Interesting solution! Although you are very patient in your presentation and like to go to very details, however this makes it somewhat cumbersome. Note that the viewer of this level of math must already be familiar with advanced math concepts like complex numbers. Thus, just stating the Euler's identity is enough and going and substituting 1,2,3,4 for 4 to show it will be equal to 1 is kind of dull. Then later in your solution procedure you want to find the equivalent of ln(-3) and you go at length to substitute -1 with another Euler entity. But is that necessary? Why not just simply substituting -1 with i^2? You get: ln(-3) = ln (3*i^2) = ln(3) + ln(i^2) = ln(3) +2 ln(i) which is much simpler and cleaner approach.
@@hajstra1307 Not sure which mistakes you are referring to but bear in mind that math problems may be approached from different angles. Thus different solutions could exist although one may be the shortest and most efficient one. If you think he has made explicit mistakes you need to mention it in your comment with details.
@@Emry11 My mistake---this mistakes are not interesting. I was trying to be polite. #1 error is neglecting the identity property of multiplication. 1^x = 1*1^(x-1) 1^x = 1 = 1*1 = 1*1^(x-1) but there is 1^x = 1 that contradicts 1^x = -3. The identity property of multiplication shall be neglected to continue with this nonsense. Following consequences are not dealt with: a) 1 shall be kept explicit in all calculations since 1*x is not equal x anymore and multiplying or dividing by 1 could change the value. Example is an exponential form z=re^(iθ) when r=1. There shall be 1*e^(iθ), not e^(iθ) b) ln(1) = 0 is no longer: 1^x = -3 ln(1) = ln(-3)/x c) one can proof for example that 0 = -1. This goes as follow: 1^x = 1*1^(x-1) 1^x = 1^(x-1) x = x-1 0 = -1 (I respected the identity only when it is convenient, following the approach used in the video) #2 error is eliminating the principal value of complex numbers by eliminating k=0. It is impossible and baseless. Conversion to the general form of complex number is always possible so 1*e^(i2kpi) = 1 + i*0 = 1 for any k. #3 error is not testing the result for x by using it in 1^x = -3 for any selected k. #4 error is producing videos that prove 1^x = 5 , = 3, = -3, = -4 and not REMOVING them from RUclips #5 error is attempting to become new Euler
@@hajstra1307 Your points are valid only in the Natural numbers domain not in Complex number domain. Your very first assumptions that 1^x=1 contradicts 1^x=-3 is only correct in N domain.
@@Emry11 x is complex in 1^x = 1*1^(x-1) 1^x = 1 = 1*1 = 1*1^(x-1). Yes it contradicts 1^x = -3. This is the POINT! The complex numbers domain includes the real numbers. We are not talking about the Natural numbers. All my points are about the Complex numbers (even #5)
Writing the solution for x in terms of general values of integers k & m is not mathematically sound. In order for your solution to work, 1 on the left-hand side of the original equation would need to be in the complex plane defined by arguments in the domain ((2k-1)π,(2k+1)π] w/ k≠0, whereas -3 on the other side would need to be in the complex plane defined by arguments in the domain ((2m-1)π,(2m+1)π]. To avoid this kind of multi-valued inconsistency, fix m=k≠0.
But fixing m=k≠0 removes the principal value of the complex number. This is not mathematically sound--is there such a thing like a complex number w/o the principal value? The presentation is just a series of 'interesting' mathematical mistakes
@@hajstra1307 x itself can take its principal value, even for k≠0. The argument of x, θ, can satisfy tanθ=-ln3/[(2m+1)π] w/ θ∊(-π,π] & in the appropriate quadrant. 1 can not take its principal value, but -3 & x can.
@@kyintegralson9656 You are for selective approach. Not use the principal value only when needed. The same goes for the identity of multiplication. This way yo can prove anything. For example: As long as there is THE IDENTITY of multiplication: 1^x = 1*1^(x-1) = 1^(x-1) Now there is NO IDENTITY of multiplication (as is done selectively in the video) 1^x = 1^(x-1) -> x =x-1 -> 0 = -1 Do you really believe that 1^x = 5 = 3 = -3 = -4 (?)
