how is i^x=2 possible?
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- Опубліковано 9 тра 2024
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Can the power of the imaginary unit i ever give us 2? We know i^1=1, i^2=-1, i^3=-i, i^4=1, and so on. So is it possible for us to find x so that i^x=2? Well, the answer is yes but it is not real. #ComplexNumbers #Exponentiation #ImaginaryNumbers #Mathematics #blackpenredpen
0:00 We know i^1=1, i^2=-1, i^3=-i, i^4=1, and so on. So is it possible for us to have i^x=2?
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pls Can 2^x=0 in the field of complex num ?
@@ayoubbenchetioui6481 x is negative infinity
Please solve this equation: (-2)^x=2
Thank you
Where can I apply this useless knowledge?
We need more τ. e^-(τ/4 + nτ).
Complex world is crazy.
I wil be find new world
Name is fantastic number
By saying crazy, you are underestimating the craziness of complex numbers.
Gerçekten bu inanılmaz. Kuantum mekaniğinde de çok önemlidir.🙂
I am not even good in real numbers
Because the world is complex by itself
I really love it when you say, " I don't like to be on the bottom, I like to be on the top"🤗🤗🤗🤗
🤨
A very normal, totally not suspicious, comment
🤨📸
@@thexavier666 yea absolutely, no complications there right? 🤨
@@noreoli true, no complications 👌👌👌
Of course!
In fact, a^b can always = c, for all Complex numbers {a,b,c} where {a,b,c} not = 0 or 1.
I’m scared to look up that proof
@@DroughtBee The proof is left as an exercise for the reader
@@maximilianarold I think there isn't really a proof. I think it's just an assumption. Like 5^3 will always be equal to a value. Lets call this value c. so 5^3 = c. As we know, c = 125, so 5^3 is 125. We knew what a and b were, but we didn't know what c was, but there was an answer. So what if we know the value of a and c, but not b. so i^b = 2. Last time we didn't know a varible, there was an answer, so there must be an answer again. That is my assumption on how he got it, but he might actually have some reason behind it...
Let a^b=exp(b*ln(a)), since exp: C -> C* is surjective, there is a complex number z such that exp(z)=c.
@@TheEGod. The proof is actually really easy. Suppose we want to find b such that a^b=c with a,c not equal to 0 or 1. Take a natural log of both sides to get b*ln(a)=ln(c). So b=ln(c)/ln(a) works. That's it.
Note that this is the same answer from this video. Here c=2 and a=i, and ln(i)=i*pi/2, so b=-2*i*ln(2)/pi.
The other case is to suppose we want find a such that a^b=c with b,c not equal to 0 or 1. This one is even easier: a = c^-b.
"Complex number is a pathway to many abilities some consider to be unnatural." - Chancellor Palpatine said during his complex analysis lecture
"Please don't say 90 degrees, we are all adults here"
Please don't say 90 degrees, as we are all adults now...I think I laughed a little more than I should have lol
😆
This one is fun. I actually found your e^(e^x) = 1 video first.
I took a different approach to solving this. I tried computing the log-base i of 2, which using log quotient rule, is equivalent to ln(2) / ln(i), the bottom part of which I was able to solve by getting the angle (or family of angles) that has a cosine of 0 and a sine of 1, which is pi/2. Thanks!
Awesome, thanks!!
@@blackpenredpenhello
I have been watching your videos for about 2 years now, now that Im in UNI, and im learning more and more about math, I can follow the videos much better, and I just love to see the progress, and the interesting things you show here on youtube
please upload more, I really enjoy your videos
Great video... much appreciated. Your info shared and your style... and your nice manner.
Very nice explanation about why the +4n part is needed to complete the answer. i raised to any multiple of 4 will always result in 1.
this was such a fun video lol i love how happy you get
Your videos are amazing, thanks professor!!! 🤗🤩🥳
Great, that is fascinating 👍
I'd add a simplification. You can pull the 2 inside the ln as an exponent, so 2*ln(2) = ln(2^2) = ln(4). It makes the result a little prettier :D
Make π τ again.
