x^x ≠ 0 ⟹ x^x > 0 Show that x^x is not bounded above Suppose BWOC x^x is bounded above, that is, there exists M > 0 s.t. x^x ≤ M for all x > 0 By the property that x^x is continuous (Proof is left as an exercise) ⟹ lim(x^x, x→∞) ≠ ∞ which is false (Again, skipping some stuff here that the reader may complete) Therefore x^x is not bounded Finding the minimum of x^x min(x^x) = y ⟹ d[x^x]/dx = 0 d[e^(xlnx)]/dx = 0 x^x(ln(x) + 1) = 0 ⟹ x^x = 0 (No solutions) Or ln(x) + 1 = 0 ln(x) = -1 e^(ln(x)) = e^(-1) x = 1/e ⟹ y = (1/e)^(1/e) = 1/e^(1/e) ∴ x^x ∈ [1/e^(1/e), ∞) ∪ {(-1)^n/n^n | n ∈ ℤ⁺} Edit: Forgot that negative integers still work with the real x^x
Okay, so... If you insert a negative number into the function x^x, you end up with (-k)^(-k) which is 1/(-k)^(k), where k>0. So, if -k is -m/(2n-1), you would have 1/(-m/(2n-1))^(m/(2n-1)) Observe that, if you have an odd root 2n-1 of a number -k (where k>0), then: (2n-1)\/(-k) it's just -(2n-1)\/(k) (we can take the minus sign outside of the root). Then, 1/(-m/(2n-1))^(m/(2n-1)) = -1/(m/(2n-1))^(m/(2n-1)). We can interpret that function as a function of x, where g(x) = -1/x^x = - x^(-x). If we change the sign of the negative domain of x^x, it has to be a subset of the positive domain of g(x), so we can study the function g and interpret from there. Let's derive g: g = - x^(-x) -g = x^(-x) ln(-g) = -x*ln(x) If we differentiate: -g'/-g = -1 - ln(x) g'(x) = - x^(-x) * (-1 - ln(x)) We can find the critical points: -1 - ln(x) = 0 => ln(x) = -1 => x = 1/e We can easily check that it's a minimum, with value -1.444667861... For any odd number 2n-1, we can find a number m such that m/(2n-1) < 1/e < (m+1)/(2n-1), and we can make the number 2n-1 as big as we want, essentially getting as close as we want to the value of 1/e (without ever being it). So, my conclusion is that the inf(x^x) is the value we get with 1/e in function g and so the range of x^x is (-1.444667861, +∞). We would have to check if we can get all values in that range. My intuition says no, but I don't have a clue on how to prove it. I hope that what I wrote was easy to follow!!
not that it fixes any problems, it just creates more. equation like f(z)^g(z) = h(z) doesn't even make sense on a complex plane, as you have possibly infinite number of values on the left hand side for each given z and only one value on the right hand side
MUCH smoother this time! And even correctly spelled 'video'. I bet you can make it more rigorous, proving that this gets every possible x and does not include any false input or overlap. I mean, we can see it intuitively; you nailed this answer. But mathematicians are often pedantic. Especially math teachers are unwilling to accept "it's intuitively obvious."
More or less whether or not 0 gets included in the Naturals depends on the context of the author. Authors who are used to dealing with the Naturals being derived from set theory as the set of finite cardinalities include 0 because that’s the size of the Empty Set. On the flip side, if you’re used to coming at the Naturals from factorization and Number Theory you’ll often exclude 0 because it can create oddball corner cases in theorems where you have to keep treating 0 as a special case. This also means that different countries or schools teach new students about the Naturals differently. That’s why there are comments in this video saying “my country includes 0” and “in my country I learned it doesn’t include 0”. And thus there is no generally accepted standard, your best bet is when you’re doing math involving “the Naturals” to define them in the beginning for the reader so they know whether you’re including 0 or not.
