Its unsolvable because we can rewrite it as e^(e^x)=e^0 so we can simplify the base and find e^x=0 which doesn't have any solution neither real or complex
You have two different functions that converge to 0. But do they converge to 0 at the same rate? what happens if you use those two functions to work out the value of 0/0? [Edit] Sorry, I posted this under the wrong video. I blame autoplay! I meant this to be a response to the 0^0=0 video.
@@hybmnzz2658yes, you are right, just that many ignorant people cannot get a clue, that limit of the function is not the same as the value of the function, they don't have to coincide, they even don't have to exist simultaneously
I’ve been watching a lot of your Lambert W function videos lately and they’re awesome! Randomly yesterday I realized that I could use the Lambert W function to solve for the equation of a separatrix of a system of ODEs in my Numerical Analysis class. So cool to see it randomly pop up in my studies and I knew how to solve it because of your videos. Thank you!
You made a horrendous mistake, at 1:38, two factors' product equals zero IF AND ONLY IF one of them is equal to zero, while the other is DEFINED! You can't literally multiply zero by smth nonexistent and get smth existent (zero, in this case) . Where I've studied math you would either get zero ( point) for the solution, or get some points subtracted from your score ( depending of the severity of the examiner) . The same is with 0 to the 0th power, it is undefined, and that's the explanation why, there's not "no agreement" but the result of this operation CAN'T be defined, because of fundamental rule not to devide by zero , you literally can't pick ANY number real or complex that soild be a reasonable value of 0^0.
@@TeFurto777 Oh there is one good example in another comment. (assuming we are working under principal branch) on one hand: ln(e^(2pi i))=ln(1)=0, on the other hand 2pi i ln(e)=2 pi i. But clearly 0 does not equal 2 pi i! So this is a valid example showing ln(a^b) does not equal b ln(a). Of course, this isn't really ln(x^x) but it gives you an idea (if I give an example with ln(x^x) there will be a lot of computations, but if you insist, I'll try). The issue is branch cut and the definition of log. It turns out if we want to generalize log into the complex plane in a nice way (differentiability/analyticity), the most natural definition would leave a ray of undefined points starting at 0. Turns out there are multiple possible way of defining log based on where the ray is pointing at. In fact, near the ray, the function has a gap. Take a look at the right side of the picture. functions.wolfram.com/ElementaryFunctions/Log/visualizations/5/02/imagetext/0031/text31.gif (the left graph is the real part of the log function, and right graph is the complex part) This gap causes issues. And (this part i am not so sure), ln(x^x) and xln(x) will differ by a constant multiple of the "gap" size (which turns out to be 2pi i).
@@adiaphoros6842 He did do it for the right hand side, log(0)= 2\pi i n for some fixed n. But he used the incorrect identity log(a^b)=blog(a) on the left side. While he is considering some branches, it isn't a rigorous justification, he didn't justify he is considering all branches and all possible answers. -If Log is defined on the branch (pi/2, 2pi+pi/2], Log(i^{1/4})= (2pi+pi/4) i while 1/2 Log(i)=(pi+pi/4) i. Note that Log(i^(1/2))-1/2Log(i)= pi, and is -*-not in the form of 2pi i n-* . Edit: the above example was wrong, it turns out to be the case that log(x^x)-xlog(x)=2pi i n. Without justification, it could be the case that Log(x^x)=xLog(x)+ci where c is some random real constant, and that BPRP is not justifying why he is only considering the case of 2pi i n. and not proving that log(x^x)-xlog(x)=2 pi i n for some integer n. Moreover, even with that, say log(x^x)=xlog(x)+2pi i m and log(0)=2 pi i n. Then xlog(x)=2pi i(n-m). Since both n and m are dependent on log, one need to justify that all possible n-m over different log is the set of integers.
I love how easy it is to prove bullshit in math, like it’s so easy to “prove” 1=-1 with more advanced math and yeah there are rules you’re breaking almost all the time but nobody catching that shit.
To be fair, there are many cases where 0⁰ = 1 is acceptable, like discrete math, combinatorics, or Taylor series. It is also consistent with the empty product definition, and many programming languages calculate 0⁰ as 1. As a _limit,_ though, it is indeterminate.
thanks for your hard work sir, really elevated my spirit to study calc 2 even more. i have an integral question for you, i think this is quite hard, since wolfram alpha can't gave any indefinite form of this integral: Integral of Sqrt(t^2 + 3Sin[t] + 4) dt
Hey man, I really love your channel and your content, I watched your hours long integral and series videos to get myself through calculus 2 and I was wondering if you have or plan to do anything of the same sort for differential equations, just massive videos solving a bunch of different types of ordinary differential equations.
