1^x = −4 x · ln(1) = ln(−4) In real number, any logarithm of 1 is 0. But here we can cleverly use the identity ln(−1) = π · i: x · ln((−1)²) = ln(−1) + ln(4) 2x · ln(−1) = ln(−1) + ln(4) x = [ln(−1) + ln(4)] / [2 · ln(−1)] x = 1/2 + ln(4) / (2π · i).
I didn't want to log neg. number, so I instead took -4 = 4 (-1) = e^{ln(4)} e^{(2m+1)pi i} = e^{(2m+1)pi i + ln(4)} for m \in Z (from Euler's equation as well). As exp is monotonic, arguments must be equal to get equal func. values, so 2k pi i x = (2m+1) pi i + ln(4), which gives you the result (without dealing with complex logarithms).
Complex numbers are not complex but calculations with complex numbers definitely tend to get complex. A lot of writing gets involved and you easily make a sloppy mistake.
Hi, I really like all of your videos. You do a GREAT job of explaining your steps! I learn so much every time I watch your videos. I would love to see you work through Schrödinger's Equation if you ever want to tackle it?
@@hajstra1307 I hear you. There is definitely some debate on this one, but math has an interesting way of creating seemingly impossible tricks, like raising 1 to a power that equals -4 or finding divisible components of zero! haha. I don't know how, but it seems to happen. Yikes
Mister new Euler, just one example of one of your errors and its consequences. In in minute 10:00 you have: Ln(e^(i2kpi*x) = Ln(-4), (note the capital L for the complex log.) Applying the proper formula for the logarithm of complex number you will have: ln(1)+ i(2kpi*x + 2mpi) = Ln(-4) Do not apply ln(1) = 0. This is no longer true according to your algebra. The very 1st line of your presentation is 1^x = -4, after applying ln, and dividing by x, you have ln(1) = ln(-4)/x, and this is your algebra, erroneous, but you are bound to use this value of ln(1) all the time in your calculations. Also remember that in your algebra 1*1 is not equal 1, and 1*x is not equal x, and probably 2*x is not equal x + x. After substituting for ln(1) according to your algebra you have following to solve: ln(-4)/x+ i(2kpi*x + 2mpi) = Ln(-4) Also you can't remove k = m = 0. There is no complex number w/o its principal value.
In this video, the field is complex analysis. _Everything_ gets redefined. In particular, functions may have multiple values. The complex logarithm is log(z), not ln(x). Writing the equation as 1^x = -4 instead of 1^z = -4 is click bait, but perhaps necessary. People don't make a living off these videos. log(w) is the set of all z ∈ ℂ such that e^z = w. log(1) is the set of all z ∈ ℂ such that e^z = 1. Zero is not the only member of log(1). The conventional definitions include: ℤ = { …, -2, -1, 0, 1, 2, …} i² = -1, x, y, u, v, r, 𝜃 ∈ ℝ, w, z ∈ ℂ. Euler's formula: e^(i 𝜃) = cos 𝜃 + i sin 𝜃. Complex exponentiation: w^z = e^(log(w)z). Rectangular decomposion: z = x + i y, w = u + i v. Complex conjugate: z* = x - i y, z z* = x²+y² Complex absolute value: r = Abs(z) = +√(x²+y²). Argument: 𝜃 = Arg(z) = {𝜃 ∈ ℝ | z = r(cos 𝜃 + i sin 𝜃)}, the set of all real angles 𝜃 such that z = r(cos 𝜃 + i sin 𝜃). Addition: w + z = (u + x) + i(v + y). Multiplication: w z = (u x - v y) + i(u y + v x) Also, w z* = (u + x) + i(v - y) w z = Abs(w)Abs(z) e^( i(Arg(w)+Arg(z)) ) Division: w / z = (w z*)/(z z*) = ((u + x) + i(v - y))/(x²+y²) Polar decomposion: z = r e^(i 𝜃) = Abs(z) e^(i Arg(z)). Complex logarithm log(z) = {w ∈ ℂ | z = e^w} The complex logarithm w = u + i v = log(z) has multiple values. It is the inverse of z = e^w, that is, the set of all w such that z = e^w. w is similar to the polar form of z. z = e^w = e^(u + i v) = e^u e^(i v). u is a real natural logarithm, and v is a real angle. u = ln(Abs(z)), v = Arg(z) log(z) = w = ln(Abs(z)) + i Arg(z) z = e^w = e^ln(Abs(z)) e^(i Arg(z)) = Abs(z) e^(i Arg(z)), as required. Complex log(1) requires values of Abs(1) and Arg(1). Abs(1) = 1. Arg(1) is the set of all angles 𝜃 such that 1 = 1 (cos 𝜃 + i sin 𝜃) This is true if and only if cos 𝜃 = 1 and sin 𝜃 = 0. The second condition is redundant. Arg(1) = {𝜃 ∈ ℝ | 𝜃 / 2π ∈ ℤ, i.e., cos 𝜃 = 1} log(1) = ln(Abs(1)) + i Arg(1) log(1) = 0 + i {𝜃 ∈ ℝ | 𝜃 / 2π ∈ ℤ}. 1^z = e^(log(1) z) 1^z = e^(z i {𝜃 ∈ ℝ | 𝜃 / 2π ∈ ℤ})