@@JSSTyger But fixing k≠0 removes the principal value of the complex number. This is not mathematically sound--is there such a thing like a complex number w/o the principal value? The presentation is just a series of 'interesting' mathematical mistakes
@@Emry11 X is complex base on "If x is an any complex number". If x is an any complex number x = a+ib, we have 1^(a+ib) = e^((a=ib)ln1). Since (a=ib)ln1= 0 so e^0 =1, therefore 1^(a+ib) = 1, so 1^x =1
If I've been professor at the University of Illinois then my level of understanding math surpasses people like you who lack even basic knowledge and come here posting spam like a troll. Go learn math and then come here falsely accuse others. I'm done with rude illiterate people like you.
In the 10:45 , x= log(-3)/i2kπ , I think it is wrong... While i2kπ•x•log(e)/i2kπ = log (-3)/i2kπ = x log (e) = log(-3)/ i2kπ = x= log (-3)/ i2kπ• log(e)
I wonder what e^(i*2pi*k) is when k=5? It might be very very interesting! Why did he stop at k=4? Someone should help this guy by cutting out the relevant 2 minutes of it. I pause it and advance through it using the cursor right key ->. He won't get many stars for the video production skills.
En mathématiques, le logarithme complexe est une fonction généralisant la fonction logarithme naturel (définie sur ]0,+∞[) au domaine ℂ* des nombres complexes non nuls. Plusieurs définitions sont possibles. Aucune ne permet de conserver, à la fois, l'univocité, la continuité et les propriétés algébriques de la fonction logarithme. (IN FRENCH) en particulier ln(1)=0, donc ln(1^x)=ln(1)*x=0 You have a big problem when you take a natural complex log of this equation... because natural complex log is not following any great rules of algebra.
this isn't true. by definition of exponentiation, 1^z = exp(ln(1)z) = exp(0z) = 1 for all z complex. if you wanna start talking about complex multivalued exponentiation then 1^x is not well-defined so the question is invalid. put your solution into 1^x in any computation engine and it will evaluate to 1 for this reason
i suppose i should explain what the video is tryna say as well ( (2m+1)π - ln(3)i ) / 2nπ is the solution x to the equation e^xw = -3 subject to e^w = 1 (therefore w is a “complex logarithm” of 1, hence -3 is a “complex exponential” of 1 by x) a correct (well-defined) solution would be: first we know the general formula for complex logarithm, e^z = y e^Re(z) e^Im(z)i = |y|e^(iarg(y)) z = ln|y| + (arg(y) + 2πn)i, where n is any integer. this is because e^(a+b) = e^a e^b holds for all complex a,b (the proof is not trivial tho) hence e^w = 1 w = 2πni and e^xw = -3 xw = ln(3) + (1+2m)πi whence the solution follows by dividing xw by w (so n is not 0)
@@jimdotz You are wrong. Ln(e^(0+i2kpi)) = ln(1) + i2kpi = 0 + 0 for k = 0. You shall always consider the principal value of any complex number, and never remove it. You can also return to the general form of complex number: e^(0+i2kpi) = 1 + i*0 = 0 (for any k), so you have ln(1) = 0 again. Remember, complex numbers expand arithmetic (for example by providing extra solutions of equations), they include real results (if they exists), but not contradict results of real numbers math.
The whole point is not everything is possible. I mean some statements are simply false. So not possible. Like someone else said they mentioned Godel too.
Why doesn't this give me the value I expected? >>> import cmath >>> x = lambda m,k: ( (2*m+1)*cmath.pi - 1j*cmath.log(3) ) / 2*k*cmath.pi >>> 1**(x(301,28)) # for m=301 and k=28 (1+0j) Should this not be (-3 +0j) instead?
It all depends on what is log(3) [and log(1)]. For a real value of it, you get a real value of the power, that is 1 [and 0, resp.]. And the real value of log(3) [and log(0)] is the principal logarithm of the general, complex logarithm function. Sometimes the principal is written as Log(3) [Log(0)], but sometimes the general logarithm is Log and the principal is log. The complex log function has inf. many values. Let k=m=1. 1^((3*pi -i*log(3))/(2*pi)) = exp((3*pi -i*Log(3))*(log(1)+2*i*pi*n)/(2*pi)) = exp(n*Log(3)+3*i*pi*n) = 3^n*(-1)^n = (-3)^n, where n is determined by the branch chosen for complex logarithm of 1). For n=1 you get -3, for n=2 it's 9, for n=3 - -27 etc.
