Calculating i to the power i the right way. Why every proof you have seen is wrong
ฝัง
- เผยแพร่เมื่อ 21 พ.ค. 2024
- The imaginary number i is equal to the square root of negative 1. But i^i is a real number, approximately equal to 0.208. Why is this true? It is an incredibly tricky calculation if you want to justify the steps properly. All of the videos I've seen on i^i use an unjustified step, including mine. So what is the proper way to calculate i to the power of i? Special thanks this month to: Daniel Lewis, Kyle, Lee Redden, Mike Robertson. Thanks to all supporters on Patreon! / mindyourdecisions
0:00 calculation
2:11 complex numbers
3:30 "proof" 1
6:21 "proof" 2
8:20 complex exp
9:12 complex log
12:32 the right way
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Complex Variables and Applications, James Ward Brown, Ruel V. Churchill, 7th edition, Chapter 1 Section 8, Roots of Complex Numbers
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I really like the canceling digits method example 😂
Truly😂😂
It’s called anomalous cancellation for those who are curious
It killed me
I like using them in my math class to illustrate how the wrong method can still get the right answer on occasion, and show the common error of improper "cancellation" to simplify fractions.
Just came by to drop x^2 = 25
See ya
(a^b)^c is indeed not always applicable to complex numbers, however, your example is *wrong*.
1 = e^2PI*i
then you raise both sides to the power of 1/2, effectively taking the square root. The square root function is not injective, you should have added +- to one of the sides, which would have made the equation correct.
Basically you've shown that both 1 and -1 are square roots of 1 (which is not a mistake)
Agree, but in my opinion, (a^b)^c is applicable in case of purely 1 dimensional calculations. For example, here, We are purely using y-axis, so using this formula gives correct answer.
When it comes to 2D calculations, like vectors (in complex number when both y axis and x axis are involved), this formula should become invalid, or should be generalised to get proper answer.
@@raffaeleraia2223 The thing is, square root of 1 truly have two values 1 and -1. But, we defined it as only positive 1.
In the equation it will always give two values.
And, a mathematical solution doesn't necessarily follow what we defined, unless we change the solution to make it follow what we defined.
I am not good at words here it seems.
with complex numbers, the log function and fraction exponents are multivalued. in the steps where you "disproved" rules of math for complex numbers you conveniently decided to ignore the multivalued nature of these operations, which you can not do.
The video is a nice explanation of branch cuts and all, but I think a lot of it can be avoided by simply noticing that a complex number has more than one polar representation. Then the first “proof” works trivially.
I mean, ℂ is a characteristic 0 field for crying out loud. That’s a pretty nice algebraic structure as far as I’m concerned.
Thank you, I was getting so mad watching those "disproofs." Conveniently ignoring the multivalued nature of roots and logarithms to say things are wrong.
Literally every video (and in english like he mentions) I've seen showing i^i is real either explains that they're taking principle values of everything meaning that ln(e^2ipi) would _indeed_ be 0, or they specifically mention that it's multivalued and get the final result with n somewhere indicating its multivalued nature.
There is nothing wrong with the first two proofs.
Those are functions that give multiple branches, which is why you can produce an apparent contradiction, but in each case the calculation does still give _A_ branch, and that branch is a valid evaluation.
However, choosing a specific branch without giving a reason why, is still wrong. The first proof in fact technically falls into the fallacy mentioned with canceling digits - for example, canceling digits works as long as the number of trailing zeros in the numerator is greater than or equal to the number of trailing zeros in the denominator, and the denominator is a power of 10, and the only digit being canceled is 0. But still it's an unjustified way to prove for the others.
The second one is more like you're saying, but still, choosing a branch requires justification, otherwise you can never be sure that you chose the correct branch. So it is technically wrong (but incomplete is a better description)
@@vr10293That's basically the principal branch, and taking the principal branch by default is reasonable to me.
I would say these proofs are to introduce the concept of index form rather than slam dunking the audience terms of complex analysis which is complex as u see by its name.
If there are multiple branches the function isn't bijective though.
@@d3ds1r There are two topics. For solving equation, it's totally fine to have multiple answers.