Let's be clear here. 1^x=1, 1^x does not equal -4 under any conditions. The author of the video found the solution to x for e^(i*2*π*k*x)=-4, not 1^x=-4. Any solution for e^(i*2*π*k*x) = -4 would involve complex numbers, but the original equation 1^x = -4 is a contradiction for any real or complex x. The expression 1^x always equals 1 for any x, so there is no solution to 1^x = -4 in the real or complex number systems. Substituting 1 with e^(i*2*π*k) leads to complex solutions that do not satisfy the original equation, as 1^x remains 1 regardless of x.
To put it more simply (I have a simple mind), in the original equation 1^x=-4, you can't change fundamental laws of math: 1 raised to any power is 1 and if that is true, then 1^x is NOT equal to -4 in any case, you matter how you try to manipulate the math.
@@JohnFlathead He did, actually. See @2:14 to @8:41. He spent a whole section proving that: e^(2*i*PI*k) = 1, when k is restricted to integer values. It's a proof based on the periodicity of the cosine and sine functions, and their values 0, 2*PI, 4*PI, etc. (that is, integer multiples of 2*PI). And also using Euler's identity, e^(i*Theta) = cos(Theta) + i*sin(Theta). In the Complex numbers, these are true statements. Not if you limit yourself to Real numbers, only if you allow expansion into Complex numbers. As a test, you could simply plug in any integer value for k and m, such that k not equal to 0, and, using that value for x, if you work it out, you will find that the result will evaluate to -4. Again, you have to use the 'complex unit' construction that he proved at @2:14, but when you do that, it will evaluate to -4.
@higher_mathematics, would you say that math has an interesting way of creating seemingly impossible feats, like raising 1 to a power that equals -4 or even finding divisors of zero?
This video is wrong. Per definition of complex exponents with positive real base (en.m.wikipedia.org/wiki/Exponentiation#Complex_exponents_with_positive_real_bases): b^z = e^(z Log(b)) where Log is complex log Further: Log(r e^(i t)) = log(r) + i t + 2 k pi i where k in Z Log(1) = 2 k pi i where the usual log(1)=0 is just the primitive log branch, and thus 1^z = 1 is based on the primitive log branch So the complete equality is 1^z = e^(z Log(1)) = e^(z 2 k pi i) But z = x + y i, then 1^z = e^((x + y i) 2 k pi i) = e^(x 2 k pi i - y 2 k pi) = (e^(2 k pi i))^x / e^(y 2 k pi) Using Euler: e^(i t) = cos(t) + i sin(t) e^(2 k pi i) = cos(2 k pi) + i sin(2 k pi) = 1 + 0 i = 1 1^z = 1^x / e^(y 2 k pi) = 1 / e^(y 2 k pi) This is a real-valued multi-valued expression. And it can never become negative. So 1^z ≠ -4 Note: 1^(x + y i) = 1^x 1^(y i) = 1^(y i) Note: 1^(y i) ≠ (1^y)^i Note: 1^(y i) = (1^i)^y
Mistakes in video: @9:40 Complex Exponents (e^(i 2 k pi))^x ≠ e^(i 2 k pi x) x is complex number and this action is not legit Ex: (1^i)^i ≠ 1^(i i) = 1^-1 = 1 @10:00 Complex Log Log(e^(i t)) = i t + 2 m pi i Log(e^(i 2 k pi x)) = i 2 k pi x + 2 m pi i Missing the period shift, can't be absorbed into k and again forgetting that x is complex, so action is not correct Also not a big deal but all Z parameter ranges include the negative integers too. But with the first and second mistake, it is hopeless that the final answer is useful at all. It is easy to show the answer shouldn't have a real part since 1^r = 1.
