Solving x^i=1 vs i^x=1
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- Опубликовано: 11 июл 2024
- Learn more math on Brilliant via brilliant.org/blackpenredpen/
Today we will be solving a complex power equation x^i=1 and a complex exponential equation i^x=1. You will see that my answer is a bit different than the answer from WolframAlapha.
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0:00 Hello
0:25 Solving x^i=1
2:12 Solving i^x=1
8:41 check out Brilliant
9:37 bonus part
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blackpenredpen
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The disappointment on his face after realizing that no one paused the video to solve the equation
😆
@@blackpenredpen
The disappointment on Oreo's face after realizing that no one paused the video to solve the equation🐰
I did! I at least attempted to do the second one
i did and it was fun(?
No, I paused to solve it, but still incorrect lol
This is the greatest video of a math debate, not like the 6÷2(3) stuff.
😆
By the way, the answer is 9 right? :P
@@pbj4184 the answer is 0
@@pbj4184 you multiply first because of the bracket
You mean Mind you Decision?
Got to love those edge cases where the commutativity of operations that works perfectly fine for real numbers becomes a trap in the complex world.
You pronounced "Euler's formula" quite well! Greetings from Germany :)
Thanks.
I'd like to know the average wage in Germany. Could you tell me that?
@@Mnemonic-X €22 per hour.
Is pronounced wheeeeler
Hello fellow LP fan
thank you for all the content you put out, man! You keep me interested in math
I return to this channel nearly after a year and damn you got a nice beard now!
4n is included in 4n/(4m+1) because 4n is just when m is equal to 0
True
@@JB-ym4up 4n/x for any integer x != 0 arguably produces solutions, and for x = 1 unarguably produces solutions. Any operations on an integer m that leave it as an integer won't effect that, so 4n/(4m-1) is just as valid as 4n/(4m+1) or even 4n/(3m^2-26)
i just started my math studies this fall and you are a source of constant inspiration for me
I don't know if you upped the res or I changed the settings on my phone, but that's one high quality vid. Great stuff as always!
Thanks. I actually have been using my phone to record in 1080P, 60FPS. My old videos (when I was recording in classrooms) were recorded on my MacBook Air, 720p!
wow I remember when you did your 100k video, glad to see you're getting close to a million
Great freaking video again! Soo close to 700k
Thanks.
I see you in Michael Penn as well
THANK YOU SO MUCH THIS MADE SO MUCH CLICK IN MY BRAIN
The main problem with polar coordinate (which pops up often in your videos) is that you can assign multple values to get the same results. The form x = (4n) / (4m+1) is equivalent to taking the (4m+1)th root of i^(4n), for which 1 is a possible solution (but you know there are 4m+1 answers in total, so another 4m possible answers, and those will not be 1, rather complex numbers).
A more strict definition of polar coordinates restricts the argument to be in the [-PI, PI) range (or some other 2-PI length interval, e.g [0, 2PI) which is perhaps more practical) and avoids somewhat whacky results like this.
but also in a 2pi range multiple solutions exist e.g. for x^5=1. so doesn't the ambiguity remain? i.e. 5th root of 1 can be mapped to different values in polar coordinates even in a 2*pi range..
even if you define such a restriction so only one answer pops up. in my eyes, the info that other answers exist, should be stated. Not for compitionisms sake, but because, such an interval is not the same for every problem, or even have a formula for every different problem. So the only reason to do it is for visualization in a class or smthng. Such debates should exist if this is not your Calc 1,2 etc. For me its just that the question is poorly stated. There should be an interval in the question or further explanation. Its not something new. Poorly stated questions are asked all the time in math. And create fun or not so, problems.
blackpenredpenbluepen
Nice work! You destroyed Wolfram Alpha! 😄
😆
congrats 1M subs :)
Seeing you stumped on this mystery was unnerving but kind of exciting.
thanku for introducing us from Brilliant
I think i^x just raises more questions when you forget that exponents have different branches in their function. Also, by looking at branches I think you can get infinitely many answers from i^(4/pi)...
Pi-rt(I) 💀💀💀💀💀
Great video! I don't really know how we should define complex numbers raised to a non-integer exponent, as if we just treat them like usual then we could say that i^ANYTHING=1 as we could write the ANYTHING as (ANYTHING/4)*4 so we'd have (i^4)^(ANYTHING/4) which supposedly is 1.
