Can e^e^x=1? Sol here kzbin.info/www/bejne/mZzGanlmi8-Nac0
@69Gigantosaur Жыл бұрын
Hello😊
@JustAPersonWhoComments Жыл бұрын
You can take the natural logarithm (ln) of both sides: ln(e^(e^x)) = ln(1) Using the property that ln(e^a) = a: e^x = 0 Now, you have e^x = 0, which has no real solutions because you cannot raise a positive number (e) to any power and get 0.
@dragondg6412 Жыл бұрын
e=0 x=1 easy 0^0^1=1
@Mono_Autophobic Жыл бұрын
@@dragondg6412bro studied from kfc toilet 💀
@ajb16384 Жыл бұрын
@@dragondg6412bro graduated from the 15 year old marker board in social studies
@Isometrix1163 жыл бұрын
Imaginary numbers are the math equivalent of going into the shadow dimension to get through obstacles
@albertoaltozano83543 жыл бұрын
Fourier transforms are more like that
@seankrkovich28693 жыл бұрын
This
@situationnormal22173 жыл бұрын
I like that analogy
@youngsandwich99673 жыл бұрын
@@albertoaltozano8354 Fourier transforms are within the shadow dimension of imaginary numbers, to be fair
@neoxus303 жыл бұрын
It is straight up that)
@brenn77543 жыл бұрын
When blue pen gets in involved, you know it's serious...
@bombdog39733 жыл бұрын
Good one xd
@WerewolfLord3 жыл бұрын
Really serious is the purple pen.
@GianniCampanale3 жыл бұрын
@@WerewolfLord you don't want to see the purple pen
@godson2003 жыл бұрын
@@GianniCampanale purple is for thanos
@brenn77543 жыл бұрын
@@godson200 this was blackpenredpen infinity war. Next is blackpenredpen endgame...
@jamesfleming49193 жыл бұрын
“360, but we are adults so we use 2pi” I felt that
@jamesfleming49193 жыл бұрын
@Memes shorts 1 like for you from me
@Joserider1233 жыл бұрын
Nobody like James Fleming’s comment. It needs to stay this way
@fatitankeris63272 жыл бұрын
What about tau?
@meowcat71242 жыл бұрын
360 likes
@fantiscious2 жыл бұрын
420 likes
@McNether3 жыл бұрын
Actually Wolfram-Alpha is correct. Too understand why we will need some function-theory/complex analysis (for example: Complex Analysis, Elias M. Stein S. 97-100). At first we will need a definition of z^w with w,z in C. For any z in C\(-∞,0] we can define a function z^: C --> C by z^w:=exp(log(z)•w) where log is the principal branch of the logarithm (that means that log(1)=0). Of course you can choose another branch but in this case the definition does not match with the exponetialfunction with a real basis. Using this definition we get: 1^x =exp(log (1)•x)=exp(0•x)=1 which states that the equation 1^x = 2 got no solution. Now we take a look at the Question: "Can we finde a x in C such that 2^(1/x)=1?" Using the definition we get 2^(1/x)=exp(log (2)•(1/x)) which is equal to 1 whenever log (2)/x=2πi•k, for any k in Z. This gives the solutions you are getting too. After clearing this we should talk about the "contradiction" at 06:23. What you are writing there is correct but its not a contradiction to the above: 1^x=2 => 1=2^(1/x) means "every solution of the first expression is also a solution of the second Expression" (which is correct cause the left expression got no solutions). The other direction 1^x=2 x²=1 is correct but x²=1 => x=√1 is wrong (the solutions -1 "gets lost"). And this gets even "worst" when complex numbers are involved...
@writerightmathnation94813 жыл бұрын
The main error is conflation of the notion of a function with the notion of a relation, by using what are sometimes referred to in complex analysis books as "multi-valued" functions. There is no such thing. A function, by definition, is decidedly NOT "multi-valued". This leads to an aberration involving a failure to understand how equal signs work in a coherent presentation of mathematics. There are two fundamental ways they can be used coherently, and all other coherent uses are definable from these. (a) The main fundamental semantic use of an equal sign is to write a mathematical statement that is interpreted as true in the context of some given model of some theory because it pertains to the facts in that context, AND that statement that is interpreted as true in that context BECAUSE the one and only thing that is described to the left of that equal sign is exactly the same thing in that model as the one and only thing described to the right of that one and the same equal sign. This is written to convey clearly to the reader some accurate information about the context provided by the given model. Mistakes can be forgiven, but persistent incoherent abuses of notation should be excised from mathematical vocabularies. (b) The main fundamental syntactic use of an equal sign is to write a mathematical statement that is to be tested for truth in the context of some given model or class of models of some theory because it pertains to the facts in that context, AND that statement that is an hypothetically testable assertion in that specific context BECAUSE the descriptions to the left and right of that particular equal sign are interpretable in the context of the model or class of models to be considered, and the question of whether some model or class of models satisfies that particular statement is a coherent question in that context. Such a syntactic use of an equal sign is written to convey clearly to the reader some problem (i.e. it is a mathematical query) aimed to elicit some accurate information (a clearly formulated and completely explained solution of the problem and answer to the question) about the context provided by the given model or class of models. Mistakes can be forgiven, but persistent incoherent abuses of notation should be excised from mathematical vocabularies. Sensationalism should be rooted out and excised, just like incoherent abuses of notation should be. Otherwise, we our logic system will prove absurdities like 0=1 in the real number system. A logical system that allows such nonsense is not useful, because from such things, the notion of "provable equation" and "equation" are indistinguishable. By a variant of Occam's Razor, we should not invent a terminology that pretends to distinguish things that are indistinguishable.
@nicktravisano71523 жыл бұрын
bruh what
@stochasticks3 жыл бұрын
@@nicktravisano7152 There's a distinction between log applied as a function and the relation called the inverse image of a point in a space via a function. Both are relations between elements of sets but a function has the property that if x=y then fx=fy . The relation named "inverse image" has not in general such property. What blackpen writes on the board is formally incorrect. You cannot use the equal sign if applying something which is not a function on both sides of the equation, such writing down "log" but in fact meaning inverse image of the exponential function in the complex plane. The apparently revolutionary results you find in the complex plane are not so much revolutionary but abuses of the inverse relation treated as a function when it should not be. Still fun though.
@HDitzzDH3 жыл бұрын
tf am I reading
@degeestvanpeterrdevries33663 жыл бұрын
@@HDitzzDH University students having a discussion
@alex_marr Жыл бұрын
Notice that 1^x = 2 and 1 = 2 ^(1/x) are, actually, two diffferent equations with different domain of x. You solved the second equation and not the first one. Edit: that is EXACTLY why wolfram can solve the second one.