1^x = -4 can be defined as an attempt changing 1 to -4 using a complex power of x. In the complex plane every complex number can be changed to any other complex number by changing the argument (rotation). and by changing its magnitude. This can be done by addition, multiplication, and by using complex power. There is an exception for number 1. The same as in the real numbers domain, the number 1 is the identity of multiplications so the result of multiplication by 1, and taking 1 to any power, is 1. Also an exception in number 0 that could be changed only by addition. Converting 1 to (-4) using multiplication can be done in two steps: multiply by (-1), then multiply by 4. The multiplication by (-1), in the complex form, is rotation by π. (-1) = e^iπ The multiplication by (4) in the complex exponential form is the change of magnitude by multiplying by e^ln(4): 1*(-1)*4 = -4 , OR (e^i0)*(e^iπ)*(e^ln(4)) = -4 So, multiplication works. But in the video the intention is to do it by power. For 1 = e^(0+i0) nothing can be done to change it using any complex power--we have multiplication by 0: 1^x = (e^(0+i0)) ^x = e^(0*x+i0x) = e^0 = 1. The trick used in the video is multiplying 1 by e^(i2πk) before applying power. Therefore, the author of the video uses a combination of the multiplication and the complex power. In the complex plane the multiplication by e^(i2πk) rotates the complex number by 2πk: 1*e^(i2πk) = (e^(0+i0) * (e^(i2πk)) = e^(i2πk); k = N∗={1,2,3,…} The power 0 is changed to the power of i2πk. Now we can calculate complex power to get any given complex number. Particularly to get real numbers 3, 5, -3, -5 presented in several ‘Higher Mathematics’ videos. Is 1 the same as e^(i2πk) where k = N∗={1,2,3,…}? Definitively not. 1 = e^(0+i0) is the principal value of e^(i2πk). We removed 1 from e^(i2πk) by removing k = 0. Returning to the video (from minute 9:50), just to solve e^(i2πk*x) = -4, but not to solve 1^x =1. The x can be calculated quite easily in 3 simple steps w/o using logarithms. 1st: Express (-4) as an exponential complex number: (-4) = (-1) *4 = e^(i(π+2πm))*e^(ln(1/4)) = e^(i(π+2πm) + ln(4)) 2nd: Substitute converted (-4) into the equation: e^(i2πk*x) = e^(i(π+2πm) + ln(4)) 3rd: Compare powers and extract x: i2πk*x = i(π+2πm) + ln(4)) --> x = (1+2m)/2k - i*ln(4)/(2πk) where k = N∗={1,2,3,…} (and this is the same result as presented in the video) Testing the result: e^(i2πk*x) = e^(i2πk*((1+2m)/2k - i*ln(4)/(2πk))) = e^(iπ(1+2m) + i*ln(4)) = e^(iπ+iπ2m) *e^(ln(4)) = (-1)*4 = -4
Wrong! But it will certainly fool a lot of people. I shouldn't even be commenting, because I know that the intention of this video is to attract people (particularly those who know that the so-called "solution" doesn't make any sense). But, I repeat, the so-called "solution" is wrong! The equation presented has no solution, either in real or complex numbers.
wow. You have certainly caught my attention. I noticed the other day from looking at these maths questions, that log(-2) has an infinity of complex values. "I don't remember learning that at school". Maybe I misremembered what I read the other day. but whatever it said, it has prepared me have a look at your post for this question. Now I'm stuck on 1^x = -4 take logs of both sides x log 1 = log (-4). I was prepared from what I read last week, to accept that log(-4) has exotic values. But how can x log 1 be anything else but 0?? After noticing that open ai agrees with you, I concede defeat!!
log(1) 1 = e^(i.2nPi) where n is an integer = i.2nPi If n = 0 then log(1) = 0 But for any other random integer value of n log(1) = i.2nPi i.e. log(1) = ....., -i.4Pi, -i.2Pi, 0, i.2Pi, i.4Pi, .... For those who are "trapped" mentally to the world of real numbers log(1) = 0. But for those who have successfully negotiated the transition into the world of complex numbers log(1) has an infinite number of complex values as outlined above. (Pi represents the number 3.1415...)
I think the k = 3 and 4 parts are unnecessary -- you don't really need to establish the pattern this far. If you must, use k in {1,2}, but that's all. Going to 3, 4, 1000 is unnecessary.
