case 6 works with something akin to the squeeze theorem tl;dr x^x as x approaches -infinity is in the form re^itheta where r approaches 0 if we rewrite x as re^itheta, as x approaches -infinity thats the same as x equals the limit as r approaches infinity of re^ipi (re^ipi)^(re^ipi)=(re^ipi)^-r=(r^-r)(e^(i(-rpi))) r^-r is a positive real number that approaches 0 as r approaches infinity, meanwhile e^(i(-rpi)) in the form e^itheta meaning its just polar form with a radius of r^-r since its always on the circle on the complex plane with a radius of r^-r, and because the circle gets compressed to a point specifically down to 0, this means limit as x approaches -infinity of x^x should be considered 0 this is not the most rigorous proof but it works for me
@blackpenredpen Hey bprp. Can u do a formula for a tetrated by x plus b exponentiated by x plus c times x plus d equal to 0. Love ur videos. Bcos of u I learnt and mastered calculus at the age of 10.
If x is a complex number other than 0 we can write x=r*e^iθ and then we see that x^x=(re^iθ)^x=r^x e^ixθ In order for this expression to be 0 we need either r^x=0 or e^ixθ=0. But since r>0 hence r^x != 0. Likewise e^ixθ != for any x. Thus x^x=0 has no solution in the complex numbers
@@derda3209 If a != 0 then a^x=e^(x*log(a)) which is never 0 (exponential function is never zero over real numbers and complex numbers). Note that log(a) is defined for none zero complex numbers
I think, the easiest proof that the solution doesn't exist in the complex world is that there aren't any zero divisors in the complex world, so for x^a to be equal to 0 , x=0 , however contradiction.
Nope. a^b = c means multiplying any constant K by the factor a for a number of b times, is equal to multiplying the same constant K by the factor c. Hence, 0^0 = c means multiplying any constant K by the factor 0 for a number of 0 times, is equal to multiplying the same constant K by the factor c. The former side gives K (because multiplying K by 0 for a number of 0 times, gives K), the latter side gives K*c (because multiplying K by the factor c gives K*c), so K = K*c , valid for any value of K ==> c = 1 Therefore, 0^0 = c = 1 Q.E.D.
how can 0^0 be 1/2 ??? If we say that "it is defined", and these 2 zeros are identical, then the solution must be 1, because it assymptoticaly goes to 1. But more strictly it is not defined. So in none of case it could be 1/2.
I'd argue -inf works without djscreetization because the negative ultimately only changes the complex sign and the magnitude still goes to 0 and 0 with any complex sign is still 0
I understand that mathematician say : "It has no solution, because negative Infinity is not a real or complex number"... But, it is obvious that assymptoticly -inf (in integer) converges to solution. It could be really interesting to make some physical experiment, where the nature say if the -inf is the solution of this equation.... For me, from technicial - engineering aspect, it is solid solution (but not ideal); The only question for me is, if the nature supports the infinity .. :)
im pretty sure even in complex this works because complex exponentiation doesnt change magnitude, so if it converges to 0 on integers it should converge for the remaining values in between that are bounded by surrounding integer values.
I think the lim as x-> negative infinity is valid. You can see that if you try x= -10, -100, -1000 then x^x surely is approaching zero. It could be argued that the function is discontinuous in the negatives, but for going to -infinity at every point the function exists x^x approaches zero as x gets more negative.
This is where new math is to be found. Idk why no one looks here or ever talks about numbers like these. Juat as we thouught i couldnt exist, it does. So this probably does too. Same with |x|=-1
I believe as long as x is real, case 6 with the limit works. lim x->-inf (x^x) = lim x->inf ((-x)^(-x)). (-x)^(-x)=(-1)^(-x) * x^(-x) The latter factor always approaches zero as x approaches infinity. The former factor is cyclical and the absolute value is always equal to one, so (-x)^(-x) always approaches 0 as x approaches infinity and thus x^x approaches zero as x approaches negative infinity. I think there’s an extension of this argument if you allow x to approach negative infinity from paths in the complex plane, but I haven’t fully formulated it yet.
There are problems trying to define a^x for non-integer values when a is negative. E.g. (-1)^(1/3) should equal (-1)^(2/6) since 1/3 = 2/6. The first expression would give the cube root of -1 which is -1 (note that the cube root function *is* defined for negative real values). However, the second expression would give the 6th root of (-1)^2, which equals 1. See en.wikipedia.org/wiki/Exponentiation#Real_exponents for more information.
@@DutchMathematician you have to simplify exponent fractions first. Values that you can get when it's not simplified don't matter. He showed this in one of his "1=2 proof" videos.
@@zachansen8293 There is no need to simplify fractions first. If f(x) = a^x is a well-defined function, then f(1/3) = f(2/6) *should* hold, no matter how we supply the same argument to the function f. If we give a function two equivalent expression then the outcome should be the same.
@@DutchMathematician Go watch the video, you absolutely do. That's how he proves that 1=2 or 1=-1 or whatever it was. You have to simplify fractions in an exponent (or you can do them in a specific order if it isn't). If you don't you are likely to get incorrect results. edit: Go search for the video named "Hate to be that guy but I need the extra credit! " - it's the last chapter starting at 3:48. He shows that i=1 because i = i^(4/4) = 1 edit2: I believe if it's not simplified you can do the denominator first, but you cannot do the numerator first. But the easiest thing to remember is to just always simplify fractions that are exponents. last edit: and of course you should run it through wolfram alpha to see that it doesn't get your "solution"
I think 💭 the first video I watched from was pi^e vs e^pi. Back then I think was entering masters in math/stats and today I’m a year away from PhD in statistics. Always inspiring! Btw did you ever did a video on Riemann-Stieltjes integration???? I recall it but can’t find it
Love how postmodern math is caught in a fantastic sceptical philosophy. He shows using 10 different reality based ways how x^x cannot be zero but proceeds with conclusion that « in a complex world » it « might ». Bake skepticism into the choice of words to really screw with all brains for 200 years and expect good results.
Well, it's easy to prove that I cannot find a solution. It's way harder to prove that there is definitely no solution... Or weirder: that even if a solution exists it is impossible for someone to find it.
@@Kleyguerth the one with the burden of proof is the one making a claim, an assertion. If you say it is possible, what proof do you present? Until you do, all of the refutations of this statement stand and you must shut up.
Am I misunderstanding your comment or are you categorizing complex numbers as "postmodern math?" Because they have some quite interesting applications in "reality"
Your teacher's argument doesn't make sense. By the same logic, 0^1 is undefined, because 0^1 = = 0^(2-1) = (0^2)/(0^1) = 0/0 = undefined but of course we know that 0^1 = 0 is _not_ undefined.
@@yashrajtripathi4832 Oops! My bad. I think what the original commenter's high school teacher meant, is that in general, a^b = (a^(b+1)) / a which, in the case of a=2 and b=2 , would lead to 2^2 = (2^(2+1)) / 2 = (2^3)/2 = 8/2 which is correct (both sides equal 4); so nothing wrong there. With a=0 and b=0 , this formula would lead to 0^0 = (0^(0+1))/0 = (0^1)/0 = 0/0 which on the righthandside is undefined (and therefore, according to his teacher, 0^0 should be undefined). However, the rule a^b = (a^(b+1)) / a is not valid for a=0 , because for example with a=0 and b=1, it leads to 0^1 = (0^(1+1))/0 = (0^2)/0 = 0/0 which is clearly wrong, because the lefthandside is _not_ undefined; 0^1 equals 0. Therefore, the high school teacher's argument as to why 0^0 is undefined, is wrong, as it doesn't hold water; he applied a rule that shouldn't be applied in this case.
subtract 1 from x^x and from 0, square both sides, take x^x as t, solve the quadratic equation and get that x^x = 1 +- i . multiply x^x and 1+- i by x^-1 and get that x = 1 +- i
I expressed the complex x in the base using euler's form aka re^iphi and the one in the power in the a + bi form, then I got the equation e^(a*ln(r)-bphi)*e^i(aphi + b*lnr), for that to equal zero the term in the first power (that is a*ln(r)-bphi) has to go to negative infinity, so for exact values it is impossible, but if we take the limit of 0 + bi as b goes to infinity, it does equal zero because phi is bounded and the real part of 0 + bi is exactly zero.
z^z = exp(z*Logg(z)) in complex numbers, where Log(z) is the principal branch of the complex logarithm. However, the modulus |z^z| = exp(Re(z*Log(z))) = exp((x/2)*ln(x^2 + y^2) - y*ArcTan(y/x)) The expression in the exponent seems to have a local minimum for x = -1/e and y = 0, so it seems |z^z| >= 1/e, for all complex z = x + 1j*y
This is probably the best video you have ever done. It’s every way to approach the solution to an equation disguised as a search for the answer to x^x=0. Bravo. Beautiful.