@@Jaahquubel I suppose I am not following your explanation, then. There must be something wrong with the way I tried to calculate this value, otherwise it would have produced "-3". The log() function in the cmath module /is/ a complex-valued function.
@@Roarshark12 cmath.log computes the principal value of the logarithmic function and for a real argument it's just the real logarithm. Complex logarithm of a real number is of the form Log(x)+ i*2*n*pi, where n is any integer and Log is the real/principal logarithm. cmath.log computes the value for n=0. And you're not doing anything wrong. If you want to see -3 as the output, you need to use the version of the problem I mentioned in my first reply, the one that uses "cmath.exp()" instead of "1**". Type cmath.exp(cmath.log(3)+3j*cmath.pi). That is the n=1 case.
Interesting solution! Although you are very patient in your presentation and like to go to very details, however this makes it somewhat cumbersome. Note that the viewer of this level of math must already be familiar with advanced math concepts like complex numbers. Thus, just stating the Euler's identity is enough and going and substituting 1,2,3,4 for 4 to show it will be equal to 1 is kind of dull.
Then later in your solution procedure you want to find the equivalent of ln(-3) and you go at length to substitute -1 with another Euler entity. But is that necessary? Why not just simply substituting -1 with i^2?
You get: ln(-3) = ln (3*i^2) = ln(3) + ln(i^2) = ln(3) +2 ln(i) which is much simpler and cleaner approach.
No any interesting solution--just a series of interesting mathematical mistakes.
@@hajstra1307 Not sure which mistakes you are referring to but bear in mind that math problems may be approached from different angles. Thus different solutions could exist although one may be the shortest and most efficient one.
If you think he has made explicit mistakes you need to mention it in your comment with details.
@@Emry11 My mistake---this mistakes are not interesting. I was trying to be polite.
#1 error is neglecting the identity property of multiplication.
1^x = 1*1^(x-1) 1^x = 1 = 1*1 = 1*1^(x-1) but there is 1^x = 1 that contradicts 1^x = -3. The identity property of multiplication shall be neglected to continue with this nonsense. Following consequences are not dealt with:
a) 1 shall be kept explicit in all calculations since 1*x is not equal x anymore and multiplying or dividing by 1 could change the value. Example is an exponential form z=re^(iθ) when r=1. There shall be 1*e^(iθ), not e^(iθ)
b) ln(1) = 0 is no longer: 1^x = -3 ln(1) = ln(-3)/x
c) one can proof for example that 0 = -1. This goes as follow:
1^x = 1*1^(x-1) 1^x = 1^(x-1) x = x-1 0 = -1
(I respected the identity only when it is convenient, following the approach used in the video)
#2 error is eliminating the principal value of complex numbers by eliminating k=0. It is impossible and baseless. Conversion to the general form of complex number is always possible so 1*e^(i2kpi) = 1 + i*0 = 1 for any k.
#3 error is not testing the result for x by using it in 1^x = -3 for any selected k.
#4 error is producing videos that prove 1^x = 5 , = 3, = -3, = -4 and not REMOVING them from RUclips
#5 error is attempting to become new Euler
@@hajstra1307 Your points are valid only in the Natural numbers domain not in Complex number domain. Your very first assumptions that 1^x=1 contradicts 1^x=-3 is only correct in N domain.
@@Emry11 x is complex in 1^x = 1*1^(x-1) 1^x = 1 = 1*1 = 1*1^(x-1). Yes it contradicts 1^x = -3. This is the POINT!
The complex numbers domain includes the real numbers. We are not talking about the Natural numbers.
All my points are about the Complex numbers (even #5)
He manages to write L so that it is identical to e. That's very artistic!
So L=e thus Ln e = e*n*e = n*e^2 = 1 and from there on...
Writing the solution for x in terms of general values of integers k & m is not mathematically sound. In order for your solution to work, 1 on the left-hand side of the original equation would need to be in the complex plane defined by arguments in the domain
((2k-1)π,(2k+1)π] w/ k≠0,
whereas -3 on the other side would need to be in the complex plane defined by arguments in the domain ((2m-1)π,(2m+1)π].
To avoid this kind of multi-valued inconsistency, fix m=k≠0.