And yet, even after using all these examples, some people in the comments still did not see the pattern. ¯\_(ツ)_/¯ So, I think him using extra steps is fine, since it makes it very very clear that he's trying to show a counter-intuitive pattern, requiring more than just a couple of steps to communicate the idea to a wider audience of people.
Despite giving 5 examples, he neglected to use non-positive values for k . He should've written e^(i*θ) = cis(θ) = cos(θ) + i*sin(θ) , and said θ and θ + 2*π*k yield the same value for k ∈ ℤ because sine and cosine have a periodicity of 2*π . Then, if he wants to be clear, he can say θ, θ±2*π, θ±4*π, ... all yield the same result.
well... if someone asks, what is 1^(1/2 - i/pi ln(2)) would you really says it's -4? Because 1^x = 1 for all complex x, not only for real x. only of you add branches from 1^x=exp(x ln(1)) you CAN get a multivalued function - without any branch cuts!!. but usually 1^x = 1 and nothing else
I asked open ai and it said (I can't copy verbatim what open ai said as I don't know how to cut and paste it successfully) Yes! complex solutions for the equation 1^x = 2 do exist. The solutions take the form x = (ln 2)/ (2 p i n) + k/n, n = 0, k in Z . Z is the set of integers, positive, negative, and zero. "I don't remember learning any of this at school".
Isn't this the same as the Perpetual Enigma i.e. 1 + 2 + 3 + 4 + 5 + …. = -1/12 ? I think I can show 1^x = -(any positive number if I substitute the appropriate numbers in the solution given). Just like the above I can show 1 + 2 + 3 + 4 + 5 + …. = anything I like.
Your writing is not clean and proper in math syntax. You cannot write => when meaning = You would not even get full points in a test at any university or high school. Neither is this a good explaining.
This is so tediously repetitive and boring, yet you've neglected to state that k can be negative, too. Also, your pronunciation is bad: "natural" isn't pronounced as "nature-el"; "cosine" isn't pronounced as "kah-sign". You should just state the identity of: e^(i*θ) = cis(θ) = cos(θ) + i*sin(θ) Since sine and cosine have a period of 2*π , then this applies to e^(i*θ) = cis(θ) = cos(θ) + i*sin(θ) . Thus, for k ∈ ℤ , θ and θ + 2*π*k yield the same result. Next, you can mention that cis(0 + 2*π*k) = 1 , and cis(π + 2*π*k) = -1 . That is, the cis of even and odd multiples of π are 1 and -1 , respectively. Now, apply them to the problem: 1ˣ = -4 (1)ˣ = (-1)*4 (e^(i*(0 + 2*π*n)))ˣ = (e^(i*(π + 2*π*m)))*2² ; m, n ∈ ℤ x*i*(2*π*n) = i*(π + 2*π*m) + 2*ln(2) x = [i*π*(2*m+1) + 2*ln(2)]/[i*π*2*n)] ; n ≠ 0 ∴ x = (2*m+1)/(2*n) - i*ln(2)/(π*n) ; m, n ∈ ℤ , n ≠ 0 One solution with m = 0 and n = 1 is 1/2 - i*ln(2)/π .
1^x = -4 e^(i.2nπx) = 4e^(i.(2m+1)π) n, m ∈ Z i.2nπx = ln(4) + i.(2m+1)π x = ln(4)/(i.2nπ) + (2m + 1)/(2n) n ≠ 0 x = -i.ln(2)/(nπ) + (2m + 1)/(2n) i.e. For 1^x = -4 it follows x = -i.ln(2)/(nπ) + (2m + 1)/(2n) where m, n ∈ Z and n ≠ 0
1^x = −4
x · ln(1) = ln(−4)
In real number, any logarithm of 1 is 0. But here we can cleverly use the identity ln(−1) = π · i:
x · ln((−1)²) = ln(−1) + ln(4)
2x · ln(−1) = ln(−1) + ln(4)
x = [ln(−1) + ln(4)] / [2 · ln(−1)]
x = 1/2 + ln(4) / (2π · i).
WolframAlpha believes that if you raise 1 to the power you got in your answer, you get exactly 1. Not -4
as would be expected for any value of x.
I didn't want to log neg. number, so I instead took -4 = 4 (-1) = e^{ln(4)} e^{(2m+1)pi i} = e^{(2m+1)pi i + ln(4)} for m \in Z (from Euler's equation as well).
As exp is monotonic, arguments must be equal to get equal func. values, so 2k pi i x = (2m+1) pi i + ln(4), which gives you the result (without dealing with complex logarithms).