Since the set of reduced form rational numbers with odd denominators is dense, you could look at the graph of the continuation of the function in the negatives by setting the function value to the limit of its odd-denominator domain at every point.
x^x is really only continuous if only the negative integers are real, although using only odd denominator rationals principally give real solutions, there are branch cuts of exponation where they don't which are needed for continuity here
Sometimes desmos shows the negative x part of the graph y=x^x but you have to avoid zooming in with anything but the +/- buttons. I wish I could show it here but I have a screenshot of y=x^x where the negative x points are shown on desmos. I used the downloadable app extension on an old Android tablet
6:00 why is 2n - 1 better than 2n + 1 for the definition of odd? Is it just so when you set it equal to something you are adding instead of subtracting?
Some people prefer that n=0 be the first term, while others prefer that n=1 be the first term. That, or adding instead of subtracting might reduce sign errors
Depends on if n includes 0 or not if it does not then you actually need 2n-1 to represent every odd natural number, if yoi use 2n+1 the smallest value you could get is 3 and not 1
If you look at the concept of the principal nth-root in complex world, you should get a complex value for all nth-roots (except 1st) of negative (both even and odd roots), because the chosen root must be at arg(negative)/n which is 180°/n by convention. For example the principal 3rd-root of -8 or (-8)^(1/3) is not -2 but rather 1 + i√3 because it is located at 180°/3 = 60°, but -2 is located at 180°. This holds for other odd roots. For even roots are obvious though. Wolfram Alpha and Mathematica also take odd (principal) roots of negative in this way. Also if you take odd roots of negative to be negative, the x^x graph at the negative domain will be not continous because if we take (-1/3)^(-1/3) which is approximately (-0.333333)^-0.333333 (by taking 6 digits behind the comma for example) in that way, it won't be connected to its neighbors like (-0.333332)^-0.333332 and (-0.333334)^-0.333334 (which are both complex valued). (-1/3)^(-1/3) = ((-1)^(-1/3))/(3^(-1/3)) = (3^(1/3))/((-1)^(1/3)) ≈ 1.442249/((-1)^(1/3)). Since 3^(1/3) has no problem, we should take a look for the 3rd-roots of -1 or (-1)^(1/3). The three 3rd-roots of -1 are -1 (the real root), 1/2 + i√3/2 (the principal root) and 1/2 - i√3/2. Here we will test just for the principal root (1/2 + i√3/2 ≈ 0.5 + 0.866025i) and the real root (-1). If we assume (-1)^(1/3) = 1/2 + i√3/2 (the principal root), then (-1/3)^(-1/3) ≈ 0.721126 - 1.249024i (complex valued) If we assume (-1)^(1/3) = -1 (the real root), then (-1/3)^(-1/3) ≈ -1.442249 (real valued) But if we take the principal value of (-1/3)^(-1/3) instead which is approximately 0.721126 - 1.249024i, you can see the smooth transition for the input -0.333332, -0.333333 (or -1/3) and -0.333334 in the function x^x: (-0.333332)^-0.333332 ≈ 0.721130 - 1.249022i (-0.333333)^-0.333333 ≈ 0.721126 - 1.249024i (-0.333334)^-0.333334 ≈ 0.721122 - 1.249026i But, if you take (-0.333333)^-0.333333 ≈ -1.442249, the graph will be disconnected hence discontinous.🙂
x^x = e^(x ln x) f(z) = z, f(z) = ln z, f(z) = z ln z are all holomorphic functions. f(z) = e^z is also holomorphic, and the composition of holomorphic functions is a holomorphic funciton. I think it's a theorem that when you combine holomorphic functions, worst case you get the intersection of their respective domains. f(z) = ln z is holomorphic on the punctured plane, and that's what you get in this case. There are exceptions to this. You can cancel poles. Consider f(z) = z and g(z) = z^-1. The composition is an entire function even thought g(z) has a pole a z = 0 + 0i Again; domain: the punctured complex plane. Unless you care about the range being real and singular valued.
no, odd denominator doesn't work either. first of all, (-1/3)^(-1/3) is not the same as cbrt(-3). second of all, if we assume, that raising to 1/n power is the same as taking n-th root, we would expect, that (-1/3)^(-1/3) = (-1/3)^(-2/6), because -1/3 = -2/6, but that's not the case. on the left hand side we have cbrt(-3) = -cbrt(3). on the right hand side the have (-3)^(2/6) = cbrt(3). -cbrt(3) is cleary not equal to cbrt(3). that's the main issue with exponential functions with negative base of a real variable and that's the main reason why we say that base should be greater than zero
of course it could be too easy a thing , but why not give ourselves an idea about the continuity question via the simple series of f(-1), f(-2), f(-3), .. the range containing -1, +¼, -1/27, +1/256, .. lying on the two curve parts of the functions g(x) = 1 / ( | x | ^ (|x|) ) and h(x) = -1 / ( | x | ^ (|x|) ) , for the domain part x < 0 , exclusively . both g(x) and h(x) should be real and continuous in this domain . ?