@zenedhyr761211 ай бұрын
To simplify: {1^x-2=y, y=0} ≠ {1-2^(1/x)=y, y=0} Similar example: x^2-1=0 and x-sqrt(1)=0 will give different graph. [Search on google] x^2-1=0 x^2=1 x=sqrt(1) x-sqrt(1)=0 ■
@vaarmendel165710 ай бұрын
1) x real -> No solution. OK 2) Let's suppose x complex -> x = a + ib (a, b Real numbers !) -> 1^x = 1^(a + 1^ib) = 1^a . 1^ib = 1 . 1^ib = 1^ib Applying ln on complex numbers: ln(z) = ln|z| + i arg(z), we get : ln (1^ib) = ln|1| + ib = 0 + ib = ib Hence ln(1^x) = ib = ln(2). Pure imaginary = pure real -> Impossible. x is not a complex number Solution: NO SOLUTION. What about that ?
@xinpingdonohoe39787 ай бұрын
@@vaarmendel1657 close, but the issue is you've used the wrong definition of arg(z). arg(z) requires base e, not base 1. So ln(1^ib)=ln(e^(2πni×ib)) =ln(e^2πkb)=2πkb. Then 2πkb=ln(2) has solutions for b.
@rodabaixo135 ай бұрын
@@zenedhyr7612bro forgot the plus or minus. Ofc x^2-1 is different from x-1. Its not just about degree. The solutions arent even the same like equivalent equations. Thats just blatant nonsense
3 жыл бұрын
My man starting to look like an ancient philosopher who lives on a mountain, I dig it
@HourRomanticist3 жыл бұрын
Like a sage or a monk or something
@sophiacristina3 жыл бұрын
He keeps on a cave meditating about the marvels of math...
@mastrammeena3283 жыл бұрын
Nah He's just a muslim
@ManjotSingh-sf2ri3 жыл бұрын
With a sacred pokeball
@papajohn60813 жыл бұрын
Thus spoke BlackPenRedPen
@stephenbeck72223 жыл бұрын
“We need two things. The first thing is the distance. The next thing is to erase the equal sign better. The third thing is the angle.”
@blackpenredpen3 жыл бұрын
Lol
@lordmomstealer3 жыл бұрын
@@blackpenredpen I have question for you 4^x+6^x=9^x FIND THE VALUE OF X
@lordshen30783 жыл бұрын
@@lordmomstealer haha dude this is video by mind ur decisions
@talkgb3 жыл бұрын
@@lordmomstealer very basic u = 2^x and v = 3^x substitution
@TheHashimq3 жыл бұрын
@@lordmomstealer this is from mind your decision He would do it in minutes
@randomyoutubecommenter43 жыл бұрын
"Can 1^x = 2?" - No. *video ends*
@abhaysharmafitness3 жыл бұрын
x=log base 1 of 2
@Pirater666l3 жыл бұрын
@@abhaysharmafitness log base 1 of 2 = indefinite, so no
@korayacar14443 жыл бұрын
@@abhaysharmafitness no such thing as log base 1
@ramg46993 жыл бұрын
@@Pirater666l *indefinite in real numbers
@seroujghazarian63433 жыл бұрын
@@korayacar1444 yes there is, in the complex world.
@dr.downvote3 жыл бұрын
Mathematicians whenever they wanna look complicated : Let's talk about complex numbers Physicists whenever they wanna look complicated: let's talk about Quantum physics. Chemists whenever they wanna look complicated : let's talk about chemistry!
@KBMNVLpNdLumkstz2 жыл бұрын
To be fair, modern chemistry based on quantum physics
@magmar-wt5on2 жыл бұрын
And QM involve hilbert space so everybody talks about complex number 😂
@aabahdjfisosososos2 жыл бұрын
Chem is not hard
@avy12 жыл бұрын
Chemistry is applied quantum mechanics, quantum mechanics is applied mathematics. And as always, math is king.
@pf329002 жыл бұрын
Complex numbers? Why not quaternions, octonions, sedenions and the Clifford algebras?
@AethernaLuxen Жыл бұрын
I like how this whole time he was holding a poke ball and half of us were too busy having our brains crushed to realise
@IONProd Жыл бұрын
It's actually his mic (in case you didn't notice)
@j03man44 Жыл бұрын
WTF 😮
@takvacs Жыл бұрын
Isn't that usual though? He has it in every video
@Shikogo Жыл бұрын
I clicked the video because of the Pokeball lmao
@The_Red_Scare Жыл бұрын
I actually noticed immediately and soon figured that it must be either his mic (somehow) or a random thing he holds as a gag for all his videos.
@mrhatman6753 жыл бұрын
You are evolving into one of these chinese big beard philosophers lol
@zaheersuhabuth26773 жыл бұрын
@Sierox same xD
@ADeeSHUPA3 жыл бұрын
@@zaheersuhabuth2677 Pakistani
@Potato20173 жыл бұрын
@Sierox me too
@zaheersuhabuth26773 жыл бұрын
@@ADeeSHUPA 🇵🇰
@ManishGupta-hb4iu3 жыл бұрын
kzbin.info/www/bejne/rIbGf3ePoaenpcU
@iQKyyR3K3 жыл бұрын
That felt like a mathematical crime.
@twakilon3 жыл бұрын
Nah. It's perfectly legal, as long as you DON'T consider the wrong logarithm branch.
@bagochips12083 жыл бұрын
More like exploiting loopholes
@twakilon3 жыл бұрын
@@bagochips1208 it's not a loophole though. The problem lies in the argument function.
@angelmendez-rivera3513 жыл бұрын
@KiwiTV It doesn't defeat itself. It just happens to be inconvenient for humans. Mathematics has never been intuitive, though. Human brains didn't evolve to be able to easily deal with complex numbers. They evolved so that we could do 3rd grade elementary school arithmetic. Everything else is just us making ourselves more miserable against our own evolution for the sake of additional benefits.
@WorkinDuck3 жыл бұрын
@KiwiTV Complex Numbers allow us to solve real word phenomenons, like apparent/reactive power in electrical systems, pretty elegant. It doesn't defeat itself, it only offers multiple perspectives of a problem
@Kdd1603 жыл бұрын
"I don't like to be on the bottom, I like to be on the top."
@samarth.suthar3 жыл бұрын
Now that's what everyone wants to be... Underrated comment...
@Barocalypse3 жыл бұрын
"i don't like to be on the bottom, i like to be on the top."*
@MrDerpinati3 жыл бұрын
*mm nice*
@akmalfaiz70943 жыл бұрын
That phrase leads to two different endings
@gani91763 жыл бұрын
We got the same surname😂
@robinbrowne54193 жыл бұрын
I would guess 2 possibilities: 1. No. 1^x cannot equal 2. 2. If 1^x can equal 2, then 1^x can equal anything, because there is nothing special about 2.
