Breaking the rules of math π = 0. Where is the mistake?
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- เผยแพร่เมื่อ 28 มิ.ย. 2023
- Very few people will be able to identify the wrong step. We start with Euler's identity and derive several equations until we get to π = 0. Complex exponentiation is tricky.
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No mistake, Circles are zeros
So are donuts...
I love this
tow = 0, pi = half 0
@@skytech2501So pi = ∪?
@@smylesgy e s
I always found it fascinating that anytime you have an exponent with i in it, things get 'weird' really fast. Beginners often try to factor things and move 'i' around just like any other variable. But of course you've just shown that isn't always the case. Wonderful explanation of how 'i' is NOT always 'just another variable'. Thanks for this.
For imaginary numbers, they're hard to imagine.
I is not a variable. In fact, it isn't really a constant too.
i* smh auto caps
@@jonjuathanjoathan2426How is it not a constant?
i is an ordinary number, really. The same mistake can be made using ordinary integers, you can say 2² = (-2)² therefore 2 = -2 - that's not valid! The same with exponents of complex numbers.
Most of the confusion in the algebra of complex numbers comes down to the branches of the logarithm. If one is simply consistent with a branch cut of the natural logarithm (such as the principle branch), everything works out without discrepancy. Here, I'll make one critique of the video. In the video, it was expressed that ln(exp(2(pi)i+1))=1. This, however, is only true given that it was specified that ln(z) is the principle value of the natural logarithm (i.e. Log(z)). Without context, one could've easily and 'correctly' written that ln(exp(2(pi)+1))=2(pi)i+1, which was indirectly done initially. In general, the natural logarithm is not a function and can be expressed as ln(z)=Log(z)+2(pi)ki, where "k" is some arbitrary integer. So if anything, this video only adds to the confusion surrounding this number system, further mystifying it. Admittedly, I doubt this comment will clear anyone of confusion, but then again, I'm not a math educator.
I was thinking it was because going from polar back to rectangular we don't have an angle representing what domain we came from in powers of n (this case power of 2). So 0 to 2π range by squaring moved to angles now in 0 to 4π. So 2π and 4π both now represent e^i(2π). His correction makes sense but e^i∆ = cos∆ +isin∆ real and imaginary conversion of i∆ now has to consider e^u = cos(iu) + (or minus I forgot) sin(iu) that now is not an angle but an imaginary angle. I think this is the interpretation of his correction meaning. 🤔 All my courses never came up with a thinking of wrong thinking contradictions in complex numbers analysis.
Based on this comment I think he should have said in the video:
"π may or may not be equal to 0."
Yeah, I found it pretty disappointing that the video was just like “you can’t do this” without giving any real justification for why you can’t do it
I am bad at math but at some point its written that e^2ip+1 = e.. so 2ip+1 = 1 so 2ip = 0 so i = 0 ??? Why is that not commented ?? Anyone knows where is my mistake?
@@vsm1456 You can't remove the e because of the fact that it's using a generalized version of exponentiation that works with imaginary numbers and uses trigonometry. (cos(x)+isin(x)). In this form you can easily see that taking logarithms of stuff like this just doesn't work. This is also why e^(x*i*pi) is a circular function.
You could have just gone from e^{2pi i}=e^0, “so” 2pi i=0 so pi=0. This makes the same fundamental mistake.
no it makes a completely different fundamental mistake
Not exactly, e^2pi i does equal e^0 without 2pi i equalling zero
@@battlesheep2552 Yes, I know that. The fundamental error is in assuming complex exponentiation is unambiguously defined, which is an equivalent error to thinking complex exponentiation is injective
The exponent rule fails for the same reason complex exponentiation isn’t injective; arguments are not unique
@@battlesheep2552 You wrote this wrong. You need grouping symbols: e^(2pi*i).
A simpler one. At 1:00, we have e^(2i x pi) = 1. Now taking natural log on both sides gives us 2i x pi = 0, which means pi = 0.
It’s an complexe exponentiel u can’t
@@jonasvuillemin9412well, it's *supposed* to be wrong.
@@jonasvuillemin9412
We can do it without taking natural log.
e^(2i x pi) = 1
e^(2i x pi) = e^0
If they have to be equal, the exponents have to be equal. Which means, 2i x pi = 0, which means pi = 0.
FYI. Those who are still confused. For a function that is cyclic, there will be many points on the domain which has the same value/image. Equating two different points which has same value and then saying those points are equal is the issue.
f(x) = f(y) does not necessarily mean x = y.
We can understand it better with an example.
cos(2 x pi) = cos(0)
2 x pi = 0 (wrong here!)
Hence, pi = 0. :)
@@therealsachin well TBF that *_still is_* "taking natural log." just no longer explicitly so. 🙃
And to the OP:: yeah, that's exactly what i expected to happen at that point. Turns out, the trick was more subtle with complex exponentiation than it would've been with just the log.
This whole thing boils down to the fact that some complex functions (like the power function) are not actually functions unless you include branch cuts. This creates a lot of need for careful watching as you go. For example, when you took the natural log of both sides at the end of your demonstration, you really should have included the coterminal answers also. It's still already incorrect, but that's besides the point.