I'll become the -pokemon- mathematics master!
Very fun video. I surelly enjoyed it
Despite being a medical student, I really enjoy your videos! ❤️
Thanks, future dr.!
I'm in the same situation Tazrian T!!!! His videos are awesome!!!
tbh you don't need that much to enjoy most math. You can learn up to calc 3 in a few months if you have good resources and aren't bad at learning new concepts in general. What's hard is being a good problem solver because that needs a lot of practice to be able to recognize a lot of patterns in all sorts of different topics.
As awesome as always dear teacher🥰😍
Having some problems with limit comparison test in series. Can u do a video on how to find the perfect series for limit comparison test so that I get the desired result?
Black pen red pen, yay!!!
Amazing Video. Tomorrow Is My Physics Test. But My Mind Is Focusing More On Your Video. 😰😅😂
only if m tends to infinity the error or difference from zero and 8m + 2 (reduced to the range 0 to 2 * Pi) in each turn could eventually give values close to zero and only if m tends to infinity will the error be zero. in a continuous circle of dots
🤔 Wait a minute! When both m and n are negative, the fraction 4m/(4n+1) turns into a 4p/(4q-1), where p=-m and q=-n. That means fraction can be generalized to 4m/(2n+1).
The best maths teacher who catches pokemon also 😆
This is very interesting.
I have just cleared my 10th standard but I am watching your videos for about 1 year and I can now solve most of the calculus and complex no. Problems that students upto class 12th can't do.
Tan(x) a lot
Maybe the ambiguity comes from the fact that taking a root of one can yield multiple values. Take the square root of one: technically, there are 2 real answers to this: 1 and -1. Negative one times negative one also equals one. By just saying that 1^(1/pi) = 1, you may be missing other complex solutions to the pi'th root of 1. If you put 1^(1/3) into wolfram alpha, you get 3 complex solutions (and maybe more if you add 2pi, idk).
an equation is well defined if it is required that each of its solutions make the equality true: an equation can have multiple solutions but not solutions that solve the equality or not in an arbitrary way. So in the second equation the only acceptable solution is x = 4n. The polar representation of complex numbers, in equations, must be used when it does not give rise to ambiguity, that is, when it effectively allows to identify all the solutions, not when it produces expressions with multiple results that give rise to "partial" equalities.
Very nice bro
I mean, I saw the derivation fine, but why isnt the denominator any odd number? Any odd root (2m+1) of 1 should give 1, not just 4m+1.
And if you want to accept the multiple answer paradigm, why not all integers for denominator? Sqrt(1) will give 1 and others.
Why x=4n/4m+1 and not x=4n/2m+1 or even x=4n/m?
if we consider (4n)/(4m+1) to be the set of solutions, shouldn’t we also consider (4n)/(4m+3)?
It’s included in that expression. Since n and m are integers. For example, to get 4/3, you just pick n=m=-1 in my expression.
Then take (4n)/(4m+(2k+1)) as a set of solutions where k=integers😂😂
I've got a question, in the end m could be any real number as 4n=1 if n is integer.
@@blackpenredpen Thank you !
The confusing part is the fact that if you compute it as (i^4)^(1/anything), you will always get 1 because i^4 is 1 and 1 to the power of anything is 1. So isn't the answer (4n/b) where n is an integer and b =/= 0?
The property doesn't apply to multivalue power function
consider
let p q φ real numbers
exp(iφ)^p = exp(iφp + 2πinp)
(exp(iφ)^p)^q = exp(i(φ+2πn)pq + 2πimq)
It depends on order of applying power. For φ=π/2 we have 1 when n=0 m=-1 q=1/π and can't have 1 for q=4.
Another examples for the nice and astonishing effects of complex numbers.
I agree with WFA. The best way to avoid any confusion is to implement principal values and principal functions for multivalued situations. It's the best way to communicate your intentions in math.
This means we have to sacrifice the relation (z^a)^b = z^(a*b). It is no longer guaranteed in the complex world, unless you take proper precautions! The proper way to multiply exponents is, unless z is purely a positive real number, first rewrite z = |z| e^(Arg(z) i) where the principal argument Arg(z) lies in the interval (-pi, pi] THEN multiply the exponents. WFA, wikipedia, and my math textbook agrees.