@speedyx34932 жыл бұрын
The 1st one is correct :) 1^x will ALWAYS equal 1, even if x is a complex number. The video is just tricking you, it’s like those old 1=2 videos when the guy slily hides the fact that he is breaking the math axioms somewhere
@glitchy9613 Жыл бұрын
@@speedyx3493 Both possibilities are wrong, the 3rd correct possibility that 1^z can have a countably infinite amount of solutions (not indeterminant like 0/0), but only when z is not rational.
@xinpingdonohoe3978 Жыл бұрын
@@glitchy9613 I feel you're misunderstanding his point. He's saying 1^x can equal anything given a sufficient value of x. And by branches, it's correct. 1^x=y, take the right branch and you can get x=ln(y)/2πni. Maybe take different branches of the ln(y) and you'll get even more solutions. The exception will be 0. Branches don't matter, taking a logarithm of 0 will cause some sort of issues. Maybe on a Riemann sphere you can argue it, but even then not necessarily.
@glitchy9613 Жыл бұрын
@@xinpingdonohoe3978 he literally says "1^x will ALWAYS equal 1" I doubt that was his point
@xinpingdonohoe3978 Жыл бұрын
@@glitchy9613 I reread my thing, and I can only assume I wasn't referring to Speedy Gonzales here. I think I was referring to BPRP, just from where I stand. Sure, for each x it is true that 1^x can equal 1, but for complex x, 1^x may be something else too.
@danielyuan98623 жыл бұрын
Okay if you solve for x in 1=2^(1/x) using wolfram alpha, you indeed get x=-i*ln(2)/(2pi*n), but as I have learned in math competitions, you should -always- usually plug it in to the original equation. And plugging x=-i*ln(2)/(2pi) as 1^x in wolfram alpha, you get 1. So x=-i*ln(2)/(2pi*n) are all extraneous solutions, which is why they are not solutions to the original equation 2^x=1.
@extraterrestrial4611 ай бұрын
How exactly extraneous, wdym by that
@sashimidude328811 ай бұрын
@extraterrestrial46 the original equation has a domain of all real x, and a certain range. The equation 1 = 2^(1/x) has a more restricted domain, and a different range. This changes the solutions, producing new solutions that do not work for eqn 1.
@blackpenredpen3 жыл бұрын
Can 0^x=2?
@namantenguriya3 жыл бұрын
Love you from India 🥰🥰
@davinderSingh-zr1hu3 жыл бұрын
Nah
@robrazzano91683 жыл бұрын
No. Ln 0 is undefined, and r=0 on the complex plane, so you are always stuck dividing by zero.
@redstoneplayz093 жыл бұрын
If you do it the way you did it, I get: x = ln(2)/(ln(0)) and ln0 is ln(0) + i*n Maybe if there is a different way it's possible but not with how you did it in the video.
@LuVD9903 жыл бұрын
I came back to your channel. It is so funny the topics related.
@dr.kelpshake45733 жыл бұрын
2:59 I was gonna say 360 degrees like the child that I am. I can't wait to be an adult and say 2 pi!
@ojaskumar5213 жыл бұрын
Radians 4 adults 😎
@shakeztf3 жыл бұрын
2 7.18808272898 funny factorial joke haha (or rather abuse gamma function for a factorial joke joke)
@jmhpt3 жыл бұрын
Or you can take it a step further and say tau!
@josefmuller863 жыл бұрын
Whaddaya mean 360°? 2pi? I only know 400 GRAD
@melonenlord27233 жыл бұрын
@@josefmuller86 Every thing, that don't give you a round number, if it's a right angle, is stupid. ^^
@HourRomanticist3 жыл бұрын
You know what. This came in my recommended, and man let me tell you, my algebra 2 teacher must be doing a great job because I don't know how I willingly clicked on this and was genuinely interested lol
@mihailmilev99093 жыл бұрын
ikr lmao
@geometrividad77163 жыл бұрын
This is very similar to the equation sqrt(x)=-1. If you put that into Wolfram, it will tell you that it has no solutions. You can try to argue that well, actually, one of the square roots of 1 is -1, but the thing is that's not what sqrt(_) actually is. The same is true under complex exponentiation: the principal branch is used by definition and as such, 1^x=1 no matter which x you plug in. As others have pointed out, this does not contradict that 2^(1/x)=1 does have solutions in C (even when we are taking the principal brach). So no, Wolfram's right here.
@MrBauchnabbel3 жыл бұрын
I think wolframalpha is right here. 1^x=2 has no solution but 1=2^(1/x) does. Grinding this down to fundamentals you see that (a^b)^c is not equal to a^(bc) for complex numbers, exactly because the change of branch of log you expertly portrayed in the video. Another instance that messes with this is my all time favourite: a = e^(log a) = e^((2\pi i / 2\pi i ) log a ) = (e ^ (2\pi i) ) ^ (log a / 2\pi i) = 1 ^ (whatever) = 1.
@hippityhoppity6573 жыл бұрын
"ok so we'll just take a logarithm and set the base to 1" this triggers me
@InfiniteQuest863 жыл бұрын
@Tessellating Tiger Lol then you can divide. He had to hide it lol.
@nanamacapagal83423 жыл бұрын
Log base 1 is so cursed
@prashant26503 жыл бұрын
log base one isn't defined in mathematics
@МаратМаранкян3 жыл бұрын
@@prashant2650 in complex numbers too?
@BenedictMHolland3 жыл бұрын
It is a rule that 1 to any power is 1. I assume this is true for all numbers but whatever tricks he is doing, you should always get a div by zero error.
@DepFromDiscord3 жыл бұрын
“Yes, but not all the time” YOU’VE BROKEN MATH
@lykou18213 жыл бұрын
The math police have issued a warrant for your arrest.
@ultrio3253 жыл бұрын
He went from: guys I have pen and I do math to: 筆子曰:「無實數解既找虛數解」。
@yehe2973 жыл бұрын
"no real number solution then go find complex number solution"
@shinobi51898 ай бұрын
@@yehe297doing gods work
@danielc5313 Жыл бұрын
Answer from google: Logarithm is not defined for base 1.
@arandomelf30503 жыл бұрын
As a random 13 year old, my mind imploded and exploded at the same time
@ManishGupta-hb4iu3 жыл бұрын
kzbin.info/www/bejne/rIbGf3ePoaenpcU
@jondo76803 жыл бұрын
So it cancels out and nothing happen to your mind.
@vibaj163 жыл бұрын
@@jondo7680 nah, it just disintegrated in place
@jondo76803 жыл бұрын
@@vibaj16 hey, I'm just someone trying to make a 13 years old feel dumb. Don't come with such high level stuff to me.