I've always loved this channel, for years. I always watch the vids, but I also pause the vid early, and solve the problem for myself, then restart the video and get confirmation of my solution. This is my new favorite video, because I completely fell into the trap. I was plain WRONG and needed the video to correct me; and I really like that.
Nice one!
I would be careful in defining z^w as e^(w ln z), which is not quite accurate, as ln z is not well-defined for non real values of z. In other words, ln z is not a function from C to C. It is a multi-valued function. For ln z to be well-defined you need to use an extended notion of the complex numbers called a Riemann surface.
Defining ln z as a function would cause all sorts of unintended consequences. For example it cannot be continuous. To see that note that ln 1=0. If you approach 1 from the fourth quadrant of the complex plane the natural log approaches 2 i pi, which is different from zero.
Agreed. I've found that on TH-cam, there is a lot of confusion regarding complex exponentiation and that's because many math content creators fail to properly communicate that not all functions are invertible. Ideas such as branch cuts and principal values need more attention. Hell! Most people don't even fully understand that quadratic equations only have two solutions because quadratic functions cannot be inverted by a function. Forget about periodic functions! This problem is far more deep-rooted. I'd go as far as to say that the school system is primarily to blame.
Yes, it's unfortunate Presh didn't mention the fundamental point here. The reason you get into difficulties is because logarithms of complex numbers are multi-valued. It isn't that log(e^(2πi+1)) doesn't equal 2πi+1. It does. It just doesn't *only* equal that. It also equals 1. And it turns out 1 is the value we need in this scenario. It is like square roots - there are two square roots and if you pick the wrong one you can easily get to contradictions.
To clarify further, the “correct” derivation at 7:10 makes the same mistake as the original derivation, and you could “prove” that pi = 0 by repeating the same steps, since the multiplied exponent is one possible value for the iterated exponent. The real problem is using “ln e = 1” as a step in proving equality, since that is only one possible value of the log at e. We can go all the way back to 1:25 to see that the real error lies in obfuscating the taking of the log on both sides.
Stepping in to ask question since you guys explained it better than the video. If e^(2iPi) = 1 = e^0, as seen at 0:51, why can’t we set the exponents of the same base equal to each other to get 2iPi = 0? I realize this can’t be true since it leads to contradiction. But I haven’t taken any higher math classes on complex numbers to know what the issue with this logic is. Is this an illegal step? Or is the problem one of the steps before it, like squaring both sides, which I know is notorious for creating dubious roots, for instance when x = -3, squaring both sides allows x = 3 and x = -3. But this Pi thing feels different. Help?!
@@aaronliss8713 you got it right in the last couple sentences.
The reason is the same as the reason for why we can’t go from 3^2=(-3)^2 to 3=-3. In math language, exponential function e^x on C is not one-to-one.
I think the origin of all trouble lies in the simple fact that the polar representation of a number is not unique. Two numbers in different sheets may be the same but you cannot equate their nat logs (like you would instead equate two real exponents of the same real base) because they in fact differ by a multiple of 2pi.
No! He misleads us in his instruction teachings. He forgot in the real number plane the same error happens. We can't use the unity bases of {0, 1} in the real number plane also for envoking y^m = y^n base powers equality implies m = n. He danced around the obvious also in real numbers it fails in base 1 (or 0 to any power let alone itself).
From the first few seconds of the video, I know where this is going and the answer is: the exponential function is not injective over C, so you can't cancel it.
exp(x) = exp(y) implies x-y is an integer multiple of pi, so not necessarily equal.
I don't care what you have to say. I still believe that Pi equals 0...
An integer multiple of iπ, but I get your point.
@@ShaunakDesaiPiano An even integer multiple of i pi, but I get your point.
@@RGP_Maths oops yes I stand corrected! 😂
@@ShaunakDesaiPiano Isn't it an integer multiple of 2*pi*i?
Presh did not explain *why* this careful handling of complex exponentiation is necessary. Complex exponentiation is basically trigonometric: e^iϑ = cos ϑ + i sin ϑ. Therefore we have e^iϑ = e^(iϑ + 2ikπ). This means that the inverse function log z is multivalued.
The faulty proof really boils down to:
e^2ikπ = e^0 ==> 2ikπ = 0 ==> π = 0.
Also, it's best to simplify to avoid these annoying mistakes
Reminds me of a favorite Tom Swifty: "Taking the natural log of -1 is as easy as pi," Euler imagined.
Around 7:00 it is also important to point out that ln(e^z)≠z in the general case for complex z. The logarithm of a complex number in the general sense has an infinite number of values (just as the inverse tangent function has for real arguments), but like arctan, we define it to produce its principal value whose imaginary part is in the range (-π π] (or perhaps [0 2π), I can't recall) by removing multiples of 2π. Either way ln(e^(2iπ+1)) is 1, not 2iπ+1.
I was wondering about this too, but he does actually point it out at the end of the video.
I agree.. Mind masturbation with i, not maths
Thank you! Even though he did point this out at the end of the video, I agree that he should have addressed it earlier.
@@vivada2667 ... He still pointed out a wrong hypothesis. It is also a problem with real number planes also that he doesn't mention. Anytime the polar complex angle of the unity exponent is manipulated is just as the case in real numbers. I watched this video again and again getting so confused why e^(-4π^2) was not balancing out. It is because in unity calculations exponent equality fails. In fact, in real numbers, any base not equal to {0, 1} has the powers relationship y^m = y^n implies m = n with the powers so that we can solve those mathematics problems. For example finding x in 3^x = 3^(x^2 - 1) and other powers equality. We can't do this equality for unity 1 nor 0 in the real numbers also.