@@kristianbojinov6715 He wants to say that if he got 0 as a solution then it is wrong because ln[0] is undefined as well as the original question of x^x we will get 0^0 after putting it which is also undefined
Actually I think you can't justify like that. ln(x^x) won't always be xlnx when talking about complex numbers, lneⁿ = n neither. So it should be a different and accurate way to handle with it, granting the right answer as well. Even if just by justifying the use of this rule. For exemple: let 1 = e^i2pi and ln both sides it'd be ln1 = ln(e^i2pi) i2pi = 0
【The Same #LOGICAL_DISASTER As In Chinese Version】 Is your complex analysis professor majored in PE ?? Do you know ln(a^b) = b ln(a) in complex variable is generally wrong ? If you didn't know, but I've told you in Chinese version, why the logic error is still here? Have you taken response on your major ? or only views , subscribes , or profits ?? 1. Is is true that 0=ln(1) = ln(e^{2pi i}) = 2 pi i ?? 2. If you say Branch Number 1 makes ln(e^{2 pi i}) = 2 pi i. Then according this branch, x^x is obvious NOT 1. I know you knew you've made a mistake. And you don't correct it and post it as another video. this is the worst commercial activity, I think I doubt your video is not for dilivering math but for money, which is of course not the value of math communities.
Why I'm feeling so disturbed with such a deliberate dealing with complex functions, especially logarithms and exponents which don't provide a single value for each value of complex argument. Especially when he puts multiple values in rhs, but simply cancels exp with log on the left side, while there are multiple values there as well. To me all this videos are to be considered as some type of mental exercise, rather than rigorous solution having anything common to mathematics as a science and just an educational subject, aimed to learn people think critically and outside of the box. But when you omit such crucial details in your solutions that questions even this goal, for you can do something like that in your real life, and it might have much more dramatic effect in your life, than just a bad mark on math 😮
x^x= e^(x.ln(x)). Using L’Hôpitals rule you can easily show that x.ln(x)-> 0 as x goes to 0, namely by writing it as ln(x) / (1/ x). Both numerator and denominator go to (minus) infinity as x goes to zero and so you can replace them with their respective derivatives resulting in (1/x)/ (-1/x^2) = -x which goes to 0 as x goes to 0. Therefore 0^0 = 1. More interesting would be x^x = 2. There we get x = e^W(ln 2) as a possible solution, where W is the mentioned Lambert function.
If I plot the function y(x) = Re(x^x), I get an almost continuous graph that seems to intersect the y-axis at y = 1. This seems to imply in some sense that x^x approaches 1 as x approaches 0.
For complx numbers, ln(z)=ln|z|+iArg(z) where Arg(z)∈(-π,π]. If you ignore this rule as you have done you can have things such as 0=ln(1)=ln(e^2πi)=2πi which is clearly wrong. Everything after your log calculation is meaningless.
My last 3 videos have been the Lambert W function and seeing this, I was able to get the solution without even knowing precisely what the Lambert W function is
But you can. Perhaps it's mostly circumstantial, but this can be seen when you consider the complex number in euler's form, and rewrite it in polar form. e^theta(i) * e^beta(i) = (cos(theta)+sin(theta)i) * (cos(beta)+sin(beta)i). This is further simplified to cos(theta)cos(beta) + cos(theta)sin(beta)i + cos(beta)sin(theta)i - sin(theta)sin(beta) = cos(theta+beta) + sin(theta+beta)i. We can let (theta+beta) = zeta for simplicity and treat it as a parameter. Meaning the above expression may be simplified to e^zeta(i). Therefore, e^theta(i) * e^beta(i) = e^zeta(i), which means this specific piwer rule is applicable. The same may apply to the other rules I pressume. Please correct me if I am incorrect in this.
Tbh, I still don't understand why there is "no agreement". When limits are never actually finding the correct answer. They only find the approaching number on a function line and literally anything could be it depending on the function. While a series LITERALLY adds to the number and CAN'T be dependent on a function, because there is only one function there and you just plug a number in and get the correct answer. e^x series clearly proves 0^0 does equal 1. It makes zero logical sense to try and say take the "limit" when trying to get the CORRECT answer to the EXACT number. Is there another series that I don't know where it can give you 0^0 exactly and be a different answer than 1? We also have the definition of anything to power of 0 is equal to 1, why does that stop at 0? Cause you can think its somehow 0/0 or something?
Can you give a more detailed response to why e^x series proves 0^0 equals 1? Sort of confused because series is also defined as a limit. But I would also say the limit does find actual answer...The formal definition of limit: Epsilon-Delta definition proves the uniqueness and existence of a single limit for every converging sequence, so taking limit does give a unique answer (or I may be confused on what you are talking about). I believe there is no rigorous justification of what 0^0 is, it is more of a debate on which definition is nicer. There are many places where 0^0 is *defined* to be 1 since it is more convenient in many cases.
x^0 for any number besides 0 is 1. 0^x for any number besides 0 is 0. The limit isn't consistent depending on how you approach 0,0 for x^y. That alone is enough for me to think 0^0 is ambiguous.
