach zero from different directions, some of them complex numbers. This is only one reason why 0^0 cannot be defined. For other values
I know all of this, I just presented a way to "prove" 0^0=1 with a false proof, as the binomial theorem needs a and b to be non-zero numbers. As for 0^0, there are many ways to get different answers. Take the limits of x^0, 0^x, x^x where x approaches 0+, the first and last will equal 1, the second will equal 0 - basically the same as what you wrote.
We assume that (-5)^2=25 and 5^2=25 then (-5)^2=5^2. then if we enter the square root in both sides we get : sqrt((-5)^2)=sqrt(5^2). if sqrt(x)=x^(1/2) then ((-5)^2)^(1/2)=(5^2)^(1/2) (-5)^(2/2)=5^(2/2) so -5=5. How do you explain that?
That is because x^(1/2) actually has two solutions while sqrt(x) is defined to be only the solution that has the the same angle on the complex plain as x. That's why you often should use ±sqrt(x) if you're not sure you only need the one solution for your specific application.
Hey, It has nothing to do with the video but can there be something like a 0.5th or any rational derivative? I just thought a bit about it and I thought like it could work in any way. Of course it would be useless but... Math for fun... Maybe you can find like an answer for my thoughts because you are much smarter than me.. I'm just a 10th grade pupil xD
iTzTien things like 1^i are known as indeterminate forms, where just plugging in the numbers gets you a result that cannot be evaluated itself, but some more math makes it possible to give a meaningful value to the limit. For example, a common indeterminate form is i*0. For a simple example of how this works, imagine the limit as x goes to infinity of x*1/x. Naturally, no matter what value you plug in for x, you get 1, and the expression being limited can easily be evaluated as 1, so the result is 1. However, plugging in infinity gets you the indeterminate form 0*i. Do the same thing with ax*1/x, where a can be any number you want, and the result is a, but they all have that same indeterminate form. This form essentially represents cutting up the ultimate value of the limit into an infinite number of pieces, all of which have to be 0 since any greater number multiplied by i is i; it could mean anything without tracking how the total value behaves as it is cut up into more and more pieces. Now try the same with (a^(1/x))^x; this is similar, except instead of dividing a number into an increasing number of pieces and adding them back together, it uses roots to divide it into pieces and multiplies them back together. It basically does with multiplication what 0*i did with addition. This turns out to be the form 1^i; since it is an indeterminate form, this can mean anything without knowing the limiting expression.
![The most beautiful equation in math, explained visually [Euler’s Formula]](/img/1.gif)
0:50 Blackpenredpen keeping it real!
"Differential symbols make me happy"
_Blackpenredpen, Jul 2018_
Dark Mage : )
@@blackpenredpen najeeb choudhury
1:50: To solve for 0^0, you need to apply L'Hopital's rule, if possible.
When y=x, x^y = x^x
lim x->0 (x^x) = 0^0 = indeterminate. Take the derivative of both parts (base & exponent) and try again.
lim x->0 (1^1) = 1
Therefore, the limit exists at (0, 0).
L'Hospital's rule does not apply to 0^0 situations. It's not as simple as you described in your comment unfortunately
Off topic, but I tried seeing if integrating (e^sinx - e^-sinx)/2 was possible. Couldn’t find a way, so I decided to plug the function into desmos, and weird enough, it looks exactly like a sin function. In fact, it looks EXACTLY similar to 1.175 sin x. There are very VERY small differences between f(x) = (e^sinx - e^-sinx)/2 and f(x) = 1.175 sin x. Each function had the same maxima and minima, and also the same x-intercepts.
ALL I WANT TO KNOW, is how and why does the number 1.175 relate to (e^sinx - e^-sinx)/2
NoNeedToBeUpset try de moivre (google it if you don't know) and you'll know.
This is sinh(sin(x)). The number 1.175 comes from when you look at the peak of the sine function, which is when x=pi/2 and thus sin(x)=1, the function you're looking at becomes (e - 1/e)/2 = 1.175. If you look at the power series of sinh(y) you'll get y + y^3 / 6 as the leading term and since y is only ever between -1 and 1 in this case, that will be a relatively good approximation as well.
I was thinking about this in my head all week.
For no reason.
I got the trivial cases y=x, (4,2) and (2,4) , the asymptotes and no negatives but needed a guide to take me home.
Thank you.
