Solving A Weird Exponential Equation | Any Solutions?
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- Опубликовано: 12 фев 2024
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Y=1 is a horizontal asymptote. This implies that e^1/x approaches 1 as x increases or decreases without bound.
the "minus" sign can be omitted in the final result, as n can be both negative and positive! Just make a substitution: m=-n, m≠0
How did you type the not equal symbol? Did you look it up?
By holding the equal button in the numbers section
One can notice immediately that there is no real solution. Hence , one has to look for complex solutions . If one knows that e^(i 2*k*π )=1 ,
k=1,2 .....,the solution is simple .
e^(1/x) = 1
If we use the normal method, the solution we get is 1/x = 0, because n⁰ = 1
However, 1/x = 0 is unsolvable.
Instead, we need to use Euler's formula, e^ix = cosx + isinx
e^(0+2πn)i = cos2πn + isin2πn = 1 + 0i where n is a positive integer
e^(2πn)i = 1
Therefore, 1/x = 2πni
x = 1/2πni
It is generally good practice to convert the denominator to a real number
1/2πni = 1/2πni × i/i = i/(-2πn)
x = -i/2πn
take complex logarithm: 1/x = 2nπi, n(e)Z
multiply both sides by x: 1 = 2nπix, n(e)Z
swap: 2nπix = 1, n(e)Z
divide both sides by 2nπi: x = 1/(2nπi), n(e)Z, n ≠ 0 (in the standard complex plane)
Standard complex plane? Really now? Is it a Cartesian coordinate system?
Great...Sir
Thanks
Taking the xth power on both sides
e=1^x
In one of your other videos didn’t we agree there are no real or complex solutions to this equation?
thanks for the video, and please forgive my curiosity but where are you actually from? you sound a lot like a Turkish guy
Ne demek. Ankarada dogdum büyüdüm ☺️
Or we can get by complex numbers
This problem also shows why L'Hopital's Rule can't be set up to do a 0/0 or infinity/infinity analysis. Taking the ln on both sides shows at 1/x = 0 has no x because 0/(1/x) only produces a 0/(infinity) result at x=0 that your e^(1/x) graphs shows the disjoint of equalling 0 from 0- and approaching infinity from 0+. The disjoint of real solutions and the x=-i/(2πn) for n not 0, integer are complex solutions that aren't specific or unique nor 0. The problem with going to the complex world to solve a e^(1/x) = 1 type equation means yes the complex numbers solved an equation from no real world applications math that never created e^(1/x) evaluated to be "1" 😮
In Electrical Engineering -i (1/(2πn)) sort of looks like a -i (1/(2πf)) dampening or attenuation of frequencies type problem of a capacitor of integer only frequencies. No real world applications capacitor relatable problem exists either since anything between 2nπ and 2(n+1)π is garbage and with no 0 frequency representing DC analysis shows how impractical the e^(1/x) = 1 set of complex solutions can work for any real world physics problem!! 😬👎
There is no finite answer, the simple way to do it is to take the natural log of both sides (since e and the natural log [ln] are inverses they cancel each other, and the natural log of 1 is 0) so what you get after you do ln(e^1/x)=ln(1) you'll get 1/x=0 So the answer would be expressed in terms of a limit as x-->±∞ 1/x=1
e^(iθ)=cosθ+i*sinθ=1
1/x=iθ or θ=-i/x or x/i=-1/θ or x=-i/θ
θ=±2πj, j=1,..., so x=±i/2πj, j=1,2,...
2:17 the equation is changing the whole graph just to avoid hitting 1.
Let me know your thoughts on it.
We often “ complexity” 1 as e^2PIi and manipulate in the complex world. If the exponent was multiplied by an integer k then n=0 covers all answers. If n is divided by k then n goes from 0 to k-1 providing k number of solutions. If 2PIni wasn’t divided by any number in the complex world only n=0 applies. For this reason you can’t have n in the denominator in this example. So there are no real or imaginary solutions.
Let z = a + bi, with A, b real. e^z = (e^a)(cos b + i sin b), where |e^z| = |e^a|. In particular, if e^z = 1, then a = 0. And sin b = 0 while cos b = 1.
I.e., b = 2 pi * k, for some integral k.
To solve e^(1/x) = 1, we solve (1/x) = 2 pi k,*i for k in Z. I.e., x = 1/(2 pi k*i) for k in Z, excepting k = 0.
Or, since 1/i = -i, you could enumerate the solutions as x = i/(2 pi k), for k in Z, k non-zero.
Just from the thumbnail, I'd say there are at least countably infinite solutions.
Begin with 1/x = 2n(pi)i and go from there. (n = integer not = 0)
There are exactly countably infinite solutions.
@@rickdesper
Well, Aleph Null to you, buddy!
(Yeah, I knew that answer, but I didn't want to get into an argument about it.)
Edit: My answer was still correct, BTW.
x -> +♾️ or -♾️
So, there are only science fiction solutions. Congrats.
You mean complex solutions? 😀
@@SyberMath Yes. Sci-fi solutions.
@@pelasgeuspelasgeus4634 more real than real solutions!