You're exactly right.
Math educators on TH-cam should really begin to stress this particular nuance of principal values of complex multivariable functions, because I see people make incorrect statements all the time.
It should be fairly easy to teach. Start with the familiar fact that the square root is multivariable, but has a defined *principal value* denoted by the symbol "√", then extend that definition for the nth root by introducing the principal branch (-pi, pi], and finally properly define the principal natural log, among other things.
Only then should you start to teach how to manipulate and solve complex equations like:
e^x = -1
=>
e^x = e^(Log(-1) + i 2pi n)
=>
x = Log(-1) + i 2pi n
where "Log(-1) = i pi" is the principal natural log of -1, and the last term encapsulates all the other natural logs of -1.
Now we can use the above to get all the solutions of:
x⁴ = -1
=>
x⁴ = e^(Log(-1) + i 2pi n)
=>
x⁴ = -1 * e^(i 2pi n)
=>
x = ⁴√(-1) * e^(i 2pi n/4)
where "⁴√(-1) = e^(i pi/4)" is the principal fourth root of -1, and the other factor encapsulates all the other fourth roots of -1.
1:07 in the last equation at this time stamp
Exp raised to some power yields same exp means the power must be 1
So 2ipi+1=1
So either pi = 0 or i = 0
I learned something about complex exponents -- nice way to motivate that.
Good video, I thought it had made a big mistake by having an additional fallacy of using two different branches of the natural log at 3:38, but I just realized that was just the real natural log function and that the only fallacy was from the power rule misuse, very clever setup.
You have got e=e^(2ipi+1), why don't you let 1=2ipi+1 directly? Then you can derive pi=0 immediately, although the result is ridiculous.
I have no idea what you are doing in the next steps and make it so complicated...
He probably thought doing it in a convoluted way is supposed to make it harder to spot the mistake
e^x = e^y does not mean x=y for complex numbers
@@dasla9492 of course not so that the result is ridiculous. Since he just wants to prove pi = 0 in a unreasonable way, this can be much faster.
Incredibly nice and simple, as always.
Thanks !
This was a great video and introduction to the power rule of complex numbers for me, but I got confused with the formula. Does the e in (e^x)^y = e^(y ln(e^x)) refer to the base, like z, or always Euler's number? In the future, it might be better to use a less ambiguous variable, or to specify which one e is
The mistake is @1:40 e^((2ipi+1)*(2ipi+1)) does not equal e. it equals e^((1 + 2 i π)^2). the rules of exponents are slightly different when dealing with e.
It was a very important point that you mentioned it well
The truth is, I had not paid attention to this point until now.
Thanks dear Prof.❤
I found the wrong step, but I was quite confused at first. The steps seem legit. But yeah, (e^x)^y isn't always e^xy, basically because ln(z) is not a continuous function for all z. In particular, there's a big discontinuity when z is a positive real number.
If you interpret ln(z) as a Riemann surface, or something like that, or perhaps an infinite pile of values (with multiples of 2*pi*i), you might be able to salvage the continuity and keep some of the rules.
I managed to figure this one out by seeing 2i(pi) as 0, and then when you started multiplying the complex numbers, I knew something was up.
Step 3: Squaring an equation can change the solution set of the equation.
W
Finally, thank you! Yours should be pinned. If anyone doesn't get it: think of pi as a variable in the initial equation, and we're solving for its value. Once we square, we've introduced the potential for extraneous solutions. (It's not clear to me that we lose the correct solution at this step, but it's moot because having taken this step, we have to check our solutions against the original equation regardless.)
@@sethkingman2118 this is not the problem
Except it's not an "equation" and therefore there's no "solution set" for it...
@@irrelevant_noob Wikipedia says "The word equation and its cognates in other languages may have subtly different meanings; for example, in French an équation is defined as containing one or more variables, while in English, any well-formed formula consisting of two expressions related with an equals sign is an equation."
Also think the mistake is different from the one that was shown. When we write "e^z" we usually mean some exponential function defined as either:
1) exp(z) = e^x * (cos(y) + i*sin(y))
2) exp(z) = 1 + z + z^2 / 2 + z^3/6 + ...
Of course, we can show that those definitions are equal.
You can notice that if we actually use proper notation, we cannot make the same mistake:
exp(2i*pi + 1) = e -- easy to see from 1st definition that it's true
However how we cannot make a statement similar to [e ^ (2i * pi)] ^ (2i * pi) = e.
On contrast to definition of the exponentioal function, If we want to raise a to the power of b in Complex numbers we shall use definition
a^b = exp(b * Ln(a)) , where Ln(a) = ln|a| + i*arg(a) + 2i*pi*k, k any from Z.
By that definition e^(2i*pi + 1) = exp((2i*pi + 1) * (1 + 2i*pi*k)) = exp(2i*pi - 4*pi*pi*k + 1 + 2i*pi*k) = / by def (1) / = exp(2i*pi - 4*pi*pi*k + 1).