But fixing m=k≠0 removes the principal value of the complex number. This is not mathematically sound--is there such a thing like a complex number w/o the principal value? The presentation is just a series of 'interesting' mathematical mistakes
@@hajstra1307 x itself can take its principal value, even for k≠0. The argument of x, θ, can satisfy
tanθ=-ln3/[(2m+1)π] w/ θ∊(-π,π] & in the appropriate quadrant.
1 can not take its principal value, but -3 & x can.
@@kyintegralson9656 You are for selective approach. Not use the principal value only when needed. The same goes for the identity of multiplication.
This way yo can prove anything. For example:
As long as there is THE IDENTITY of multiplication:
1^x = 1*1^(x-1) = 1^(x-1)
Now there is NO IDENTITY of multiplication (as is done selectively in the video)
1^x = 1^(x-1) -> x =x-1 -> 0 = -1
Do you really believe that 1^x = 5 = 3 = -3 = -4 (?)
What application can we have of this equation on science or engineering?
none. There are multiple errors.
Mientras no haya luz,..... busquemos que no haya luz.
It would be nice if you could show integer values of m and k where the solution actually works.
2kπi must always be zero (thats how he rewrote ln(1)) so the division by zero is what is making the solution not work.
@@JSSTyger But fixing k≠0 removes the principal value of the complex number. This is not mathematically sound--is there such a thing like a complex number w/o the principal value? The presentation is just a series of 'interesting' mathematical mistakes
1^x=-3 x=((2m+1)π-Ln(3)i)/((2Kπ)) K ≠ 0 K = Z Log[1,-3]=Undefined
If x is an any complex number, we have 1^x = e^(xln1).
Since xln1= 0 so e^0 =1, therefore 1^x = 1
Not correct! Where is complex number i in your equations?
@@Emry11 X is complex base on "If x is an any complex number".
If x is an any complex number x = a+ib, we have 1^(a+ib) = e^((a=ib)ln1).
Since (a=ib)ln1= 0 so e^0 =1, therefore 1^(a+ib) = 1, so 1^x =1
If I've been professor at the University of Illinois then my level of understanding math surpasses people like you who lack even basic knowledge and come here posting spam like a troll. Go learn math and then come here falsely accuse others.
I'm done with rude illiterate people like you.
You could improve your presentation massively by not repeatedly saying identically the same or variations of the same thing several times over.
In the 10:45 , x= log(-3)/i2kπ , I think it is wrong...
While i2kπ•x•log(e)/i2kπ = log (-3)/i2kπ
= x log (e) = log(-3)/ i2kπ
= x= log (-3)/ i2kπ• log(e)
I wonder what e^(i*2pi*k) is when k=5? It might be very very interesting! Why did he stop at k=4? Someone should help this guy by cutting out the relevant 2 minutes of it. I pause it and advance through it using the cursor right key ->. He won't get many stars for the video production skills.
Apparently, you’re not familiar with Gödel's incompleteness theorems. Not EVERYTHING is possible in mathematics.
Actually Gödel's help is not needed. Mistakes in presentation are numerous and easy to be shown
1^3=1*1*1*=1 (x ➖ 1x+1).
ln(1) is automatically zero and does not need to be rewritten as 2kπi. Thats where the error lies. Your denominator at 10:35 is zero.
Sophism
Please note that erroneously ln(1) = (ln(-3))/x (after conversion of 1^x = -3). There is more errors.
@@cazacumihail3671 Real sophists cold be upset...and author could be proud
En mathématiques, le logarithme complexe est une fonction généralisant la fonction logarithme naturel (définie sur ]0,+∞[) au domaine ℂ* des nombres complexes non nuls.
Plusieurs définitions sont possibles. Aucune ne permet de conserver, à la fois, l'univocité, la continuité et les propriétés algébriques de la fonction logarithme. (IN FRENCH) en particulier ln(1)=0, donc ln(1^x)=ln(1)*x=0 You have a big problem when you take a natural complex log of this equation... because natural complex log is not following any great rules of algebra.
Complex roots again?
It’s in my head.
1^x = −3
x lg(1) = lg(−3)
x = lg(−3) / lg(1)
x = lg(−3) / 0.