I have actually found that there *are* negative real numbers that satisfy the equation, they just produce complex numbers in the process, but x and y are still real numbers (for example, x=2 produces not only y=2 and y=4, but also y≈-0.76666469596212…)
The main problem is that there is not a continuum of these numbers, hence they would only be disconnected points and not a curve (even though as x or y grows, the points get closer and closer)
@@mytriumph can you explain what method you used to get that? I know that blackpenredpen used lambert w function to get -0.76666469596212..., but are there other ways?
Its 3 AM,but I cannot sleep without seeing the graph of this thing xD
Boypig24 So true 😂
I can relate
Yes
I was hoping for some great content from this one, after seeing part one - and you delivered! I had to stop and rewind a few times to make sure I fully understood what you had done, but I got it in the end. Thank you.
You are the best maths teacher I have ever seen
It's worth substituting t=1/u. You'll find that when the parameter is greater than 1, the x and y values are the y and x values for 1/t. So you don't need to work out that region even if you're interested in where the curve is for values of the parameter.
I see that you've only managed to consider the first quadrant, but that leaves you short of some solutions to x^y = y^x.
For example, the function f(x, y) = x^y - y^x will have the solutions to x^y = y^x whenever f(x, y) = 0.
But if y is an even positive integer, then f(x, y) is large and negative when x is large and negative; and f(x, y) is large and positive when x is large and positive. Given that y is an even positive integer, f(x, y) is clearly continuous, and therefore the intermediate value theorem provides that there will be an odd number of solutions to f(x, y) = 0.
At 19:30 I can see two solutions for y=2, but where's the third one? Likewise for y=4, 6, 8 ... Are those third values for x inevitably negative? Are they all close to -0.8? Is there a generating function for them?
Two quick comments:
1. Substitution with t=1/s algebraically shows the symmetry of x and y
2. Sometimes when L'Hôpital's rule is applied, the fraction is just the definition of derivative at a point
You're a really genius in maths. It's incredible how can you know all of that XD.
At around 13min, t→1, you get x = t^(1/[t-1])→e.
Then to get y = t^(t/[t-1]), you can shortcut the calculation by noting that
y = xᵗ → e¹ = e
Fred
Ah!!!!!!! I missed that! Thanks, Fred, as always!
blackpenredpen and for the first part you can substitute t=1+z where z approaches 0, and the resulting expression for x is the literal limit definition of e (well with a function limit rather than the usual series limit, but whatever)
very nice....
much easy if you root both sides by (x*y)
it makes LHS and RHS separable
power(X,1/X) = power(Y,1/Y) =t...........1
Also, there are only 2 real solutions here.
If Y > e then the other solution is X < e
If Y < e then the other solution is X > e
and at Y =e both roots are e only
You can use the binomial theorem to expand
(X-Y)(X-X2)*F(X,Y), X2 is real
where
F(X,Y) has only complex solutions
If x^y=y^x, it follows that e^[y*ln(x)]=e^[x*ln(y)], and y*ln(x)=x*ln(y), and ln(x)/x=ln(y)/y. If you consider the function f(t)=ln(t)/t, it follows that f'(t)=[1-ln(t)]/t^2, therefore f(t) will be increasing when t is less than e, and decreasing afterward. So each pair of x and y will have one greater than e, and the other less than e!
see
0^0=1
and 0^0=1
so they are equal and the line should go through
At 12:18, couldn't you just use the fact that y = tx to solve everything more quickly?
lim(t->1) tx = 1 * lim(t->1) x
And since we already know that limit is e, then y must also be equal to e.
don't you mean y=x^t? same answer comes out though
sami labidi y is both equal to x^t and tx in this case
@@user-en5vj6vr2u last video he defined y=tx
But as you said y=x^t
0:48 blackpenredpen keepin it real
Some interesting insight on the 0^0 "debate"...
Using an arbitrarily-small number b to represent "almost 0", beginning with a value of 1 and graphing y=x^b, and y=b^x, we get the lines y=x and y=1, which disagree in what to return at x=0 to begin with. As we decrease b, they both approach different lines, but their y-intercepts both remain firmly at 0 and 1 respectively. Funnily enough, y=x^b approaches y=1, while y=b^x approaches y=0 (they seem to have swapped places, is the joke), and y=x^b conforms to y=1 much more quickly than y=b^x conforms to y=0
(at b = 2^(-10), 0.1^b ~=0.998, and b^0.1 = 0.5, as an example of how quickly the two lines conform)
And, Desmos says that y=0^x has two values for x=0, both (0, 0) and (0, 1). (Though, y=x^0 only gives (0, 1))
However, I think 0^0=0^0 simply due to the fact that it looks strikingly like an identity. Anything, even when undefined, _should_ be equal to itself, right?