Complex numbers are not complex but calculations with complex numbers definitely tend to get complex. A lot of writing gets involved and you easily make a sloppy mistake.
Hi, I really like all of your videos. You do a GREAT job of explaining your steps! I learn so much every time I watch your videos. I would love to see you work through Schrödinger's Equation if you ever want to tackle it?
This time you have learn wrong staff.
@@hajstra1307 I hear you. There is definitely some debate on this one, but math has an interesting way of creating seemingly impossible tricks, like raising 1 to a power that equals -4 or finding divisible components of zero! haha. I don't know how, but it seems to happen. Yikes
Mister new Euler, just one example of one of your errors and its consequences.
In in minute 10:00 you have:
Ln(e^(i2kpi*x) = Ln(-4), (note the capital L for the complex log.)
Applying the proper formula for the logarithm of complex number you will have:
ln(1)+ i(2kpi*x + 2mpi) = Ln(-4)
Do not apply ln(1) = 0. This is no longer true according to your algebra. The very 1st line of your presentation is 1^x = -4, after applying ln, and dividing by x, you have ln(1) = ln(-4)/x, and this is your algebra, erroneous, but you are bound to use this value of ln(1) all the time in your calculations. Also remember that in your algebra 1*1 is not equal 1, and 1*x is not equal x, and probably 2*x is not equal x + x.
After substituting for ln(1) according to your algebra you have following to solve:
ln(-4)/x+ i(2kpi*x + 2mpi) = Ln(-4)
Also you can't remove k = m = 0. There is no complex number w/o its principal value.
"Mister new Euler" ©
Ahah😂 Golden!
Beautiful explanation 😀
In this video, the field is complex analysis. _Everything_ gets redefined. In particular, functions may have multiple values. The complex logarithm is log(z), not ln(x).
Writing the equation as 1^x = -4 instead of 1^z = -4 is click bait, but perhaps necessary. People don't make a living off these videos.
log(w) is the set of all z ∈ ℂ such that e^z = w. log(1) is the set of all z ∈ ℂ such that e^z = 1. Zero is not the only member of log(1).
The conventional definitions include:
ℤ = { …, -2, -1, 0, 1, 2, …}
i² = -1, x, y, u, v, r, 𝜃 ∈ ℝ, w, z ∈ ℂ.
Euler's formula: e^(i 𝜃) = cos 𝜃 + i sin 𝜃.
Complex exponentiation: w^z = e^(log(w)z).
Rectangular decomposion: z = x + i y, w = u + i v.
Complex conjugate: z* = x - i y, z z* = x²+y²
Complex absolute value: r = Abs(z) = +√(x²+y²).
Argument: 𝜃 = Arg(z) = {𝜃 ∈ ℝ | z = r(cos 𝜃 + i sin 𝜃)}, the set of all real angles 𝜃 such that z = r(cos 𝜃 + i sin 𝜃).
Addition: w + z = (u + x) + i(v + y).
Multiplication: w z = (u x - v y) + i(u y + v x)
Also, w z* = (u + x) + i(v - y)
w z = Abs(w)Abs(z) e^( i(Arg(w)+Arg(z)) )
Division: w / z = (w z*)/(z z*) = ((u + x) + i(v - y))/(x²+y²)
Polar decomposion: z = r e^(i 𝜃) = Abs(z) e^(i Arg(z)).
Complex logarithm log(z) = {w ∈ ℂ | z = e^w}
The complex logarithm w = u + i v = log(z) has multiple values. It is the inverse of z = e^w, that is, the set of all w such that z = e^w. w is similar to the polar form of z.
z = e^w = e^(u + i v) = e^u e^(i v).
u is a real natural logarithm, and v is a real angle.
u = ln(Abs(z)), v = Arg(z)
log(z) = w = ln(Abs(z)) + i Arg(z)
z = e^w = e^ln(Abs(z)) e^(i Arg(z))
= Abs(z) e^(i Arg(z)), as required.
Complex log(1) requires values of Abs(1) and Arg(1). Abs(1) = 1. Arg(1) is the set of all angles 𝜃 such that 1 = 1 (cos 𝜃 + i sin 𝜃)
This is true if and only if cos 𝜃 = 1 and sin 𝜃 = 0. The second condition is redundant.
Arg(1) = {𝜃 ∈ ℝ | 𝜃 / 2π ∈ ℤ, i.e., cos 𝜃 = 1}
log(1) = ln(Abs(1)) + i Arg(1)
log(1) = 0 + i {𝜃 ∈ ℝ | 𝜃 / 2π ∈ ℤ}.