Therfore, e^(2i*pi + 1) has actually infinitelly many values one of which is indeed equal to exp(2i*pi + 1) = e.
On the complex plane, every number x+iy (except zero) can be written as r*exp(i*t) where r>0 and t is a real number. It also happens that for every integer n we have x+iy = r*exp(i*t + 2*pi*n)
In some sense, it's as if the exponential has infinitely many inverses at each point, one for every integer n. There isn't a unique complex logarithm that can be defined everywhere.
However, the "main", "standard" or "usual" complex logarithm is defined as:
log(x+iy) = log(r*exp(i*t)) = log(r) + i*t
where r>0, log(r) is the real logarithm of r and t is in the interval (-pi, pi)
Now, after choosing a logarithm, you can define the complex exponentiation z^w by the rule:
z^w = exp(w*log(z*exp(2*pi*n))) where n is an integer such that z*exp(2*pi*n) falls inside the domain of your chosen logarithm.
And then after rewatching his erroneous talk about this being a complex only plane problem. It is total nonsense. Anyone equating exponent results in the real number plane also don't find equality in the exponents implied with {0, 1} because the y^m = y^n means the exponents m = n equation is true only if the base is not 1 (nor 0 let alone the undefined 0 with itself).
5:15 I felt a wave of power flush over me. As he said where I thought something went wrong, actually was the part when it went wrong.
The mistake was in the assumption that (a^b)^c always equals a^(bc). This statement holds true for real numbers, but in the case of complex numbers, the actual formula is a little messier:
(a^b)^c = a^(bc) * exp(τci*round(-Im(bln(a))/τ))
Where Im denotes the imaginary part, and round denotes the function which rounds numbers to the nearest integer (rounding up if it's a tie). I'm also using τ to denote 2π, not out of preference but to avoid using a confusing number of parentheses.
When c is an integer, this is just the same as a^(bc). That's also the case when a and b are real (unless a is negative or zero).
If we use logarithm defined for complex values ln(e) = 1 + 2πni where n is integer. In definition z^w = exp(w*ln(z)) the logarithm is a multivalue function. You can write as well z^w = exp(w *(ln(z) + 2πni))
Just learned for the first time that the power rule for complex numbers isn't the same as that for real numbers. Thanks!!!
Can we not come up with the contradiction much earlier in the process? Like:
e^(2*i*pi)=1=e^0. Therefore, 2*i*pi=0. Since 2i is not 0, therefore pi=0. The underlying lesson still holds, tho.
In fact, at 1:33 the equation is already wrong, wolfram alpha also likes to make mistakes, but if you write a query (e+e-e)^(2*pi*i+1) it will find other values besides e, same for ((e+e-e)^(2*pi*i+1))^(2*pi*i+1). The bottom line is that there is a difference between raising the number e to some power and using the exponential function. The first equation is true if the exponential function is on the left, and the number e is on the right. Thus, replacing one with another in the third line makes no sense and, of course, leads to the mistake.
The mistake is in 0:40
As you can see, squaring both side which have different sign (which e^iπ is positive, and -1 is negative) will give a contradiction proof. In other word, imagine when there is 1 = -1 and then you square them up, you will got 1 = 1. Now this is fishy that it is no way that 1 = -1 can be true. You may square both side if :
- 1 from 2 equations has square root or (1/2n), n for all integer.
- both of them have square root or (1/2n), n for all integer.
- 1 from 2 equations, or both of them have absolute equation.
(Example for absolute equation :
|n + 1| = n and |n + 1| = |n|)
In an Equation one has to do the same operation on both sides. In minute 1.31 in the video effectively only the left side was raised to the power of (2i phi +1), the right side is e and has to be raised with the same power but stayes the same. This is the mistake. Presh inserted in a recursive way e only on the left side! This is not allowed in manipulating equations.
Euler’s Lemma: Euler’s identity
Euler’s Theorem: π = 0
Euler’s Corollary: Circles don’t exist
in real life probably not. circles are too perfect to exist in reality
@@rohangeorge712 Circles don’t exist in real life almost surely. The probability of drawing a perfect circle is 0.
I knew it was going to be when logarithms got involved! It's kind of interesting that we can use the periodicity of complex exponentiation to wind up with very strange situations w.r.t. inverse functions. For example, we can "kind of" define a principal solution to sqrt(x)= -1 with x=e^(2*pi*i), as _distinct_ from one, which is pretty weird if you ask me.
I'm at 2:09 and it looks like the problem is that pi is cylclical and you're about to make the mistaken assumption that 1 is the only value for which a^x = a whereas this isn't actually true with imaginary numbers because they function almost like degrees, repeating every two i pi
4:08 your error is that you assumed one constant was equal to zero, and not the other… Everyone knows the value of pi therefore the value of -4 must have been zero
1:27 i already get that result without the next step. Since you have a common base on both sides, simply equate the 2 exponents:
2i* pi +1 = 1
2i*pi = 0
i*pi=0
pi = 0
Same
no that proves either i or pi is = to 0. what if u divided both equations by pi instead? u would get i = 0. so there u could be "dividing by 0" if u will. its kind of goofy.
@@rohangeorge712 but pi isn't 0 so it isnt division of 0. pi=0 and i=0 are extraneous results, they're not true.