Isn't there a set of irrational numbers that can be written as infinite series of rational numbers with odd denominators? If so, shouldn't those work aswell?
That proof is in error. Infinity - Infinity = undefined or infinite numbers so ✓(x + 1) - √x does not have a limit going to zero necessarily. ✓(x + n) - √x also goes to zero as x becomes large for any integer n of finite value!!
6:40 Isn't the set of natural numbers supposed to be the set of natural counting numbers? So they're not supposed to include 0 because you don't normally count from 0. 0 is, however, included in the whole numbers set. Correct me if I'm wrong.
That will be horrendous. Or will it? z^z=exp(z ln(z)) If we fix a branch of ln(z), then given that we can write d(z^z)/dz=z^z (1+ln(z)) it would appear to be a differentiable, hence holomorphic, except at 0. I don't know if we have to fix a specific branch, such as ln(1)=0, or if any branch will work, but I expect most of the complex plane will work for some branch. We may then get a domain of C\(Branch cut U {0}) And if we take the branch cut to be the usual R-, then the domain for a continuous function of the form z^z would be C\(-∞,0]
All negative integers are in the domain, not just even ones. Moreover, they are all included in -m/(2n-1) (m,n in Z+) that he has in the video. Just set n=1.
@@xinpingdonohoe3978the issue is that using -1/3 vs -2/6 as exponents creates a lot problems when the base is negative. Like (-1)^(1/3) = -1 but (-1)^(2/6) will either give you 1 if you do [(-1)^2]^(1/6) or a complex number if you do [(-1)^(1/6)]^2 The easiest way to address it is just to say that you cannot apply the x^(m/n) = [x^m]^(1/n) rule if x is negative and m/n is not in simplest form
@@xinpingdonohoe3978, well by this we remember what problems emerge when working with negative powers, negative bases, negative or broken exponents on school , right before the introduction of logarithms with “all” kind of bases . and, the rationals -1/3 and -2/6 may of course have the same numerical value but as an operator, e.g. as exponents they are surely really different . just say odd and even .
@@keescanalfp5143 even as exponents, they're the same. p/q as an exponent is the same as it is in irreducible form, and that can be verified by turning it into an exp equation and using Euler's identity, for whatever branch of logarithm.
@keescanalfp5143 they're the same even as exponents. The exponent p/q is the same as the exponent ap/aq, and that can be checked by going into exponential form and using Euler's identity, for whatever logarithm branch.
i think that a function joining the functions g(x) = |x|^x, x∈ℝ+ ∩ x = (-2m)/(2n-1), m,n∈ℤ+ and h(x) = -|x|^x, x = -(2m-1)/(2n-1), m,n∈ℤ+ describes the x^x
@@Tricky313chess two points: first, this would have a lot of multivalued shenanigans. Second, even as principal values, they have the same value. f(-1/2)=f(-2/4)=-i|√2|
y = x^x = e^{xln(x)}>> ln(y) = x ln(x), Let u=lnx then ln(y) = u e^u then u=W(ln(y)), Substitute back ln(x) = W(ln(y)) so x=e^{W(\ln(y))} Final Result, The inverse function of for f can be expressed as: f^{-1}(x) = e^{W(ln(x))} ( without forgetting the domain of x which is (0;1/e] or [1\e;+oo[ because we have to check tat th function is one to one)
Except Z+ does in Europe. Because here 0 is considered to be both a positive and a negative number and not neither. So Z+ is the same thing as N which also includes 0 here.