Just asking how did you become so good in maths? I saw your previous videos for doubts and you make questions easy.
Practice solving problems, read college textbooks, participate in competitions, watch and learn proofs. Over a few years you'll rack up so much intuition and knowledge if you practice right snd consistently.
Can you do a video on how to change the pens in your hand? Thanks for your wonderful videos
4:17 me too dawg glad we got one thing in common 💯
You are the best ❤
Hey, nice video, I've got a fun challenge for you: Determine all positive integer pairs (p, q) for that p^q + q^p is prime. Not what you usually do, but it has an interesting solution.
oh hello there lol
Nice question. xD.
Since you could say that p is equal to a number 2ⁿ-2 and q is 1 therefore p^q+q^p is 2ⁿ-1. Since i know that it isnt determined whether there is an infinite amount of mersenne or not, answering this question, would be either quite impossible or just impossible.
A couple mersenne examples would be
(2,1): 3
(4,1) : 5
(6,1) : 7
.... You could generally say, that there are infinitely many tupels where p is a prime -1 and q is 1 so that (p -1)^1 + 1^(p-1) = p
So my answer would be that there are infinitely many tuples, as many as there are Prime numbers...
(2,3)...
wlog assume p=2 and eval mod3
Very easy ngl
@@zoomlogo lmao hi
**For those who want the tl;dr explanation:**
i^x = 2, so x = log base i of 2 = ln(2)/ln(i) by base-change. This is just ln(2)/(pi/2 i) = ln(4)/(pi * i).
I'm so prone to clickbait thumbnails when it comes to maths, usually they don't work on me but your thumbnails always get me😂
Our genius is back we amazing questions 😀
you could also take the ilog of 2 and rewrite it as ln(2) / ln(i) = ln(2) / 0.5ln(-1) = 2ln(2) / pi i if im not mistaken
Please make a video on how to solve any kind of ∑ problem...
I need to learn..
its a pleasure watching you. thanks
Broo this is insaneee 😵
this kind of math is so interesting to me
I never took precalc or a calculus class
just college algebra
we only got a slight introduction to imaginary numbers so all of this baffles me
glad I don't need calculus for my degree 😅
Thank you, sir
Very nice video wow I'm a really huge math fan and keep it up !
Thank you!
Stunning proof👍👍👍
This is my humble request to whomsoever is reading, please consider my problem:::
By Euler's identity
=> e^iπ + 1 = 0
=> e^iπ = -1
Square both the sides
=> (e^iπ)² = (-1)²
=> e^2iπ = 1
take natural log of both the sides
=> ln(e^2iπ) = ln(1)
=> 2iπ = 0
Please explain😢😢😢
By the way I have a couple more such demonstrations that kinda contradicts the identity which I am unable to recall rn, although I also have worked with this a lot, it feels not a peaceful identity at somepoints, But I do remember that the problem in the eq comes right after squaring both the sides. I am not sure with all this as I did this a long time ago but whatever I wrote is what I remember rn, sorry if I wasted any time, please consider atleast replying 🙏🙏🙏
Here is what you wrote:
Square both the sides
=> (e^iπ)² = (-1)²
=> e^2iπ = 1
^^This is not true. For complex inputs, a^b^c is not always equal to a^(bc). The rule of complex analysis dealing with log and exponent branches says so. Or else, one can prove anything we want. Here's one of my personal favorites with this fallacy:
Suppose we have a real number a. Then, a = e^(ln a) by definition.
It follows that,
a = e^(1 * ln a)
= e^[(2iπ/2iπ)*(ln a)]
= e^[(2iπ)*(ln a/(2iπ))].
Applying our fallacy, we see that the above expression equals [e^(2iπ)]^[ln a/(2iπ)].
But e^(2iπ) = 1 by Euler's identity. Thus, we get:
a = 1^[ln(a)/(2iπ)]. But 1^x = 1, (unless you use the same fallacy!) so all a = 1. Thus, all real numbers are 1 (obviously not true).
