The trick here is avoiding the exponent equal to zero. Instead, we can use 1 = e^(2*pi*i*k), where k is a non-zero integer.

@Aelcyx

Жыл бұрын

Should there be an i in the exponent?

@gdtargetvn2418

Жыл бұрын

​@@Aelcyx yes

@Alians0108

Жыл бұрын

@@Aelcyx 2*pi*k comes from a unit circle rotation while i makes the length = 1

@Aelcyx

Жыл бұрын

@@Alians0108 the radius comes from the coefficient of e. You need i or otherwise it's not periodic

@mcwulf25

Жыл бұрын

1^x = exp(2mπxi) -1 = exp( (2n+1)πi ) Compare exponents and cancel πi to get: x = (2n+1)/2m

The problem here is your powers are not well-defined. When you use complex powers, you need to specify which branch of of the logarithm you are using. 1^x is equal to e^(x* log(1)) where log (1) can be any branch of the log. If you start defining the powers correctly, your solution will make a lot more sense. For example, if you use the branch of the log where -Pi/2

It all depends on what complex logarithm you use. If the chosen logarithm satisfies ln(1) = 0, then there are no solutions. But if ln(1) is any other integer multiple of 2πi, then you get infinitely many solutions.

@ladymercy5275

Жыл бұрын

I like to use ln(1)/2πi to symbolize integers.

@mcgamescompany

Жыл бұрын

the thing is that ln(1)=0 only for the main value of the logarithm, and you cant really stick to that since you are using the infinitely many other angles that represent both 1 and -1, so using just 1 for the logarithm makes no sense in the first place. It would be more correct to say "no solution if x is a real number, and there are infinitely many solutions is x is a complex number"

@grrgrrgrr0202

Жыл бұрын

@@mcgamescompany hard disagree on everything you said.

Let a and b both be element of R and z element of C: then (a^z)^b a^(bz). In general (a^z)^b is not defined. In order to be well defined, b has to be an integer. See also De Moivre‘s theorem. Wolfram Alpha is correct, there is no solution.

@dragunityx12

Жыл бұрын

Average discrete math enjoyer

@thomaslangbein297

Жыл бұрын

@@dragunityx12 ???

@markener4316

Жыл бұрын

Everything is fine you did a great job with the example

The inverse function of x^2 is defined on a 2-sheeted Riemann surface. In the complex plane only one of the two branches of sqrt(z) can be defined.

My first thought was to use the multivalued-ness of the exponential function. 1 can be written as e^(2*pi*i), so to get -1, we can take x = 1/2. But then 1 = e^(2*k*pi*i) for any integer k, so x could be of the form 1/(2*k), except k = 0. Now, could we've guessed this? Sure, one of the square roots of 1 is -1, since (-1)*(-1) = 1, so the x = 1/2 solution is rather "obvious". The other ones are too, once you realize an even power of -1 is +1. But of course, this hinges on using those "non-principal" branches of exponentials. The canonical answer, in my book, is "no solution".

This is a little like Numberphile and 1 + 2 + 3 + 4 + ... = -1/12. WA and Desmos are not wrong because they are following the rules of real number mathematics and ignoring extraneous roots. The question remains whether this is a case where the non-principal roots are not just possible answers but really relevant ones. By the same token I can assert that the square root of 4 is -2, and I am kinda right but not really. So you can fairly tell me I was thinking of a number which squared gives me 4, which is a different thing than the square root of 4. But if I said I was talking about the square root of 4 + 0i I would be off the hook. This is what I am gathering is your argument here. ;)

@tollspiller2043

Жыл бұрын

I mean, don't you have that a root only has one solution, and a quadratic equation has two, meaning you have to take +- in order to not lose information. By that, by just taking 1^(1/2) you are simply taking the root of 1, which is still only 1, because you are not solving a quadratic. 1^2 on the other hand would have 2 solutions, 1 and -1