1^z = e^(log(1) z)
1^z = e^(z i {𝜃 ∈ ℝ | 𝜃 / 2π ∈ ℤ})
Let's be clear here. 1^x=1, 1^x does not equal -4 under any conditions. The author of the video found the solution to x for e^(i*2*π*k*x)=-4, not 1^x=-4. Any solution for e^(i*2*π*k*x) = -4 would involve complex numbers, but the original equation 1^x = -4 is a contradiction for any real or complex x. The expression 1^x always equals 1 for any x, so there is no solution to 1^x = -4 in the real or complex number systems. Substituting 1 with e^(i*2*π*k) leads to complex solutions that do not satisfy the original equation, as 1^x remains 1 regardless of x.
To put it more simply (I have a simple mind), in the original equation 1^x=-4, you can't change fundamental laws of math: 1 raised to any power is 1 and if that is true, then 1^x is NOT equal to -4 in any case, you matter how you try to manipulate the math.
Your statement that one to the power of a complex number must equal one is where you have assumed a result without proof.
This is how people get fooled by politicians as well
@@dougr.2398 The burden of proof lies with the author of the video. He conveniently didn't verify his solution because he can't.
@@JohnFlathead He did, actually. See @2:14 to @8:41. He spent a whole section proving that: e^(2*i*PI*k) = 1, when k is restricted to integer values.
It's a proof based on the periodicity of the cosine and sine functions, and their values 0, 2*PI, 4*PI, etc. (that is, integer multiples of 2*PI). And also using Euler's identity, e^(i*Theta) = cos(Theta) + i*sin(Theta).
In the Complex numbers, these are true statements. Not if you limit yourself to Real numbers, only if you allow expansion into Complex numbers.
As a test, you could simply plug in any integer value for k and m, such that k not equal to 0, and, using that value for x, if you work it out, you will find that the result will evaluate to -4. Again, you have to use the 'complex unit' construction that he proved at @2:14, but when you do that, it will evaluate to -4.
No need for variable "m".
Substitute left side with e^ (i2kπx), substitute right side with: 4e^(i(π+2kπ)).
well the solution has two degrees of freedom, no? so you do need two variables
The most general solution allows for k and m to be different. Try it with k=1, m=2, for instance.
0=log(x)^4
@higher_mathematics, would you say that math has an interesting way of creating seemingly impossible feats, like raising 1 to a power that equals -4 or even finding divisors of zero?
Very good
This video is wrong.
Per definition of complex exponents with positive real base (en.m.wikipedia.org/wiki/Exponentiation#Complex_exponents_with_positive_real_bases):
b^z = e^(z Log(b))
where Log is complex log
Further: Log(r e^(i t)) = log(r) + i t + 2 k pi i where k in Z
Log(1) = 2 k pi i
where the usual log(1)=0 is just the primitive log branch, and thus 1^z = 1 is based on the primitive log branch
So the complete equality is
1^z = e^(z Log(1)) = e^(z 2 k pi i)
But z = x + y i, then
1^z = e^((x + y i) 2 k pi i)
= e^(x 2 k pi i - y 2 k pi)
= (e^(2 k pi i))^x / e^(y 2 k pi)
Using Euler: e^(i t) = cos(t) + i sin(t)
e^(2 k pi i) = cos(2 k pi) + i sin(2 k pi) = 1 + 0 i = 1
1^z = 1^x / e^(y 2 k pi)
= 1 / e^(y 2 k pi)
This is a real-valued multi-valued expression. And it can never become negative.
So 1^z ≠ -4
Note: 1^(x + y i) = 1^x 1^(y i) = 1^(y i)
Note: 1^(y i) ≠ (1^y)^i
Note: 1^(y i) = (1^i)^y
Mistakes in video:
@9:40 Complex Exponents
(e^(i 2 k pi))^x ≠ e^(i 2 k pi x)
x is complex number and this action is not legit
Ex: (1^i)^i ≠ 1^(i i) = 1^-1 = 1
@10:00 Complex Log
Log(e^(i t)) = i t + 2 m pi i
Log(e^(i 2 k pi x)) = i 2 k pi x + 2 m pi i
Missing the period shift, can't be absorbed into k and again forgetting that x is complex, so action is not correct
Also not a big deal but all Z parameter ranges include the negative integers too.
But with the first and second mistake, it is hopeless that the final answer is useful at all. It is easy to show the answer shouldn't have a real part since 1^r = 1.