Wolfram Alpha does not assume x in R. Wolfram Alpha assumes the principal branch of exponentiation. In the principal branch, 1^x = 1 for all complex x. In order to get any other value, you have to assume the non-principal branch of Log(1).
@robdenteuling32703 жыл бұрын
@@angelmendez-rivera351 This man complexes
@abdulkabeer73133 жыл бұрын
what is this principle and non principle branch?
@tupoiu3 жыл бұрын
@@abdulkabeer7313 Principle just means restricting our polar form to have an argument between 0 and 2pi.
@volxxe3 жыл бұрын
@@tupoiu isn’t it from -pi to pi?
@edgardojaviercanu47403 жыл бұрын
You have just extracted juice from a stone. It´s beautiful.
@PicaroPariah3 жыл бұрын
is this an idiom in your country?
@edgardojaviercanu47403 жыл бұрын
@@PicaroPariah Not really. The expression in Spanish is: "sacar agua de las piedras". I preferred to use "juice" instead of "water". A failed poet, as you can see.
@porfiriodev3 жыл бұрын
@@edgardojaviercanu4740 in portuguese is "tirar leite de pedra" which means extract milk from stones lol
@mihailmilev99093 жыл бұрын
@@yoonsooham3261 f
@mihailmilev99093 жыл бұрын
@@porfiriodev in Bulgarian it's water too
@JamesHesp2 жыл бұрын
There are many things wrong with this video, but they serve as good indicators of why we need the concept of branches in complex analysis. Let us first see that some things are certainly not right: Suppose the reasoning by BPRP works, and that one can indeed write ln(1) = ln(e^(2pi*i*n)) = 2*pi*i*n, for any integer n. Then we run into the problem that ln(1) = ln(1^2) = 2ln(1). This immediately implies that ln(1) = 0, because that is the only solution to x = 2x. So either BPRP is wrong, or ln(1^2) = 2ln(1) is wrong. However, this same property of logarithms is used by BPRP himself when he writes ln(e^(2pi*n*i)) = (2pi*n*i)*ln(e), thus in any case BPRP's argument is not self-consistent. This property of logarithms should be familiar and we would certainly want this to be true. Let us now get to the heart of the problem. BPRP did not consider that in the case of inverting the complex exponential, you may not use all properties that we are used to when dealing with logarithms of real numbers. To 'invert' the complex exponential, you need to choose a specific branch, precisely to deal with problems like the one that we see in the video, namely that 1= e^2pi*i = e^4pi* i = e^6pi*i = e^(2pi*n*i) for any integer n. What does choosing a branch mean? Well from these equalities we see that there is no single inverse value to the complex expontential for the value 1: We need to choose one of the values 2*pi*n*i to get a step closer to defining something like an 'inverse' function to the complex exponential. Making such choices in order to always know which value to pick is (crudely speaking) what mathematicians call choosing a branch. The natural logarithm for complex numbers is an example where such a choice of branch has been made: The natural logarithm is defined for complex numbers by choosing the principal value branch, which restricts to the interval (-π, π]. This means that even though 1 = e^2pi*n*i for any integer n, when we use ln(e^(iθ)), we choose the value inside (-π, π] (even if θ is outside this interval!). In the case of ln(1) = ln(e^(2pi*n*i)) for any integer n, the natural logarithm then simply gives 0. This last point stresses that ln(e^(2pi*n*i)) = 0, thus the argument in the video does not work. Then one might be tempted to defend BPRP's argument by saying that he implicitly chose a different branch for the natural logarithm, precisely by asserting that ln(1) = 2pi*n*i for some integer n other than 0. However, even then one encounters the problem we discussed above: ln(1) = ln(1^2) = 2ln(1). This does not mean that one cannot take different branches for logarithms. Instead it means that when we do take a different branch, we cannot expect precisely the same rules to hold for our logarithm. In particular the rule for logarithms that log_a(b^c) = c*log_a(b) does not hold anymore if we choose a branch corresponding to ln(1) = 2pi*n*i for n ≠ 0. Of course this is a cherished property of logarithms, and motivates even more why mathematicians prefer to choose the principal value branch: It is the only branch in which this property holds. Thus in conclusion, BPRP's algebraic gymnastics to solve the equation 1^x = 2 is not correct, and upon further inspection Wolfram Alpha turns out to be exactly right: There is no solution to this equation. However there are some solutions to the equation 1 = 2^(1/x), which is NOT an equivalent equation. But my comment is long enough as is. If anyone is interested, I can elaborate more on this later.
@kevinruiz56592 жыл бұрын
Pretty interesting comment. I would like to see why 1 = 2^(1/x) its not an equivalent equation to 1^x=2. I'm guessing that it is because we also have to choose a branch of the function f(z)=z^(1/x) to apply on both sides?
@NowhereMan56912 жыл бұрын
interested
@aura-audio2 жыл бұрын
Thanks for bringing this up! In an engineering course I'm taking, the instructors were very up front about telling us to give our answers in terms of this interval. Now I'm starting to see why. Time to do some more research/learning.
@General12th2 жыл бұрын
Does this video do a disservice to the field?
@Lightn0x Жыл бұрын
It's much easier than that... the video is wrong from the first minute where they take log base 1 of both sides. You can't do that, that's equivalent to dividing by 0.
@ejb79693 жыл бұрын
Question: Do you need the negative sign in the numerator? If n runs through all positive and negative integers, it's the same "solution set" with or without the negative sign ... ... isn't it ?? PS - Your gentle emotional delivery here (moments of disappointment, exasperation, near-defeat) is a refreshing new contribution to the art of math videos!
@carultch17 күн бұрын
It's a similar issue as when you rewrite the following diffEQ solution: y = e^(-x + C) as: y = C*e^(-x) The C came from a constant of integration that was completely arbitrary, and when we rewrite the final solution, the C is still an arbitrary constant. However, it is NOT the same arbitrary constant as you originally generated. Some professors insist you either assign subscripts or a tilde on top of one of them to tell them apart, or use a different letter entirely. If it were my choice, I'd accept a reuse of C, as long as write "Reassign C", or something else that means the same thing, to indicate that it isn't the same letter. Likewise, with this solution of: -i*ln(2)/(2*pi*n) The value of n is an arbitrary integer, so you could just as easily assign a different letter like m, and write it as: i*ln(2)/(2*pi*m) Heehee....pi m. Dr Peyam, this one is for you.
@vellagang6783 жыл бұрын
Wait a minute, Log base 1 is undefined Anyways, The pokemon in his hand is more important
@Kokurorokuko3 жыл бұрын
So is division by 0, but he managed to get past it
@galgrunfeld99543 жыл бұрын
@@Kokurorokuko L'hopital would be proud.