@fotnite_

Жыл бұрын

I don't think this is really true. The result from the video happens because the complex root is assumed to be a multivalued function when solving for some variable unless otherwise specified. If you take the rth complex root, you will always get r answers, where in this case -1 is one of those answers, and that's possible because of this assumption. On the other hand, 1+2+3+... doesn't have a value at all under any circumstances, because it doesn't converge under any domain. Now, it's true that ζ(-1) = -1/12, but ζ(s) = Σ1/n^s _if and only if_ Re(s) > 1, so ζ(s) ≠ Σ1/n^s. ζ(s) is an analytic continuation of Σ1/n^s. The "numberphile result" is just not a true result in _any_ case, whereas the result in the video _is_ true under a complex domain. So WA is completely correct when it tells you about 1+2+3..., but not correct for a complex domain when it tells you 1^x = -1 has no solution. Of course, it assumes a real domain, and it would be correct for that domain, as you pointed out. But I really don't think this is the same kind of thing.

@MDarioF

Жыл бұрын

mmmm 1^(x), x€R or x>=0 equal 1 ever.

The solution is in conflict with the definition of the square root function. The result is always positive, otherwise it would not be unique and in the consequence the square root no function.

@daanskiel1236

Жыл бұрын

That's exactly what i thought, this solution only works if you neglect the rules of the square root.

@RexxSchneider

Жыл бұрын

The square root function is not defined in the same way for complex numbers. In fact, it's often considered a multivalued function, and the solution -- x = (2m+1)/2n -- will always contain -1 as one of its values.

@leif1075

Жыл бұрын

@@RexxSchneider butbthst implies what I was thinking tomsolve this you can take the log of both sides..say natural log..but then you get x times ln 1 which ln 1 equals zero so there doesn't seem to be a solution that way since ln of (-1) is not zero we know ln of (-1) equals i*pi by Eulers formula..so this problem is indeed ill posed..see what I mean because even if you say x is a complex number like a plus bi when you multiply by ln 1 it always comes oubtto zero..unless now you tell me you can't take log of a negstive number..well why not since as you say log is a multi valued function in complex world..

@RexxSchneider

Жыл бұрын

@@leif1075 In the complex plane, all numbers have multiple representations for the same value. For example 1 = e^(2nπi) where n is any integer. Similarly -1 = e^(2mπi + πi) where m is any integer. So we could write 1^x as e^(2nπix), and we could re-write the equation 1^x = -1 as this: e^(2nπix) = e^(2mπi + πi) Now we can happily take natural logs of both sides, giving: 2nπix = 2mπi + πi So x = (2mπi + πi) / 2nπi -- as long as n ≠ 0 We can now cancel top and bottom by πi and we get: x = (2n + 1) / 2m Q.E.D. Incidentally, that should show you how to take the natural log of -1 by writing -1 as e^(2mπi + πi), which immediately gives the complex natural log of -1: ln(-1) = (2m+1)πi for all integer values of m, which shows that it is indeed a multivalued function. Therefore, I respectfully disagree that the problem is ill-posed.

@wolfbirk8295

Жыл бұрын

​​@jash21222 why for nonreal numbers...? You must define 1^x for nonreal numbers...; else this has no sense...! There is a definition for real numbers x ....! ( There are different ways to do that.!)

The final value of x = 1/2k for all values of integer n provided that k is non zero; The reason is that 1 raised to the power (2n+1) is always unity for all integer n;

@livedandletdie

Жыл бұрын

1 raised to any power is 1, doesn't matter what the exponent is. If it's positive, imaginary, negative, nothing turns 1^c into -1 or any other number for that matter. 1 is after all the multiplicative idempotent, no matter how many times you multiply it, it will remain itself forever.

We can do in simplee way that we will write 1^x as (1+ i(0))^x and -1 as -1+i(0) this will become (cos2pi+ isin2pi) ^ x = cos pi+ isinpi by using demoviers theorem and comparing real parts cos2pix=cospi. Cos gets and pi gets canceled we get 2x=1 .: x=1/2

I took y = 1/x, then raised both sides to the power of y, giving the equation 1 = (-1)^y which gives us y=2, from there we get x = 1/2.