The magnitude of x^y for real numbers y, is |x|^y, and if we have the magnitude of a limit approaches 0, then we can safely say that the limit is zero. the limit as x-> -\infty of |x|^x, is also equal to the limit as x->\infty x^{-x} = 0, meaning that the magnitude is 0, therefore the limit as x-> -\infty x^x = 0
0:50 From one side, 0^0 is 0 because 0^a is 0, from the other, 0^0 is 1, because a^0 = 1. So if we multiply the rules it will be (a^0)(0^a)/(0^a)(a^0)=0/0=1. For a big part of people, 0^0 is equal to 1, for a smaller fraction of them it's 0 or/and undefined, for the smallest one it's superposition between 1 and 0 or/and 0,5 (Corresponds with the mathematical "joke" that says 1+2+3+4...=-1/12). I count that 0^0 is undefined, because taking the power of a number is just taking it to the power of 1/(nth root power), so 2nd power is 1/0.5=2. If it's to the power of 0, then it's 1/0, which is undefined, and any operation with undefined gives just undefined in the output.
my intuition before I watch the video, not sure if this will get anywhere: I learned early on in calculus that if you have an equation f(x) with no constants, you can rewrite it as e^ln(f(x)) to solve that issue.
Well, if you consider x>0, then we can use x^x= e^{x*lnx}, and e^y is never 0. Same goes for the Complex Exponential; 1 is the only value e^z doesn't take. I think this does it for X^x=0.
There is no finite complex solution If z = r* e^it, r and t real z^z = r^z * e^itz So the equation becomes r^z=0 or e^itz=0 Case 1 r^z=0=>|r^z|=0 =>r^r =0, which has no solution as r is a positive real number Case 2 e^itz=0 => |e^itz|=0 =>e^-ity=0 Since t is between - pi and pi, y has to be - inf, thus no finite complex solution This also requires
(a+bi)^(a+bi) can be rewritten as (a+bi)^a*(a+bi)^bi now we know that either (a+bi)^a=0 or (a+bi)^bi=0 then (a+bi)^bi can be expressed as e^(bi*Log(a+bi)) where Log is the complex log (if f(z)=e^z then f^-1(z)=Log(z)) let's bring the formula for the complex Log: Log(z)=ln(|z|)+arg(z)i now now let's simplify e^(bi*ln(|a+bi|)+arg(a+bi)i) simplify a little bit more (e^bi)^(ln(|a+bi|)+arg(a+bi)i) now simplify e^(bi*ln(|a+bi|))*e^(bi*arg(a+bi)i) then just solve the equation e^(bi*ln(|a+bi|))*e^(bi*arg(a+bi)i) = 0 we get (cos(b)+sin(b)i)^ln(|a+bi|)*e(-b*arg(a+bi)) now (cos(b)+sin(b)i)^ln(|a+bi|) can't equal 0 because |cos(b)+sin(b)i|=1 for any b (= R and and for this to equal 0 |cos(b)+sin(b)i| must equal 0 now the second part e^(-b*arg(a+bi)) |e| =e so this is not equal to 0 for the same resons now let's come back to our original problem: either (a+bi)^a=0 or (a+bi)^bi=0 but we proven that (a+bi)^bi/=0 so (a+bi)^a=0 then |a+bi| = 0 then a+bi=0 then a=0 and b=0 then we get 0^0 which is either 1, undefined or has infinitly many values but for all cases 0^0/=0 so (a+bi)^(a+bi) has no solution
Extra: let's see if equation z^z has any solutions in set denoted as C(epsilon, phi) (I will type j for epsilon and k for phi) where j^2=0 where j /=0 j/j=1 , k=1/j and k^2 is undefined firstly let's define exponents: e^z=sum from: n=0 to: infinity z^n/n! then e^jz=sum from: n=0 to: infinity (jx)^n/n! = 1+jx+0+0+0+... =1+jx now the ln: ln(a+bj)=c+dj then e^(c+dj)=a+bj then simplify (e^c+(e^c)dj) = a+bj then form two equations e^c=a and e^c(dj) = bj solve the first one c = ln(a) substitute a(dj) = bj now ad=b now d = b/a so ln(a+bj)=ln(a)+(b/a)j
X= -infinity seems like a good solution to me. Even if we have to use complex numbers to define powers of negatives, they still approach 0 in absolute value
Intuitively I knew that before your attempts to find a solution. My Calc teacher may have gone over this with me in college and I am just recalling the lesson.
They should as the definition of the power is 0 multiplications from the original definition of 1. This is a formulation for the inductive proof of every single number. The fact that every other positive power of 0 is 0 is irrelevant to what the zeroth power of zero is.
@@mhm6421 When the definition of power by simple integer is set up with 1 as the basis, it makes little sense to claim that 0^0 is not 1. You'd be putting in a branch for no reason. In other words, it is simpler.
@@ronaldking1054That definition was not just randomly defined, it comes from a pattern seen with exponents. For example, 2^4 = 2*2*2*2. 2^3 = 2*2*2. Based on this, we could assume that x^n = (x^(n+1))/x. By this logic, 0^0 = 0^1 / 0. Division by zero is undefined, and so this has no answer. We could define one, but not everyone will agree on the definition.
@yurenchu it's the Banach limit. I'm not actually sure you can apply it to 0^0. It applies to sequences. The Banach limit of {0,1,0,1,...} is 1/2. This sequence does not converge under the usual sense of limits, but Banach limits extend the usual limits to non-convergent sequences. Not all sequences have uniquely defined Banach limits.
Thanks for your reply. I didn't receive a reply notification because your reply wasn't posted within the same thread, but luckily I happened to revisit this comment section and noticed it by accident. Interesting to read about the Banach limit, I'm not familiar with it, so thanks for that. It leaves me puzzled though how it can be applied to a function as described in your previous post (i.e. a real function f(x) that's 1 for rational x , and 0 for irrational x, and then show that the Banach limit of f(x) equals 1/2 for x approaching infinity).
As for more (fruitless) ways to solve it, there were a couple I came up with (assuming you're using the extended complex plane) The first is pretty simple: if z^z = 0, then e^(z ln z) = 0, and thus z ln z = -infinity (we're in the extended complex plane so this is fine) The only value this works for is z = -infinity, so that's our solution The second is just substitution; set z = x+1, so (x+1)^(x+1) = 0 Then (x+1)^x * (x+1) = 0, then x * (x+1)^x + 1 * (x+1)^x = 0 So x * (x+1)^x = -(x+1)^x, meaning ln(x) + x ln(x+1) = ln(-1) + x ln(x+1) Assuming x =/= -1, we can remove x ln(x+1) from both sides, giving us ln(x) = ln(-1), which lets us conclude x = -1, which is a contradiction of our assumptions and thus the equation has no solutions
Why would z*ln(z) = -♾ necessarily imply z = -♾? If we assume ln(-♾) = ln(-1) + ln(♾) = iπ + ♾, then we have -♾*ln(-♾) = -♾*ln(-1) + -♾² = = -i*♾ - ♾ We end up with complex infinity here, or rather, unsigned infinity, not -♾
actually for no. 6, the 2n limit, it even works for x = n for n integer, but it approacges 0 from up then down, up the down, and so on (much like sin(x)/x as x approaches infinity
Funnily enough if you put in Desmos x = 0 and 0^(1/x), then you end up with 0^(1/x) = 0 I suppose from a coding perspective, Desmos sets priority that 0^anything overwrites the calculation that 1/0 is undefined. Just an interesting mention!