BPRP actually did a video on this a while back, seeing if 1^x = 2 is possible. Long story short, that equation had no solutions. But, there was a solution to the equation 1 = 2^(1/x). Raising both sides to the power of (1/x) caused some issues with domain and range restrictions, so the solutions obtained were technically "extraneous."
@@ianzhou3998 Thank you very much for correcting me or rather teaching me the actual reason.
I just took a look at some articles, I didn't noticed the actual law which was a^b^c = a^(b×c) for these elements should belong to the Real world.
Well I was on the right path to find my mistake on my other such demonstrations that had the same mistake, when I said "the problem in the eq comes right after squaring both the sides".
Half knowledge is more dangerous than no Knowledge
Anyways thank you for explaining that much and telling about the video of 1^x=2 and sorry If I were to be silly. 😊
Can you make a video about the levi civita symbol, more specifically about some identities? Anyways, nice video, your work is appreciated!
I just discovered your channel and it IS GREAT!
Your enthusiasm is amazing, I had a math texher like that 25 years ago, you really remind him. (his name was Gustave😂)
I'm 38, love maths and I stopped at this level (high school math option in my country)
But because of life and the obstacles on the way, I never was able to pursue in polytechnic.
But I always kept a close link with mathematics and especially analysis. I litteraly do integrals during my free time, it's so beautiful..
My favorite is to trick arrogant people in the STEM fields with a "simple ∫ 1/(x^4 + 1) dx
Most people fall in the trap.
Anyway, I love your content and I'm gonna be watching a lot of it to stay sharp!
Thank you
Thank you very much!!
What trap?
@@DPME820 people think it's an Ln
Professor please make a video on tricks used to solve limits
I shudder to think of the complexity of any maths problem that would require the use of a green pen in addition to the red, black and blue!😄
5:09 "Check this out" watch the word "note" at 144p, trippy.
Great job!
Is there a use to solving this for the more general form of e^i(pi/2+2npi) and show that you can produce an x that satisfies all infinitely many integer values of n?
I looked at that expression and rewrote the theta value as (pi+4npi)/2 to make it more comfortable, then I set (e^i(pi+4npi)/2)^x = 2 and followed the same process to isolate the x from the equation.
(e^i(pi+4npi)/2)^x = 2, n ∈ Z
x * i(pi+4npi)/2 = ln(2)
x = 2ln(2)/i(pi+4npi)
x = -2iln(2)/(pi+4npi)
Someone could ask "but what if we don't start with e^i(pi/2)? What if we start with e^i(5pi/2), or e^i(13pi/2)?" I asked that while watching, but I realized that other polar values of i can still be raised to a power that makes i^x = 2.
Edit: Thanks to another discussion I thought about possible problems here and the reason why the video's focus on the principal value was there, based on where ln(x) isn't well-defined. If anyone has input on where this does and doesn't work, that would be appreciated!
If you wanna make it work for non-positive complex numbers, just change the 4npi part.
The solution of i^x = -2 is 2n - 2iln(2)/π
Will you do hypercomplex numbers or quarternions?
You are great teacher
Hey bprp, thank you for showing us all this crazy stuff, i am reeeeeeally enjoying while watching you
I learnt lambert w Function with you and i improved myself a lot, and for so long you didnt make a video about lambert W Function, i have little bit hard problem for you about this stuff; Can you solve x^(1/W(x))=y? I miss the w Function videos btw. I hope u see, thx again! (If you put in wolframalpha it may not show the solution but it is actually solveable, and the domains are suitable enough)
I have known a lot about complex from your video ✨
HI STEVE, CAN I GET A PROMO CODE FOR YOUR LAMBERT W FUNCTION PURPLE TSHIRT BECAUSE I WANT TO ORDER 100 OF IT AND THE SHIPPING PRICE IS 800 DOLLARS
Yes so true
i agree, the lambert w function shirt looks so good haha
bprp merch ftw
makes you feel smart
Could you send me an email blackpenredpen@gmail.com and let me know why you are ordering 100 t-shirts? I will see what I can do for you.
I took log base i on both sides and got x=logi(2)
I would love to see you go further than complex numbers and solve an equation with quaternions instead - maybe something like x^x = 2?