@prime934

Жыл бұрын

But x=1/2 doesn't satisfy this equation

@akshatpratapsingh5476

Жыл бұрын

@@prime934 it does satisfy watch the full video

Square root of a positive integer is positive

I agree with @felsenHome:. This solution show a big mistake about the definition of square root. Because without loss of generality, 1^(m/n) with m odd and n even always can be write as 1^(1/2) [square root of 1] which is always a positive number, it is in the fundamentals of mathematics.

@Francu8942

Жыл бұрын

In the complex numbers all roots have as many solution as their index, i think

@Schockmetamorphose

Жыл бұрын

@@Francu8942 Yes. And any even root has the property that it includes -1

I could understand every step ... but I have difficulty to accept that the found solutions are all real numbers...there must be something that's not adding up properly...

A game where everyone is playing by different rules.

You don't need all of that. We know that i^4=1 and i^2=-1, so: (i^4)^x=i^2 i^4x=i^2 Then we have: 4x=2 *x=1/2*

Thanks for answering my question!

Perhaps another video could be make to explain the solution shown at the end of this vidoe?

Mathematicians would write an enormous paper to prove that the sun will rise tomorrow.

The answer for x could be anything, as it will still come out as 1. Both the negative and positive of the same number, will give a positive answer, when squared.

@paolarei4418

Жыл бұрын

1x1 = 1.(Absolute infinitely many zeroes later)1 = how CPUs counts= yo bad at math = good^-1 + you x are = word^quality^ok^-1^-2 + human^yo + he+haves = math of words? Wth

Sorry, this time you are wrong!

very well made, keep it up

@SyberMath

Жыл бұрын

Will do! Thanks

Q: Determine complex nums analytical continuation for Zeta Function at Ramanujan's sigmaN=-1/12

equation is unsolveable tho. if x=1/2, then 1^1/2=-1, which is wrong, 1^1/2=+-1

I’m not convinced. If our final answer is a real number then 1 to a real number is 1. I’m not sure how to reconcile this.

@Archimedes_Notes

2 ай бұрын

😂😂😂

Correct me if I'm wrong but from what I can tell any fraction will work as long as the denominator is odd and the nominator is even. also we can still think about it in terms of real numbers if we tweek it a little bit: 1^x=-1 ==>> 1=(-1)^(1/x) and since we know that (-1)^2 is 1 by that mehod we deduce the 1/2. Going from here 1^(1/2) which is √1=+-1 which in it selfe contains -1 so the solution is good, moving forward we can also wright this as (-1^1)/(-1^2) so -1/1=-1 and we can expend upon this whith the logic that if we want the -1/1 with the -1 basic than -1^z=-1/1 than z=(-1)^n/(-1)^m so -1^(n/m)=-1/1 so n/m is even/odd and since -1^(1/x)=1^x than z=1/x so 1/(n/m) which is m/n so odd/even

@zirco77

Жыл бұрын

On your very first line, you raise both side of the equality to the power 1/x. While the new result might be equal, it does not prove the original equality. Simple counter example: let's try to prove 5 = -5 5^(1/x) = (-5)^(1/x) : true? let x = 1/2 as in your example 5^2 = (-5)^2 : definitely true while the last line is true, I don't think this proves that 5 = -5 to start with ;)

@paolarei4418

Жыл бұрын

Im from the 6th grade and this breaks da mind

8:28 1^(1/2) is the principal branch. Meaning it’s only going to be 1. All real powers give positive numbers when applied to positive numbers.

1^x = -1 Using exp(2*pi*i*k) = 1 And exp(pi*i*(2*k+1))= -1 Then the equation becomes: exp(2*pi*i*m*x) = exp(pi*i*(2*n+1)) And since the exponential function on complex domain are an entire function with period 2*pi*i, their arguments inside are equal to the module 2*pi*i 2*pi*i*m*x = pi*i*(2*n+1) x =( 2*n+1) / (2*m), where m and n are integers. When m=0 the equation are not defined. (x goes to Infinity) With n=0 and m=1 the solution are x=1/2, which is the second branch of the square root: Sqrt(1) = -1

@PTETdorf

Жыл бұрын

dont care

@King1Z7

Жыл бұрын

​@@PTETdorf I think you have a spelling error in your comment. It was obviously meant to say "wow, amazing"

Just intuitively, it seems like x=1/2n where n is an integer?