@@kobalt4083 yea ik im just saying why that wouldnt work. the proofs are wrong anyways so it doesnt really matter. cause yea ur right pi and i arent 0
You can't equate the complex exponents with same base like e^0 = e^iπ doesn't impliy 0 = π that argument is wrong.
When we proved (a^b)^c = a^(bc), we probably used natural numbers. If the rule was proven for natural numbers, that doesn't mean it works for the integers, or reals, or complex numbers. It may or may not.
But it does always work for integers.
For real rational numbers, as long as the fraction we get from doing bc doesn’t simplify we are good
That formula works for all complex numbers
@@benjaminojeda8094 The point of the video was that this rule does not work for all complex numbers.
The fundamental problem is exp(2ipi) =exp(0) so when you raise 0^0 you have an undefined result.
The fundamental problem is we as a species long for both equality and wealth, which are mutually exclusive ideals. But I agree with the rest.
@Qermaq Wealth is not incompatible with equality, limitless overconsumption and abhorrent greed is.
@@timmydirtyrat6015 Those two are a pickle, I agree. But the desire for wealth is by definition the desire for more than others can have, which is the antithesis of equality.
@@Qermaq Desire for wealth =/= desire for more than others can have. That would be more true if you specified desire for relative wealth. And even then it's not contradictory for everyone to be equally wealthy. I don't think your statement is defensible.
@@Qermaq what a low level materialistic thinking
If that reasoning were true, then 1 = anything you want:
1^r = (e^(i2*pi))^r = e^(i2*pi*r) = cos(2*pi*r) + i*sin(2*pi*r)... (where 'r' is a complex number),
which is absurd.
Thanks for the video.
I know my "proof" is also false, but I would show the following:
At 1:10 we have e^(2i+1) = e. | ln
2i+1 = 1 (bcs e = e^1) | -1
2i = 0
And as neither 2 nor i equals 0 it must be pi = 0.
e^3ipi = -1 = e^ipi take the ln(), and ignore that ln(-1) is undefined
3ipi = ipi; next divide by ipi
3 = 1; next subtract 1
2 = 0; next multiply by pi/2
Pi = 0 QED
Very interesting math here. Is this basically trying to apply a rule of exponents to sines and cosines --or phasors?
this is really neat! however, there's one thing i think you glossed over a little too quickly: when taking the natural log of a complex number, what's especially important is to pay attention to the imaginary part and make sure it's within the right range (i think it's -π, π? i can never remember LOL). otherwise at around 6:55 you show ln (e^[2iπ+1]) which, without this important explanation, might mislead someone into just saying "well, ln is the reverse of the exponential function, so they cancel out" and get back to the initial result.
This video reminds me of another fallacy (the explanation of which I will leave as an exercise for the reader!) June 28 was Tau Day (6.28), analogous to Pi Day (3.14), but celebrated by those who think that the more natural "circle constant" would be the ratio of a circle's diameter to its RADIUS (rather than its diameter). They call this new circle constant TAU and it is mathematically equal to 2π.
There are arguments pro and con regarding tau , but the fun begins when you apply Euler's formula to both of these circle constants. It its more familiar π form we have e^(π*i) = -1. If we take natural logs of both sides we have π*i = ln(-1). This is actually a very useful relationship and can actually serve somewhat as a DEFINITION of logs of negative numbers. For example we could say that since -2 is equivalent to (-1)(2), the log of -2 could be considered as being log 2 + πi. Multiplying using logs (which is what we had to do back when I was in high school, as calculators had not yet been invented!) would allow us to substitute addition for multiplication. A problem like X = 2*3 would become log X = log 2 + log 3. If we were to use the above definition of logs of negative numbers, X = (-2)*(-3) would be solved as follows: log X = log (-2) + log (-3) = log 2 + πi + log 3 + πi = (log 2 + log 3 ) + 2πi = (log 6) + 2πi. Taking "antilogs", this would become 6 +0i (since the e^ix function is cyclical) and give us the correct answer of 6.
But what about the equivalent tau formula? It can be written as e^(tau*i) = +1. Taking natural logs of both sides we have tau*i = ln 1 = 0. But if a product of numbers = 0, then at least one of the numbers must also = 0, which implies that either tau = 0, or i = 0!
i'm very confused but ok...
huh didn't know tau = 2
tau = 2π @@commoncoolchannel8588
0:40: Squaring is no equivalence transformation. Which means that you are generating new pseudo-solutions at this point.
I would assume that problems can arise if we have a variable as in exponent equation and then, on evaluation, find that the variable is complex. Presumably you have to feed that back in and find out if the steps were valid for complex exponents. Maybe not something you are ever likely to come across, but I do wonder what other general rules we were taught at school that fall apart where complex numbers are involved.
nb. I confess I used Wolfram to find out what step was invalid, but I had no idea why it wasn't.
What's really intersting here is that the mistakes comes from a property of the log we're supposed to be able to use thanks to an other property of it, which is that on the set of real numbers, the log • exp (x) = x, and that's possible for exp and log are both bijections from one set to an other.
However, this doesn't work on the complex field for exp(i2n×pi) = 1 for any given integer n, which shows exp is no longer a bijection and as such cannot verify the first property.
7:20 doesn't the natural logarithm become an ambiguous multiple valued function in complex plane? So this might cause some problems, I think
Edit: maybe it works out if you define ln(z) for complex z as the principal value of complex natural logarithm?