4:17 "i dont like to be on the bottom, i like to be on the top" xddd
@Blackpenredpen , look up this book called "
(Almost) impossible integrals, sums, and series" and do a video on it.
I just found this platform. So how does one start Calculus? Which video first?
lovely. Thank you. I will visit here whenever I got freetime like now.
i^i. . . . .my favorite
Very kewl video....love the info
The "secret" is always the same: put everything in base *e* (Euler's number)
we can use ln(z)= ln(|z|) + i(2npi+theta) too
can u do vid solve equation
erf(x) = 2 find x?
I didn't get why we have to write i^4n ? thank you !
I got the principal value for x just fine, but for the general solution I somehow ended up with 4n in the denominator.
Same :(
Splendid.
Bro you look so much better without a beard, no kidding
At school
Teacher: Whats your favorite number?
A random kid: 3
Another kid: 7
This guy: *i*
brother i took both side to the power i then replaced i^i by e^-pi/2
then after something i got x = i^-4lni/pi
pls explain how much wrong i am
Before watching video
i^x = 2
x=log_{i}(2)
x=ln(2)/ln(i)
ln(i)=iπ(1+2n)/2 for any integer n
x=(2ln(2))/(iπ(1+2n))
x=(-2 i ln(2) )/( π (1 + 2n) )
focusing on the principle value, we have -2i ln(2) / π
Edit: shoot I didn't notice the extra answers with raising i to the fourth power
I did it this way too before watching the vid lol I also didn't notice the extra answers, also I think you made a slight mistake, ln(i) = iπ(1 + 4n)/2 after you combine the π/2 with the 2πn into a single fraction and factor out the π. It didn't affect the principal answer tho
Request: is there a complex number x such that 2^x = x?
2^x = x
e^(x ln 2) = x
x e^(-x ln 2) = 1
-x ln 2 = W(-ln 2)
x = W(-ln 2) / -ln 2
Now there's a couple cool things of note here.
Any number such that n = W(-ln x) / -ln x can be represented as an infinite power tower. I'll post the proof in a separate reply underneath this one.
The other cool thing is that for 2^x = x, you can do 2^(2^x) = x or 2^2^x and following the chain, you're solving for the infinite power tower of twos
n^x = x
xln(n) = ln(x)
ln(n) = ln(x)e^(-ln(x))
W(-ln(n)) = -ln(x)
e^(-W(-ln(n))) = x
Identities of the W Lambert function tells us now that
x = W(-ln(n))/-ln(n)
For
n^x = x
So that's pretty cool, it's a shortcut to solve any convergent power towers.
Sir, what's about taking log both sides...
Can you do 100 related rates please
Can u do a video where to use blackpen and redpen
my man's hoarding whiteboard markers like they're Hagaromo chalk
😂
can u solve ln(4x+5)=4x-2016 ?
1:37 is i multiply by i allowed with index law(power to multiply) in complex world?
i = e^(ipi/2), so (e^ipi/2)^x = e^x(ipi/2) = 2, take the ln
of both sides: we have x(ipi/2) = ln(2) => x = 2ln(2)/(ipi)
In case of the derivation of y= i^x to y´=x*i if x=2: i^2(Y)= 2(y´)
"We are adults now, so say 'pi over 2.'"
Thank you for this. From the bottom of my tired heart.
Où avez-vous trouver votre t-shirt ?
In step two where you wrote (e^ipi/2)^x, isnt it incorrect to multiply the powers as one of them is complex? I saw it in another video where pi was falsely proven to be = 0 due to this mistake.
Could you please help me with a curiosity of mine?
Is there a way to create the exact formula for the following definite integral?
Integral between 0 and a:((((x^2)(b^2))/((a^4)(1-((x^2)/(a^2))))+1)^1/2) dx
Wolfram Alpha says that this is an indefinite integral - and the solution contains an Elliptic integral of second kind.
Yes, it can. Take the natural log of both sides to get the exponent out, then divide by the ln i. X=ln 2 / ln i.