@amineelbahi25283 жыл бұрын
base 1 is undefined in R , he's working in C
@DanPolo30003 жыл бұрын
base 1 is undefined in any Field, what he's doing is messing with the fact that log function on C is not a function by definition (one value of z leads to infinitely many values of log(z)), we have to use the principal value of log, the Log function, instead, which locks the n value to 0, and is bijective.
@DanPolo30003 жыл бұрын
Literally doing that, if z1 = z2 then log(z1) = log(z2) in complex field, which is false (log in lowercase is not bijective!!!).
@akshitsingh59123 жыл бұрын
Teacher : 1 to the power anything is 1 BPRP: Hold my M A R K E R
@gabrielfoos93932 жыл бұрын
another way to solve this is 1=i^4n, n being an integer different than 0, you then have ln(i^4n)=4nln(i)=4nln(e^pi/2*i)=4n*pi/2*i, so ln(2)/ln(1)=ln(2)/4n*pi/2*i=ln(2)/2n*pi*i=-i*ln(2)/2n*pi, same result with a slightly different method
@Lightn0x Жыл бұрын
All good until "ln(2)/ln(1)". ln(1) = 0, so you are juat dividing by 0. Which is the same mistake in the video (just that in the video it's packaged differently). Wolfram alpha is correct, the equation has no solutions (including complex ones). The only way you get solutions is by doing something illegal (like dividing by 0).
@gabrielfoos9393 Жыл бұрын
Yeah you are right but breaking the law is fun
@MagicGonads Жыл бұрын
@@gabrielfoos9393 ln(1) is not 0 for other branches of the logarithm in C, so it can be divided by for those, it's only the one that aligns with the real-valued logarithm (the principal logarithm) that has ln(1) = 0, it's usually the most convenient to use but *not the only one*
@gabrielfoos9393 Жыл бұрын
Oh okay well I’m pretty new to complex logarithms, thanks for clarifying !
@flowingafterglow6293 жыл бұрын
I think this is probably my favorite problem you've done. It just shows how completely messed up things can get when you get to the complex plane, to the point where 1 raised to a power does not equal 1. Complex analysis is just bizarre. But in the end, this problem is so dang easy. You can get to things like -1^x = 10 x = i ln 10/pi How crazy is that?
@chem7553 Жыл бұрын
Both beautiful and horrifying😆😆😆
@W.2026 Жыл бұрын
Why tf is he holding a pokeball is he trying to get sued by nintendo
@molybd3num8235 ай бұрын
it's his mic
@alkankondo893 жыл бұрын
The most EPIC beard in all of the KZbin Mathematics community! Also, LOL at that look-of-disappointment at 2:31! 😆
@blackpenredpen3 жыл бұрын
LOL thanks!
@ejb79693 жыл бұрын
Love your username! The double-sharps in your logo caught my eye. Is that where the triple-sharp is in the Quasi-Faust?
@alkankondo893 жыл бұрын
@@ejb7969 Oh, wow! Thanks for noticing the reference. NOBODY has EVER noticed it before you! I extracted these notes from the 3rd movement of the Concerto for Solo Piano (Op.39, No.10). In addition to Quasi-Faust, there are 2 occurrences of a triple-sharp in this movement, after it modulates to the parallel major, F-sharp major. Again, thanks for noticing!! 👍😀
@ejb79693 жыл бұрын
I didn't know that about the Concerto movement, and I've been over that score many times! (As a listener, not a player.)
@alkankondo893 жыл бұрын
@@ejb7969 Yeah, for the longest time, I also only knew about just the triple-sharp of Quasi-Faust. That’s the example that’s widely used in mentioning Alkan’s use of triple-sharps. I just happened to hear about the ones in the Concerto. They’re very easy to overlook in the torrent of notes in the score!
@fetchfooldin32523 жыл бұрын
😂😂😂 omg. I love the fact that you're searching for interesting equation to solve. That's amazing 👏🏻 keep going . This is your folower from Morocco ❤️
@molybd3num8235 ай бұрын
hello fellow moroccan
@ibrahimmassy27533 жыл бұрын
All depends of the branch of logarithm chosen because k^x=exp(x*"log"(k)) where you need specify the 2*pi magnitude interval of the imaginary part of function "log" how is defined; if it contains 0 there are no solutions. For example, the case of principal Log doesn't work because Log(1)=0; thus, for powers of principal branch 1^x is always 1
@docteuressciencemathematiq8461 Жыл бұрын
In general, we can find the solution of x 1 = P(n) where P(n) is a polynomial in R or C the solution is: -i(2 *Pi* k + Log ||(P(n)||)/2*Pi with ||(P(n)|| the modulus of P(n) and k a relative integer. it is the magic of complex numbers that allows this in particular the possibility of writing: 1= exp(2*Pi*k)
@twilightfox69483 жыл бұрын
*blackandredpen: writes log1(2) to the board* Me: *wait, that's illegal!*
@Compasseye3 жыл бұрын
I came here to see how many people undestand this, because boi i really weak in math. It took me hours to understand this, and open bunch of books about ln, log, and how they works. By the way i'm in 8th grade, and your vids helped me to understand lots more of how to calculate. So i thanked you for that 💙
@eekumbokum67703 жыл бұрын
I hate to do this but......r/imverysmart
@Safouan03 жыл бұрын
@@eekumbokum6770 Not at all lmao. Lots of people are motivated to make positive changes in their lives...
@Lightn0x Жыл бұрын
The funniest thing is that the video is just wrong :). So if you really claim you understood it, then you didn't understand it enough :P. Hint: think about what it means to take log base 1.
@xinpingdonohoe39787 ай бұрын
@@Lightn0x you don't have to take log base 1 though. Just express 1=e^(2πni) and take the natural log.
@conrad53423 жыл бұрын
Thank you. This is one of the cases my Math Prof. warned me about. Does the imaginary unit really enable such relations or are the exponential / logarithmic laws more limited in the complex world than one might first time think?
@alexaisabird3 жыл бұрын
the exponential function in the complex plane is not biejctive, and ln(z) is not the inverse of e^z
@frfrchopin Жыл бұрын
This is hard to accept since it's zero division of exponents
@helo38273 жыл бұрын
The first thing I noticed is he don't have glasses.
@Nino-eo8ey Жыл бұрын
After finally having learnt complex numbers, it feels so good being able to understand these types of videos! Keep up the great work.
@chzhao Жыл бұрын
A complex number raised to a complex power has multiple values. If we treat 1^x as such a multi-value expression, and the question is to find x so that 2 is one of values of 1^x, then x=-i*ln2/(2*pi) as stated in the video is a solution. Actually there are multiple solutions: x=-i*ln2/(2*pi*k) with k being any non-0 integer.