@@qwerty_ytrewq4452 This my logic, a more mathimatical explaination I would watch "The Most Controversial Number In Math" by BriTheMathGuy. It may be defined with a limit to make it infinite terms, but it its not actually "approaching" anything, it actually adds to or we say converge to, not approaching. It actually adds terms to the other terms continuing and WILL get to such number in the end. Like geometric series are infact just limited amount of area of a shape and WILL have a limited area or one correct area no matter. If a series does infact converge it would be no difference to an limited area of a shape just more abstract compared to a geometric series, which is called geometric for a reason, actual shapes. If you can rigorously prove "e^x" 's series is correct, then e^0=0^0/0!. And we know for a fact e^0=1. So if e^0 is equal to one, then 0^0 has to equal to one.
@@chitlitlah But it only approaches a number on a function line and is never the exact number itself. Limits can't prove what the exact number is. It can happen to be the same as the exact number, but it doesn't actually prove to be it. Limits SEEM to get that number, not actually get to that number.
My two cents here: I believe saying x=0 is not a solution because x^x has no agreement is absolutely stupid in the context of this particular function since it approaches 1 at x=0 no matter which side in complex plane you approach it from Source: Wolfram Alpha
The points lie on 2 curves, one for negative n and one for positive n. The curves are almost straight lines, and they are complex complements from each other.
We can also see that x=0 is an extraneous solution by plugging it back into the multiplicative equation. That gives us an ln(0) term, which is undefined.
In my opinion:: X^X = 1, then X = 1 or X ~ 0 (nearly to 0) since X^0 = 1. Example: For X = 0.00001, then X^X = 0.99988487. For X = 1e-10, then X^X = 0.99999999769 Smaller X will give X^X close to 1. Using ghraphic aid, I also get X = -0.999992, with X^X = 0.999998, (but I don't know how to interpret it)
That exponent subtraction property of exponents doesn't work for 0, since that just results in 0 ÷ 0 every time. The best argument I can come up with for x⁰ = 1 is just that it is an empty product, meaning that it is the product of no elements, which defaults to 1, since that is the multiplicative identity and you would want the product of a 1-element list to evaluate to said element. This extends to 0⁰, which then becomes 1. Going the other direction with 1/0 for the argument of a product, you still haven't multiplied any (1/0) terms to cause the product to evaluate to undefined. • Most programming languages accept 0⁰ = 1 as well. • Taylor expansions of functions like cos(x) or eˣ rely on this. • It is useful in combinatorics.
IMO, if 0! is defined as 1, it seems logical that 0^0 could also be defined as 1. After all, we KNOW that 0 times any number is zero, but 0! isn't 0???
I don't understand your first solution. From what I understand, with any complex number multiplied by any complex number you multiply the magnitudes and add the angles. If you square any complex number, then the square of the magnitude has to equal 1 in this case. How is the magnitude of 2.213534 + 3.1139999i squared equal to 1? Wait, we're not squaring. It's to the power of itself. Still, I can't visualize it. Are there any videos showing complex numbers powered by complex numbers?
I believe saying x=0 is not a solution because x^x has no agreement is absolutely stupid in the context of this particular function since it approaches 1 at x=0 no matter which side in complex plane you approach it from
Because it is a special function. And not just any but a pretty specific one and is rarely used so it's not put on calculators, similar to other special functions. It is available in different computer algebra systems, in Wolfram alpha it is known as "W", "LambertW" and "ProductLog" Alternatively you can compute it using a few iterations of newton's method by using the fact that W is an inverse of x*exp(x)
@@kodirovsshik Oh, thanks for your exhaustive answer! If I may further ask: There is nothing special about W() that would make it impossible or illogical to put it on a calculator, right? It's just uncommon so they did not implement it.
Yes sir, exactly. That is to say, if some of the special functions were implemented on a calculator, I would guess that one would be much more likely to see something like (poly)gamma, zeta, elliptic integrals, Si, Ei, li, hypergeometric function, and maybe only then the Lambert W. I might be wrong though, I'm just guessing from what I've seen. I myself sometimes lack W in tools I use.
The equation y=x*e^(x) isn't invertible in elementary function (i.e. arithmetic, powers, roots, logs, trig, & inverse trig), so we define the Lambert W function to be the inverse to this function. If you know the value of y, the LambertW function solves for the corresponding value of x. Really multiple values, since it is a multivalued function, and there is a particular region of LambertW, where you need to pick a branch to get an answer. Similar to how y=x^2 has two branches of its inverse function, both y=sqrt(x) and y=-sqrt(x).
The equation y=x*e^(x) isn't invertible in elementary function (i.e. arithmetic, powers, roots, logs, trig, & inverse trig), so we define the Lambert W function to be the inverse to this function. If you know the value of y, the LambertW function solves for the corresponding value of x. Really multiple values, since it is a multivalued function.
Can you justify that Natural log of complex numbers. Because 'mindyourdecision' show some problem with that method in his video dedicated to solving i^i.