But if you want to define a function, You usually apply a polar form and construct a Riemann Sphere to try to make an only output.
it depends on how you define complex logarithms. if you want to have a function, you need to allow a principal value of the logarithm. if you want to find all solutions, it isn't a normal function anymore because it has multiple outputs. method 2 isn't a "wrong" proof, it just glosses over some small detail with +2*k*pi*i in the logarithm function which does cancel out when you raise it to e. It's obviously different from the fraction coincidence at the beginning because this is a good method that always works for any similar problem. you can proof euler's theorem with taylor series or you could just assume the result. doesn't make it wrong.
I'm a wrong believer in two wrongs make a right. Let me repeat that, I'm a wrong believer in two wrongs make a right.
Actually the 2kπi won’t cancel out because you have an extra i, so raising it to e would give you a factor of e^(-2kπ), which contributes to the multi-valued nature of i^i itself.
@hapedisedivide1980 What are you sayin' ?
@@samueljehanno Please simply refer to the video and compare.
At 6:12, the reason you get a nonsense answer here is independent of the (a^b)^c = a^(bc) step. You've stated that 1^(1/2) = 1, which is correct, used (a^b)^c = a^(bc) to find that 1^(1/2) = -1, which is also correct. The problem is that 1^(1/2) has two solutions and you have incorrectly equated them to one another.
If you are careful to always note that e^f(x) = e^(f(x) + 2 i n pi), where n is an integer, then that avoids the problems associated with (a^b)^c = a^(bc). Of course, using this method isn't helpful when the exponent has a base of 1, because using a base of 1 is generally problematic and breaks a lot of general rules. For example, 1^(infinity) is undefined rather than infinite.
There are other examples of (a^b)^c = a^(bc) not holding for complex numbers. The point in the video is still true.
1^(1/2) = 1 It doesn't have 2 solutions, you're mistaking it with x^2=1 which would be |x|=1^(1/2) the x is inside a module so there you have 2 solutions.
@@Max_Power_ That only holds for real numbers. For complex numbers, functions can be multivalued.
@@Max_Power_ No, you're thinking of sqrt(1), which is positive by definition
"The problem is that 1^(1/2) has two solutions and you have incorrectly equated them to one another." - ya, because the sqrt function has 2 branches, one of which is the principle one
doesnt the 1=-1 proof fail because you decided that
squareroot 1 = 1
instead of
squareroot 1 = +-1 ?
For complex numbers, if you account for all branches the identity (a^b)^c=a^(bc) should hold.
likewise in proof 2 counterexample, the natural log needs to be defined respective to a branch of the natural log which effectively reduces it mod 2πi. if you want to avoid that you need to choose a branch which includes 0 properly, so you can let πi =-πi and then you wouldnt get 0=2πi rather 0=0
(a^b)^c=a^(bc) does hold if you treat it as a multivalued function. But I think the point in the video is that typical presentations do not do that.
@@TheEternalVortex42 That's actually not true, although Presh's example doesn't make that clear. There is an example due to Clausen where the formula fails, even when considering the multi-valued case. (See the Wikipedia page "Exponentiation" and search for "Clausen").
@@TheEternalVortex42I agree that that is the point of the video. But I disagree that that is an actual typical problem. I feel like he was assuming presentstions have the problems he's saying, but they just don't.
Yep. Squareroot is a multi-valued function, which results in exactly the same problem that taking the log has.
Your counterexample to proof 1 is finding out that 1 has two square roots, one of which is 1, and the other of which is -1. This is not wrong. That's always something you need to consider when undoing a root with a square! You're focusing on an extraneous root.
In general though, the core of this is that when you're using multivalued complex functions, yes, you do in fact need to be careful because the principal value isn't always the one you're looking for! This is why the proof 2 got the nonsensical result of 0 = 2pi*i
So yes, the careful analysis is accurate.
@@sylverfyreThe simplest solution is to note that polar representations are not unique. The issue is not the algebraic operation in the exponent as he makes it seem. That is perfectly fine. And roots are taken care of by applying De Moivre’s formula. (Actually even complex roots are no trouble since ℂ is a field and we can always rewrite a rational function in i as a linear polynomial in i.)
I love complex analysis! In the case of the natural logarithmic function, we can make a branch cut and define the principal branch as log z = ln r + iθ, where r > 0 and θ is in (-π, π]..
How would you define log(-1) then? It has no argument in the range you stated.
@@MichaelRothwell1 I erroneously excluded π because I couldn’t find a
Cutting a locally holomorphic function kinda defeats the pupose of it being locally holomorphic in the first place. Path dependence is just the way to go in complex analysis. If cutting is required, you would want to choose the cut freely anyway.