@@Xnoob545 maybe not in every single Europe in country. But I am a math student and I have had the occasion of exchanging with people from several European countries about math and all of those people were taught the same things.
@@vascomanteigas9433 well, you *could* cut them any way you like that works, but if we follow the standard logarithmic cut then yes, over the negative real axis.
As the negative rationals which are in the domain are dense in the negative reals, we can still ask for which negative values in the domain the function is continuous.
0 is neither positive nor negative. That is much less disputed than whether 0 is a natural number or not. That is why “positive” and “non-negative” are not the same because the former does not include 0 but the latter does
@Ninja20704 how do you quantify how much something is disputed ? Also I can say that "positive" and "strictly positive" are not the same because the first one includes 0 and not the second one
![แฉกลโกงงานวัด 2024 [ โกงมั้ยครับ ep.106 ] | DOM](http://i.ytimg.com/vi/taC5dpP3HXA/mqdefault.jpg)
If you want a super challenge, then find the real range of x^x
x^x ≠ 0 ⟹ x^x > 0
Show that x^x is not bounded above
Suppose BWOC x^x is bounded above, that is, there exists M > 0 s.t. x^x ≤ M for all x > 0
By the property that x^x is continuous (Proof is left as an exercise) ⟹ lim(x^x, x→∞) ≠ ∞ which is false (Again, skipping some stuff here that the reader may complete)
Therefore x^x is not bounded
Finding the minimum of x^x
min(x^x) = y ⟹ d[x^x]/dx = 0
d[e^(xlnx)]/dx = 0
x^x(ln(x) + 1) = 0
⟹ x^x = 0 (No solutions)
Or ln(x) + 1 = 0
ln(x) = -1
e^(ln(x)) = e^(-1)
x = 1/e
⟹ y = (1/e)^(1/e) = 1/e^(1/e)
∴ x^x ∈ [1/e^(1/e), ∞) ∪ {(-1)^n/n^n | n ∈ ℤ⁺}
Edit: Forgot that negative integers still work with the real x^x
@@Dravignor -1^-1=-1.
@@Dravignor
That's only for the positive part, we know f(-1) = -1, so -1 ∈ range(f(x))
The real challenge is to find the range when x is negative
Okay, so... If you insert a negative number into the function x^x, you end up with (-k)^(-k) which is 1/(-k)^(k), where k>0.
So, if -k is -m/(2n-1), you would have
1/(-m/(2n-1))^(m/(2n-1))
Observe that, if you have an odd root 2n-1 of a number -k (where k>0), then:
(2n-1)\/(-k) it's just -(2n-1)\/(k) (we can take the minus sign outside of the root).
Then,
1/(-m/(2n-1))^(m/(2n-1)) = -1/(m/(2n-1))^(m/(2n-1)).
We can interpret that function as a function of x, where
g(x) = -1/x^x = - x^(-x).
If we change the sign of the negative domain of x^x, it has to be a subset of the positive domain of g(x), so we can study the function g and interpret from there.
Let's derive g:
g = - x^(-x)
-g = x^(-x)
ln(-g) = -x*ln(x)
If we differentiate:
-g'/-g = -1 - ln(x)
g'(x) = - x^(-x) * (-1 - ln(x))
We can find the critical points:
-1 - ln(x) = 0 => ln(x) = -1 => x = 1/e
We can easily check that it's a minimum, with value -1.444667861...
For any odd number 2n-1, we can find a number m such that m/(2n-1) < 1/e < (m+1)/(2n-1), and we can make the number 2n-1 as big as we want, essentially getting as close as we want to the value of 1/e (without ever being it).
So, my conclusion is that the inf(x^x) is the value we get with 1/e in function g and so the range of x^x is
(-1.444667861, +∞).
We would have to check if we can get all values in that range. My intuition says no, but I don't have a clue on how to prove it.
I hope that what I wrote was easy to follow!!
@braisrg5 great work, but you should exclude 0 from the range
We surely need a 10 part series on this for the full domain
Major motion picture.
@@robertpearce8394 yeah
We’ll need an infinite series that (hopefully) converges!
@@mreverything7056 now that is a smart reply!
@@mreverything7056 best reply lol!
This is why I like complex analysis. The domain of z^z is all of C, except for a branch point at z=0.