4:44 me during any presentation ever
"The Ellen Teeth Power"
Thank you BPRP for Keeping It Real!
You should do case (2) and (3) first and calculate just x. For y, instead of doing complicated calculations, you can just use "y = t x" and get the y values right away.
Case (1) is by symmetry with case (3).
Also, this is not all the solutions in real space.
y^x = x^y ==> y^(1/y) = x ^(1/x)
The advantage of the latter form is that it is symmetric and has variable separation between x and y.
Now you can go to desmos and plot "y = x^(1/x)". Drawing lines parallel to the X-axis (y = c), that intersect this curve in more than 1 location yield (x, y) pairs that are solutions to the original equation. Here the two intersection points with "y = c" would be "(x1, c)" and "(x2, c)". The (x, y) pairs that are solutions to original equation correspond to the x values of the intersection points (x1, x2).
Note: The curve "y = x^(1/x)" rises from (0, 0) through (1, 1) to (e, e), like an S-curve, then peaks at (e, e) and finally monotonically decreasing for higher values of x, with the y value being asymptotic to 1 (going towards (infinite, 1)). All positive (x, y) solution pairs will lie on this curve in the range ((1, 1), (infinite, 1)) with (e, e) being the only unpaired solution in this range.
There are discrete valued solutions for negative x values that are real. In the range x = [-e, -infinite], if y is real and positive, it monotonically increases in the range from about [~0.692, 1]. All of these values pair up with exactly one positive value from the range [(~0.755, ~0.692), (1, 1)]. The two x values (one positive and one negative) are all solution pairs for the original equation.
In the x value range [-1, -e], the graph is again decreasing and some positive real y values are created in range [-1, ~0.692], which pairs up with exactly one positive x value solution in the range [(0, 0), (1, 1)], which is the rising part of the S-curve. These can also potentially pair up with one of the y values generated in the x value range [-e, -infinite], creating 3 x values with the same y value, meaning (x, y) solutions to original equation of the form (x1, x2), (x1, x3), (x2, x3).
There are some positive solutions in the x value range [0, -1], that have real y values between 1 and e. These will pair up with two positive solutions creating 3 solutions to the original equation similar to the three point case above.
There are many negative y value solutions corresponding to all odd negative integers less than -2 of the form "(-n, - n^(1/n))", where n is an odd positive integer greater than 2. The y-values monotonically increase from ~ -0.693 to -1. All of these potentially pair up with negative x-values in the range [-1, -2], yielding more solutions.
Finally, there are solutions in the complex plane, where y takes on the same complex value for two or more negative values of x. One very interesting solution of original equation with negative integers and complex y values is, e.g. (x, y) --> (-2, -4) => (-2)^(-4) == (-4)^(-2) == 1/16
Best RUclips channel
please do a video on x^x=y^y./ I dont un derstand that graph
One of the best maths guys on RUclips no question 🙌🙌 #yay
Thanks!!!
Color consistency, #yay
You don't need to find that t=1 point is useful just by chance. You know it is asynptotic to x=1 and y=1and is symmetric about x=y. Plug in x=y, so t^(1/(t-1))=t^(t/(t-1)), which gets you t=1. Then you plug in 1 and find your x or y, whichever is easier, and you have a point perpendicular to x=y at that point.
So hash it to also show where x^y < y^x and x^y > y^x
I have a pretty heird question about that thing of "0^0 = 0^0", it's "ok" to say that "1/0 = 1/0"? I know that it may not exist something like "it's ok" sort of thing here, but it made me think a lot
I think I'm missing something basic. If x is plotted on the x axis, and y is plotted on the Y axis, then you get the diagonal curve y=x. Are you plotting t on the z axis?
We can count on you for always "keeping it real" ;)
Of course it could be also x=t^(t/(t-1)) and y=t^(1/(t-1)) if you want go top to right instead of right to top for increasing t.
This becomes so much easier when you accept that 0^0=1
Can't wait to see the video about the derivatives!
It's already done here: ruclips.net/video/604njpZ-bbA/видео.html
blackpenredpen Yay!
Let n = 1/(t-1) then x =((n+1)/n)^n ; y = ((n+1)/n)^(n+1) … is much easier and rational solutions can also be found
for n = 1, 2, …
Could you please explain the Gödel’s incompleteness theorems?
I fully understand what you're doing, but I can't for the life of me grasp how you come up with the various strategies, but I guess it comes with practice ;-)
Keep it up mate, I love it.