It's treated as a different function than natural log, but there is a complex log function, that takes the natural log of the magnitude, and then adds on the angle (plus any integer multiple of 2*pi) times the imaginary unit. Such that: log(z) = ln|z| + i*(angle(z) + 2*pi*k) where k is any integer The way we can derive it, is as follows: Let z = r*e^(i*t), and let L = a + b*i, where a, b, r, and t are all integers Define L such that: e^L = z Carry out e^L, based on its polar form breakdown: e^(a + b*i) = e^a * [cos(b) + i*sin(b)] Since the magnitude of [cos(b) + i*sin(b)] will always equal 1, no matter what be equals, this means that the magnitude of z is equal to e^a. |z| = e^a Which means: a = ln(|z|) cos(b) = real(z) sin(b) = imag(z) This means that be equals any angle that is a coterminal angle to z, since cos(angle(z)) also equals real(z), and likewise for sin(angle(z)) equaling imag(z). Thus: b = angle(z) + 2*pi*k where k is any integer. Since e^L = z, this means if we solve for a and b, the components of L, the log of z, we show that L = ln(|z|) + i*(angle(z) + 2*pi*k).
how is e^e^x=1 solvable? th-cam.com/video/ckc9F0VjZ3k/w-d-xo.htmlsi=m91azalG4twF0nTo
I solved a x^x^x=2
If you want I can give you the answer
Its unsolvable because we can rewrite it as e^(e^x)=e^0 so we can simplify the base and find e^x=0 which doesn't have any solution neither real or complex
@@saliryakouli1260 We can not rewrite as x=e^x
But we can say x=e^ln(x)
@@leonardobarrera2816 no I said that e^x is equal to 0 because we can simplify when it's the same base
@@saliryakouli1260 But the equation is other, it does not contain 0 in it original expretion
ill stick to the 1
What's wrong with 2.213534+3.1139999i? Don't be such a hater against complex numbers.
BODMAS 9. KIDDING 😅
more like the 0
@@Sir_Isaac_Newton_HUH
@@Sir_Isaac_Newton_0⁰=1 ?
You have two different functions that converge to 0. But do they converge to 0 at the same rate? what happens if you use those two functions to work out the value of 0/0?
[Edit] Sorry, I posted this under the wrong video. I blame autoplay! I meant this to be a response to the 0^0=0 video.
That's the problem, doing this method will result in many different answers, one for each fumtion that you choose, that's why it's undefined.
0/0 is not a value but an informal saying for lim f/g where f and g go to zero. Anyways, it can equal anything.
Lim ax / x = a
Lim lnx / x = inf
@@hybmnzz2658yes, you are right, just that many ignorant people cannot get a clue, that limit of the function is not the same as the value of the function, they don't have to coincide, they even don't have to exist simultaneously
I’ve been watching a lot of your Lambert W function videos lately and they’re awesome!
Randomly yesterday I realized that I could use the Lambert W function to solve for the equation of a separatrix of a system of ODEs in my Numerical Analysis class. So cool to see it randomly pop up in my studies and I knew how to solve it because of your videos. Thank you!
You made a horrendous mistake, at 1:38, two factors' product equals zero IF AND ONLY IF one of them is equal to zero, while the other is DEFINED! You can't literally multiply zero by smth nonexistent and get smth existent (zero, in this case) . Where I've studied math you would either get zero ( point) for the solution, or get some points subtracted from your score ( depending of the severity of the examiner) . The same is with 0 to the 0th power, it is undefined, and that's the explanation why, there's not "no agreement" but the result of this operation CAN'T be defined, because of fundamental rule not to devide by zero , you literally can't pick ANY number real or complex that soild be a reasonable value of 0^0.
You are the reason I love calculus and algebra so much. Thanks for the great videos!
ln(x^x) does not necessarily equal to xln(x) when we extend to the complex numbers, more care in handling the left hand side.
Why?
@@TeFurto777 Oh there is one good example in another comment. (assuming we are working under principal branch) on one hand: ln(e^(2pi i))=ln(1)=0, on the other hand 2pi i ln(e)=2 pi i. But clearly 0 does not equal 2 pi i! So this is a valid example showing ln(a^b) does not equal b ln(a). Of course, this isn't really ln(x^x) but it gives you an idea (if I give an example with ln(x^x) there will be a lot of computations, but if you insist, I'll try).
The issue is branch cut and the definition of log. It turns out if we want to generalize log into the complex plane in a nice way (differentiability/analyticity), the most natural definition would leave a ray of undefined points starting at 0. Turns out there are multiple possible way of defining log based on where the ray is pointing at. In fact, near the ray, the function has a gap. Take a look at the right side of the picture.
functions.wolfram.com/ElementaryFunctions/Log/visualizations/5/02/imagetext/0031/text31.gif (the left graph is the real part of the log function, and right graph is the complex part)
This gap causes issues. And (this part i am not so sure), ln(x^x) and xln(x) will differ by a constant multiple of the "gap" size (which turns out to be 2pi i).