Take two complex numbers z and w. If z,w≠0, then z^x=w can be solved for x. If z=0 and w=0, then x>0. If z=0 and w≠0, x has no solutions. If z≠0 and w=0, x has no solutions. x^x=0 should have no solutions. Sure, if we take the limit as x→-∞, where -∞=inf R travelled along the negative real axis and not the general complex ∞, then we'd get a complex number of magnitude decreasing to 0, but we want numbers, not extended numbers.
x^x=(re^(i phi))^(a+bi) where a=r cos(phi) and b=r sin(phi) leads to x^x=e^(r ln r cos(phi) - r phi sin(phi)) e^i(r phi cos(phi) + r ln r sin(phi)) Since sin(theta) and cos(theta) are never simultaneously 0, this can approach zero only as phi-->infinity. (The magnitude approaches zero.)
For the original #6: it feels like the argument should be makeable based on that the magnitude of said expression feels like it should approach zero. If we restrict to rational values, then it should be trivial to argue - for irrational values, perhaps we can still show any branch approaches zero, and argue that way?
With x = a + ib we have z = x^x = (a + ib)^(a+ib) = exp([½ a ln(a²+b²) - b theta] + [½b ln(a²+b²) + a theta] i), where theta = arg(a,b). We want │z│= exp(½ a ln(a²+b²) - b theta) = 0. To me it seems this is the case if a goes to minus infinity independent of b. (Or possibly if b goes to plus infinity ???). So my tentative answer would be x = - infinity + i b with b arbitrary. I checked (-50)^(-50) is very small, i.e. close to zero. (-50)^(-50) = 1.126 · 10^(-85). So it seems plausible.
Maybe we should try making up numbers to solve this. Previous big updates were caused by addition, multiplication and exponentiation, while the problem in the video is caused by tetration. And the solution doesn’t have to be -♾️, it could be an orthogonal unit just like i
To me 0^0 is undefined even if we take a seventh grade approach. 5^0 = 5^(m-m) = 5^m / 5^m. If we take 0 instead of 5 we get: 0^m / 0^m m can be any number yet we still have undefined fraction of 0/0 which could be be approached if we have some context, like in sinc function: sin(x)/x when x -> 0. But here we don't have that context, so it remains undefined. P.S. I also was amazed that in English this limit doesn't have a special name. For example in Russian this limit is called "Первый замечательный предел" or "first remarkable (or wonderful) limit". And the second such limit is (1+1/x)^x when x-> infinity.
Simply, x^x=0 ⇒ x ⋅ x = 0 ⇒ (x ⋅ x)/2 = 0/2 ⇒ x = 0 but , putting it back into the equations → 0^0 , which is indeterminable, which means the only solution, which is x is an extranuos root(fake) and thus this equation has no solution at all :)
I suspect it would be some kind of complex number. I’d have to look up the exact formula since it’s been a few years, but raising something to a complex power breaks down into the cos(x) + sin(x)i or something like that. Seems like the direction you will need to go to find a complex solution.
Yeah, we arrived at the same conclusion. Lim -> - infinity but I didnt think of rational numbers, saying x belongs to Z solves it the best we can do. Now I have a question, cam we do lim in Complex numbers?
In my mind 0⁰ is undefined because take 2, divide it by 2 you get 2/2 which is also equal to 2¹‐¹ =2⁰ (sorry for the poor notation I'm on mobile) the same can then be said for 0, take 0 divide it by 0 and get 0/0 which is undefined if I remember correctly and can be rewritten as 0⁰. That's my two cents but I may be wrong
I know as x->0, x^x goes to 1, it takes a dip to x=1 then shoots up past that. I know it only exists at negative integers for real solutions. The only way it could possibly get to 0 is at x=-inf. Right? How does it behave in the complex plane?
x^x = e^(x*(ln(-x)+iπ)) = e^(x*ln(-x) + xiπ) So as you go into increasingly negative numbers along the real line, you get an ever-shrinking spiral around the x-axis, which... yes, I'm pretty sure converges to 0 as you approach x=-∞.
Why does function y=x^x does not have any values on the left side of the graph? I mean for x=-1, f(-1)=(-1)^(-1), which would be equal to -1, and so on for other arguments?
It does have values, For x that is in the form of even/odd, its |x|^x For x in the form of odd/odd, its -|x|^x For x in the form of odd/even or irrational, it’s imaginary, if you allow imaginary numbers, forget everything above and just do complex calculations using e^(xlnx). Anyway, overall, the function jumps between positive, negative and imaginary/undefined infinitely many times in any finite region, so it couldn’t be graphed using normal methods, I would graph it is x^x for x = 0 or x>0, ±|x|^x for x < 0
There are problems trying to define a^x for non-integer values when a is negative. E.g. (-1)^(1/3) should equal (-1)^(2/6) since 1/3 = 2/6. The first expression would give the cube root of -1 which is -1 (note that the cube root function *is* defined for negative real values). However, the second expression would give the 6th root of (-1)^2, which equals 1. See en.wikipedia.org/wiki/Exponentiation#Real_exponents for more information.
0^0 is exactly as defined as 0+0. That is: we made it up, and you can choose to agree with our definitions or you can choose not to. But at the end of the day, whether, and how, something is defined is up to the presenter. Definition isn't a matter of consensus, it's a matter of context.
I would think that it depends on whether you consider 0 a digit or an un-digit. If it is an un-digit then in base 0 the only digit is the un-digit 0. So 1 in base zero is the the un-digit 0. If however you think that 0 is a digit then base 0 has no digits and therefore there is no way to represent 1 in base 0 which I suppose is undefined. That might suggest that base 0 is not a base with no digits but rather a an unary base like base 1, except its only digit is 0 (nonary?). If we represent the base by prefixing the number (like in coding) with 0n (n for nonary) then 0n0 is nonary 1, 0n00 would be nonary 2, and 0n000 would be nonary 3 but would still evaluate to 0 in decimal because the value of the nonary value would be 0 times the number of digits which will always be 0. I think that at the very least one hast to define what a base 0 system is before using it in expressions.
A negative real number to the power of itself gives a number with a definite magnitude, if not a definite angle in the polar form. I think if the magnitude goes to 0 as x goes to negative infinity, that is sufficient to say that the limit is 0.
There is another way to solve this. Solutions for x^x = 0 have to be solutions of either x^a = 0 or a^x = 0 with a = x, as x^x = 0 is a special case of both of these equations a^x = 0 has no solutions (as it implies that x = log a (0), which is undefined), and x^a = 0 has 1 solution: x = 0. Trying out x = 0 in x^x = 0 returns 0^0 = 0, and 0^0 is undefined, so x = 0 is not a solution for x^x = 0. This means that there are no solutions for x^x = 0.
Why can't we use logarithmic definition to conclude there is no solution? Loga (b) = c a^c = b So x^x = 0 logx (0) = x And since log of 0 is undefined, then there are no solutions.