19:26 Geogebra! YAY!
thank you for watching
I still don't understand why we can assume y=tx in the first place. It seems like a logic step, but is it proven that there aren't more solutions?
I'm not super strong with parametric functions, but I don't think he's actually assuming anything by saying that. He was initially inspired to set up that equation by the case of 2^4 = 4^2, for which t happens to be an integer, but he's evaluating the function for all values of t, including non-integer ones. So all it's really saying is that for each pair of values x and y, there exists some number t by which x can be multiplied so that it equals y, and that's true of any pair of real numbers
That's a good explanation. I was fixed on the assumption that t has to be a natural number and was concerned about pairs of prime numbers.
Actually, he derived the equations with t by plugging y=tx into the original equation and solving stuff. Since we’re not going to be working with 0s or any such with this equation, you can be confident that this is always meaningful for any numbers you could plug into the equation, so solutions won’t disappear this way.
I'm positive you are correct !
Does it have a non parameterized form?
hey we know how to flip some curve around the plane but do you think about rotating it with some degree use parametric equations and try to plane the rotaed graph
and do video about the Batman equation
show my comment
But still there are a set of solutions missing right? For example, 2 has 3 solutions: [-0.766665, 2, 4]
I wonder what would the graph look like if we even include those ones!
I enjoy it and thank you👏
You're amazing ! Thanks for your videos !
I remember messing around in desmos and I put it
yth root of y equals xth root of x
It graphed something similar.
Ach, I've been looking into all possible values that satisfy x^y = y^x if you allow complex x and y... this video was good revision but... please... RPBP... are you going to consider the (admittedly multivalued) case of (a+bi)^(c+di) = (c+di)^(a+bi)?
I hope you make a video to answer this question, if limit((1+1/n)^n,n->infinity) equal to e, then limit(product(1+1/(n/2+k),k from 0 to n),n->infinity) is equal to?
Can you convert the parametric equation into a cartesian one? In the form of y=....
I like you keepin' it real
Wait, have you always been wearing glasses? I can't remember seeing you wear them.
Also, that's a pretty awesome t-shirt you're wearing there.
so is this technically not a function since it fails the vertical line test? Awesome video btw!
Why is e the intersection ?
Soufian 27 Bc both x and y have the same value e
Try this, if x = e, solve for t based on the parametric equation. The solution for t should give you e if you plug into the parametric equation for y
just tried that and it doesn't work lol. The solution for t is 1 in that case. It's a limit situation wherein 1 to the infinity is e. Not sure how to prove that
You can use the classic (1+1/t)^t, which in the limit as t->inf is equal to e. I know that BPRP has a video showing this, and that this form (and it's relative) are the definition of 'e'.
Warrick Dawes But how is that related to the parametric equations?
*russian anthem starts playing*
You r doing a great job sir. But i have a small question, in here for the same x we r getting two values of y, but we know functions dont do that. So i am a little confused. Please explain this little ambiguity!!
PLEASE TUTOR ME
But there are infinitly many solutions for x=y. Why the graph only intersects with (e,e)?
What is the foci of this hyperbola?
I saw it on Desmos, but you cannot type it in on your graphing calculator since it’s impossible to solve for y.
The five significant points being (1, inf) (2,4) (e,e) (4,2) and (inf,1). Always with 2 and 4, so funny.
Can you do a video on why you can use L'Hôpital's rule when you have 0/0 or inf/inf for the exponent of e?
It's cause you are getting the value of e^(f(t)) while "t" goes to 0 and f(t) being the equations in the second blue point, that is because using some algebra you can move the limit from the base to the exponent, so you are essencially getting a limit but this time the limit is the exponent (sorry about my english)
You should have plotted t=1/2 (4,2) and t=2 (2,4) before plotting the limits.
How do you obtain the THIRD solution for 2^x = x^2 besides (2,4) and (4,2) ?
If it exists, then it must be a complex solution.
@@physicsphysics1956 yep, you use the Lambert W function to obtain a third complex solution
does it cross or approximate to (y-1)(x-1)= (e-1)^2
You are amazing!
If we plug (1+0) into the binomial theorem, 0^0 has to be 1, or (1+0)^n would be 0, while it clearly is 1.
Any opinions on this?
Gergő Dénes
Let z = x^y
Then z = 1 for y=0 and x is not zero but approaches zero.
Also z = 0 for x = 0 and y is positive, approaching zero.
So far, we have two line sgments approaching (0,0) leading us to expect two different values.