In this context, it doesn't quite matter, since it's in the exponent and the +2kπi turns out to not have an effect on the result, doesn't it?
@@irrelevant_noob It does matter especially if the original exponent has a non-zero imaginary part, such as here, I think. And non-integer real exponent might truly change the result if multiplied by 2kpi
For me this is indicative of a defect of our teaching of rules for exponentiation and the various definitions of the exponentiation operation itself. We teach the multiplicative rule (a^b)^c=a^(bc) (as well as other rules for exponents) in a context where b and c are (usually positive) integers, and then later we expand the definition of exponentiation to allow complex exponents. Never do we stop and consider whether the usual rules for exponents remain valid when we do so; it turns out that (sticking to positive real bases as the expanded definition requires, although this too is often not said explicitly) the additive rule a^b*a^c=a^(b+c) remains valid in general, but the mentioned multiplicative rule is simply no longer valid. To save it one needs to add the condition that the inner exponent b is _real_; however I have never seen a text that states this restriction explicitly. We are simply supposed to understand how complex exponentiation works, and refrain from using the multiplicative rule where it is no longer valid because it is obviously wrong (as the video shows). This oversight is witnessed by the numerous proud presentations of a calculation of i^i, despite the facts that (1) it falls outside the domain of the expanded definition of exponentiation (not a real base) and (2) the computation invariably invokes the multiplicative rule in a case where it is not valid.
2pi = 0 in polar coordinates. But then division is not so simple. You can't simply split tau, otherwise -(2pi)^2 = -0*0 is correct. To divide by 2 correctly, we should think on the whole set of points that represent the same angle [0, 2pi)+2kpi, k integer. 0 is the entire set {2kpi}, divide by 2 => {kpi}. As many others have stated, in the complex plane, we are working with a multivalued function.
We could take the logarithm of both sides on 0:52 as another path.
Another problem is that e^{x}=1 doesn’t imply x=0 over the complex numbers. It implies that x=2ipi n as n ranges over the integers. The logarithm “function” over the complex numbers isn’t a function (it’s a so called multivalued function). It is defined as Log(x)=ln|x|+i(arg(x)+2pi n) where ln here is just the normal logsrithm over the reals, and Log is just ro highlight the fact we are taking Log over the complex numbers.
The way of looking at it is that the function sending a complex number x to e^x isn’t bijective.
It's the same as proving that 1 = -1. Just square the both sides and see the results!
(x^y)^n=x^(ny) is actually valid when n is an integer, including in our case of (e^(iπ))^2=e^(2iπ). This is because x^n is not sensitive to the branch cut. Indeed, you can just multiply repeatedly without ever taking a logarithm. To the extent that a branch cut still applies to (x^y)^n=x^(ny), it’s determined solely by x which is equal on both sides. This result is much more general than one little coincidence.
In 7:40 it seems like we got the same result but with extra steps: (e^(2ipi+1))^(2ipi+1)=e. So how is it wrong in the first variant and true in the second?
I just followed each step after starting with e^ipi(1+2n) = -1 where n is an integer and came to the conclusion that this only works if n = -1/2 which cant happen
Videos like these confirm that math is not certainly not one of my strengths.
beta male fixed mindset
Neither are double negatives...
i^i = e^(-π/2). Here we convert i=e^(iπ/2). Then we have to multiply the two powers
It strikes me that near the 1 minute mark you already have e^2ipi = 1, which already naively implies pi = 0
3:16 less confusing way of doing this step is saying e^(4iπ) = e^(iπ) * e^(iπ) * e^(iπ) * e^(iπ) = (-1) * (-1) * (-1) * (-1) = 1
Mistake actually begins from where you started to square both sides.
Nah, everything was still good up to that point, and for quite a while after that too.
Is there a mathematical formula to find a number with the most factors? In the case of a math problem there would be constraints, like "what number has the most factors between 1 and 100?" Is there a way to solve that question with something like a formula without taking every number and counting the factors? (The answer in that question is 60, with 12 factors, but I had to check every number to make sure, there has to be a more streamlined way.) The reason I have this question is because my family has an obsession with prime numbers, so I want to be able to find the "least prime" number.
Your story is interesting
@@goldensnitch5492 they full on microwave in only prime numbers, set alarms in prime numbers, etc. its a contest for them, seeing who can do it for the longest
I don't know of a formula, but this wikipedia page should be up your alley: en.wikipedia.org/wiki/Highly_composite_number
This question looks not obvious in general. But you should note that the number of divisors is given by the produit of the k_p+1, where k_p is the exponent of p in the prime decomposition of n
You can construct what you want for small values. But then it seems difficult
restaurant bill: 53 $
meanwhile presh : let me show you how 53$ equals to zero
I Like your funny words magic man.
but really this video has renewed an interest in mathematics in me. great content.
At 7:00
The usual log rule says u can take out the exponent and write it outside the log. Ln x^y = y Ln x
U saying "we know that is e" just means ur undoing the thing that is supposed to prove it wrong, basically saying we know its e so its e.
That said. I cant really find where the contradiction starts. Except if taking out the exponent isnt something u can do in these situations. Given the calculator said this is where the value changes.