It is also what my advanced calculator does. If I type sqrt[ e^(-pi i) ], it doesn't return (-i) = e^(-pi/2 i).
It returns i = e^(pi/2 i). WFA does the same.
This is because they both rewrite e^(-pi i) = e^(pi i) using the principal argument -pi < Arg(z) ≤ pi before multiplying the exponents. This is of course just a convention, but it's a good convention, and everyone should know about it.
Then, the principal value of i^(4/5) is [e^(pi/2 i)]^(4/5) = e^(pi/2 i * 4/5) = e^(2/5 i pi) = 0.31 + 0.95i.
i^(4/5) =/= (i^4)^5, and is not a valid solution to i^x = 1.
The mistake that leads to the extraneous 4n/(4m+1) solutions is that you didn't express i in its principal polar form before multiplying the exponents. The proper way to do it would've been:
i^x = 1
e^(pi/2 i x) = e^(2pi i m), m is an integer
x/2 = 2m
x = 4m
Since the angle 8m + 2 is multivalued within 2 * Pi, then by increasing m it will cover points at the angles from 0 to 2 * Pi. then when m tends to infinity we will have a reduced angle equal to 0 or 2 * Pi. only then Pi is irrational.
been subscribed for two years now, great video as always. i'm graduating next week with a BS in electrical engineering. as for the answer to your question:
I'm an engineer, i will just round it to 1 anyways lmao.
actually though, i don't think it is wrong to say both solutions are correct, as long as it is understood why the 2nd method can produce multiple values.
cheers
That's exciting Brian! Congrats!!!
I think the second method in the bonus part is the proper way to calculate irrational complex roots. More explicitly, converting the complex number to exponential form then simplifying the exponents, not calculating each exponent separately.
There should be some notation to distinguish whether i^x= 1 means that any value of i^x should equal 1, or that the principal value of i^x should equal 1.
Nice :)
Hello, if I get your point , I think m =0 in the beginning, but it’s change to n= 1 when you let n=m , so in the last equation( m i #/2 x )= 2i# n / yo start from #=0, not in rad not 2# .. I hope this will help, if doesn’t, please let us know the answer. Not( my English language is so bad) . Maybe I don’t understand what you say .
If 4n/4m+1 is a solution isn't then 4n/m m0 also a solution? Because 1 is then always one solution
How about using the Taylor series e^z = 1 + z + (z^2)/2 + (z^3)/6 + ...
Let a = m/n, and insert z = i*a*pi*(4k+1)/2 into the formula.
So that we can get only one anwser for a k value?
We miss the pim (Peyam) part of those sets of solutions, such as the first one e^2pin.
awesome one!!! Thankfully not too advanced for me😂
Damn you're so good....
The main thought-error happend at 2:06. exp(2·PI·n)=x with x==1 works only for n=0, but not for all integers. E.g. the expression provides for n=1 an x=23.1407, etc. The resolution is quite simple: Eulers polar form of complex numbers only work in the C-world. In the moment, i is removed from the exp(i)-term, everything is shifted back to R. In R even sqrt(-1) is undefined. Things will become clearer, by using complex math strictly. While it is technically legal to eliminate any number by ist inverse (e.g. k·(1/k)=1), in the C-world context matters.
Hello sir, is there posible for this?
x^y+bx+c=0 when y,b,c are real numbers and letter y is a fraction form
Because raising e to a complex exponent is not 1 to 1, you cannot really say that if e^(ix) = e^(iy) then x=y
This is the step at 4:30
Your video fantastic
I need to know what is true at the last part!
on the other hand, if I perform the operation first for the exponent 4. the result converges directly to 1. but operations with irrational factors will only be achieved at infinity.
Not quite, since 1 has 5 5th roots, the number 1, and 4 more complex numbers, all of which can be represented in polar form with many solutions, hence 4m+1 solutions in either case, if you don't limit your domain :)
There's a problem I've been banging my head against a while that might interest you... what do all the solutions of "X to the power Y equals Y to the power X" look like if you allow X and Y to be complex?
i didnt have clue so i just looked at the second one. i = sqrt-1, so i^2 = -1 . -1 * -1 = 1 . Since i has to be times by itself twice (i mean ^2) to get -1, and then -1 is squared to get 1, then surely i^4 = 1
Wouldn't you necessarily need to take the values of x that satisfy the equation otherwise the multi valued nature of the function interferes with your solution? I think it's maybe a situation where you have to be careful to choose a solution that avoids the branch cuts of the complex function or else you get contradictory answers like this, but I'm not sure either. This is interesting!