@violet_broregarde3 жыл бұрын
This answers a question I've had for a long time: does some math treat e^2 and e^(2+tau*i) differently? I learned something today. Thank you so much :D
@angelmendez-rivera3513 жыл бұрын
e^2 = e^(2 + 2·π·i), but this does not imply 2 = 2 + 2·π·i. z |-> e^z is not injective if z is a complex number.
@tfdtfdtfd3 жыл бұрын
4:15.....why would he assume that WE know "HE" likes to be on the top rather at the bottom?
@MrAlRats3 жыл бұрын
He considers all his viewers to be his rabid fans and expects us to have done the necessary enquiries so we are prepared if we ever meet him.
@Simqer11 ай бұрын
This proves it, complex numbers were invented by mathematicians who were on some extremely good weed.
@jamespetersen21211 ай бұрын
There should be more solutions. You can also split 2=2*exp(2*pi*m*i) and use the additive rule of the logarithm. The general solution is x=-iln(2)/2/pi/n+m/n,n is not 0,m,n are integers.
@sherkoza Жыл бұрын
You can change ln(2) also to ln(2e^(i(0+2pi*m))) This way you can get more answers X=(ln(2)+2pi*m*i)/(2pi*n*i)
@duggydo Жыл бұрын
I think this might be why Wolfram won't give an answer. Too many possibilities.
@yichen63133 жыл бұрын
Cancelling log and exponent so casually gives me anxiety about messing up with branches....
@writerightmathnation94813 жыл бұрын
Good.
@mr_cod33 жыл бұрын
Hello there is a mistake from the beginning when u wrote X=log(2) because log(1)=0 so basically X is multiplied by zero
@dissonanceparadiddle3 жыл бұрын
Moving math into higher or adjacent dimensions is always a very neat thing to see. Things you think that are impossible become commonplace
@official-obama3 жыл бұрын
no, jerry, you’re forgetting quaternions. there _is_ a solution for sin(e^(7x^2))=53x-25,000 -elementary schools in the year 900000
@EebstertheGreat Жыл бұрын
In general, if 1^x = z, with x,z ∈ *C* and z≠0, then x = k + [(ln z)/(2πm)] i for some k,m ∈ *N* with m≠0. In this case, the values of 1^x are the principal values of z^(-n/m) for all integers n. So the principal value of 1^x is always z^0 = 1, but the desired value is always in there. Specifically, 1^x = z on the branch n = -m.
@slavsquatsuperstar3 жыл бұрын
3:02 I’m pretty sure I heard “we are adults now, so we use 2 pi” xD
@d4slaimless3 жыл бұрын
Table at 6:03 it should be Ln, not Log I guess? Btw, like this you can get any number by replacing ln(2) with ln(k) where is the number (or it's power) that you want.
@wdyuyi5 ай бұрын
actually "log" could have different meanings in each category we dealing with. Therefore, denote as "Ln" might be more clear👍
@hopewelltv83413 жыл бұрын
All I could see was the beard.
@betabenja3 жыл бұрын
all I could hear was "stop looking at the beard! concentrate! concentrate!"
@riyadamin1913 жыл бұрын
And his mic
@riyadamin1913 жыл бұрын
This is also a clickbait 😂
@itskarudo3 жыл бұрын
wait, that's illegal!
@robobuilder133511 ай бұрын
I would like to note that this problem has some additional weird properties. For example, if you attempt to solve this via a limit, you find that if you try to solve for 2^x=1, you end up with this being true as x approaches 1/infinity, essentially as the infinitieth root. Or just use the identity where raising something to the power of 0 makes it into 1 However on the other hand, you can prove that for all real exponents of x, 1^x is 1. Imaginary numbers are useful, yes, but they also kinda messy and can pretty commonly give non practical solutions(like in the real world, there is no real operation to do i rotations or whatever) Long story short, solutions like this never really see use except when i is representative of some specific axis. Like the phase of an AC circuit. And even then, the most you really see of i is e^i(angle) or a basic complex number. Or at least, in the classes ive been in so far. College junior rn
@The_Commandblock11 ай бұрын
You dont need a limit. You just need a zero. Anything to the zero is equal to one
@robobuilder133511 ай бұрын
@@The_Commandblock I didn't use the 0 power property since the video didn't really go over them and I wanted to use a way other people on the comments didn't use. I will add that in though
@slytan15063 жыл бұрын
4:15 "I don't like to be on the bottom I like to be on the top"
@nicknice78393 жыл бұрын
The pokemon in the pokeball in his hands probably learnt more maths than me
@W1ckedT0astr3 жыл бұрын
Math itself must've felt violated after the problem was solved
@viktorramstrom37443 жыл бұрын
Wh-what are you doing, redpenblackpen-chan?
@Usuario4593 жыл бұрын
@@viktorramstrom3744 what are doing step brpr
@OptimusPhillip Жыл бұрын
Yes, if x = (-ln(2)*i)/(2nπ), where n is any non-zero integer. But only for non-principal values of 1. The principle value of 1 is e^(0*i), or e^0. And raising e^0 to any power will yield e^0, which is 1. So entering 1^x into a calculator will yield a result of 1, even if you use this x value. But if you use this value of x when entering 2^(1/x) into a (sufficiently sophisticated) calculator, you will, in fact, get 1. EDIT: Cool, I didn't miss any details. I actually solved this slightly differently from how you did. I actually jumped straight into polar form, restating the equation as (e^((0+2nπ)*i))^x = 2 right off the bat.
@LK90512 Жыл бұрын
For increased effect you should have asked if 1^x can equal e. Then following the same reasoning you would have concluded that 1^(-i/2pi) = e, which is quite beautiful, and correlates directly with euler's formula: [1^(-i/2pi) = e]^(2pi*i) -> 1 = e^(2pi*i) I know the math is ultimately not correct, but it's just for good fun.
@xinpingdonohoe39787 ай бұрын
It's not "ultimately" correct, but it's not particularly wrong either. The principal value of 1^x is 1. Always. But if x is not an integer, you will be getting either finitely many or infinitely many more values. 1^(ln(2)/2πni) has principal value 1, but can take on other values such as 2. 1^(1/2) has principal value 1, but can also take on -1.
@sarthakchavhan3 жыл бұрын
log functions are not defined for base 1 . in fact its in the definition that base of log is positive and can't be equal to 1 . so how can you take log with base 1 on both sides 🤔
@spiderjerusalem40092 жыл бұрын
Don't you acknowledge it in the first place? Suppose look up how "i" was invented? looking up for complex value, then one may have to go through complex solution just as doing things with √(-1)
@aura-audio2 жыл бұрын
I love this! I'm taking an EE class right now which revolves around complex numbers, and your videos are super helpful/inspiring.