@@qwerty_ytrewq4452 That’s why BPRP wrote ln(exp(2nπi)), which equals 0 when n = 0. So he’s not working with the principal branch only.
@@adiaphoros6842 He did do it for the right hand side, log(0)= 2\pi i n for some fixed n. But he used the incorrect identity log(a^b)=blog(a) on the left side. While he is considering some branches, it isn't a rigorous justification, he didn't justify he is considering all branches and all possible answers.
-If Log is defined on the branch (pi/2, 2pi+pi/2], Log(i^{1/4})= (2pi+pi/4) i while 1/2 Log(i)=(pi+pi/4) i. Note that Log(i^(1/2))-1/2Log(i)= pi, and is -*-not in the form of 2pi i n-* .
Edit: the above example was wrong, it turns out to be the case that log(x^x)-xlog(x)=2pi i n.
Without justification, it could be the case that Log(x^x)=xLog(x)+ci where c is some random real constant, and that BPRP is not justifying why he is only considering the case of 2pi i n. and not proving that log(x^x)-xlog(x)=2 pi i n for some integer n.
Moreover, even with that, say log(x^x)=xlog(x)+2pi i m and log(0)=2 pi i n. Then xlog(x)=2pi i(n-m). Since both n and m are dependent on log, one need to justify that all possible n-m over different log is the set of integers.
@@TeFurto777 Log(z^w) = w * Log(z) ONLY IF -pi < Im( w * Log(z) )
Accidentally proved 0^0 = 1 🤯
I love how easy it is to prove bullshit in math, like it’s so easy to “prove” 1=-1 with more advanced math and yeah there are rules you’re breaking almost all the time but nobody catching that shit.
Not really. Cuz ln0 is undefined. Even if u define it as -infinity. 0*-infinity is also an undefined expression
not really
Actually you can also “prove” it by taylor series but Ofc it is wrong
To be fair, there are many cases where 0⁰ = 1 is acceptable, like discrete math, combinatorics, or Taylor series.
It is also consistent with the empty product definition, and many programming languages calculate 0⁰ as 1.
As a _limit,_ though, it is indeterminate.
thanks for your hard work sir, really elevated my spirit to study calc 2 even more.
i have an integral question for you, i think this is quite hard, since wolfram alpha can't gave any indefinite form of this integral:
Integral of Sqrt(t^2 + 3Sin[t] + 4) dt
who let him cook this time because this is some good 2 AM maths
Hey man, I really love your channel and your content, I watched your hours long integral and series videos to get myself through calculus 2 and I was wondering if you have or plan to do anything of the same sort for differential equations, just massive videos solving a bunch of different types of ordinary differential equations.
Here's one video: th-cam.com/video/e-cTygNbEUE/w-d-xo.html : )
cheers!
Worst mistake is taking Ln both sides
xlnx=0
And picking 1 solution as x=0
how is it equal to 0 when it's equal to 2pi*n*i, that's precisely the point of "finding all the solutions"
Could you argue your point ?
Ln(1) = 0 wym
@@kristianbojinov6715 He wants to say that if he got 0 as a solution then it is wrong because ln[0] is undefined as well as the original question of x^x we will get 0^0 after putting it which is also undefined
Actually I think you can't justify like that. ln(x^x) won't always be xlnx when talking about complex numbers, lneⁿ = n neither. So it should be a different and accurate way to handle with it, granting the right answer as well. Even if just by justifying the use of this rule.
For exemple: let 1 = e^i2pi and ln both sides
it'd be ln1 = ln(e^i2pi)
i2pi = 0
【The Same #LOGICAL_DISASTER As In Chinese Version】
Is your complex analysis professor majored in PE ??
Do you know ln(a^b) = b ln(a) in complex variable is generally wrong ?
If you didn't know,
but I've told you in Chinese version,
why the logic error is still here?
Have you taken response on your major ?
or only views , subscribes , or profits ??
1. Is is true that 0=ln(1) = ln(e^{2pi i}) = 2 pi i ??
2. If you say Branch Number 1 makes ln(e^{2 pi i}) = 2 pi i. Then according this branch, x^x is obvious NOT 1.
I know you knew you've made a mistake.
And you don't correct it and post it as another video.
this is the worst commercial activity, I think
I doubt your video is not for dilivering math
but for money, which is of course not the value of math communities.
Lambert w function is the biggest W
The values you've found are unreal!
Could somebody make a playlist of all the BPRP videos where he uses the Lambert W function? I want to binge them.
Never heard of this Lambert function, but now I'm curious.
I don't understand why this man would keep shoes in camera frame😅
Please , geometry session full complete simple to andvanced lebel and olympiad questions
As amc ,india ioqm,rmo,inmo,imo etc.