My favourite aspect about mathematics is that any two people independently discovering the same subject should derive equivalent definitions.
@@caspermadlener4191 In this case, the reason why we define a branch cut is because the logarithmic function is multiple-valued.
@@JaybeePenaflor Thanks for the clarification! (here is the ≤ symbol, in case you need it in the future).
Than explains so much! Thank you! No matter if it's i^i or whatever else, the fact you made me pay more attention while carrying out supposedly obvious operations is priceless.
I mean, the second proof was basicaly showing that e^(-pi/2) is an answer. The only thing it did differently from the final proof was considering just 1 answer to ln(x), but this answer is still valid so it should also get only one answer to i^i (and it did)
The lesson here is: when in doubt trust Wolfram not TH-cam.
This has to be one of your best ever videos, if not the best - bravo!
Dam this definitely blew my mind🤯
I have been so used to doing these problems normally that I almost forgot the limits of the basic mathematics rules like this indices one!
This would be great if Presh included the part about e^z also being multivalued and different from exp(z)
I've seen "log" used for the complex log function along with "ln" for the real natural logarithm.
No, "log" is used for log-base-10 (though some weirdos at ISO have decided that "lg" means base-10 when that is more commonly used for base-2...the idea is that "log" means the generic logarithm where you need to provide the base, making the binary logarithm "lb.")
Are you thinking of "Log", with a capital L?
@@dhwyll In the Wikipedia entry for complex log, they use "log" as a generality, and "Log" for the principle value.
@@dhwyllit depends on the field of study. In some texts that focus on graph theory and computer algorithms, they write log and it is assumed to be based 2. In some complex analysis texts, log is base e. You have to adopt the convention of the author and the field of math you are working in.
@@dpgoulette Exactly 😊
At high school, we used the names ln, log and log_a, for e-based, 10-based and a-based logarithms.
At the university we used (almost) only the natural logarithm and named it log.
That is: Check the definition of log in the text you are studying, and if the definition is not provided you have a natural choice 😂.
@@dhwyllAnd as we all know, the generic logarithm is the logarithm with base e, aka. natural log.
you have taken 2nd principal value for 1 which will give you -1 its no surprise there as 1^1/2 may equal to -1. but for 1=e^0 1st principal value it will remain 1. Well new method is also correct but the previous method also works we just need to be careful. as we have applied the the method to find roots of unity x^5=1.
At 6:00 it was you that messed up. 1 = e^(2*pi*i) you were fine, but when you took the square root of both sides you messed up. -1 is also equal to 1^(0.5). You played a trick with the multi-valued nature of the square root function. The issue has nothing to do with the "multiply the exponents" rule - it was another mistake altogether.
To put it another way, when you introduced the 1/2 exponent, you took the principle value on the left but the other value on the right.
The whole cancelling digits thing is really interesting. I want to know what process can lead you to find a general solution for what fractions that happens to
Brute force of Chuck Norris 😅
In 4:43 The identity *_(a^b)^c = a^(bc)_* is not only true with *_b_* and *_c_* as real numbers. It is also true as long as the variable *_a_* is a positive real number and that the variable *_b_* is in the form *_Log z_* or is multivalued with respect to *_2πni_* where *_n_* is any integer. The variable *_c_* could be any non-multivalued complex number. This is proven in 8:56 where Presh showed that *_z^w = (e^Ln z)^w = e^(w Ln z)_*
I was also confused about the step at 8:56. As the example at 5:50 shows, it is not sufficient that "a" is real: Using complex numbers for "b" and "c", we get 1=(-1) for "a"=e. And e is definitely a real number. However, it seems that the key is the definition of the complex exponentiation; the problem is that this definition is introduced at 12:48 but it is already used at 8:56.
This is not correct. (OK, you could define both sides to be multi-valued, but even then you could say only that all valid values for the RHS are also valid values for the LHS. I'm using the Principal Argument idea from the video as the definition (although I would exclude negative numbers).) For example, let a=e, the base for natural logarithms, let b=3 pi i/2, and let c=i. Then a^{bc} = e^{-3pi/2} < 1. However, a^b = -i, and (-i)^i = e^{pi/2} > 1.
Slight correction on proof 2:
Since e^{2i\pi} = e^{0}, and in fact any angle shifted by 2\pi is the same angle on the unit circle, it can be said that 0 is in fact equal to 2i\pi. Or more precisely, this is a number system in which all numbers in the exponent are modulo 2\pi, and mod(2\pi, 2\pi) = 0.