@@SyberMath Complex number theory is fake invented math because
(1) the definition of a complex number contradicts to the laws of formal logic, because this definition is the union of two contradictory concepts: the concept of a real number and the concept of a non-real (imaginary) number-an image. The concepts of a real number and a non-real (imaginary) number are in logical relation of contradiction: the essential feature of one concept completely negates the essential feature of another concept. These concepts have no common feature (i.e. these concepts have nothing in common with each other), therefore one cannot compare these concepts with each other. Consequently, the concepts of a real number and a non-real (imaginary) number cannot be united and contained in the definition of a complex number. The concept of a complex number is a gross formal-logical error;
(2) the real part of a complex number is the result of a measurement. But the non-real (imaginary) part of a complex number is not the result of a measurement. The non-real (imaginary) part is a meaningless symbol, because the mathematical (quantitative) operation of multiplication of a real number by a meaningless symbol is a meaningless operation. This means that the theory of complex number is not a correct method of calculation. Consequently, mathematical (quantitative) operations on meaningless symbols are a gross formal-logical error;
(3) a complex number cannot be represented (interpreted) in the Cartesian geometric coordinate system, because the Cartesian coordinate system is a system of two identical scales (rulers). The standard geometric representation (interpretation) of a complex number leads to the logical contradictions if the scales (rulers) are not identical. This means that the scale of non-real (imaginary) numbers cannot exist in the Cartesian geometric coordinate system.
Why do you have desmos on high contrast though?
you mean dark mode?
Why don't you have a life?
Do your own videos on whatever TF demos contrast you want, you whiner.
A bit of a dubious answer, but e^1/x approaches 1 as x goes to either infinity or -infinity. Not sure if x = +/-infinity is accepted as an answer though,
Unfortunately not
@@SyberMath Can I hence say that there is no real solution for x, but for what it's worth, the function e^1/x approaches 1 as x goes to positive or negative infinity?
or ±∞
First look. Using limits, I’d say that x -> -♾️ or x-> +♾️.
I divided my comments in to 3 and this is a classic example to back up the second comment.
e^1=e*1
e^1=e^1(e^2PIni )
e^1=e^(1+2PIni)
Taking logarithms you get 1=1+2PIni
This is true only when n=0
If you use this method to calculate 4th root then n can go from 0 no to 3 that completes unit circle. But when you are dealing with 2PIn 0 by itself completes the unit circle.
The equation 1^x= anything other than 1 has complex solutions only when n is not equal to zero. For solutions to exist k must be zero. So there are no real or complex solutions to that equation.
That’s why your other video on this subject makes perfect sense.
There are many others publishing 1^x=2, e, 3,etc. All there solutions essentially involve division by zero and all of them are wrong. You are the only one to withdraw such video and my hat is off for that.
For this reason
x -> INFINITY
😜😮
Well you need 1/x = 0, so obviously you solve for x and get x = 1/0! Not sure what all the fuss is about.
Problem with this equation is there is no solution as the exponent has to be zero.
One divided by any number will never be closed to zero.
Keep in mind that any base to the power of zero always gives you one.
No real solutions
X-->infinite
=> e^1/ ♾️ e^0 =1
S={1000 -♾️}
you can't just plug in infinity. infinity is a property of limit.
×=1/0 😂
anything over zero is infinity
@@DarsheelAE not defined, it's not infinity
@@DarsheelAEinfinity is not a number
@@bhoju_ it is, when denominator approaches zero the fraction approaches infinity
You can't use 1/0 as you get an undefined number.
I'm confused...
1 = e^(1/x)
(1 = e^(1/x))^x
1^x = e
So... What's the difference between e = 1^x and 1 = e^(1/x)🤔🤔
(a^b)^c = a^(bc) does not always work for complex numbers
x =i/(2kpi)
No need to keep the - sign since k is in Z
That's true!
-i/2npi
USE GEOGEBRA
Why?
No, there aren't any solutions (based on the last video in aplusbi).
there clearly are solutions
More precisely, no real solutions. Go into the complex solutions and u will find them
Have you seen the demonstrantion in this video? you did it? Do not you agree? So, you have two choises.
1. You have to say WHY, the proof is wrong, WHERE it is wrong.
OR
2. you can say: I don't understand anything about this matter... but deep of my heart I feel that this demonstration is wrong.
Second thing:
"based on the last video in aplusbi"
what video?
No, I predicted there would be no solutions before watching it based on what was then the most recent vid in aplusbi. I thought e^(1/x)=1 would be equal to e=1^x which Syber found no solutions for in that channel.@@ZannaZabriskie
Check again
Be simple, please... there is a easiest way to solve.
Multiply top and bottom by NEGATIVE i?? Sir, how dare you. 😲
I don't understand, why couldnt he do it? Im in 8th grade so forgive me if im wrong.
That's how it's supposed to be done 👊🔥
@@UrosLetic in algebra, you are always permitted to multiply any quantity by "one"... for example (i/i) equals one; similarly, (-i/ -i) also equals one. In this case, multiplying by (-i / -i) was helpful to simplify the expression because the new denominator is positive.
@@dixonblog thank you. I thought so too.
I was just busting his chops. I agree, his way is better.@@UrosLetic
Isn't this just 1^x = e all over again?
Nope
@@SyberMath hmm. Both sides to the xth root?
@@mcwulf25 nope
Gents, please minus sign must be in front of fraction line so need to be accurate when writing....
1 / x = 0 => no soluzioni .
lim 1 / x = 0
x -> infinito
😊👍👋