Can you do x=1/Sq root (x)
x = 1/√x
Multiply by √x on both sides
x√x = 1
Square both sides
x².x = 1
x³=1
x = ³√1
x = 1 (real)
x = -√3/2 ± i/2 (complex)
Not quite! Here we use some number theory at the end sort of.
x = e^lambert w function(2 pi i n)
n is any integer!
If n = 0:
x = e^W(2iπ0)
x = e^W(0)
W(0)→ W(x.e^x) = x → 0 = 0.e⁰ → W(0.e⁰) = 0
x = e⁰
x = 1
So n can be any integer
Btw, e^W(x) is just x/W(x). So if you don’t want to deal with nested powers, you can write it as 2pini / W(2pini)
right cuz Wx.e^Wx = x
genius!
Why I'm feeling so disturbed with such a deliberate dealing with complex functions, especially logarithms and exponents which don't provide a single value for each value of complex argument. Especially when he puts multiple values in rhs, but simply cancels exp with log on the left side, while there are multiple values there as well. To me all this videos are to be considered as some type of mental exercise, rather than rigorous solution having anything common to mathematics as a science and just an educational subject, aimed to learn people think critically and outside of the box. But when you omit such crucial details in your solutions that questions even this goal, for you can do something like that in your real life, and it might have much more dramatic effect in your life, than just a bad mark on math 😮
Not, gonna lie I wish this video came out before my Methods 34 exam this year as it could of helped with one of the questions.
x^x= e^(x.ln(x)). Using L’Hôpitals rule you can easily show that x.ln(x)-> 0 as x goes to 0, namely by writing it as ln(x) / (1/ x). Both numerator and denominator go to (minus) infinity as x goes to zero and so you can replace them with their respective derivatives resulting in (1/x)/ (-1/x^2) = -x which goes to 0 as x goes to 0. Therefore 0^0 = 1. More interesting would be x^x = 2. There we get
x = e^W(ln 2) as a possible solution, where W is the mentioned Lambert function.
If I plot the function y(x) = Re(x^x), I get an almost continuous graph that seems to intersect the y-axis at y = 1.
This seems to imply in some sense that x^x approaches 1 as x approaches 0.
It does which makes sense.
Eddie Woo's video on 0^0 goes into this in more detail, I recommend it
For complx numbers, ln(z)=ln|z|+iArg(z) where Arg(z)∈(-π,π]. If you ignore this rule as you have done you can have things such as 0=ln(1)=ln(e^2πi)=2πi which is clearly wrong. Everything after your log calculation is meaningless.
It's 3am why am I watching this?
Isn’t the product log of 2nπi just 2nπi since 2nπi * e^(2nπi) = 2nπi * 1, so the product log of 2nπi would just be that. And e^(2nπi) would just be 1
My last 3 videos have been the Lambert W function and seeing this, I was able to get the solution without even knowing precisely what the Lambert W function is
Could you do
x
∫ (ln[t]/[t^x])dt = 0
1+1/x
The answer is nice
@@anshumanmondal8317It's the golden ratio
@@anshumanmondal8317how did you manage to get complex numbers from a real-valued function integral 🤨
Can you do some problems on Rolle's theorem and how to decide on the primitive function that should be used?
6:19 the normal exponential power rule cannot be used in case of complex number right??
But you can. Perhaps it's mostly circumstantial, but this can be seen when you consider the complex number in euler's form, and rewrite it in polar form. e^theta(i) * e^beta(i) = (cos(theta)+sin(theta)i) * (cos(beta)+sin(beta)i).
This is further simplified to cos(theta)cos(beta) + cos(theta)sin(beta)i + cos(beta)sin(theta)i - sin(theta)sin(beta) = cos(theta+beta) + sin(theta+beta)i.
We can let (theta+beta) = zeta for simplicity and treat it as a parameter. Meaning the above expression may be simplified to e^zeta(i).
Therefore, e^theta(i) * e^beta(i) = e^zeta(i), which means this specific piwer rule is applicable. The same may apply to the other rules I pressume. Please correct me if I am incorrect in this.
As a 15 yr old highscooler in Britain (we dont get taught calculus until post 16 btw) I can confirm that you my friend are a magician
X^X=2 try this
Can you do more differential equations solutions?
Second: what branch w lambert we get?
Good question. I think that question would be where the real fun starts. And by "real" I mean complex.
but isn't i to the power of any multiple of 4 1 so shouldn't you add +4n where n is any integer
my casio fx-991 cw says the solve for this eqn is- 1e-50
I'm in highschool pre-calculus. This is all gibberish.
you r a genius..taught us another and more general solution.
Does the Lambert W function come from hell? :)
Please, amc for best books
Tbh, I still don't understand why there is "no agreement". When limits are never actually finding the correct answer. They only find the approaching number on a function line and literally anything could be it depending on the function. While a series LITERALLY adds to the number and CAN'T be dependent on a function, because there is only one function there and you just plug a number in and get the correct answer. e^x series clearly proves 0^0 does equal 1. It makes zero logical sense to try and say take the "limit" when trying to get the CORRECT answer to the EXACT number. Is there another series that I don't know where it can give you 0^0 exactly and be a different answer than 1? We also have the definition of anything to power of 0 is equal to 1, why does that stop at 0? Cause you can think its somehow 0/0 or something?