I think the values of x that give multiple solutions should be allowed. I tend to think of branch cuts as arbitrarily destroying the beautiful Riemann surface structure of complex numbers (and complex exponentials in particular).
"namely: multiples of n" -> "namely: multiples of 4" ~ @ 2:30
Hm maybe you can take some limit of maxima(cos(8x+2), x is integer) or something
Also, maybe you can't really cancel out the pi with the 1/pi in that expression?
The issue with the second question is that it is ambiguously phrased.
*For* *fractional* *x* , i^x is multi-valued, and since in the complex world there is no good way to define principal root [e.g., which is the principal root among the first question's answer e^(2nπ)?], the second question "i^x = 1 " is as dubious/incorrect as "1^(1/2) = -1" (the latter because ±1 ≠ 1).
So, the second question ought to be changed to “For what values of x is 1 *_a_* value of i^x ?".
On the other hand, because the first question’s LHS turns out to collapse into a single value 1, its LHS indeed properly equals its RHS. So, the first question is coherent and valid as it is.
I would say that 4n/(4m+1) is technically correct because it accounts for all possible solutions. However, using the form of (i^4n)^(1/(4m+1)), like you did with (i^4)^0.2, justifies the WFA answer of just x=4n because the 1/(4m+1) is redundant.
Doesn't this logic mean that i^(4/m)=1 is one solution for all m as you would have the mth roots of 1? Or am I missing something becuase there are a lot more integers than ones of the form 4m+1?
Are n and m in x = 4n / (4m + 1) really independent variables? It'd seem to me, intuitively at least, that they aren't, since I presume the angles on both sides of an "=" sign to be linked. In that case, m = n + 2πk + π/2, which would result in x = n / ( n + 2πk + π/2 + 1/4 ). Ouch...
I figured something out. If you take 4n/q where n and q are integers, it still works. Because qth root of 1 will give q solutions out of which 1 will always be one of them.
Thanks, even I didn't thought such
Counterexample: let n=1, q=8, then 4n/q = ½, but what is i^(½)?
In fact {4n/q | n,q are integers} is just Q, so you are saying that _i_ raised to any rational power is 1.
@@JivanPal Considering your point. Can we add the statement “where q is not divisible by 4”??
@@VedantArora, yes, I believe that would then work! I reasoned this out to myself as follows, thinking geometrically about the n'th roots of unity and how you can decompose any of the n'th roots of any complex number into the product of the number's principal n'th root and any of the n'th roots of unity:
Let n be any positive integer. We'll consider the n'th roots of unity, as well as the principal value of i^(1/n); that is, the n'th root of _i_ that has the smallest positive argument, namely the number z = e^(iπ/(2n)).
Then raising z to any multiple of 4 gives a complex number which is equal to one of the n'th roots of unity; let integer m be given, then this number is y=z^(4m). Since y is equal to an n'th root of unity, there must be another n'th root of unity whose angle/argument "fills the gap" between y and 1.
Judiciously choosing _w_ to be this other root of unity (that is, _w_ is the n'th root of unity such that w^(4m) has argument equal to 2π-arg(y)), we get y·w^(4m) = 1.
In particular, observe:
y·w^(4m)
= z^(4m) w^(4m)
= (zw)^4m,
and _zw_ is any of the n'th roots of _i,_ so the above equals i^(4m/n).
I am now trying to think about what it is in this chain of logic that n being a multiple of 4 doesn't work with.
@@JivanPal damn, love your calculations dude!
Why is this video invisible? I only found it with the link from “Introduction to Power Series for Calc2”, but it doesn’t appear on your channel
LVE STREAM STARTING!