The term in the denominator is sort of kicking the problem back at you. 'e^x = i' is not at all clear to me.
Nice. Can you please also cover possible applications? I find it easier to remember that way!
One application is to exercise your math skills...
@@esajpsasipes2822 That's given right he's an effective communicator. I'm just asking for actual applications.
@@lucidx9443 I was more wanting to say that there might not be any (not sure), but even if there aren't any direct applications, it at least serves as an exercise (or entertainment i guess).
More generally, quantum mechanics uses this type of math
What if we are looking for solutions to this type of equations: i^x = k
Can there still be solutions if k is a negative number? I'm asking this since we are using the natural log to find the solutions, while the function is not defined for inputs smaller than zero.
Just wondering.
They are using the extended version of natural log. The usual exponential function can be extended to exp: C -> C\{0}, which is a surjective but not injective function. Since exp is not injective, it doesn't have an inverse, but given any nonzero w in C, the solutions to exp(z)=w always differ by a factor of the complex exponential of an even integer multiple of the imaginary unit. Therefore, we can instead define inverse branches of exp, analogous to natural log in the positive real numbers. In this case, since exp is surjective, we can rewrite i^x in terms of exp, and there would be infinitely many solutions as long as k is nonzero.
BPRP: x=4n-(2iln2)/π
My mindset: x=log_i(2)
4:17 "I don't like to be on the bottom. I like to be on the top."
Is there a particular reason for writing 2ln2 in the solution rather than ln4? I assume since that's the answer on WA, there must be a reason.
it's the standard form, because it's "more beautiful" to have the lowest possible number in the ln. Otherwise you could also write "ln 4^i"
@@Engy_Wuck Thank you.
Why wolfpharm alfa give the answer with log not ln ?
great video
(4×ln2)/2πi is also a solution
I got everything but why on 5:54 we don't multiply everything on the right side by 4 instead😅
Can you have exact result for x, in x^i = i^x ?
Yes. x is either i or -i. Sorry if this is disappointing.
How many integer values of parameter m are there for the function y = x^4 - 4x^3 +(4−m)x+1 to have three extremes?
To have three extremas, the derivative must have zeros at 3 different points.
y' = 4x^3 - 12x^2 + 4 - m
Now setting this to zero
4x^3 - 12x^2 + 4 - m = 0
From here I trust that you can factor this cubic using whichever formula you desire to find how many integer values exist under the given conditions.
@@gamerpedia1535 Isn't it given by the cubic discriminant b²c² - 4ac² - 4b³d - 27a²d² + 18abcd ?
Hi Dr. Pen!
Good one.
Good morning.
Is it possible for you to post your long videos in the video section of Facebook?
I would unload them from UA-cam but I have no credit card.
I would love to study each of your solutions now tht I have plenty of time for it (living at the foot of the northwestern argentinian mountains
-iLn(2)/pi
How do you know these are all solutions?
i^x=2 is the same as e^(½πix)=e^(ln2). So x=ln2⁄(½πi), or −i⋅ln4∕π
Impressive
Can you please give a link to the video you mention at 0:09? What pops up in the video isn't linking.
NM I found it. ua-cam.com/video/9wJ9YBwHXGI/v-deo.html
It should be in the description
@@blackpenredpen You know what it was? UA-cam oh-so-wisely formatted that hyperlink so it would not look like a hyperlink at all. So it was UA-cam's fault for failing as usability once again grrr....
0:07 😮
the complex world is crazy
“Don’t say 90 degrees cause we are all adults now”🤣🤣
Cool!
I wrote the answer as ln((2)^(2)(-i/pi)). I wonder if the whether general answer of i^x = a, would always be ln((a)^(a)(-i/pi)). It looks quite nice as well.
Try an induction proof
Tried it lol. Doesn't work. I did find after starting again that the actual general form is x = ln(a^(-2i/pi)) which I could prove by induction.
Thanks for the hint. @Simon N I don't get to do maths much these days as I have left sixth form and uni course doesn't have any advanced maths in it really alway fun when bprp uploads.
"Teachers, feel free to use this on your next test"