(1)^2=1(x ➖ 1x+1 ). 2^2 (x ➖ 2x+2).
1^x = -4 can be defined as an attempt changing 1 to -4 using a complex power of x.
In the complex plane every complex number can be changed to any other complex number by changing the argument (rotation). and by changing its magnitude. This can be done by addition, multiplication, and by using complex power. There is an exception for number 1. The same as in the real numbers domain, the number 1 is the identity of multiplications so the result of multiplication by 1, and taking 1 to any power, is 1. Also an exception in number 0 that could be changed only by addition.
Converting 1 to (-4) using multiplication can be done in two steps: multiply by (-1), then multiply by 4.
The multiplication by (-1), in the complex form, is rotation by π. (-1) = e^iπ
The multiplication by (4) in the complex exponential form is the change of magnitude by multiplying by e^ln(4):
1*(-1)*4 = -4 , OR (e^i0)*(e^iπ)*(e^ln(4)) = -4
So, multiplication works. But in the video the intention is to do it by power. For 1 = e^(0+i0) nothing can be done to change it using any complex power--we have multiplication by 0:
1^x = (e^(0+i0)) ^x = e^(0*x+i0x) = e^0 = 1.
The trick used in the video is multiplying 1 by e^(i2πk) before applying power. Therefore, the author of the video uses a combination of the multiplication and the complex power.
In the complex plane the multiplication by e^(i2πk) rotates the complex number by 2πk:
1*e^(i2πk) = (e^(0+i0) * (e^(i2πk)) = e^(i2πk); k = N∗={1,2,3,…}
The power 0 is changed to the power of i2πk. Now we can calculate complex power to get any given complex number. Particularly to get real numbers 3, 5, -3, -5 presented in several ‘Higher Mathematics’ videos.
Is 1 the same as e^(i2πk) where k = N∗={1,2,3,…}? Definitively not. 1 = e^(0+i0) is the principal value of e^(i2πk). We removed 1 from e^(i2πk) by removing k = 0.
Returning to the video (from minute 9:50), just to solve e^(i2πk*x) = -4, but not to solve 1^x =1. The x can be calculated quite easily in 3 simple steps w/o using logarithms.
1st: Express (-4) as an exponential complex number:
(-4) = (-1) *4 = e^(i(π+2πm))*e^(ln(1/4)) = e^(i(π+2πm) + ln(4))
2nd: Substitute converted (-4) into the equation:
e^(i2πk*x) = e^(i(π+2πm) + ln(4))
3rd: Compare powers and extract x:
i2πk*x = i(π+2πm) + ln(4)) --> x = (1+2m)/2k - i*ln(4)/(2πk) where k = N∗={1,2,3,…}
(and this is the same result as presented in the video)
Testing the result:
e^(i2πk*x) = e^(i2πk*((1+2m)/2k - i*ln(4)/(2πk))) = e^(iπ(1+2m) + i*ln(4)) = e^(iπ+iπ2m) *e^(ln(4)) = (-1)*4 = -4
Wrong! But it will certainly fool a lot of people. I shouldn't even be commenting, because I know that the intention of this video is to attract people (particularly those who know that the so-called "solution" doesn't make any sense). But, I repeat, the so-called "solution" is wrong! The equation presented has no solution, either in real or complex numbers.
wow. You have certainly caught my attention. I noticed the other day from looking at these maths questions, that log(-2) has an infinity of complex values. "I don't remember learning that at school". Maybe I misremembered what I read the other day. but whatever it said, it has prepared me have a look at your post for this question.
Now I'm stuck on 1^x = -4
take logs of both sides
x log 1 = log (-4).
I was prepared from what I read last week, to accept that log(-4) has exotic values.
But how can x log 1 be anything else but 0??
After noticing that open ai agrees with you, I concede defeat!!
log(1) 1 = e^(i.2nPi) where n is an integer
= i.2nPi
If n = 0 then log(1) = 0
But for any other random integer value of n
log(1) = i.2nPi
i.e. log(1) = ....., -i.4Pi, -i.2Pi, 0, i.2Pi, i.4Pi, ....
For those who are "trapped" mentally to the world of real numbers log(1) = 0. But for those who have successfully negotiated the transition into the world of complex numbers log(1) has an infinite number of complex values as outlined above.
(Pi represents the number 3.1415...)
Been 50:years since I did my maths degree but I always thought 1 raised to any power is 1
At 5:43 I pointedly ask why anyone would not expect the thame anther for all k.