(e^(2·i·pi+1))^(2·i·pi+1) to e^(-4·pi^2+4·i·pi+1) changes the log-branch. The correct general form would be e^(-4·pi^2·n+2·i·pi·n+2·i·pi+1) which for n=0 reduces to e.
Actually in general, ln(exp(2iπ+1)) can equal 2iπ+1. It can equal 2kiπ+1 for any integer k. What this "paradox" showed is that in this particular problem, k=0 was the only valid branch
I wouldn't exactly call this a contradiction so much as a misalignment of which branch you use. The rigorous definition guarantees you use the principal branch and is like the "actual" answer, but you could probably find an infinite number of ghost solutions that are less rigorous but satisfy the equations.
The neat thing is that z^w = e^(w*ln(z)) works for real numbers too. It just so happens that raising a number to a multiple of a logarithm with its base will always give you the exact same result as exponentiating the logged number by the coefficient if all numbers involved are real. The rule is basically just a simplified form of the full equation.
1,000³ = 10^(3log10(1000)) = 10^(3 * 3) = 10⁹ = 1,000,000,000
The same thing happens with e and the natural log too, it's just a lot less intuitive to read
Other perplexing results that i found in my Complex Analysis course were:
Cos and sin are not bounded.
ln is not holomorph on the whole plabe
e^1/z = a has an infinity of solutions (i think)
You can solve real integrals using complex analysis
Every time I see someone squaring both sides in a proof I immediately get concerned
For good reason
I think the following question leads to similar issues:
What is i^(4i) ?
Firstly, we develop this general formula and result: x^i = e^(i.lnx) = cos(lnx) + i. sin(lnx). Thus 1^i = 1.
So is it = (i^i)^4 = [e^(-π/2)]^4 = 0.2078… ^4 = 0.016… ?
OR
(i^4)^i = 1^i = 1 ???
OR is it both?
Simple rule whenever you find something not possible that means some of the real number rule not applied you can easily find them by analysing your process of doing.
1:07 Should we not be able to, by means of logarithms, already state that 2iπ + 1 = 1 2iπ = 0, and then either i or π should be zero? How do we explain and correct that?
The problem is more related to that the logarithm is not an analytic function over the entire complex plane but it has a branch cut between zero and infinity. When you evaluate (a^b)^c = e^(c*ln(a^b)) you have to consider what branch you evaluate the logarithm on.
In modular arithmetic, where the characteristic modulus is divisible by 2, pi can be defined to be a nontrivial solution to 0/2. So, one other 'problem' is trying to say that 2pi=0 implies pi=0
It was so easy to find the mistake. It was there all along. All I had to do was look in the mirror.
Whenever you have e (or any other real number) raised to some complex exponent z of the form z = a+ib, where a and b are real, a will dictate the modulus (size) if the complex number, while b will dictate the argument (rotation in the complex plane).
Because any angle of θ and θ+2π, are equivalent, this can give results where the exponentioation of e to a complex number z=a+ib is equal to the exponentiation of e to a different complex number x=a+i(b+2πn), and so it looks like the complex numbers are equal when they are not. This in my opinion is the core issue of the proof, or at least the root thereof.
I think there is an earlier mistake at 1:27
We can't substitute an equation into itself we will end up with both sides of the equation to be equals
Like if we have the following equation
y=x+2 equ1
Then x=y-2 equ2
Then we substitute equ2 in equ1
y=y-2+2
---> y=y
1:11 From here, we could also do this:
e^2ipi=e^1
So 2ipi=1
i×pi=1/2
i×pi=0.5
so he gives a mathematical understanding of the issue, but i wanted to give a more intuitive one:
complex exponentiation for powers of e can be understood as rotation around a circle in the complex plane. for e^(a+bi), e^a gives the magnitude of the output vector, or the radius of the circle, and b is how many radians you move around the circle. this is why e^(1+2pi(i)) is e, because e^a (e^1 here) is e, and b radians (2pi) just takes you back to the start of the circle.
the top equation at 5:30 is true because its just a repeating process of moving 2pi radians around a circle of radius e in the complex plane, but the bottom equation is false because, if you multiply the exponents in the way described, you effectively are mixing together the a and b coefficients. since these coefficients are what determine the radius and distance around the circle you travel, mixing them together ruins the almost iterative way that higher exponents act like they did in the first equation.
what tripped me up for a second as someone thats worked a good bit with complex powers is that i spot checked the first time he used the flawed rule he discusses at 8:00 in my head and it worked out, so it took me a few minutes to go back and consider the second time he used it. i was able to figure out the whole w ln(z) thing a few years ago on my own, but i guess since i didnt have a teacher that showed me i never consciously recognized how the regular exponent rule doesnt quite work until now.
Similar to 'proofs' turning on ignoring the ± square root: every number has an infinite number of logs. That x^n == x^(n+2πi) does not mean that 2πi is zero.
8:41
ln(e^(2ipi+1))=1
ln(e)=1
Ergo e^(2ipi+1)=e
e^1=e
Ergo 2ipi+1=1
Ergo 2ipi=0
Ergo i*pi=0
Ergo either i or pi must equal 0, and since pi≠0, therefore i must equal 0.
But what is the definition of ln(e^z) then? Thats what the question of the power rule seemed to reduce to?