Can you give a more detailed response to why e^x series proves 0^0 equals 1? Sort of confused because series is also defined as a limit. But I would also say the limit does find actual answer...The formal definition of limit: Epsilon-Delta definition proves the uniqueness and existence of a single limit for every converging sequence, so taking limit does give a unique answer (or I may be confused on what you are talking about). I believe there is no rigorous justification of what 0^0 is, it is more of a debate on which definition is nicer. There are many places where 0^0 is *defined* to be 1 since it is more convenient in many cases.
x^0 for any number besides 0 is 1. 0^x for any number besides 0 is 0. The limit isn't consistent depending on how you approach 0,0 for x^y. That alone is enough for me to think 0^0 is ambiguous.
@@qwerty_ytrewq4452 This my logic, a more mathimatical explaination I would watch "The Most Controversial Number In Math" by BriTheMathGuy.
It may be defined with a limit to make it infinite terms, but it its not actually "approaching" anything, it actually adds to or we say converge to, not approaching. It actually adds terms to the other terms continuing and WILL get to such number in the end. Like geometric series are infact just limited amount of area of a shape and WILL have a limited area or one correct area no matter. If a series does infact converge it would be no difference to an limited area of a shape just more abstract compared to a geometric series, which is called geometric for a reason, actual shapes.
If you can rigorously prove "e^x" 's series is correct, then e^0=0^0/0!. And we know for a fact e^0=1. So if e^0 is equal to one, then 0^0 has to equal to one.
@@chitlitlah But it only approaches a number on a function line and is never the exact number itself. Limits can't prove what the exact number is. It can happen to be the same as the exact number, but it doesn't actually prove to be it. Limits SEEM to get that number, not actually get to that number.
My two cents here: I believe saying x=0 is not a solution because x^x has no agreement is absolutely stupid in the context of this particular function since it approaches 1 at x=0 no matter which side in complex plane you approach it from
Source: Wolfram Alpha
I wonder what shows up if one plots all solutions in complex plane. Too lazy right now.
The points lie on 2 curves, one for negative n and one for positive n. The curves are almost straight lines, and they are complex complements from each other.
The Domain of w Lambert is real or complex?
What exactly is the lambert W function?
We can also see that x=0 is an extraneous solution by plugging it back into the multiplicative equation. That gives us an ln(0) term, which is undefined.
3:38 wait cant you only use the principal root..?
In my opinion::
X^X = 1, then X = 1 or X ~ 0 (nearly to 0) since X^0 = 1.
Example:
For X = 0.00001, then X^X = 0.99988487.
For X = 1e-10, then X^X = 0.99999999769
Smaller X will give X^X close to 1.
Using ghraphic aid, I also get X = -0.999992, with X^X = 0.999998, (but I don't know how to interpret it)
0^0 = 1*0/0 = undefined (basically 0 itself is undefined)
That exponent subtraction property of exponents doesn't work for 0, since that just results in 0 ÷ 0 every time. The best argument I can come up with for x⁰ = 1 is just that it is an empty product, meaning that it is the product of no elements, which defaults to 1, since that is the multiplicative identity and you would want the product of a 1-element list to evaluate to said element. This extends to 0⁰, which then becomes 1. Going the other direction with 1/0 for the argument of a product, you still haven't multiplied any (1/0) terms to cause the product to evaluate to undefined.
• Most programming languages accept 0⁰ = 1 as well.
• Taylor expansions of functions like cos(x) or eˣ rely on this.
• It is useful in combinatorics.
IMO, if 0! is defined as 1, it seems logical that 0^0 could also be defined as 1.
After all, we KNOW that 0 times any number is zero, but 0! isn't 0???
As you like vidios titles photo Einstein😮😮😮😮
I don't understand your first solution. From what I understand, with any complex number multiplied by any complex number you multiply the magnitudes and add the angles. If you square any complex number, then the square of the magnitude has to equal 1 in this case. How is the magnitude of 2.213534 + 3.1139999i squared equal to 1? Wait, we're not squaring. It's to the power of itself. Still, I can't visualize it. Are there any videos showing complex numbers powered by complex numbers?
(-1)
nobody wants fake numbers lol.
What W function is? I totally forgot that.
the 9999 sounds like wawawawa in FF lmao
I believe saying x=0 is not a solution because x^x has no agreement is absolutely stupid in the context of this particular function since it approaches 1 at x=0 no matter which side in complex plane you approach it from
How do I use W() on my Casio?
You don't
@@kodirovsshik Why is it not there?