It is still interesting to see where these steps can go wrong, so thank you for the informative video.
At 8:50, how did you come to that conclusion/derivation for the complex exponent? Seems like that was not well justified?
That is actually not clear at all, but it can be derived from the Maclaurin series definition of e^z with some sneaky summing. It is probably easier to start with (e^u)*(e^v) and then multiply out their series expansions. From here, collect terms of the same degree and you should see a binomial pattern which can be factored to give something like (u+v)^n in the terms of the series. With a little fudging around in the coefficients, you can get show that you have the exact same series representation as e^(u+v). Since these are unique, they have to be the same function.
8:00 The natural log of any number except zero has infinite solutions. When you take the ln of 1, it would give the following solutions: ln(1) = 0 + n*2*pi*i, where n is an integer. Similar to ln(e^(2*pi*i)), it would result in: ln(e^2*pi*i) = 2*pi*i + n*2*pi*i = (n+1)*2*pi*i = m*2*pi*i, where m is an integer. This means: ln(1) = ln(e^(2*pi*i)).
*Yeah so that's basically the same as saying "2 = -2 because they both square to 4, but that's impossible, so we can't define square roots."*
@@mr.d8747the nth root FUNCTION is defined using the principal branch. Therefore, only 1 value comes from it.
The 1/n exponent takes all branches. It's a multivalued function and you have to specify the branch
This video answered my so many questions
Brilliant.
Excellent channel
Or, you could just do it directly. The formula for complex exponentiation is:
(a + bi)^(c + di) = (a^2 + b^2)^[(c + di)/2] * e^[i(c + di) * arg(a + bi)]
Where arg(a + bi) is the angle with the positive real axis for the number.
Thus, for i^i, that is (0 + i)^(0 + i). Therefore, a = 0, b = 1, c = 0, d = 1. Thus:
(0 + 1i)^(0 + 1i)
(0^2 + 1^2)^[(0 + 1i)/2] * e^[i(0 + 1i) * arg(0 + 1i)]
1^(i/2) * e^[i(i) * (π/2 + 2πn)]
1 * e^-(π/2 + 2πn)
e^-(π/2 + 2πn)
Thanks! Where did you get this formula?
@@nitey123 My Complex Analysis class. It's a consequence of Euler's formula and the polar coordinate nature of the complex plane.
Hi, im no expert and didnt understand the video at all, for id like to get an answer to make me understand, why i^i gives such a weird number?, it is just a 1 on a different side of plane, like -1, no matter what it is, is a 1 for example, 1^1, or -1^1 or -1^-1, they will result on a 1 with a symbol that tells on which side of the plane it lays, like, I mean, the number (i) could be rewritten as 1 with another symbol, like $1, so how can it just be that $1^$1 not be either of this four: 1, -1, i, -i?
Sorry, got lost in the middle of the whole thing. IMHO you also fell in the pitfall you wanted to avoid, and this started the moment you presented i in polar form. The real problem, before you get to i^i is to explain, what it means to elevate anything to a power of a complex number.
What I mean is that sometimes the meaning of some mathematical operators become a little esoteric. For instance a+b is a + 1 + 1 + 1 ...+1 (b times) a * b is a + a + a ... (b times) +a. But when it get to exponentiation, and we start having a^1/2, it starts being something else than a * a (half a time?). The notation starts taking a special meaning. And this is a simple point that could be better explored in the video.
Thanks. Great video
Isn't this the method that Blackpenredpen uses? So you wouldn't be the only one that uses a justified method.
Wait so what happened to the infinite series at 9:00? Why show that if it wasn’t used?
Putting 1 as √1 is slightly deceptive (Though theoretically correct) since there are two numbers whose square is 1 I e 1 and -1
he didn't put 1 as root(1), he put root(1) as 1, not the same thing
I posit there is a gulf between spuriously cancelling digits in a fraction, and unjustifiably applying the third law of exponents, ***in this case***.
You state that this law "should [have] some conditions like...", and give the simplest and most restrictive form of those conditions.
Wikipedia's "Exponentiation" page gives the "next level up", but once again qualifies it with a disclaimer:
"*** In general ***, (b^z)^t b^zt, unless z is real or t is an integer".