@EEEEEEEE11 ай бұрын
E
@denissudarev3 жыл бұрын
Is base 1 for logs exist?
@SupaGut20013 жыл бұрын
I have the same questions...
@SupaGut20013 жыл бұрын
But seems that exist in complex
@pendragon76003 жыл бұрын
No. Also, 1^x = 2 has no solutions.
@neon_trotsky3 жыл бұрын
When watching these videos I feel like I am fcking up whatever little maths I know
@hamsand25273 жыл бұрын
@@pendragon7600 then how did he do it using the laws of math?
@MichaelGrantPhD2 жыл бұрын
This is basically a division by zero issue. After all, 1^x = e^(ln0 * x) = e^0 = 1.
@Lightn0x Жыл бұрын
Exactly. Taking log base 1 of both sides is equivalent to dividing both sides by 0, which is why I am very confused why the comments are so positive. I thought in 2023 we would be over the "divide both sides by 0 to show that 1 = 2" type of "maths"...
@Shikogo Жыл бұрын
Dude your flicking motion to flip between the markers is so smooth.
@Electric_Bagpipes3 жыл бұрын
When Wolfram alpha breaks you know your fkd.
@marvenkorte34373 жыл бұрын
The result you calculated is only correct for the equation 1=2^(1/x). The taking of complex roots is not bijective when you consider values with higher imaginary part than pi. This means that 1^x=2 does not have any solutions like wolfram alpha says. And your reasoning for why the number also satisfies the equation 1^x=2 is flawed because you assume the identity ln(e^x) = x holds for any complex number. Which is not true because the complex version of the exponential and Log are again not bijective on the entire complex plane only on the principle branch.
@H336-p1v3 жыл бұрын
How about "x tetration i = 2" :0
@angelmendez-rivera3513 жыл бұрын
x^^i is not a well-defined operation.
@nanamacapagal83423 жыл бұрын
How does that even work
@Endrit719 Жыл бұрын
if a guy holding a pokeball tries to prove something, he is trying to prove nothing
@axbs48633 жыл бұрын
“But you know me I don’t like to be on the bottom I like to be on the top” bruh 4:16
@helloitsme75533 жыл бұрын
Depend on which branch of 1^x you take
@shambobasu15793 жыл бұрын
The title should be "10 ways to (not) write zero"
@nilsgoliasch2443 жыл бұрын
I have no idea why I'm watching this on a Saturday evening, but here I am
@chayakumarsedutainment47993 жыл бұрын
Can you solve this equation: X^(log2 to base 3) = X^(1/2) + 1 Can you find the value of X here? It should be 9. I found this question on internet without solving steps but had a final solution
@MattMcIrvin Жыл бұрын
The way I thought about it is that i is a non-principal fourth root of 1, and you can get all kinds of real numbers by raising i to an imaginary power, so the answer is "sort of, if you abandon computing principal values."
@xinpingdonohoe39787 ай бұрын
Yes, the principal value of 1^x is always 1, but other values may exist, and can cause some fancy shenanigans. Even as simple as 1^x=-1, you'd get solutions of the form x=1/(2n). Take the principal value of any of these 1^(1/(2n)) and you get 1, but other possibilities include -1 for each.
@blackpenredpen3 жыл бұрын
2^x=-1 vs. (-1)^x=2 but in ONE minute kzbin.info/www/bejne/pnPRhGqBYtp1Y8k
@secavara3 жыл бұрын
Regarding this video, the two answers from wolfram alpha are consistent. b^z when b is real and positive and z is complex, is being taken in both equations as Exp[z Log[b]], where Log is a branch in which Log[b] is real. Hence, you get different answers in Wolfram depending on whether you ask Solve[1^x == 2, x] or Solve[2 \[Pi] I x m + 2 \[Pi] I n == Log[2], x]. Given the fact that you have the possibility to reach big audiences, you could make a bigger effort to present these topics in a more precise fashion, and use this topic to explore its nuances, and you are intentionally choosing not to do this.
@angelmendez-rivera3513 жыл бұрын
@@secavara This is dishonest criticism, as the purpose of these videos are not to present the topic rigorously as understood by mathematicians. The purpose of these video are to showcase what happens when one is not careful, and to present topics heuristically, which is necessary before a student can begin to approach a complex topic with rigor. I have my disagreement with BPRP regarding how some subjects should be presented, but implicitly accusing him of being dishonest via your tone and wording when the criticism is not even applicable is itself dishonest and hypocritical.
@j.hawkins87793 жыл бұрын
i love how you hold a pokeball, because i love pokemon
@ManishGupta-hb4iu3 жыл бұрын
kzbin.info/www/bejne/rIbGf3ePoaenpcU
@Kdd1603 жыл бұрын
@@secavara No! Bprp’s just having fun for the viewer’s entertainment!!
@yoursole68173 жыл бұрын
You should make a video on x^x=x^2 Obviously 1 and 2 are the only real solutions but I did this with some code I wrote, it would be interesting to see how the actual math works. Also what about complex solutions
@blackpenredpen3 жыл бұрын
Love your quick explanation! You would have to rule out 0 since 0^0 is not defined. Also 0^0 isn’t 0^2.
@MathElite3 жыл бұрын
@@blackpenredpen Hey thanks for the feedback! I appreciate it so much. I edited the title and description I'm trying the whole math youtuber thing, I like it so far (inspired by you and others)
@esajpsasipes2822 Жыл бұрын
@@blackpenredpen i think it is defined for algebra purposes, but even with that it's not a solution as 0^0 =/= 0^2 => 1 =/= 0
@zeroitedono25473 жыл бұрын
any tips on how to increase self esteem in maths? i was always interested in maths but people are just so toxic and competitive man
@iCarus_A3 жыл бұрын
In what environment are you encountering toxicity? Personally, I just didn't really find maths fun until I got to higher maths like Mathematical Reasoning, Graph Theory and Group Theory. It really pays off when you finally see your work in one subject area began helping you understand other areas (ie, some understanding of computer datastructure helps a lot with Graph Theory, and Vector/Matrix stuff is all in Group Theory). That just never happened for me when I did high-school stuff. I also don't usually talk with other people about maths and instead rather figure stuff out from scratch myself with help from textbooks or forums, so that might be why I never felt any peer-pressure or competition
@zeroitedono25473 жыл бұрын
@@iCarus_A like I remember doing a ton of stupid and easy mistakes on a maths paper, people thought oh I must be a genius or something and then I overheard people saying that they thought I was smart and it really hurt to be honest everyone is just so competitive idk why and i feel my passion for maths getting dimmer-its not the maths thats hard its just overcoming the mental barrier when doing maths, for some reason when I do maths I just assume I can't do it since thats what some of my teachers said
@iCarus_A3 жыл бұрын
@@zeroitedono2547 Something I learned is that the more passion you hold for a specific hobby/profession/pursuit, the more likely it is for you to get stuck in a mentality of perfectionism. At some point, you just become so paralyzed with the fear of failure that even taking a single step feels difficult. The first thing to know when recovering from something like this is to realize that you're not perfect and the only way to get better is to keep doing it. I'm usually fairly good at maths and tend to lead math-related projects in small groups, but I also make tons of small mistakes regularly. A mistake fixed is not a mistake at all, and it's far more important to get the overarching goal right -- try to get a friend or someone else to talk your stuff over without getting too involved and see if you can explain your thought progress clearly to someone who's not even deeply familiar with your field.