Because it is a special function. And not just any but a pretty specific one and is rarely used so it's not put on calculators, similar to other special functions. It is available in different computer algebra systems, in Wolfram alpha it is known as "W", "LambertW" and "ProductLog"
Alternatively you can compute it using a few iterations of newton's method by using the fact that W is an inverse of x*exp(x)
@@kodirovsshik Oh, thanks for your exhaustive answer! If I may further ask: There is nothing special about W() that would make it impossible or illogical to put it on a calculator, right? It's just uncommon so they did not implement it.
Yes sir, exactly. That is to say, if some of the special functions were implemented on a calculator, I would guess that one would be much more likely to see something like (poly)gamma, zeta, elliptic integrals, Si, Ei, li, hypergeometric function, and maybe only then the Lambert W. I might be wrong though, I'm just guessing from what I've seen. I myself sometimes lack W in tools I use.
I’ve been working on a similar problem, but I can’t seem to solve it. It’s x^x^x^x=9
Hey anyone what is this W fxn......
Asking as I am in highschool(Class 12)
The equation y=x*e^(x) isn't invertible in elementary function (i.e. arithmetic, powers, roots, logs, trig, & inverse trig), so we define the Lambert W function to be the inverse to this function. If you know the value of y, the LambertW function solves for the corresponding value of x. Really multiple values, since it is a multivalued function, and there is a particular region of LambertW, where you need to pick a branch to get an answer. Similar to how y=x^2 has two branches of its inverse function, both y=sqrt(x) and y=-sqrt(x).
Whenever I see one of these insane problems where the only real solution is 1 or 0, I know W can’t be far behind…
Are you growing a beard again?
Bro wearing the Lambert but still I am dumb enough as a lamb to notice it until he mentioned it.
he doesnt solve he simplifies
I don't know why I'm watching this even tho I don't understand a single thing😅.
What is the W function?
aap kha se ho. m Bharat se hu. ap acha padata hai
The answer is 1 my guy
It's easy to show that for all n exist an m where both are integers
w(m,2nπi)=2nπi
Is it possible to find a relationship between m and n
Solution: x isn't real, therefore x doesn't exist. so x is DNE
wtf?
Find all solutions to x^x =x including real and complex
That shirt would be so fire with the angry fish instead of x
Bro is stuck in a time loop making the same videos ten trillion times
These Tshirts are awesome
this dude can figure out why the chicken crossed the road
What about x^x^x=1. ?
x = 1 🤩
Btw Lambert W doesnt work on this kind of equation (x^x^x = y)
Yes i can use it. Solution equation x^x^x =a is recurency x(n+1)=exp(W(W(x(n)*ln(a))) x(0)=1. For a =2pi*i we have 2.34604680777561+0,67808152886982*i
I want that shirt of Urs with the W f(x)😊,how can I get it?
This looks unreal
Make me understand the 'w' thing
Anyone?!
The equation y=x*e^(x) isn't invertible in elementary function (i.e. arithmetic, powers, roots, logs, trig, & inverse trig), so we define the Lambert W function to be the inverse to this function. If you know the value of y, the LambertW function solves for the corresponding value of x. Really multiple values, since it is a multivalued function.
What is this W( ) function? Spell it for me so I can look it up.
th-cam.com/video/Qb7JITsbyKs/w-d-xo.html
great thumbnail
Can you justify that Natural log of complex numbers. Because 'mindyourdecision' show some problem with that method in his video dedicated to solving i^i.
It's treated as a different function than natural log, but there is a complex log function, that takes the natural log of the magnitude, and then adds on the angle (plus any integer multiple of 2*pi) times the imaginary unit.
Such that:
log(z) = ln|z| + i*(angle(z) + 2*pi*k)
where k is any integer
The way we can derive it, is as follows:
Let z = r*e^(i*t), and let L = a + b*i, where a, b, r, and t are all integers
Define L such that:
e^L = z
Carry out e^L, based on its polar form breakdown:
e^(a + b*i) = e^a * [cos(b) + i*sin(b)]
Since the magnitude of [cos(b) + i*sin(b)] will always equal 1, no matter what be equals, this means that the magnitude of z is equal to e^a.
|z| = e^a
Which means:
a = ln(|z|)
cos(b) = real(z)
sin(b) = imag(z)
This means that be equals any angle that is a coterminal angle to z, since cos(angle(z)) also equals real(z), and likewise for sin(angle(z)) equaling imag(z). Thus:
b = angle(z) + 2*pi*k
where k is any integer.
Since e^L = z, this means if we solve for a and b, the components of L, the log of z, we show that L = ln(|z|) + i*(angle(z) + 2*pi*k).
Have you thought about using triangular notation?
Uno a la uno es uno..
Too...much....letters....not....enough....numbers.......AAARRGGGHHH
damn you love that function, don't you haha
fish function?
First
Not last
What is the w?
найди чему равен i
x = 0 and 1 😅
Where is the Wilbet function fish 😳😳😳😳
Good morning
Very tricky - great👍
*solutions
i^4=1 ...
me : x = i^4