It has complete algebraic closure
not that it fixes any problems, it just creates more. equation like f(z)^g(z) = h(z) doesn't even make sense on a complex plane, as you have possibly infinite number of values on the left hand side for each given z and only one value on the right hand side
My brain relaxes when I watch your videos. Come back soon.
Same
MUCH smoother this time! And even correctly spelled 'video'.
I bet you can make it more rigorous, proving that this gets every possible x and does not include any false input or overlap. I mean, we can see it intuitively; you nailed this answer. But mathematicians are often pedantic. Especially math teachers are unwilling to accept "it's intuitively obvious."
More or less whether or not 0 gets included in the Naturals depends on the context of the author. Authors who are used to dealing with the Naturals being derived from set theory as the set of finite cardinalities include 0 because that’s the size of the Empty Set. On the flip side, if you’re used to coming at the Naturals from factorization and Number Theory you’ll often exclude 0 because it can create oddball corner cases in theorems where you have to keep treating 0 as a special case.
This also means that different countries or schools teach new students about the Naturals differently. That’s why there are comments in this video saying “my country includes 0” and “in my country I learned it doesn’t include 0”. And thus there is no generally accepted standard, your best bet is when you’re doing math involving “the Naturals” to define them in the beginning for the reader so they know whether you’re including 0 or not.
The series is forming
Again, it's always "what is the domain of x^x?" And never "how is the domain of x^x"😢
Domain expansion!
ryōiki tenkai
Since the set of reduced form rational numbers with odd denominators is dense, you could look at the graph of the continuation of the function in the negatives by setting the function value to the limit of its odd-denominator domain at every point.
Im getting PTSD of Real Analysis in undergrad.
Great video!
Came back after 4 years just to thank you for helping me understand math, and you helped more than you could ever imagine ❤
I am happy to hear this! Thank you so much.
Yeah, and my professor put this in an assignment when I was in my first semester. I couldn't find the full domain, and I now hate this function for it
x^x is really only continuous if only the negative integers are real, although using only odd denominator rationals principally give real solutions, there are branch cuts of exponation where they don't which are needed for continuity here
Zero will be a natural number on the same day that talking about "clopen" intervals becomes acceptable.
Next do the domain of f(x, y) = x^y.
6:34 this can't be real "le zero" can't be forgotten 🙏
It indeed works on desmos
But only for a second or two
It shows plenty of points
Not connected but In an array very close to e^x.
Sometimes desmos shows the negative x part of the graph y=x^x but you have to avoid zooming in with anything but the +/- buttons. I wish I could show it here but I have a screenshot of y=x^x where the negative x points are shown on desmos. I used the downloadable app extension on an old Android tablet
I'm still not convinced that *no* negative irrational numbers are in the domain. I certainly get that some aren't
I assume he'll address that in a future video (see 7:08).
@@NotBroihon hopefully.
Next, find all the zeroes of the Riemann Zeta function.
Well there are an infinite amount of them
Me, with an academic Matlab license: I can probably do that in an afternoon.
Also me: *clueless*
At least it's beautiful. And now it's matched with my version perfectly.
6:00 why is 2n - 1 better than 2n + 1 for the definition of odd? Is it just so when you set it equal to something you are adding instead of subtracting?
Some people prefer that n=0 be the first term, while others prefer that n=1 be the first term.
That, or adding instead of subtracting might reduce sign errors
Depends on if n includes 0 or not if it does not then you actually need 2n-1 to represent every odd natural number, if yoi use 2n+1 the smallest value you could get is 3 and not 1
If you look at the concept of the principal nth-root in complex world, you should get a complex value for all nth-roots (except 1st) of negative (both even and odd roots), because the chosen root must be at arg(negative)/n which is 180°/n by convention.
For example the principal 3rd-root of -8 or (-8)^(1/3) is not -2 but rather 1 + i√3 because it is located at 180°/3 = 60°, but -2 is located at 180°. This holds for other odd roots. For even roots are obvious though.
Wolfram Alpha and Mathematica also take odd (principal) roots of negative in this way.