@zeroitedono25473 жыл бұрын
@@iCarus_A thank you
@iCarus_A3 жыл бұрын
@@zeroitedono2547 Happy to help even in the slightest :D
@Rikky-the-Unstoppable Жыл бұрын
1ˣ = 2 ⇒x = In(2)/In(1) ∵ aˣ = b ⇒ x = In(b)/In(a) ⇒x = In(2)/0 [any no. divided by 0 is undefined] ∴The solution of x is undefined
@ffggddss3 жыл бұрын
First reaction to the title question: No. 1^x can't = anything but 1. Take ln of both sides: x·0 = ln2 . . . 0 = ln2 . . . contradiction ==> no solution. Now to watch, to see what it is that takes 7 minutes to do. Hmmmm, well if you want to go that way, we also have to add 2nπi to "ln2" in the numerator, don't we? Fred
@blackpenredpen3 жыл бұрын
Yea I think so. Also, some one pointed out this was just the imaginary part of the answer. So the full answer should be like a+answer in the video for any real number a.
@ffggddss3 жыл бұрын
@@blackpenredpen Good point. Adding any real number, a, to the exponent, just multiplies the result by 1ª = 1, after all. Fred
@nipunkumar11753 жыл бұрын
*I don't know if 1^x will be 2 or not,but I surely like your Pokemon mic*
@micheledejana78493 жыл бұрын
It's a bit controversial: when you solve ln(e^2pi*i) you are assuming that 2pi*i = 0 and that's true only if 2=0 or pi=0 or i=0
@blackpenredpen3 жыл бұрын
The weird thing is 1^complex has multiple values.
@lordmomstealer3 жыл бұрын
@@blackpenredpen I have question for you 4^x+6^x=9^x FIND THE VALUE OF X
@geometrydashtzolkin10253 жыл бұрын
@@lordmomstealer Its a boring lambert w function type of problem
@pravargupta62853 жыл бұрын
@@lordmomstealer Really bro? We have done these type of easy questions when we were in class 9 and 10!!
@pravargupta62853 жыл бұрын
@@geometrydashtzolkin1025 Its a simple equation when you divide both sides by 9^x and then substitute (2/3)^x as t and then it forms a quadratic, quite easy to solve
@antman76733 жыл бұрын
One of the first videos I was ahead of. Guess studying is paying off.
@ReasonMakes10 ай бұрын
Would be lovely to use tau instead of pi here
@Ostup_Burtik9 ай бұрын
I love tau
@fenghuazhou62313 жыл бұрын
1^x = 2, So x = Log(base1)2 = Log(2)/Log(1) The answer will be a complex number a + b i, where a is any rational number, and b = -Log(2)/(2 n Pi), n be an integer.
@stewartcopeland49503 жыл бұрын
I will remember this: it is enough to take off your glasses to approach the problem from another perspective and solve it !
@cpotisch3 жыл бұрын
Wait but didn’t we lose information when we took the log and moved the exponent out? I think you ended up getting an improper equality because the function is not one to one.
@pravargupta62853 жыл бұрын
Yes it is indeed a multivalued function so we can take log and put it out till it is a multiple of (0+2(i)(k)pi) , k belongs to integer
@cpotisch3 жыл бұрын
@@pravargupta6285 My point is that if a function f has a multivalued inverse f^-1, you can't f^-1(f) without losing information. Like arcsin(sinx) does not equal x. So I don't think what he did is valid.
@angelmendez-rivera3513 жыл бұрын
@@cpotisch What he did is entirely valid. The expression 1^x is inherently multivalued if x is a complex number.
@thischannelhasnocontent86292 жыл бұрын
@@angelmendez-rivera351 Exponentiation with complex numbers is *defined* by the exp() function. a^x := exp(x*ln(a)) for all a and x. The exp() function is not multivalued.
@taragnor3 жыл бұрын
"They called me mad for trying to divide by zero! But I'll show them! I'll show them all! ahahahaha!"
@Yatornado Жыл бұрын
1*1^(x - 1) = 2 | divide by 1 1^(x - 1) = 2 | but 2 also equal to 1^x x - 1 = x -1 = 0 | and we can do it for any other integer not just -1, they are all equal to 0 1^x = 2 is not the same as 1 = 2 ^ (1/x). Just like 1 = -1 is not the same as 1² = (-1)². Raisng equation to the power might change it's answer.
@apotheosys1 Жыл бұрын
Abstract: It is correct, you just need to take serious care that it is about multivalued stuff, and in the middle its easy to mistake and use normal functions properties. (Like 1^x =1 for all reals is trouble if exponential is multivalued)
@ranjitsarkar31263 жыл бұрын
4:37 My brain: starts melting
@barneyy69423 жыл бұрын
Now I'm gonna dream about this, shit. (It's 2am here)
@anushkasaini81953 жыл бұрын
But here is about 11:30 pm in India
@pravargupta62853 жыл бұрын
@@rishabhagrawal80 now 12:07 lol we are everywhere!!
@shreyan13623 жыл бұрын
@@pravargupta6285 12:09
@lordmomstealer3 жыл бұрын
Its 12:24 am in india date is 8/1/2021
@1Jarousek13 жыл бұрын
It's 8:11 PM in Czech republic. Still 07\-i^2\(sqrt(4 084 441))
@xevira3 жыл бұрын
When the white hair comes in with the beard, BPRP will evolve into a super guru with mystical powers of calculus knowing how to write the Lambert W function in its elementary form.
@daituameunagabbia73903 жыл бұрын
The definition of a^b for a, b complex numbers is a^b=e^(b*ln(a)). So 1^i=e^(i*ln(1))=1. Now, if there is a complex number z=x+iy, with x, y real numbers, such that 1^z=2, we have 2=1^(x+iy)=(1^x)*((1^i)^y)=1*1^y, which means that 1^y=2, with y a real number. This is impossible.
@apizcyril94793 жыл бұрын
this guy can definitely explain to us how gojou satoru's power really worked mathematically