I suspect the comprehensive expression of those conditions would actually allow it in the case of (e^(i.pi/2))^i
Indeed, I have seen a few videos about complex powers. Every time they start using complex powers without defining what a complex power is, and then applying all the rules of real powers without even checking whether there exists a complex function that has all the same properties as the real power function.
Same answer but that is the best proof i've seen so far. Keep up the good work.💯💯
At 10:45, why does it follow from |z|e^(i arg(z)) = e^u * e^(vi) that |z|=e^u and argz = v?
Usually, a*b = c*d doesn't imply that a=c and b=d.
It could be that |z|=e^(vi) and argz = -ui, right? Or that |z| and arg(z) are something else completely, like in 3*4 = 2*6.
|z| is always a real number.
presh sounds so done with everything
Saw some people disagreeing with the argument against the “proofs”, so here’s a bit of a take of mine.
Regarding counterexample in “proof” 1: The exponential function is single-valued, in fact holomorphic, on all of C. Therefore all the steps in the counterexample, except where the erroneous Exponentiation “rule” in question was applied, are perfectly valid.
Regarding “proof” 2: ln(2iπ) is of course multi-valued, but then the statement 0=ln(2iπ) is saying that 0 is equal to a whole bunch of different numbers like 2iπ, -4iπ, -8iπ, etc. And the fact that 0 happens to be one among those numbers doesn’t make it any less absurd.
Remark: One may also find i^i by Euler’s formula, because it is justified for all complex numbers. Note that complex sine and cosine are defined via Euler’s formula.
This blew my mind 💥
Presh, you're usually right, but this time... All the values of the "counter-proofs" are a bit of fraudulent.
In the first, you took not the principal value of one in the complex plane but another one (1=e^2πi) and you stated that is equal to one (true). But then you took the square root. 1 is the principal complex square root of the complex number 1, but you took the complex square root (which is multi-valued) of e^2πi. In other terms, you took the real square root on the LHS (which matches with the principal complex square root on complex numbers with imaginary part = 0) and the complex square root (multivalued) in RHS. If you treated 1 with the general argument (e^i(0+2πk), k is an integer) you wouldn't get to this result because it has another value which is 1.
In the other, happens the same thing. You took the real natural logarithm on the LHS (which matches the principal branch of the natural logarithm on complex numbers with imaginary part = 0) and the complex natural logarithm on the RHS (the multi-valued natural logarithm). That property has to be true, not also because you use it when solving the equation z = e^w, but because is the definition of the natural logarithm (and it has to be multi-valued because so it is the complex exponential)
In both cases, you took two different functions on the two sides, making the answer blatantly wrong. In complex numbers, is better to work with the general argument and then reduce it to the principal form if you're not sure what you're doing.
Technical misunderstandings: Ln(z) is the notation of principal branch of natural logarithm, usually not the notation for real natural logarithm. It's subtle because for real numbers is equal, but not for complex numbers. Use Log(z) and log(z) notation for complex natural logarithm (since defining based-logarithms is a bit useless in the complex world) and ln(x) for natural real logarithm.
Thank you for reading to anyone who kept reading so far and I love you Presh, this is just constructive criticism
Interesting method the other videos used. At the start of the video I solved for e^-pi/2 by doing i^i = e^ln(i)*i, since e^i*pi/2 = i, e^(i*pi/2)*i = e^-pi/2. Since at the step for substituting the ln has a family of solutions, that is why there are more solutions to i^i
The multi valued result is still ridiculous. It means that all these values are equal to i^i, but not equal to each other. This violates the postulate that if a=b and b=c the a=c.
It's amusing that all the comments point out that the counterexamples are only "wrong" because they ignore that complex log is multivalued. Yet the whole point of that part of the video is that other videos are ignoring that fact!
Thank you sir
8:04 "0 = 2*i*pi" - is there any significance that in terms of angles, 0 rad and 2pi rad are equivalent, and does 2*i*pi make it any different in terms of angles?
Imaginary/complex numbers in mathematics are like subways, overpasses, bridges, tunnels, etc. They only make sense if they connect something important with something else important. This is the only why they are important in themselves. Otherwise they would be worthless by themselves, they serve to help us move from one "useful" place to another "useful" place.
I think the thumbnail is wrong. You said i^i = e^pi/2, not i^i = e^-pi/2.
For me thumbnail is showing "?" 😂
finally someone does it in right way
Love that you would make a 15 minute video on this, and I would be absorbed watching it all. Nice job.