Let u = xln(x). Then you have e^u = 0, and this has no solution, even in complex world. However any branch of the log makes xln(x) onto (I think) so you wouldn't be able to find x in any case. Thus there is no solution For the limit case : you don't have to put x =2n as this also works in the complex world, the modulus of 1/(-inf)^inf would be 0 no matter the branch
Poser 0^0 = 1 par convention. Il n'y a pas de raison de ne pas prolonger par "continuité" en 0 la fonction f : x ---> 0^x Ainsi, 0^0 = 1^0 = 2^0 = .... = 1 On a donc posé : f(0) = lim f(x) (x ---> 0)
What I thought of is multiplying both sides by x so then we have x^x•x=0•x =>. x^(x+1)=0 but x^x is also 0 so x^(x+1)=x^x take ln both sides then cancel the lnx because x can't be one and then we have x+1=x witch definitely has no solutions
you cannot multiply both sides by x when one side is 0, if you could, nothing would make sense anymore, take x^2-1=0 and multiply both sides by x and you got x^3-x=0 and that doesnt work
You cannot take the ln at both sides when both sides equal 0 . For example, solve (x-1)^x = 0 . Clearly x=1 is a solution. But if we raise both sides to the power of 3: (x-1)^(3x) = 0 = (x-1)^x (x-1)^(3x) = (x-1)^x Take ln at both sides: ln( (x-1)^(3x) ) = ln( (x-1)^x ) 3x * ln(x-1) = x * ln(x-1) 2x * ln(x-1) = 0 2x = 0 OR ln(x-1) = 0 x = 0 OR x-1 = 1 x = 0 OR x = 2 which are clearly not valid solutions to the original equation. Instead, these are extraneous solutions that were introduced by raising both sides to the power of 3 (before taking the ln ); but that's not the problem. The real problem is that the solution x=1 that we wanted to find, is now suddenly _gone_ . And this disappearance is caused by the step of taking the ln (when both sides equal 0).
Everything exept 1, -1, i and -i to the power of negative infinity is 0 (and all those numbers to the power of positive infinity are infinity). That means the solution is negative infinity. That also means that 1/0 is infinity and 1/infinity is 0.
Just messing around on a calculator, it seems that if you let x=ni then as n approaches positive infinity, the answer approaches zero as both the real and imaginary components get arbitrarily small. Still not a solution, but it's very interesting.
@@JJ_TheGreat x^0 is only undefined for x=0, it works with anything else. 0^x is only defined for positive real numbers, I think that's why it's not taken into account that much. 0^(-1) is undefined (it would literally mean 1/0), as well as 0^i (=e^i*ln(0)=e^-i*inf=cos(-inf)+i*sin(-inf) and that's undefined) or whatever else you come up with. In contrast, (-1)^0=1 and i^0=1.
Those imaginary numbers made me think and I have a question for you: what id we define the answer? Imaginary numbers come from us defining the square root of -1 as i. What if we define a number m so that m multiplied by itself m times is 0. I mean, just at first glance, if we did that and kept the n root of 0 is zero it would mean raising something to a power and roots would no longer be whatever the math term is for their current relationship. Just a thought there. Feel free to ignore me though.
and from there it doesnt work due to 0^0 not being agreed and also that you raised both sides to 1/0, making it undefined however he did go to speak about limits which was not talked about in this "one liner"
@@keepwatching-ane no, maybe read before criticising. take the power 1/x for a given x. the result on the righthandside is necessarily 0. Then no need for useless limits as the result is already given after the first step.
@@yakaridubois3378 and the left hand side would be undefined because if you raise x^x by 1/x whilst x is 0, you just raised it to an undefined value. Im saying limits could be good for finding what x could be without making the left hand side undefined
@@keepwatching-ane Yes it seems you are right about my solution is doesn't work at x=0. 'Brithemathguy' a youtuber says that the limits are not satisfactory either because the limit never actually reaches x=0. Even more, he says that 0^0=1, which can be verified by taking exp(x) at x=0. We know that exp(0)=1. The series formula for the exponential is exp(x)=x^0+x+x^2/2+... but when x=0, all the terms except the first one necessarily vanish, so we deduce 0^0=1. The video of Brithemathguy is interesting :)
the x value needed to receive the lowest possible number with x^x, is e^(-1), and ( e^(-1) ^ (e^(-1)) ) is equal to 0.69... so x^x is impossible for 0.4, 0.5, 0.6 as well
You proven that 0^0 is not indeterminate with this. x^x = 0 has NO SOLUTION, meaning no 0^0 CAN'T equal 0 EVER. So there you can't say 0^0 has the possibility to be 0 EVER. So the answer HAS TO BE 1
You have to leave real numbers and just use ontological symbols. For 1/-infinity^infinity, you can solve to +-1=0, which can be used to symbolize all equals nothing.
This is correct, and a simple proof that e^z is non-zero for every complex z is the fact that e^-z = 1/e^z, and e^z is defined and finite on the entire complex plane, so if some complex number c were a zero of e^z, then e^-c would be infinite.
@@gamemakingkirb667 What I'm saying is that if z^c=0 for any complex number c then z^-c would be 1/0 or infinity, contradicting the fact that e^z is an entire function (no poles.)
I mean , just say there aren't any zero divisors in the complex world, so there aren't any solutions in the complex world. Also if you define that x^x =0 and make a new ring , would technically be a solution..
Finally 0^0 approaches 0:
ruclips.net/video/X65LEl7GFOw/видео.htmlsi=QRgxgdaL5If2FWnP
BPRP, how solve 1^x = 2?
@@costelnica3988 I think he has done it
0^0 = 1 so i^0 = 1
case 6 works with something akin to the squeeze theorem
tl;dr x^x as x approaches -infinity is in the form re^itheta where r approaches 0
if we rewrite x as re^itheta, as x approaches -infinity thats the same as x equals the limit as r approaches infinity of re^ipi
(re^ipi)^(re^ipi)=(re^ipi)^-r=(r^-r)(e^(i(-rpi)))
r^-r is a positive real number that approaches 0 as r approaches infinity, meanwhile e^(i(-rpi)) in the form e^itheta meaning its just polar form with a radius of r^-r
since its always on the circle on the complex plane with a radius of r^-r, and because the circle gets compressed to a point specifically down to 0, this means limit as x approaches -infinity of x^x should be considered 0
this is not the most rigorous proof but it works for me
@blackpenredpen Hey bprp. Can u do a formula for a tetrated by x plus b exponentiated by x plus c times x plus d equal to 0. Love ur videos. Bcos of u I learnt and mastered calculus at the age of 10.
If x is a complex number other than 0 we can write x=r*e^iθ and then we see that x^x=(re^iθ)^x=r^x e^ixθ
In order for this expression to be 0 we need either r^x=0 or e^ixθ=0. But since r>0 hence r^x != 0. Likewise e^ixθ != for any x. Thus x^x=0 has no solution in the complex numbers
wait what
@@TNTErick ?
probably true, but why can a^x not equal 0 for a!=0? With real numbers it's obvious, but with complex numbers I don't know
@@derda3209 If a != 0 then a^x=e^(x*log(a)) which is never 0 (exponential function is never zero over real numbers and complex numbers). Note that log(a) is defined for none zero complex numbers
I think, the easiest proof that the solution doesn't exist in the complex world is that there aren't any zero divisors in the complex world, so for x^a to be equal to 0 , x=0 , however contradiction.
He doesn't age man
Idk, I really miss his looooong beard
Maybe we age at the same rate as him 😂
He looked the same like years ago I watched him for the first time 😂
Well of course, he's asian
study math and get eternal life
For me 0^0 = 1/2, take the average value of 0 and 1, im sure everyone will be happy 😊
😂😂😊
Nope.
a^b = c means
multiplying any constant K by the factor a for a number of b times, is equal to multiplying the same constant K by the factor c.
Hence,
0^0 = c means
multiplying any constant K by the factor 0 for a number of 0 times, is equal to multiplying the same constant K by the factor c.
The former side gives K (because multiplying K by 0 for a number of 0 times, gives K), the latter side gives K*c (because multiplying K by the factor c gives K*c), so
K = K*c , valid for any value of K ==>
c = 1
Therefore,
0^0 = c = 1
Q.E.D.
@@yurenchuwhat the sigma
how can 0^0 be 1/2 ??? If we say that "it is defined", and these 2 zeros are identical, then the solution must be 1, because it assymptoticaly goes to 1. But more strictly it is not defined. So in none of case it could be 1/2.
the average value of 0 to 1 how??
If u want to know about the complex solution, u can find bprp's old video with title "the tetration of (1+i) and the formula (a+bi)^(c+di)"
You say "not my best video", but I really enjoyed it!
Naturally, let’s be rational for a moment. This is too complex to be real.