But this comment is a day old
Wait a little time treveler
@@asamenechbayissa553 NO HE IS NOT A TIME TRAVELER
@@asamenechbayissa553 Indeed I am. In the future, Steve Jobs dies of Ligma
very easy dear
I think we should only consider the principal root, and if not, it should be noticed anywher, in this situation you can have √1=1 and if you want to consider the "other solution" then it's just ±√1, and similarly, if we have x²=1 we CANT just take the sqrt on both sides, well we can but √1=1 and √x²≠x, it's just ±x.
All that stuff is because 1¹ is obviously 1, and that is also obviously 1^(2/2), so if we take multiple values we'll get 1=±1.
Also we could say 1^(2/2)=(√2)² wich in case it's multivaluated, it gives 1 also, but then i^(4/5) wouldn't be 1
I would accept (4n/4m+1) because even computing the denomitor first, 1 is always a possible answer. We know i^4n is always 1, but to add to that, i^(4m+1) is always i. That being said, if we write that out as i^(4m+1)=i and take the (4m+1) root of both sides we also find that i=i^(1/4m+1) as one of the solutions. It's just important to know its not the ONLY solution and there are 4 other ones no matter what.
I take your answer! It's absolutely obvious if you consider roots.
I think when you cancel π and 1/π in step 2, you cut the possibility of other roots.
In (i^4)^1/4, if we just cancel 4 and 1/4 we get just one value of the expression, i. And then you do it like (i^4)^1/4= 1^1/4=(1^4)^1/4=1. What i =1?
The problem lies in just taking one root at one time and then another root at another time.
Multiplication of exponents (or cancellation of factors) here can give two results, so caution needs to be taken.so (i^4)^1/4 and i^(4/4) can mean two different things
i^(4/4)=i, but (i^4)^1/4 has four different values i, -i, 1,-1.so we can wrongly say i=i^1 = i^(4/4)= (i^4)^1/4= 1^1/4=(1^4)^1/4=1.
In the last step of i=i^1 = i^(4/4)= (i^4)^1/4= 1^1/4=1, I'm taking just one value of 1^1/4 which creates the problem.
You are taking i^(4/π)= (i^4)^1/ π = 1^ 1/π
Does 1^ 1/π have just one value
Then the same expression i^ (4/π)= [i^1/π]^4=[(sth^π)^1/π]^4
Here by cancelling π and 1/ π in (sth^π)^1/π
you are taking just one value of the expression which just isn't equal to another value of the same expression.
Interensting video
Sir please solve integration of cos(x)^2
For the bonus part, I think that the second method is better. If we consider the first method, we could extend the solutions of i^x = 1 to all the complex numbers. Whichever x you choose (except 0), you'll always be able to write it with the form 4/(4/x). Therefore, when you apply the first method, you get i^x = (4/x)-th root of 1, which is 1. Maybe I'm wrong but it should work for every complex number other than 0. So, this method has something wrong in it, I'd rather use the second one.
How did he get the e^i2(pi)n at 1:27?
I'm watching this while in middle school, idk why but I want to understand what you're doing lol
Haha nice I did that too
Same here
5:20 when my answers are different then others tho it makes sense...
For x=4n/(4m+1) where n & m belong to Z,
i^x=i^[4n/(4m+1)]=i^(4*n/(4m+1))=(i^4)^(n/4m+1).
As i^4=1, (i^4)^(n/4m+1)=1^(n/4m+1).
Therefore, whatever is the value of n and m, i^x will always be equal to one and there is no other output.
How can be 4n/(4m+1) = 4/pi ?? Considering that n,m in Z...
Okay, we got solution with 4m+1 in the denominator and we checked the solution 4/5 , but what if we just check for example 4/7? i^4/7= (i^4)^1/7 =1, and 4m+1=7 has no solution in integers. Where is a mistake in my reasoning?
Finally a problem i(pun intended) am able to solve. I wonder if i raised to complex values can actually be evaluated to 1?
i^i gives real numbers so maybe with some tinkering you maybe able to actually do that
8:18 From Where U Have Got i^8=8
Maybe I'm wrong but there is something I don't understand. In De Moivre's formula, when we raise e^(i pheta) (with pheta the angle) to the power n, such that we have (e^(i pheta))^n, n must be an integer. But at 1:40 you raise both sides to the power of 1/i, which is not an integer. Therefore your demonstration must be wrong. If someone could explain me if I am wrong, I would be grateful.