So, in general, what does the plot of z^w "look like" if you plot the multi-values?
That would require 6 dimensions to plot. You can kind of get an idea by fixing one of the inputs though. If you fix z=re^(iθ+i2kπ) on the unit circle by setting r=1, then varying the real part of w=c+di rotates z around the circle. Varying the complex part of w stretches and shrinks the whole circle by an amount inversely (and exponentially) proportional to d. So if d gets bigger, the circle gets smaller. The branch cuts here i.e. the 2kπ parameter, also generate a whole integer-indexed family of other circles of different radii. So basically complex exponentiation turns the unit circle into a bunch of concentric rings.
If you don’t fix z on the unit circle, but you do fix it on the real line, θ=0, then things get a bit more difficult, but basically you get a dependence of radius on both real and complex part of the exponent. So the whole circle gets wiggled a bit as you go around the whole thing.
Great job with the details, and why the other proofs were incorrect.
Fantastic! Could you please do this with Ramanujan's sum -1/12? Please!!
most videos showing that 1+2+3+... = -1/12j ust assume a series to converge when it isn't. That's it.
Actually sum of all numbers is not truly -1/12 imo. But, using it often solves the problem, since you can't always deal with infinity.
I have learnt complex number recently so I now I can understand these proof
What an elegant video!
Have MYD just listened an university course on complex analysis?
10:19 Here you said "since u and v are real, you can say e^(u+vi) = e^(u)e^(vi)"
Why does the i not make a difference?
very cool stuff
Good job...awesome
Thank you!! This was bugging me as well!
But the supposed error at 7:56 can easily be remedied by adding 2ipi Plus 2npi Presh so technically it's not all wrong..wouldn't you agree
what is the ratio between volume of sphere and the cylinder inscribed inside the sphere and where cylinder have maximum volume
Suppose the sphere has radius r and hence volume Vs = (4*pi/3)*r^3 . Let n be the axis of the inscribed cylinder, and let a be the acute angle between n and a line through the center of the inscribed cylinder and a point on the edge of the inscribed cylinder. Then the height of the inscribed cylinder is 2*r*cos(a), the radius of the inscribed cylinder is r*sin(a), and the volume of the inscribed cylinder is Vc = 2*r*cos(a) * pi*(r*sin(a))^2 = 2*pi*(r^3) *cos(a)*(sin(a))^2 .
Determine the angle a for which Vc is maximal:
Vc(a) = 2*pi*(r^3)*cos(a)*(sin(a))^2 = 2*pi*(r^3)*cos(a)*[1- (cos(a))^2] = 2*pi*(r^3)*[cos(a) - (cos(a))^3]
d(Vc) = 2*pi*(r^3)*[1 - 3*(cos(a))^2] * d(cos(a))
d(Vc) = 2*pi*(r^3)*[1 - 3*(cos(a))^2] * (-sin(a)) d(a)
d(Vc)/da = 2*pi*(r^3)*[1 - 3*(cos(a))^2] * (-sin(a))
Setting d(Vc)/d(a) = 0:
2*pi*(r^3)*[ 1 - 3*(cos(a))^2 ] * (-sin(a)) = 0
1 - 3*(cos(a))^2 = 0 OR -sin(a) = 0
3*(cos(a))^2 = 1 OR {a = pi + 2*pi*k , for any integer k}
(cos(a))^2 = 1/3 OR {a = pi + 2*pi*k , for any integer k}
Furthermore, second derivative must be negative in order to have a maximum:
d(Vc)/da = 2*pi*(r^3)*[1 - 3*(cos(a))^2] * (-sin(a)) =
= 2*pi*(r^3)*[ 1 - 3*[1 - (sin(a))^2] ] * (-sin(a))
= 2*pi*(r^3)*[-2 + 3*(sin(a))^2 ] * (-sin(a))
= 2*pi*(r^3)*[ 2*sin(a) - 3*(sin(a))^3 ]
dd(Vc)/da^2 = 2*pi*(r^3)*[ 2 - 9*(sin(a))^2] * d(sin(a))/d(a) =
= 2*pi*(r^3)*[ 2 - 9*(sin(a))^2] * cos(a)
If (cos(a))^2 = 1/3, then (sin(a))^2 = 1 - (cos(a))^2 = 1 - 1/3 = 2/3 , and hence
dd(Vc)/da^2 = 2*pi*(r^3)*[ 2 - 9*2/3] * cos(a) =
= 2*pi*(r^3)*[ 2 - 3*2] * cos(a)
= 2*pi*(r^3)*[-4] * cos(a)
which is negative when cos(a) is positive ==> cos(a) = sqrt(1/3)
If a = pi + 2*pi*k , then sin(a) = 0 and cos(a) = -1, and
dd(Vc)/da^2 = 2*pi*(r^3)*[ 2 - 9*0] * (-1) =
= 2*pi*(r^3)*[ 2 ]*(-1)
which is negative; but a = pi + 2*pi*k for integer k is not an acute angle and hence does not lead to a physically viable solution.