Also if you take odd roots of negative to be negative, the x^x graph at the negative domain will be not continous because if we take (-1/3)^(-1/3) which is approximately (-0.333333)^-0.333333 (by taking 6 digits behind the comma for example) in that way, it won't be connected to its neighbors like (-0.333332)^-0.333332 and (-0.333334)^-0.333334 (which are both complex valued).
(-1/3)^(-1/3) = ((-1)^(-1/3))/(3^(-1/3)) = (3^(1/3))/((-1)^(1/3)) ≈ 1.442249/((-1)^(1/3)). Since 3^(1/3) has no problem, we should take a look for the 3rd-roots of -1 or (-1)^(1/3). The three 3rd-roots of -1 are -1 (the real root), 1/2 + i√3/2 (the principal root) and 1/2 - i√3/2. Here we will test just for the principal root (1/2 + i√3/2 ≈ 0.5 + 0.866025i) and the real root (-1).
If we assume (-1)^(1/3) = 1/2 + i√3/2 (the principal root), then (-1/3)^(-1/3) ≈ 0.721126 - 1.249024i (complex valued)
If we assume (-1)^(1/3) = -1 (the real root), then (-1/3)^(-1/3) ≈ -1.442249 (real valued)
But if we take the principal value of (-1/3)^(-1/3) instead which is approximately 0.721126 - 1.249024i, you can see the smooth transition for the input -0.333332, -0.333333 (or -1/3) and -0.333334 in the function x^x:
(-0.333332)^-0.333332 ≈ 0.721130 - 1.249022i
(-0.333333)^-0.333333 ≈ 0.721126 - 1.249024i
(-0.333334)^-0.333334 ≈ 0.721122 - 1.249026i
But, if you take (-0.333333)^-0.333333 ≈ -1.442249, the graph will be disconnected hence discontinous.🙂
x^x = e^(x ln x)
f(z) = z, f(z) = ln z, f(z) = z ln z are all holomorphic functions. f(z) = e^z is also holomorphic, and the composition of holomorphic functions is a holomorphic funciton. I think it's a theorem that when you combine holomorphic functions, worst case you get the intersection of their respective domains. f(z) = ln z is holomorphic on the punctured plane, and that's what you get in this case.
There are exceptions to this. You can cancel poles. Consider f(z) = z and g(z) = z^-1. The composition is an entire function even thought g(z) has a pole a z = 0 + 0i
Again; domain: the punctured complex plane. Unless you care about the range being real and singular valued.
This is exactly the definition of "harder than it looks"
The domain is x>0, although some negative values give a real number…
I'm satisfied now - thanks!
no, odd denominator doesn't work either. first of all, (-1/3)^(-1/3) is not the same as cbrt(-3). second of all, if we assume, that raising to 1/n power is the same as taking n-th root, we would expect, that (-1/3)^(-1/3) = (-1/3)^(-2/6), because -1/3 = -2/6, but that's not the case. on the left hand side we have cbrt(-3) = -cbrt(3). on the right hand side the have (-3)^(2/6) = cbrt(3). -cbrt(3) is cleary not equal to cbrt(3). that's the main issue with exponential functions with negative base of a real variable and that's the main reason why we say that base should be greater than zero
What about negative integers? They are also in the domain. And, they alternate below and above the x axis and the go to 0 as x goes more negative.
you truely are the nilered of math
6:30 my calc 2 prof always said N doesn't include 0, and specifically used N_0 to include it. Seems to me like an adequate solution.
There are several adequate solutions. There's just no agreement about which one to use and they are not all mutually compatible.
It takes a real man to admit when he was wrong 🤘
of course it could be too easy a thing , but why not give ourselves an idea about the continuity question via the simple series of
f(-1), f(-2), f(-3), ..
the range containing
-1, +¼, -1/27, +1/256, ..
lying on the two curve parts of the functions
g(x) = 1 / ( | x | ^ (|x|) )
and
h(x) = -1 / ( | x | ^ (|x|) ) ,
for the domain part
x < 0 , exclusively .
both g(x) and h(x) should be real and continuous in this domain .
?
I always thought that 0 is not in the set of natural numbers but it is in the set of whole numbers
Domain X-pansion
Can you make a video for calculate the Integral forth root of tanx Teacher? Thank you.