Did I miss something as 8 is in the formula 4n/(4m+1) if n=2 and m=0
I love how this guy hold balls!
Is there any function as
"Logarithm Integration multiplied by divination"
I guess X=4n where n is an integer is a more appropriate form as it will always result in 1.
Even in your expression X=4n/4m+1 if we keep m=1 it will result in X=4n and will return 1.
But if we take some arbitrary values like n=m=2 then we end up with X=8/9. Here if we evaluate using (i^8)^1/9 then we end up with 1 but if we take (i^1/9)^8 then we don't get 1.
If n=0 then i^X is always 1.
If we take n=1 and m=2 then we end up with X=4/9 and again if we evaluate using (i^4)^1/9 we get 1 but not in the other case.
Therefore irrespective of what values we keep in n and m there is always a possibility that we don't get 1.
So in my opinion Wolfram Alphas answe is more appropriate and generalized cause it completely eliminates the denominator by keeping m=1 in your expression.
By the way big fan of yours and love watching your videos.
The problem in your demonstration is actually when you write (1)^(1/5)=1, which is clearly true in R, but when we are in C, x^5=1 has five solutions (which are exp(2.i.k.pi/5), where k is an integer. The question you raise is about the definition of a function x---->x^q (where q is a rational) in C. The problem is particularly tough when x is not a real, so we know that there isn't any real solution. So how to choose between these five values? I haven't studied the question deeply but I don't think we can defiine a continuous function on C.
It depends of the question. If the question is "find X s.t. i^x=1", of course is correct 4n/(4m+1).
If the question is "find X s.t. i^x =1 and only 1", the answer is 4n.
I don't give a s... if i^(4/5) have multiple answers as long as 1 is one of the answers. This is the difference between "or" or "exclusive or".
7:08 By the same logic the solutions are 4n/(2m+1)
Because i^(4/7)=1...
you swapped (different) m and n (numerator and denominator) @~ 9:40.
previously: (4)n/(4)m(+1), now: m/n.
a bit confusing.
use different letters?
state that these are different ms and ns?
if you have i^(4n/(1+4m) you can rewrite it as (-1)^((4n/(1+4m))/2) then the fraction cn be converted to 2n/(1+4n) which always has the solution 1 and like some other comments pointet out that any fraction with an multiple of 4 on top and any odd nuber at the bottom is an solution
Ps: sorry if my language is a bit confusing im a 12th year math stundent in germany and english isnt one of my mayors i hope that my idea is understandabile
What about x = 4/3 or 4/7 or really 4 over anything? It's not in the solution but still works!
choose any x, choose any y such that |y|=1
i^x = i^(4*(x/4)) = (i^4)^(x/4) = 1^(x/4) = 1
now sice y*1 = y, then y * i^x = y so let z be such that i^z=y, we get i^(x+z) = y, now we define w=x-z
repeating the same trick from before, i^x = i^(w+z) = i^z * i^w = y * 1 = y
I think I have realised the mistake (without checking any comment in case anyone else has already answered this).
The correct solution to
i^x = 1 must be a rational number. So only rational numbers ( of the form
[ 4n/(4m + 1) ] with integer values of m and n ) will result in 1. Hence in bonus case of 4/pi, the solution ideally must not result in 1.
In last part both are correct because we know x⁴ must have 4 roots so, i^(4/π) =1, cos (8m+2) + i sin(8m+2) and 2 more roots.
Am I right?
WFA
without what?
thanks
From the start m and n are both integers, when u considered m as pi, you've only violated the domain value. So that's why it wasn't true
How different a debate is this from the debate that 4^(1/2) should equal 2 because it also equals -2? f(x) = x^n for most integer n values map two different x onto the same y, so taking the inverse function you will get one x mapping to two different y.
There’s nothing really special about i here, most numbers to the 4/5 will have 5 possible answers because the “to the 1/5” is the inverse of “to the 5” and the logic I said above about having multiple answers applies.
I personally think if you accept you need to plus the plus/minus when you square root both sides, you should except all values of a fractional exponentiation, but that’s more of an opinion that consistency should be maintained than a good mathematical argument. I’d be curious to hear what the real truth on how fractional exponents are evaluated in formal mathematics; is there a process for taking “one correct value” or do you need to accept them all.