Love that comment 🤓
I love the superpower of erasing the board with a knock
Hey there BPRP, been watching for some 6 years now. Good stuff man
Thank you!!
Lots of love to you and to your iconic pens❤@@blackpenredpen
Yo también lo sigo desde hace 6 años, me puse a pensar en ello al leer tu comentario.
Must be the first vid without "of course, otherwise, how could I make this video?" Love the subversion of expectations
For the limit x^x you can do:
lim x^x = lim exp(ln(x)*x) = exp(lim ln(x)*x) because exp is a continuous function. Then lim ln(x) * x = lim ln(x) / (1/x) = lim (1/x) / (-1/x^2) = lim 1 / (-1 / x) = lim -x = 0 (using de l'Hopital) and therefore lim exp(ln(x)*x) = lim x^x = 1.
I think you need to assume x>0 ( If you use x Real) , in order to write x^x= e^{xlnx}, as lnx is defined only in (0, oo) for x Real.
I'd argue -inf works without djscreetization because the negative ultimately only changes the complex sign and the magnitude still goes to 0 and 0 with any complex sign is still 0
yeah it'll be equal to 1/(x^abs(x)) * i which approaches 0 as x -> infinity
I understand that mathematician say : "It has no solution, because negative Infinity is not a real or complex number"... But, it is obvious that assymptoticly -inf (in integer) converges to solution. It could be really interesting to make some physical experiment, where the nature say if the -inf is the solution of this equation.... For me, from technicial - engineering aspect, it is solid solution (but not ideal); The only question for me is, if the nature supports the infinity .. :)
im pretty sure even in complex this works because complex exponentiation doesnt change magnitude, so if it converges to 0 on integers it should converge for the remaining values in between that are bounded by surrounding integer values.
thats not true when you write integral from b to inf it's short for lim_(a->inf) integral from b to a you don't realy use infinity
I think the lim as x-> negative infinity is valid. You can see that if you try x= -10, -100, -1000 then x^x surely is approaching zero. It could be argued that the function is discontinuous in the negatives, but for going to -infinity at every point the function exists x^x approaches zero as x gets more negative.
1:55 both arguments basically portray when an unstoppable force meets an immovable object
This is an example of a pedagogically responsible explanation.
This is where new math is to be found. Idk why no one looks here or ever talks about numbers like these. Juat as we thouught i couldnt exist, it does. So this probably does too. Same with |x|=-1
I believe as long as x is real, case 6 with the limit works. lim x->-inf (x^x) = lim x->inf ((-x)^(-x)).
(-x)^(-x)=(-1)^(-x) * x^(-x)
The latter factor always approaches zero as x approaches infinity. The former factor is cyclical and the absolute value is always equal to one, so (-x)^(-x) always approaches 0 as x approaches infinity and thus x^x approaches zero as x approaches negative infinity.
I think there’s an extension of this argument if you allow x to approach negative infinity from paths in the complex plane, but I haven’t fully formulated it yet.
This is deceptively simple but it makes sense to me.
There are problems trying to define a^x for non-integer values when a is negative.
E.g. (-1)^(1/3) should equal (-1)^(2/6) since 1/3 = 2/6. The first expression would give the cube root of -1 which is -1 (note that the cube root function *is* defined for negative real values). However, the second expression would give the 6th root of (-1)^2, which equals 1.
See en.wikipedia.org/wiki/Exponentiation#Real_exponents for more information.
@@DutchMathematician you have to simplify exponent fractions first. Values that you can get when it's not simplified don't matter. He showed this in one of his "1=2 proof" videos.
@@zachansen8293
There is no need to simplify fractions first. If f(x) = a^x is a well-defined function, then f(1/3) = f(2/6) *should* hold, no matter how we supply the same argument to the function f. If we give a function two equivalent expression then the outcome should be the same.
@@DutchMathematician Go watch the video, you absolutely do. That's how he proves that 1=2 or 1=-1 or whatever it was. You have to simplify fractions in an exponent (or you can do them in a specific order if it isn't). If you don't you are likely to get incorrect results.
edit: Go search for the video named "Hate to be that guy but I need the extra credit! " - it's the last chapter starting at 3:48. He shows that i=1 because i = i^(4/4) = 1
edit2: I believe if it's not simplified you can do the denominator first, but you cannot do the numerator first. But the easiest thing to remember is to just always simplify fractions that are exponents.
last edit: and of course you should run it through wolfram alpha to see that it doesn't get your "solution"
I think 💭 the first video I watched from was pi^e vs e^pi.
Back then I think was entering masters in math/stats and today I’m a year away from PhD in statistics.
Always inspiring!
Btw did you ever did a video on Riemann-Stieltjes integration????
I recall it but can’t find it
Love how postmodern math is caught in a fantastic sceptical philosophy. He shows using 10 different reality based ways how x^x cannot be zero but proceeds with conclusion that « in a complex world » it « might ». Bake skepticism into the choice of words to really screw with all brains for 200 years and expect good results.
Well, it's easy to prove that I cannot find a solution. It's way harder to prove that there is definitely no solution... Or weirder: that even if a solution exists it is impossible for someone to find it.
@@Kleyguerth the one with the burden of proof is the one making a claim, an assertion. If you say it is possible, what proof do you present? Until you do, all of the refutations of this statement stand and you must shut up.
@@geekonomist Yes, I'm agreeing with you, I just repeated it in simple english lol
YFAI
Am I misunderstanding your comment or are you categorizing complex numbers as "postmodern math?" Because they have some quite interesting applications in "reality"
My high school teacher once declared 0^0 is undefined because powers can be written as fractions, so 0^0 equals 0/0. I never questioned it until now.
Your teacher's argument doesn't make sense. By the same logic, 0^1 is undefined, because
0^1 =
= 0^(2-1)
= (0^2)/(0^1)
= 0/0
= undefined
but of course we know that 0^1 = 0 is _not_ undefined.
What about 2^2 😂
@@yashrajtripathi4832 What about it?
@@yurenchu definitely not about your comment bro !
@@yashrajtripathi4832 Oops! My bad.
I think what the original commenter's high school teacher meant, is that in general,
a^b = (a^(b+1)) / a
which, in the case of a=2 and b=2 , would lead to
2^2 = (2^(2+1)) / 2 = (2^3)/2 = 8/2
which is correct (both sides equal 4); so nothing wrong there.
With a=0 and b=0 , this formula would lead to
0^0 = (0^(0+1))/0 = (0^1)/0 = 0/0
which on the righthandside is undefined (and therefore, according to his teacher, 0^0 should be undefined).
However, the rule a^b = (a^(b+1)) / a is not valid for a=0 , because for example with a=0 and b=1, it leads to
0^1 = (0^(1+1))/0 = (0^2)/0 = 0/0
which is clearly wrong, because the lefthandside is _not_ undefined; 0^1 equals 0.
Therefore, the high school teacher's argument as to why 0^0 is undefined, is wrong, as it doesn't hold water; he applied a rule that shouldn't be applied in this case.
Hello sir, I just wanted to thank you for making these videos because they've helped me out a lot.
What do you mean not your best video? I enjoyed every single second. Good job
Thumbnail said this video isn't your best, but it's still one of your best to me :)
The reasoning works for when a is the Gaussian integers or Q[i] but when it is all C I have to deal with x^i =0 so yeah..
Thx for pointing out. 😊
@@AndyBaiduc-iloveu thanks 🤗
Here comments is more difficult than videos. Please get this man a medal 🏅🏅. Salute professors.
subtract 1 from x^x and from 0, square both sides, take x^x as t, solve the quadratic equation and get that x^x = 1 +- i . multiply x^x and 1+- i by x^-1 and get that x = 1 +- i
I expressed the complex x in the base using euler's form aka re^iphi and the one in the power in the a + bi form, then I got the equation e^(a*ln(r)-bphi)*e^i(aphi + b*lnr), for that to equal zero the term in the first power (that is a*ln(r)-bphi) has to go to negative infinity, so for exact values it is impossible, but if we take the limit of 0 + bi as b goes to infinity, it does equal zero because phi is bounded and the real part of 0 + bi is exactly zero.
z^z = exp(z*Logg(z)) in complex numbers, where Log(z) is the principal branch of the complex logarithm. However, the modulus
|z^z| = exp(Re(z*Log(z)))
= exp((x/2)*ln(x^2 + y^2) - y*ArcTan(y/x))
The expression in the exponent seems to have a local minimum for x = -1/e and y = 0, so it seems
|z^z| >= 1/e, for all complex z = x + 1j*y
This is probably the best video you have ever done. It’s every way to approach the solution to an equation disguised as a search for the answer to x^x=0. Bravo. Beautiful.