So the maximum inscribed cylinder volume is for cos(a) = sqrt(1/3), and this volume is
Vc,max = 2*pi*(r^3)*[cos(a) - (cos(a))^3] =
= 2*pi*(r^3)*[1 - (cos(a))^2]*cos(a)
= 2*pi*(r^3)*[1 - 1/3]*sqrt(1/3)
= 2*pi*(r^3)*[2/3]*sqrt(1/3)
= (r^3)*[4*pi/3]*sqrt(1/3)
The ratio between maximum inscribed cylinder volume and sphere volume is thus
Vc,max / Vs = [ (r^3)*[4*pi/3]*sqrt(1/3) ] / [ (4*pi/3)*r^3 ] =
= sqrt(1/3)
= 0.5773502692...
I hope that helps.
Good Video,
One Thing that still wasnt explained but was used as I suppose:
Ln ( x * y ) = Ln(x) + Ln(y)
Where x,y complex Numbers
Which is only true for
arg(x) + arg(y) < 2 * Pi.
You also forgot to define the argument of a complex number. If you want to define it properly, you can only do so on an open and simply connected subset of the complex plane (we often exclude the negative reel numbers), or else it will not be continuous; I feel like it's a counter intuitive fact people often do not know. This is also required to define properly the log, wheter you want to use the arg function or not.
kind of love on how the answer from the wrong proof's have the same answer as in the real and correct proof, its like: your right for all the wrong reasons
Thanks for the lesson. It was very understandable.
When, in "proof" 1, you substitute 1^(1/2) for 1, you must also say that 1^(1/2) has two results, namely 1 and - 1. So the power law holds for one of the values.
Exactly. It's a multivalued complex root/power.
All of the proofs you've shown are fixed by not just including the principal value.
For example:
1 = 1^½ = (e^(2πin))^½
ln(1) = ln(e^πin)
0 = πin
Which means n=0
Another
ln(e^z)
z = a+bi
ln(e^(a+bi))
a + ln(e^bi)
a + ln(e^i(b+2πn))
a + i(b+2πn)
1 = e^2πin
ln1 = 2πin
2πin = 0
So n=0
This can be extended into techniques for finding all roots of a number
Eg.
z³ = 8 => z = ³√8
³√(8) = ³√(8e^2πin) = 2e^(⅔πin)
For n={0,1,2} we get all values for which z³ = 8
how do you justify that 1=1^(1/2)? Can’t it be 1 or -1? If that is true does it break the reasoning for a^b^c not = a^bc
i^i= e^(-pi/2) for all values for ln(z), not just the principal.
There are always N Nth roots. So this does not work, (-2)^2 = 4 = 2^2. Therefore -2 = 2.
for proof 2, it’s also correct. The key point is that i=e^i(π/2+2kπ), then everything is correct. 6:43 is wrong on purpose i think.
Really amazing
Root(a) x Root(b) can be written as Root(a×b) if and only if both a,b > 0
(a^b)^c can be written as (a)^bc if and only if a is positive real number.
Every Mathematical operation that we know could eventually have a condition that follows
5:46 you just simply took the principle value of 1, but in fact for 1=e^(2niπ), n could be ANY integer but not just 1. If you raise 1 to 1/2 power, you should get 1 if n=0 and -1 if n=1. In another way to explain it , 1^1/2=±√ 1. You must figure out all values but not take the principle value and say it is wrong. For (a^b)^c=a^(bc), you may not get all values right so you must take all values
The step doesnt always work, as shown by that particular example, so you cant use it as proof. If you have cases where the step doesnt work and you cant pinpoint the reason, that makes the step unusable if you want to prove something.