You can't square both side , otherswise we can write -1=1
Squaring both sides 1=1
Interesting. I had thought the problem was substituting e^(2i*pi+1) for e in the same equation that we had derived that equality, because that would be using an equation to solve itself. Now I realize that’s not a problem since we’re not really “solving” the equation because it’s entirely constants.
In a complex unit circle, you can have different exponents (angles) pointing to the same value.
Yes that is true! But he slipped the obvious real answer and suggested the only faulty in complex numbers rule. It is also the problem in real numbers that we can't use a power base of {0, 1} to do an exponent equality enforcement. 1 (and any 0 power let alone undefined itself) in real numbers don't enforce that y^m = y^n meaning of exponent equality m=n rule that work in all bases not equal to {0, 1}!! 🤔😉
it's an extraneous solution, since you squared both sides. It's like solving sqrt(x) = -2.
This is not the problem. He wasnt solving an equation. Exp(2ipi)=1 is completely fine and he could start from there.
I came up with a simple method of resolving these X = any Y type proof: seek the point at which the entire operation is multiplied by 0. This is a reset which gives control over everything and makes the outcome determined by the design of the equation after that point, with any prior inputs being lost. Now that I have written this, I will watch the video and attempt to use my method to spot the error.
I'm not following the math very well, but already at 1:03 if e^x = e then x = 1. If we accept that i is a known value and pi is unknown, then the only way to satisfy i*pi+1=1 is i*pi=0 or pi = 0. In this particular instance, pi is essentially being used as a proxy for 0, in order to zero out the entire equation and make X = to any Y.
This is good thinking but it's not the only way to work it: sometimes you add solutions by a certain step. So π=0 is an extraneous non-solution, but if you're careful, π=π is also a solution.
@@TheReaverOfDarkness dude exp is not injective
I am being vague over here, but could this be because of the periodicity of the euler's form of complex numbers?
In my opinion, the best way to avoid such problems is to avoid writing things like a^b with complex numbers (because such things are not well defined in general) and resort to using the notation exp(z). With this rule, one wouldn't even be able to write (e^(2*pi*i+1))^(2*pi*i+1) = e.
This is a circular argument. You'd also need to explain why ln(e^z)z
But then how would you justify that (e^i*pi)^2 = e^2*i*pi? If we use the definition you said we will get e^2*ln(e^i*pi)=e^2*ln(-1) and ln(-1) has multiple complex answers.
From what i rmb we are allowed to do this if the “outer” power is an integer or real rational number that us in its simplest form without any worry.
Could someone clarify this?
So how do you find out whether ln(e^x) is x or one? With complex numbers, sometimes ln(e^x) is x, other times its 1. How do you figure out which one it is?
Same question for me 😐
subtract 2i\pi until the magnitude of the imaginary value is less than \pi, then take the natural log.
The quirk is a result of what exponentiating complex numbers really means. Euler's Identity (the equation that formed the basis for this false proof), is a specific case of Euler's Formula, a tool that makes use of Taylor Series' to equivocate cos(x) + isin(x) to e ^ ix.
When exponentiating some complex number z, what you are really doing is shifting from a cartesian representation of one complex number, to a polar representation of another. The real coordinate, a, of the first number can be dropped down from the exponent as the coefficient ln(a), while the imaginary coordinate, b, remains. This changes e ^ z to ln(a)e ^ bi.
Looking back to Euler's Formula, that can be re-expressed again as ln(a)cos(b) + iln(a)sin(b). This should look familiar if you've worked with polar coordinates, as rcos(\theta) represents the x coordinate in terms of polar coordinates, and rsin(\theta) represents the y coordinate. Since in the complex space, a cartesian coordinate pair is represented as a single number, x + iy (more commonly referred to as a + bi), ln(a)cos(b) + iln(a)sin(b) is of the form rcos(\theta) + irsin(\theta) (with r equal to ln(a) and \theta equal to b), which is simply a new cartesian complex number.
Because polar coordinates loop after a change in \theta of 2\pi (because sine and cosine have a period of 2\pi), changing the b value of your exponent z by 2\pi has no effect on the exponent, so a reduction can be performed before exponentiating without changing the result.
In short, the natural logarithm returns an infinite number of values, each separated by multiples of 2i\pi, but in computation, the one with the lowest magnitude is returned.
A much simpler 'proof' utilizing the same error would be e^(2πi)=1⇒2πi=ln(1)=0⇒π=0.
Another tricky way is to do the pi=0 much sooner. With e^(2pi*i+1)=e, you can say that the exponents should be equal on both sides. So, 2pi*i+1=1 which means 2pi*i=0 and then pi=0
That way it would be much easier to spot the mistake, because it's quite well known that e raised to different complex powers can give the same results.
1:12 - e^(2 i pi + 1) = e
or e^(2 i pi + 1) = e^(1)
Does this not suggest that:
(2 i pi + 1) = 1
[because a^b = a^c -> b = c]
Therefore
2 i pi = 0
i pi = 0
Then either i = 0 or pi = 0, so there's already been a mistake by this point.
(My background is Computer Science, not Maths; please let me know if I've done something wrong).
(Perhaps the a^b = a^c -> b = c rule doesn't apply to complex numbers?)
I knew this. I just needed you to tell me. 😊