The magnitude of x^y for real numbers y, is |x|^y, and if we have the magnitude of a limit approaches 0, then we can safely say that the limit is zero. the limit as x-> -\infty of |x|^x, is also equal to the limit as x->\infty x^{-x} = 0, meaning that the magnitude is 0, therefore the limit as x-> -\infty x^x = 0
nice exploration of techniques!
0:50
From one side, 0^0 is 0 because 0^a is 0, from the other, 0^0 is 1, because a^0 = 1. So if we multiply the rules it will be (a^0)(0^a)/(0^a)(a^0)=0/0=1. For a big part of people, 0^0 is equal to 1, for a smaller fraction of them it's 0 or/and undefined, for the smallest one it's superposition between 1 and 0 or/and 0,5 (Corresponds with the mathematical "joke" that says 1+2+3+4...=-1/12).
I count that 0^0 is undefined, because taking the power of a number is just taking it to the power of 1/(nth root power), so 2nd power is 1/0.5=2. If it's to the power of 0, then it's 1/0, which is undefined, and any operation with undefined gives just undefined in the output.
0^0 = (0^1) * (0^-1) = 0/0 = undefined
2 AM blackpenredpen? Dont mind if i do!
Woah my time when it got released was 9 am. I guess you're somewhere close to United States while I am in Europe
my intuition before I watch the video, not sure if this will get anywhere: I learned early on in calculus that if you have an equation f(x) with no constants, you can rewrite it as e^ln(f(x)) to solve that issue.
Well, if you consider x>0, then we can use x^x= e^{x*lnx}, and e^y is never 0. Same goes for the Complex Exponential; 1 is the only value e^z doesn't take. I think this does it for X^x=0.
There is no finite complex solution
If z = r* e^it, r and t real
z^z = r^z * e^itz
So the equation becomes r^z=0 or e^itz=0
Case 1
r^z=0=>|r^z|=0
=>r^r =0, which has no solution as r is a positive real number
Case 2
e^itz=0
=> |e^itz|=0
=>e^-ity=0
Since t is between - pi and pi, y has to be - inf, thus no finite complex solution
This also requires
(a+bi)^(a+bi) can be rewritten as (a+bi)^a*(a+bi)^bi now we know that either (a+bi)^a=0 or (a+bi)^bi=0 then (a+bi)^bi can be expressed as e^(bi*Log(a+bi)) where Log is the complex log (if f(z)=e^z then f^-1(z)=Log(z)) let's bring the formula for the complex Log: Log(z)=ln(|z|)+arg(z)i now now let's simplify e^(bi*ln(|a+bi|)+arg(a+bi)i) simplify a little bit more (e^bi)^(ln(|a+bi|)+arg(a+bi)i) now simplify
e^(bi*ln(|a+bi|))*e^(bi*arg(a+bi)i) then just solve the equation e^(bi*ln(|a+bi|))*e^(bi*arg(a+bi)i) = 0 we get (cos(b)+sin(b)i)^ln(|a+bi|)*e(-b*arg(a+bi)) now (cos(b)+sin(b)i)^ln(|a+bi|) can't equal 0 because |cos(b)+sin(b)i|=1 for any b (= R and and for this to equal 0 |cos(b)+sin(b)i| must equal 0 now the second part e^(-b*arg(a+bi)) |e| =e so this is not equal to 0 for the same resons now let's come back to our original problem: either (a+bi)^a=0 or (a+bi)^bi=0 but we proven that (a+bi)^bi/=0 so (a+bi)^a=0 then |a+bi| = 0 then a+bi=0 then a=0 and b=0 then we get 0^0 which is either 1, undefined or has infinitly many values but for all cases 0^0/=0 so (a+bi)^(a+bi) has no solution
Extra: let's see if equation z^z has any solutions in set denoted as C(epsilon, phi) (I will type j for epsilon and k for phi) where j^2=0 where j /=0 j/j=1 , k=1/j and k^2 is undefined firstly let's define exponents: e^z=sum from: n=0 to: infinity z^n/n! then e^jz=sum from: n=0 to: infinity (jx)^n/n! = 1+jx+0+0+0+... =1+jx now the ln: ln(a+bj)=c+dj then e^(c+dj)=a+bj then simplify (e^c+(e^c)dj) = a+bj then form two equations e^c=a and e^c(dj) = bj solve the first one c = ln(a) substitute a(dj) = bj now ad=b now d = b/a so ln(a+bj)=ln(a)+(b/a)j
Now I don't want to to waste another 30 minutes of my life solving this equation so if someone sees that please solve this problem
X= -infinity seems like a good solution to me. Even if we have to use complex numbers to define powers of negatives, they still approach 0 in absolute value
Intuitively I knew that before your attempts to find a solution. My Calc teacher may have gone over this with me in college and I am just recalling the lesson.
My unserious part had a laugh at 7:50 ... I'll go stand in the corner... 😅
I have seen people on Quora get really angry when told that 0^0 is not one.
They should as the definition of the power is 0 multiplications from the original definition of 1. This is a formulation for the inductive proof of every single number. The fact that every other positive power of 0 is 0 is irrelevant to what the zeroth power of zero is.
@@ronaldking1054
0^0 = 1 so i^0 = 1
@@ronaldking1054 0^0 is not a calculateable thing, you just define it, there is no real answer unless everyone agrees on it.
@@mhm6421 When the definition of power by simple integer is set up with 1 as the basis, it makes little sense to claim that 0^0 is not 1. You'd be putting in a branch for no reason. In other words, it is simpler.
@@ronaldking1054That definition was not just randomly defined, it comes from a pattern seen with exponents. For example, 2^4 = 2*2*2*2. 2^3 = 2*2*2. Based on this, we could assume that x^n = (x^(n+1))/x. By this logic, 0^0 = 0^1 / 0. Division by zero is undefined, and so this has no answer. We could define one, but not everyone will agree on the definition.
@yurenchu it's the Banach limit. I'm not actually sure you can apply it to 0^0. It applies to sequences. The Banach limit of {0,1,0,1,...} is 1/2. This sequence does not converge under the usual sense of limits, but Banach limits extend the usual limits to non-convergent sequences. Not all sequences have uniquely defined Banach limits.
Thanks for your reply. I didn't receive a reply notification because your reply wasn't posted within the same thread, but luckily I happened to revisit this comment section and noticed it by accident.
Interesting to read about the Banach limit, I'm not familiar with it, so thanks for that. It leaves me puzzled though how it can be applied to a function as described in your previous post (i.e. a real function f(x) that's 1 for rational x , and 0 for irrational x, and then show that the Banach limit of f(x) equals 1/2 for x approaching infinity).
As for more (fruitless) ways to solve it, there were a couple I came up with (assuming you're using the extended complex plane)
The first is pretty simple: if z^z = 0, then e^(z ln z) = 0, and thus z ln z = -infinity (we're in the extended complex plane so this is fine)
The only value this works for is z = -infinity, so that's our solution
The second is just substitution; set z = x+1, so (x+1)^(x+1) = 0
Then (x+1)^x * (x+1) = 0, then x * (x+1)^x + 1 * (x+1)^x = 0
So x * (x+1)^x = -(x+1)^x, meaning ln(x) + x ln(x+1) = ln(-1) + x ln(x+1)
Assuming x =/= -1, we can remove x ln(x+1) from both sides, giving us ln(x) = ln(-1), which lets us conclude x = -1, which is a contradiction of our assumptions and thus the equation has no solutions
Why would
z*ln(z) = -♾
necessarily imply
z = -♾?