Has anyone receive a comment by the instructor. There is no way that he can substitute the value for x and end up equaling -3.
wrong, natural logarythm is only defined in 0 to +infinity
this isn't true. by definition of exponentiation, 1^z = exp(ln(1)z) = exp(0z) = 1 for all z complex. if you wanna start talking about complex multivalued exponentiation then 1^x is not well-defined so the question is invalid. put your solution into 1^x in any computation engine and it will evaluate to 1 for this reason
i suppose i should explain what the video is tryna say as well
( (2m+1)π - ln(3)i ) / 2nπ is the solution x to the equation e^xw = -3 subject to e^w = 1 (therefore w is a “complex logarithm” of 1, hence -3 is a “complex exponential” of 1 by x)
a correct (well-defined) solution would be: first we know the general formula for complex logarithm, e^z = y e^Re(z) e^Im(z)i = |y|e^(iarg(y)) z = ln|y| + (arg(y) + 2πn)i, where n is any integer. this is because e^(a+b) = e^a e^b holds for all complex a,b (the proof is not trivial tho)
hence e^w = 1 w = 2πni
and e^xw = -3 xw = ln(3) + (1+2m)πi
whence the solution follows by dividing xw by w (so n is not 0)
Hi! 1 = exp^(i*2*k*π) and your solution looks like true! But!!! Also, 1 = n^(i*2*k*π/log(n)) and if n is not exp, your solution is wrong!
try to solve Indian exam jee advanced maths questions. And Love you from India.
This is completely wrong. There is no solution.
That’s correct. 1 to the power of any exponent will return 1.
@@Lithium59not so for complex numbers.
@@jimdotz You are wrong.
Ln(e^(0+i2kpi)) = ln(1) + i2kpi = 0 + 0 for k = 0. You shall always consider the principal value of any complex number, and never remove it. You can also return to the general form of complex number: e^(0+i2kpi) = 1 + i*0 = 0 (for any k), so you have ln(1) = 0 again.
Remember, complex numbers expand arithmetic (for example by providing extra solutions of equations), they include real results (if they exists), but not contradict results of real numbers math.
1÷0 is not possible.
First pin sir
The whole point is not everything is possible. I mean some statements are simply false. So not possible. Like someone else said they mentioned Godel too.
COS(8 pi ) = .9 .... we always thought that 1 to any pos power = 1. I think you reach too far, sorry. InventPeaceNotWar
Why doesn't this give me the value I expected?
>>> import cmath
>>> x = lambda m,k: ( (2*m+1)*cmath.pi - 1j*cmath.log(3) ) / 2*k*cmath.pi
>>> 1**(x(301,28)) # for m=301 and k=28
(1+0j)
Should this not be (-3 +0j) instead?
It all depends on what is log(3) [and log(1)].
For a real value of it, you get a real value of the power, that is 1 [and 0, resp.].
And the real value of log(3) [and log(0)] is the principal logarithm of the general, complex logarithm function. Sometimes the principal is written as Log(3) [Log(0)], but sometimes the general logarithm is Log and the principal is log.
The complex log function has inf. many values.
Let k=m=1.
1^((3*pi -i*log(3))/(2*pi)) = exp((3*pi -i*Log(3))*(log(1)+2*i*pi*n)/(2*pi)) = exp(n*Log(3)+3*i*pi*n) = 3^n*(-1)^n = (-3)^n, where n is determined by the branch chosen for complex logarithm of 1).
For n=1 you get -3, for n=2 it's 9, for n=3 - -27 etc.
@@Jaahquubel with the Python complex math (cmath) module, log() defaults to the natural log, that is, log to the base "e"
@@Roarshark12 The base of the logarithm is not the point.
@@Jaahquubel I suppose I am not following your explanation, then. There must be something wrong with the way I tried to calculate this value, otherwise it would have produced "-3". The log() function in the cmath module /is/ a complex-valued function.
@@Roarshark12 cmath.log computes the principal value of the logarithmic function and for a real argument it's just the real logarithm.
Complex logarithm of a real number is of the form Log(x)+ i*2*n*pi, where n is any integer and Log is the real/principal logarithm.
cmath.log computes the value for n=0. And you're not doing anything wrong.
If you want to see -3 as the output, you need to use the version of the problem I mentioned in my first reply, the one that uses "cmath.exp()" instead of "1**". Type cmath.exp(cmath.log(3)+3j*cmath.pi). That is the n=1 case.