Do you still have the previous video? I am curious!
In France, we say that IN = [0;+♾️[ = {0,1,2,3,...}
And to say that we don't take 0 we write it as the set IN*
That's how I learned here in Brazil.
inni da bee
Yes but you can't write it like this (normal brackets mean intervals of ℝ). For "intervals" of ℕ you use double brackets,
ℕ = 〚 0 ; +∞〚
@alexandresibert6589 ye ik it was a mistake
here we just write Z_{ \geq 0} or Z_{\geq 1}
Isn't there a set of irrational numbers that can be written as infinite series of rational numbers with odd denominators? If so, shouldn't those work aswell?
I wonder how that would look plotted.
also, we can have any negative number as long as it belongs to Z
0^0 finally approaches 0
th-cam.com/video/X65LEl7GFOw/w-d-xo.htmlsi=4zAEKTg4b3WjP0Ct
That proof is in error. Infinity - Infinity = undefined or infinite numbers so ✓(x + 1) - √x does not have a limit going to zero necessarily. ✓(x + n) - √x also goes to zero as x becomes large for any integer n of finite value!!
Mathematics 😅
No Mathematician in Nigeria will consider 0 in the set of Natural numbers but in the set of Whole numbers
I can't believe it
6:40
Isn't the set of natural numbers supposed to be the set of natural counting numbers? So they're not supposed to include 0 because you don't normally count from 0. 0 is, however, included in the whole numbers set. Correct me if I'm wrong.
In the Peano axioms for the natural numbers, 0 is included.
@@bjornfeuerbacher5514Ah, I see.
Next analyze f(z) = z^z for complex z
That will be horrendous. Or will it? z^z=exp(z ln(z))
If we fix a branch of ln(z), then given that we can write
d(z^z)/dz=z^z (1+ln(z))
it would appear to be a differentiable, hence holomorphic, except at 0. I don't know if we have to fix a specific branch, such as ln(1)=0, or if any branch will work, but I expect most of the complex plane will work for some branch. We may then get a domain of
C\(Branch cut U {0})
And if we take the branch cut to be the usual R-, then the domain for a continuous function of the form z^z would be
C\(-∞,0]
f(0)=1 or undefined
is there a way to "prove" that at least the real part of 0^0 is 1? (from all complex paths)
Stupid 0 ruining everything with "no agreement"
I want to see a graph if the negative side
There’s also x=-2n for all n in negative integers
All negative integers are in the domain, not just even ones. Moreover, they are all included in -m/(2n-1) (m,n in Z+) that he has in the video. Just set n=1.
@@TedHopp oh right
@@TedHoppwhen you set m=2 and n= 1, x=-2 which results to complex value hence the domain would not exist
@@mathmadeeasierwithdelmy953 Huh? (-2)^(-2) = 1/(-2)^2 = 1/4. No complex value in sight.
@@TedHopp my bad I was viewing it as -1/2... Thanks
I hope we will see an integral with the natural part of a number 😢
Don't try to make short videos. I love long-form!
0^0 should be undefined. I can set up an arbitrary limit problem showing 0^0 is any value. You cannot do that with other values.
What caught my attention last time was x = -2/6. Yes, this value equals -1/3, but it is NOT -1/3.
Huh? Equals simply means they're the same. -1/3 = -2/6 if they are the same, which is to say -1/3 is -2/6.
@@xinpingdonohoe3978the issue is that using -1/3 vs -2/6 as exponents creates a lot problems when the base is negative.
Like (-1)^(1/3) = -1 but (-1)^(2/6) will either give you 1 if you do [(-1)^2]^(1/6) or a complex number if you do [(-1)^(1/6)]^2
The easiest way to address it is just to say that you cannot apply the x^(m/n) = [x^m]^(1/n) rule if x is negative and m/n is not in simplest form
@@xinpingdonohoe3978,
well by this we remember what problems emerge when working with negative powers, negative bases, negative or broken exponents on school , right before the introduction of logarithms with “all” kind of bases .
and, the rationals -1/3 and -2/6 may of course have the same numerical value but as an operator, e.g. as exponents they are surely really different . just say odd and even .