I think the k = 3 and 4 parts are unnecessary -- you don't really need to establish the pattern this far. If you must, use k in {1,2}, but that's all. Going to 3, 4, 1000 is unnecessary.
And yet, even after using all these examples, some people in the comments still did not see the pattern. ¯\_(ツ)_/¯ So, I think him using extra steps is fine, since it makes it very very clear that he's trying to show a counter-intuitive pattern, requiring more than just a couple of steps to communicate the idea to a wider audience of people.
@@robharwood3538 That's a very good point!
Despite giving 5 examples, he neglected to use non-positive values for k . He should've written e^(i*θ) = cis(θ) = cos(θ) + i*sin(θ) , and said θ and θ + 2*π*k yield the same value for k ∈ ℤ because sine and cosine have a periodicity of 2*π . Then, if he wants to be clear, he can say θ, θ±2*π, θ±4*π, ... all yield the same result.
well... if someone asks, what is
1^(1/2 - i/pi ln(2))
would you really says it's -4? Because 1^x = 1 for all complex x, not only for real x.
only of you add branches from 1^x=exp(x ln(1))
you CAN get a multivalued function - without any branch cuts!!. but usually 1^x = 1 and nothing else
Your videos are very informative! Thank you for making them. Also, what is your accent?
My advice: DO NOT REMAIN MISINFORMED by this video
@@hajstra1307 I just was asking your nationality. You have a very unique accent.
@@epd807 I an not the author of this video
I asked open ai and it said (I can't copy verbatim what open ai said as I don't know how to cut and paste it successfully)
Yes! complex solutions for the equation 1^x = 2 do exist. The solutions take the form
x = (ln 2)/ (2 p i n) + k/n, n = 0, k in Z .
Z is the set of integers, positive, negative, and zero.
"I don't remember learning any of this at school".
Somehow I never learned about logarithms between high school and college. I figured that was part of this, but I don't know how to use them.
Do not warry, the author of presentation don't know how to use logarithms of complex numbers.
OMG ...... endless loops of complex numbers
Грандиозно👏
Isn't this the same as the Perpetual Enigma i.e. 1 + 2 + 3 + 4 + 5 + …. = -1/12 ?
I think I can show 1^x = -(any positive number if I substitute the appropriate numbers in the solution given).
Just like the above I can show 1 + 2 + 3 + 4 + 5 + …. = anything I like.
Your writing is not clean and proper in math syntax. You cannot write => when meaning =
You would not even get full points in a test at any university or high school. Neither is this a good explaining.
Hello i am the first.
This is so tediously repetitive and boring, yet you've neglected to state that k can be negative, too. Also, your pronunciation is bad: "natural" isn't pronounced as "nature-el"; "cosine" isn't pronounced as "kah-sign".
You should just state the identity of:
e^(i*θ) = cis(θ) = cos(θ) + i*sin(θ)
Since sine and cosine have a period of 2*π , then this applies to e^(i*θ) = cis(θ) = cos(θ) + i*sin(θ) . Thus, for k ∈ ℤ , θ and θ + 2*π*k yield the same result.
Next, you can mention that cis(0 + 2*π*k) = 1 , and cis(π + 2*π*k) = -1 . That is, the cis of even and odd multiples of π are 1 and -1 , respectively. Now, apply them to the problem:
1ˣ = -4
(1)ˣ = (-1)*4
(e^(i*(0 + 2*π*n)))ˣ = (e^(i*(π + 2*π*m)))*2² ; m, n ∈ ℤ
x*i*(2*π*n) = i*(π + 2*π*m) + 2*ln(2)
x = [i*π*(2*m+1) + 2*ln(2)]/[i*π*2*n)] ; n ≠ 0
∴ x = (2*m+1)/(2*n) - i*ln(2)/(π*n) ; m, n ∈ ℤ , n ≠ 0
One solution with m = 0 and n = 1 is 1/2 - i*ln(2)/π .
A inveja é uma merda
1^x = -4
e^(i.2nπx) = 4e^(i.(2m+1)π) n, m ∈ Z
i.2nπx = ln(4) + i.(2m+1)π
x = ln(4)/(i.2nπ) + (2m + 1)/(2n) n ≠ 0
x = -i.ln(2)/(nπ) + (2m + 1)/(2n)
i.e. For 1^x = -4 it follows x = -i.ln(2)/(nπ) + (2m + 1)/(2n)
where m, n ∈ Z and n ≠ 0