If we assume
ln(-♾) = ln(-1) + ln(♾) = iπ + ♾, then we have
-♾*ln(-♾) = -♾*ln(-1) + -♾² =
= -i*♾ - ♾
We end up with complex infinity here, or rather, unsigned infinity, not -♾
Please make a video on how to manually calculate complex answers of Lambert w function
actually for no. 6, the 2n limit, it even works for x = n for n integer, but it approacges 0 from up then down, up the down, and so on (much like sin(x)/x as x approaches infinity
Funnily enough if you put in Desmos x = 0 and 0^(1/x), then you end up with 0^(1/x) = 0
I suppose from a coding perspective, Desmos sets priority that 0^anything overwrites the calculation that 1/0 is undefined. Just an interesting mention!
Take two complex numbers z and w. If z,w≠0, then z^x=w can be solved for x. If z=0 and w=0, then x>0. If z=0 and w≠0, x has no solutions. If z≠0 and w=0, x has no solutions.
x^x=0 should have no solutions. Sure, if we take the limit as x→-∞, where -∞=inf R travelled along the negative real axis and not the general complex ∞, then we'd get a complex number of magnitude decreasing to 0, but we want numbers, not extended numbers.
I want you to discuss the hailstone sequence and Collatz conjecture and your opinion about it
Nice video, a bit in the style of your videos of a few years ago. Thank you!
x^x=(re^(i phi))^(a+bi) where a=r cos(phi) and b=r sin(phi) leads to
x^x=e^(r ln r cos(phi) - r phi sin(phi)) e^i(r phi cos(phi) + r ln r sin(phi))
Since sin(theta) and cos(theta) are never simultaneously 0, this can approach zero only as phi-->infinity. (The magnitude approaches zero.)
For the original #6: it feels like the argument should be makeable based on that the magnitude of said expression feels like it should approach zero. If we restrict to rational values, then it should be trivial to argue - for irrational values, perhaps we can still show any branch approaches zero, and argue that way?
With x = a + ib we have z = x^x = (a + ib)^(a+ib) = exp([½ a ln(a²+b²) - b theta] + [½b ln(a²+b²) + a theta] i), where theta = arg(a,b). We want │z│= exp(½ a ln(a²+b²) - b theta) = 0. To me it seems this is the case if a goes to minus infinity independent of b. (Or possibly if b goes to plus infinity ???). So my tentative answer would be x = - infinity + i b with b arbitrary. I checked (-50)^(-50) is very small, i.e. close to zero. (-50)^(-50) = 1.126 · 10^(-85). So it seems plausible.
that was some fun analysis!
Maybe we should try making up numbers to solve this. Previous big updates were caused by addition, multiplication and exponentiation, while the problem in the video is caused by tetration. And the solution doesn’t have to be -♾️, it could be an orthogonal unit just like i
To me 0^0 is undefined even if we take a seventh grade approach.
5^0 = 5^(m-m) = 5^m / 5^m.
If we take 0 instead of 5 we get:
0^m / 0^m
m can be any number yet we still have undefined fraction of 0/0 which could be be approached if we have some context, like in sinc function: sin(x)/x when x -> 0. But here we don't have that context, so it remains undefined.
P.S. I also was amazed that in English this limit doesn't have a special name. For example in Russian this limit is called "Первый замечательный предел" or "first remarkable (or wonderful) limit".
And the second such limit is (1+1/x)^x when x-> infinity.
Multiply both sides by x, and you reach a contradiction where you equate powers to get 1+x=x
Simply, x^x=0
⇒ x ⋅ x = 0
⇒ (x ⋅ x)/2 = 0/2
⇒ x = 0
but , putting it back into the equations → 0^0 , which is indeterminable, which means the only solution, which is x is an extranuos root(fake) and thus this equation has no solution at all :)
Pow(x,x ) is undefined for all Re(x)0 , will be defined ob Re(x)
no solutions in the complex, because if z is not 0 then by definition z^z = e^(z*ln z), and e^w is nonzero for any w complex
My first approach was to say, that x^x = 0 has a solution of x = - \inf over IZ, because with f : IZ -> |Q, x |-> x^x, lim_{x -> - \inf} f(x) = 0...
I suspect it would be some kind of complex number. I’d have to look up the exact formula since it’s been a few years, but raising something to a complex power breaks down into the cos(x) + sin(x)i or something like that. Seems like the direction you will need to go to find a complex solution.
Yeah, we arrived at the same conclusion. Lim -> - infinity but I didnt think of rational numbers, saying x belongs to Z solves it the best we can do. Now I have a question, cam we do lim in Complex numbers?
In my mind 0⁰ is undefined because take 2, divide it by 2 you get 2/2 which is also equal to 2¹‐¹ =2⁰ (sorry for the poor notation I'm on mobile) the same can then be said for 0, take 0 divide it by 0 and get 0/0 which is undefined if I remember correctly and can be rewritten as 0⁰. That's my two cents but I may be wrong
You can use induction to prove that any number n divided by n is equal to n⁰ so 🤷♂️
Any field has no 0 deviders. Thats all solution
I know as x->0, x^x goes to 1, it takes a dip to x=1 then shoots up past that. I know it only exists at negative integers for real solutions. The only way it could possibly get to 0 is at x=-inf. Right? How does it behave in the complex plane?
x^x = e^(x*(ln(-x)+iπ)) = e^(x*ln(-x) + xiπ)
So as you go into increasingly negative numbers along the real line, you get an ever-shrinking spiral around the x-axis, which... yes, I'm pretty sure converges to 0 as you approach x=-∞.
Why does function y=x^x does not have any values on the left side of the graph? I mean for x=-1, f(-1)=(-1)^(-1), which would be equal to -1, and so on for other arguments?
Counter-example: what do you do when x = -1/2? You're getting imaginary numbers for anything but x in Z* (I think)
It does have values,
For x that is in the form of even/odd, its |x|^x
For x in the form of odd/odd, its -|x|^x
For x in the form of odd/even or irrational, it’s imaginary, if you allow imaginary numbers, forget everything above and just do complex calculations using e^(xlnx). Anyway, overall, the function jumps between positive, negative and imaginary/undefined infinitely many times in any finite region, so it couldn’t be graphed using normal methods, I would graph it is x^x for x = 0 or x>0, ±|x|^x for x < 0
There are problems trying to define a^x for non-integer values when a is negative.
E.g. (-1)^(1/3) should equal (-1)^(2/6) since 1/3 = 2/6. The first expression would give the cube root of -1 which is -1 (note that the cube root function *is* defined for negative real values). However, the second expression would give the 6th root of (-1)^2, which equals 1.
See en.wikipedia.org/wiki/Exponentiation#Real_exponents for more information.
Actually, the negative part of the graph is just unconnected points for the integers. Because -1/2 for example.
Try graphing (|x|)^x or (|x|)^(|x|). The second one is a little weird.
0^0 is exactly as defined as 0+0. That is: we made it up, and you can choose to agree with our definitions or you can choose not to. But at the end of the day, whether, and how, something is defined is up to the presenter.
Definition isn't a matter of consensus, it's a matter of context.
Been yelling "MINUS INFINITY" two minutes in. Glad I got close enough, too bad i hate complex numbers
negative infinity?
I would think that it depends on whether you consider 0 a digit or an un-digit. If it is an un-digit then in base 0 the only digit is the un-digit 0. So 1 in base zero is the the un-digit 0. If however you think that 0 is a digit then base 0 has no digits and therefore there is no way to represent 1 in base 0 which I suppose is undefined.
That might suggest that base 0 is not a base with no digits but rather a an unary base like base 1, except its only digit is 0 (nonary?). If we represent the base by prefixing the number (like in coding) with 0n (n for nonary) then 0n0 is nonary 1, 0n00 would be nonary 2, and 0n000 would be nonary 3 but would still evaluate to 0 in decimal because the value of the nonary value would be 0 times the number of digits which will always be 0.