@@buycraft911miner2 That's... true. For rings like the real numbers. In complex numbers is usually ok to put equal sign even if something like this happens. This is because most of complex functions are multi-valued.
@@buycraft911miner2 The step doesn’t always work, but at least one of the values is correct. In previous example, 1^1/2=±1, at least one of these is correct. So the step is still useful and intuitive, unlike the 16/64’s example.
@@KewoNg-to6zj yeah, I do agree that, although funny, the example isnt even close to the same case
Tried solving using euler's formula.
i^i = exp(ln(i^i))=exp(ilni), then recall exp(ix)=cos x + isinx therefore exp(ilni) = cos(ln(i))+ isin(ln(i))
i= exp(pi/2+2pi*k) but here it breaks and we get result that i^i is complex
As always I can relax watching your videos, because you explain strict mathematics in a very pedagogical way 😊.
Just take log
I wonder because when we write 1 as e^0i then proof 1 works but when we write 2π it doesn't, and 2π is actually 0 only?
you can make some really cool fractal images from complex functions
blackpenredpen gave the correct answer using a "right" method 6 years ago
6:20 but the square root of 1 (1^0.5) can also be -1 so the answer you get is correct (not the only solution but a correct one). Your example is like saying that you can't use a square root cause you'll get 1 or -1 which means 1=-1. Imo this isn't a satisfying example for proving the problem with using power rules on imaginary and complex numbers.
Yeah, I vaguely remember Complex Analysis. Used it in grad school a few times, once or twice since. Thank you for jogging the old memory. ;-)
The existence of e^ipi = -1 cancels the entire rant about logarithms lol
Maths really is fascinating.
1=X^2 simplfies to X=1^(1/2). Which then simplifies to X can be either 1 or -1.
Ok so let's take x a real number, we have 1^x = e^[xln(1)] = e^[x(ln|1| + 2in\pi)] = e^[2inx\pi] which can be egual to any number on the complexe unit circle depending on the value taken for x. Well done for pointing out the issue but I don't get how your way of doing solves it.
So is ln(i^i) = i * ln(i) also incorrect? Every video I’ve seen on this used that step.
he used it himself without justifying where he 'defined' taking power of complex
@@Yougottacryforthis Yeah, it feels like he's just taking an opportunity to show places where complex numbers have distinct algebraic properties, but then glosses over places where the rules are similar.
I remember solving this in a signal processing course!
Great take-down of people who act as if they know what they're doing in a math video - when they don't - but I'm still a little confused on multiplying i i times.
Boy am i glad i studied economics instead of math.
Just wow!!! 👍
12:25
0
basically the first two proofs are true for any field of characteristic two provided that we can define the exponential and logarithmic functions (which we can, via the discrete logarithm) in a way that preserves all the properties the exponential and the logarithm have for real numbers
"Everybody would object using this method".
Sir, you are underestimating me xD.
The correct way is thankfully the only way I ever learned to do complex exponentiation.
What I really want to know is how you came up with those faulty fraction cancellation examples!!!
"i" like your videos.
Unfortunately, you missed a nice opportunity to discuss why one needs branch cuts and how one makes functions single valued on a subset of the complex plane. The function you describe for the logarithm is only valid on the Riemann surface of the logarithm, which also yields single valued results, but an infinite number of them given the structure of the Riemann surface. I would say it is incorrect to say we restrict to a principle value unless the branch cut has been discussed. Many branch cuts are possible, which can yield different answers. Understanding how you restrict the complex plane to make the results into continuous and even analytic functions is the important result when raising a complex number to a complex power.
There is nothing wrong about -1 being the square root of 1. We did not say that 1 is the only answer to being the square root of 1. Therefore, it is wrong to say 1 = -1, since 1 could also be exp(4πi), in which case it's 1/2 power is indeed 1.
For the second ”proof“, since you have taken logs of both sides, we are essentially looking at the arguments, in which case, 0 is indeed the same as 2π. Cos you can just keep going and say 2iπ = 4iπ = 6iπ etc etc.
"There is nothing wrong about -1 being the square root of 1" yes there is. Because it isn't.
The thumbnail is wrong. It is written π/2 rather than -π/2 😅.
Yep, I could not believe he actually got the thumbnail wrong. I looked at it twice and then a third time, figuring, how could he get that wrong?
You're right but in this rigorous context, it bothers me that you're using sqrt(-1) as an alternative definition of i, as i could also be defined as -sqrt(-1)