Now try other values of x and y both approaching zero, for example x = y = -0.000001
You will quickly find that z has many values as x,y approach zero from different directions, some of them complex numbers. This is only one reason why 0^0 cannot be defined.
For other values
I know all of this, I just presented a way to "prove" 0^0=1 with a false proof, as the binomial theorem needs a and b to be non-zero numbers.
As for 0^0, there are many ways to get different answers. Take the limits of x^0, 0^x, x^x where x approaches 0+, the first and last will equal 1, the second will equal 0 - basically the same as what you wrote.
We assume that (-5)^2=25 and 5^2=25 then (-5)^2=5^2.
then if we enter the square root in both sides we get : sqrt((-5)^2)=sqrt(5^2).
if sqrt(x)=x^(1/2)
then ((-5)^2)^(1/2)=(5^2)^(1/2)
(-5)^(2/2)=5^(2/2)
so -5=5.
How do you explain that?
I have an i-dea for that^^
That is because x^(1/2) actually has two solutions while sqrt(x) is defined to be only the solution that has the the same angle on the complex plain as x. That's why you often should use ±sqrt(x) if you're not sure you only need the one solution for your specific application.
Square root makes the absolute value
Sir, WHAT WILL BE THE VALUE IF IOTA TO POWER e
This man just says, fuck calculators
Can you make videos based on the AMC questions
Patrick Gleason you forgot the question mark
I know lol
Thankyou very much.
blackpenredpenbluepen
I do like my morning maths ;D
Why don't the negatives work?
Here's a hard problem: What is the biggest circle you can inscribe between this graph and y = (e-1)^2/(x-1) + 1?
I was really hoping you'd start with t=0, and when you did, I was like "yisss"
Can you solve x^y=y^x for y? Can you make X or y the subject?
Matt GSM I think you can, but only with the lambert function [W(x)] and other types of functions.
Hello blackpenredpen, i´ve got a challenge for you solve this one: 2^x=16x(two to the x power equals sixteen times x) I will be await.
6.75621530402476?
plz use exp() function
The equation has integer solution: 4^2 = 2^4
He covered that in his previous video on this equation.
It (along with its symmetric twin) is the *only* integer solution.
Fred
infini-t
How about x^y=2y^x
I can say unequivocally that 0^0=0^0.
X=2 y=4 ; y=2 x=4?
Wow!
Where is your RED PEN?
Soo y^x=x^y is a hyperbola?
Merci !!!
Hey, It has nothing to do with the video but can there be something like a 0.5th or any rational derivative? I just thought a bit about it and I thought like it could work in any way. Of course it would be useless but... Math for fun...
Maybe you can find like an answer for my thoughts because you are much smarter than me.. I'm just a 10th grade pupil xD
Full graph ruclips.net/video/p-R0druZiTs/видео.html
What about y=x^x, y=x^x^x, etc.? Aren't those solutions too?
Keenan Horrigan that's really interesting 🤔... So conversely x= y^y and all the other infinitely many of those?
They are not solutions, because the exponents would multiply
Connor Smith What do you mean by "they would multiply"?
Keenan Horrigan (x^x)^x is not x^x^x
Kenneth Anderson Shit, you're right
But wouldnt 1^infinity be 1 because 1 times itself is 1 no matter how many times your do it?
sry if im being dumb but im a 13 year old xd
iTzTien things like 1^i are known as indeterminate forms, where just plugging in the numbers gets you a result that cannot be evaluated itself, but some more math makes it possible to give a meaningful value to the limit. For example, a common indeterminate form is i*0. For a simple example of how this works, imagine the limit as x goes to infinity of x*1/x. Naturally, no matter what value you plug in for x, you get 1, and the expression being limited can easily be evaluated as 1, so the result is 1. However, plugging in infinity gets you the indeterminate form 0*i. Do the same thing with ax*1/x, where a can be any number you want, and the result is a, but they all have that same indeterminate form. This form essentially represents cutting up the ultimate value of the limit into an infinite number of pieces, all of which have to be 0 since any greater number multiplied by i is i; it could mean anything without tracking how the total value behaves as it is cut up into more and more pieces.
Now try the same with (a^(1/x))^x; this is similar, except instead of dividing a number into an increasing number of pieces and adding them back together, it uses roots to divide it into pieces and multiplies them back together. It basically does with multiplication what 0*i did with addition. This turns out to be the form 1^i; since it is an indeterminate form, this can mean anything without knowing the limiting expression.
#yAy
Wanted to write a comment, but cant think anything but #yay
really good video. very interesting
Brilliant! #yay
Looks like a hyperbola but ain't.