I think that at the very least one hast to define what a base 0 system is before using it in expressions.
I enjoyed this video a lot!!
Negative infinity
X=-Inf
X=0 also makes sense as a solution
X=1 also makes sense
Please cover the lambert function
A negative real number to the power of itself gives a number with a definite magnitude, if not a definite angle in the polar form. I think if the magnitude goes to 0 as x goes to negative infinity, that is sufficient to say that the limit is 0.
I understand almost nothing about these types of videos yet they interest me lmao
There is another way to solve this. Solutions for x^x = 0 have to be solutions of either x^a = 0 or a^x = 0 with a = x, as x^x = 0 is a special case of both of these equations
a^x = 0 has no solutions (as it implies that x = log a (0), which is undefined), and x^a = 0 has 1 solution: x = 0. Trying out x = 0 in x^x = 0 returns 0^0 = 0, and 0^0 is undefined, so x = 0 is not a solution for x^x = 0. This means that there are no solutions for x^x = 0.
Why can't we use logarithmic definition to conclude there is no solution?
Loga (b) = c a^c = b
So
x^x = 0 logx (0) = x
And since log of 0 is undefined, then there are no solutions.
if you put it into desmos, it will show a weird looking graph on the negative infinity side, and if you move it in any way it disappears!!! idk why
Let u = xln(x). Then you have e^u = 0, and this has no solution, even in complex world. However any branch of the log makes xln(x) onto (I think) so you wouldn't be able to find x in any case. Thus there is no solution
For the limit case : you don't have to put x =2n as this also works in the complex world, the modulus of 1/(-inf)^inf would be 0 no matter the branch
Poser 0^0 = 1 par convention. Il n'y a pas de raison de ne pas prolonger par "continuité" en 0 la fonction f : x ---> 0^x
Ainsi, 0^0 = 1^0 = 2^0 = .... = 1
On a donc posé :
f(0) = lim f(x)
(x ---> 0)
What I thought of is multiplying both sides by x so then we have x^x•x=0•x =>. x^(x+1)=0 but x^x is also 0 so x^(x+1)=x^x take ln both sides then cancel the lnx because x can't be one and then we have x+1=x witch definitely has no solutions
you cannot multiply both sides by x when one side is 0, if you could, nothing would make sense anymore, take x^2-1=0 and multiply both sides by x and you got x^3-x=0 and that doesnt work
@@nakellold if you don't include zero in the solutions Im pretty sure you can do that
You cannot take the ln at both sides when both sides equal 0 .
For example, solve (x-1)^x = 0 . Clearly x=1 is a solution. But if we raise both sides to the power of 3:
(x-1)^(3x) = 0 = (x-1)^x
(x-1)^(3x) = (x-1)^x
Take ln at both sides:
ln( (x-1)^(3x) ) = ln( (x-1)^x )
3x * ln(x-1) = x * ln(x-1)
2x * ln(x-1) = 0
2x = 0 OR ln(x-1) = 0
x = 0 OR x-1 = 1
x = 0 OR x = 2
which are clearly not valid solutions to the original equation. Instead, these are extraneous solutions that were introduced by raising both sides to the power of 3 (before taking the ln ); but that's not the problem. The real problem is that the solution x=1 that we wanted to find, is now suddenly _gone_ . And this disappearance is caused by the step of taking the ln (when both sides equal 0).
Everything exept 1, -1, i and -i to the power of negative infinity is 0 (and all those numbers to the power of positive infinity are infinity).
That means the solution is negative infinity. That also means that 1/0 is infinity and 1/infinity is 0.
The 0^x -> 0 limit is only onesided. If you look at both sides the limit doesn't exist
Suggestion: make a contour map for y = x^x. Fascinating thing. Doesn’t solve this problem, though.
For x
Just messing around on a calculator, it seems that if you let x=ni then as n approaches positive infinity, the answer approaches zero as both the real and imaginary components get arbitrarily small. Still not a solution, but it's very interesting.
1:16 Yeah, but then you also have 0^x = 0 - so don’t people also make an argument that it could also equal 0?
0^x=0 for x≠0
@@tsouphs The same thing for x^0...
x^0=1 for x≠0, which this video was discussing...
So I was making a point that you can also look at it as 0^x...
@@JJ_TheGreat x^0 is only undefined for x=0, it works with anything else. 0^x is only defined for positive real numbers, I think that's why it's not taken into account that much. 0^(-1) is undefined (it would literally mean 1/0), as well as 0^i (=e^i*ln(0)=e^-i*inf=cos(-inf)+i*sin(-inf) and that's undefined) or whatever else you come up with. In contrast, (-1)^0=1 and i^0=1.
Those imaginary numbers made me think and I have a question for you: what id we define the answer? Imaginary numbers come from us defining the square root of -1 as i. What if we define a number m so that m multiplied by itself m times is 0. I mean, just at first glance, if we did that and kept the n root of 0 is zero it would mean raising something to a power and roots would no longer be whatever the math term is for their current relationship.
Just a thought there. Feel free to ignore me though.
It's a one-liner: one can raise x^x=0 to the power 1/x on both side from which x=0
and from there it doesnt work due to 0^0 not being agreed and also that you raised both sides to 1/0, making it undefined
however he did go to speak about limits which was not talked about in this "one liner"
@@keepwatching-ane no, maybe read before criticising. take the power 1/x for a given x. the result on the righthandside is necessarily 0. Then no need for useless limits as the result is already given after the first step.
@@yakaridubois3378 and the left hand side would be undefined because if you raise x^x by 1/x whilst x is 0, you just raised it to an undefined value. Im saying limits could be good for finding what x could be without making the left hand side undefined
@@keepwatching-ane Yes it seems you are right about my solution is doesn't work at x=0.
'Brithemathguy' a youtuber says that the limits are not satisfactory either because the limit never actually reaches x=0. Even more, he says that 0^0=1, which can be verified by taking exp(x) at x=0. We know that exp(0)=1. The series formula for the exponential is exp(x)=x^0+x+x^2/2+... but when x=0, all the terms except the first one necessarily vanish, so we deduce 0^0=1.
The video of Brithemathguy is interesting :)
Hi there! 😁
May be the answer of 0⁰ is the friends we made along
the x value needed to receive the lowest possible number with x^x, is e^(-1), and ( e^(-1) ^ (e^(-1)) ) is equal to 0.69...
so x^x is impossible for 0.4, 0.5, 0.6 as well
You proven that 0^0 is not indeterminate with this. x^x = 0 has NO SOLUTION, meaning no 0^0 CAN'T equal 0 EVER. So there you can't say 0^0 has the possibility to be 0 EVER. So the answer HAS TO BE 1
Instead of fields like Q, R or C, what about using a ring containing zero divisors? Does x^x=0 have solutions there?
Nice one! What about p-adic numbers?
@rainerzufall42 Ooh you've taken that too far 😃
I guess so. Integers modulo 4 for example. Let x=2. Then...
x^x=2^2=4=0 modulo 4.
@@dogbiscuituk Yeah, this example was pretty obvious...
You have to leave real numbers and just use ontological symbols. For 1/-infinity^infinity, you can solve to +-1=0, which can be used to symbolize all equals nothing.
What actually changes from X to 2N?
0=x^x=(e^ln(x))^x=e^(xln(x)). No complex power of e can make the result zero. There's no solution.
i can bet everyone that watched this know that
This is correct, and a simple proof that e^z is non-zero for every complex z is the fact that e^-z = 1/e^z, and e^z is defined and finite on the entire complex plane, so if some complex number c were a zero of e^z, then e^-c would be infinite.
I see why e^-c can’t work, but why not e^c?
@@gamemakingkirb667 What I'm saying is that if z^c=0 for any complex number c then z^-c would be 1/0 or infinity, contradicting the fact that e^z is an entire function (no poles.)
I mean , just say there aren't any zero divisors in the complex world, so there aren't any solutions in the complex world.
Also if you define that x^x =0 and make a new ring , would technically be a solution..
Oh yeah, because 0/0 is undefined.