@@keescanalfp5143 even as exponents, they're the same. p/q as an exponent is the same as it is in irreducible form, and that can be verified by turning it into an exp equation and using Euler's identity, for whatever branch of logarithm.
@keescanalfp5143 they're the same even as exponents. The exponent p/q is the same as the exponent ap/aq, and that can be checked by going into exponential form and using Euler's identity, for whatever logarithm branch.
i think that a function joining the functions
g(x) = |x|^x,
x∈ℝ+ ∩ x = (-2m)/(2n-1), m,n∈ℤ+
and
h(x) = -|x|^x,
x = -(2m-1)/(2n-1), m,n∈ℤ+
describes the x^x
Domain expansion:f(x)=x^x
Why doesn't WA consider those other points as part of the domain?
❤
why did you say it was better as m/2n-1 and not m/2n+1 ?
Bc i put n as a pos integer. So 2n+1 would start at 3 but i need the bottom to start at 1.
@@blackpenredpen oooh makes a lot of sense, thank you :)
How can f(x) = x^x be a function for all numbers, when f(-1/2) and f(-2/4) have different values
@@Tricky313chess two points: first, this would have a lot of multivalued shenanigans. Second, even as principal values, they have the same value.
f(-1/2)=f(-2/4)=-i|√2|
How can we calculate (-pi)^(-pi)?
What is the inverse function of this function 🤔
y = x^x = e^{xln(x)}>> ln(y) = x ln(x), Let u=lnx then ln(y) = u e^u then u=W(ln(y)), Substitute back ln(x) = W(ln(y)) so x=e^{W(\ln(y))} Final Result, The inverse function of for f can be expressed as: f^{-1}(x) = e^{W(ln(x))} ( without forgetting the domain of x which is (0;1/e] or [1\e;+oo[ because we have to check tat th function is one to one)
When MATHEMATICIAN is bored❤😂
I don’t get why m/(2n-1) can’t be 0 if m, n € Z+? I thought Z+ is natural numbers with 0 being included. Maybe correctly say m is natural and n is Z+?
Please please please do x^x = -x
How did the calculus student go about his day after a breakup?
He derived with no respect to x
In desmos it shows different
It probably only shows continuous parts
Is (-6/10)^(-6/10) defined or not according to this video?
I wish N included zero definitively, because Z+ doesn't and there's no need for confusion and redundancy.
Except Z+ does in Europe. Because here 0 is considered to be both a positive and a negative number and not neither. So Z+ is the same thing as N which also includes 0 here.
@@ryznak4814 wrong
In Lithuania 0 is taught as a neutral number
@@Xnoob545 maybe not in every single Europe in country. But I am a math student and I have had the occasion of exchanging with people from several European countries about math and all of those people were taught the same things.
@@Shack263 who uses Z+ though? People who want to be pedantic.
@@xinpingdonohoe3978 being pedantic is the point of maths
It’s an entire function, it exists over the entire complex plane, right?
Yes, with an infinite number of branch cuts over the negative real axis.
@@vascomanteigas9433 well, you *could* cut them any way you like that works, but if we follow the standard logarithmic cut then yes, over the negative real axis.
@@xinpingdonohoe3978 This function are also called the Tetration Function.
why is 2n-1 better tho?
Can you solve x^6=(x+1)^6
noice
Can you solve this x^x=-1
As the negative rationals which are in the domain are dense in the negative reals, we can still ask for which negative values in the domain the function is continuous.
Lmao you used Z+ to avoid the confusion but Z+ still includes zero
0 is a positive number 0 ≥ 0, just not a strictly positive number
I think + means strictly positive
@@isjosh8064 Well it probably depends where you were taught
My point is that doesn’t clear up the confusion
Positive integer means strictly greater than zero to include zero we use non negative integers
0 is neither positive nor negative. That is much less disputed than whether 0 is a natural number or not.
That is why “positive” and “non-negative” are not the same because the former does not include 0 but the latter does
@Ninja20704 how do you quantify how much something is disputed ?
Also I can say that "positive" and "strictly positive" are not the same because the first one includes 0 and not the second one