@usdescartes

Жыл бұрын

True... except the numerator could be any odd, not just 1.

@Schockmetamorphose

Жыл бұрын

@@usdescartes It's just about taking an even root of 1

@Schockmetamorphose

Жыл бұрын

You have to exclude 0, you can't divide by 0

@SyberMath

Жыл бұрын

Why do we worry about different forms of 1 on the left hand side? Can we not just write e^(2pi*i) for 1 which would give us x=(2n+1)/2? Where does I remember you had said for the following video that it doesn't matter what form is used for i: kzread.info/dash/bejne/doN4r9yfeqTQZag.html

i just did: 1^x=-1 (i^4)^x=-1 i^4x=-1 i^4x=i^2 4x=2 x=2/4 x=1/2 but 1^(1/2)=1 there weren't any mistakes in the calculation of X so we just got a contradiction edit: it would be a contradiction if the square root wasn't multivalued

At 7:50 after squaring for the case x = ³/₂ {(1)^³⁄²}² = 1³ = 1 = (−1)² = 1 Where did you get i³ from ?! By the way i³ = −i

@RexxSchneider

Жыл бұрын

The i was simply a mistake. He meant -1.

An easier solution : 1^x=-1 ((-1)²)^x=i² ((i²)²)^x=i² i^(4x)=i² 4x=2 x=1/2

So I have two questions: Why did you include complex numbers? Why did you stop at complex numbers? At a glance, x = 1/2 is obvious because sqrt 1 = +/- 1. With a little more thought, 1/sqrt 1 will give us x = -1/2. If we couldn't find these real solutions, yes it would make sense to look at complex numbers. But we don't need them and the problem didn't hint that they were interested in going that deep. If you voluntarily look at complex numbers, why stop there? Quaternions give more solutions, I would guess dual numbers give more as well, and there are probably more to be found in more niche number systems. But we traditionally stop at real solutions unless otherwise indicated or required.

@grrgrrgrr0202

Жыл бұрын

There is nothing deep about looking at complex numbers. Complex numbers are such a household name in mathematics that you must have a very good reason if you don't look into complex numbers. Quaternions, on the other hand, are very niche and not interesting most of the time. And for this problem, if you have properly formulated the complex case (which this video has not), you have the quaternion case for free as you can replace i by any purely imaginary quaternion with absolute value 1.

@daksharora928

Жыл бұрын

the problem is sqrt function would always give a positive value kind of like x^2=1 x=1 or -1 but x=√(1) x=1 is the only solution, a square root doesnt give both values, it only gives us the positive one, that is how the function is defined

I think everyone would be pleased to know, that non-rational power exponentiation yields infinite number of results. so, x^e is absolutely cursed on a complex plane, with its compact infinite braches set

@paolarei4418

Жыл бұрын

And x^log(2)+? (except 10 multiplies)

I'm trying to open my mind up to how you solved this. The graph shows there are no real solutions, so whatever we come up with must be complex. But in your final answer, you get to 3/2 and 1/2, which are both real! It seems we are considering sqrt (1) to include the possibility of -1, which is in conflict with the graph. So to solve this, rather than using complex numbers, aren't you shifting the conventional definition of the sqrt function, which is always positive?

@SyberMath

Жыл бұрын

Real numbers are complex, too! They are just a subset

@ladymercy5275

Жыл бұрын

Maybe the graph in the video was wrong.

@paolarei4418

Жыл бұрын

Math is complex or GOOGOLPLEX

@evgtro8727

Жыл бұрын

The trick is complex functions can be multivalued in contrast to real-valued functions. So, if you treat sqrt(z) as a complex function then it has two branches. If it is a real-valued function then it must have unique value (must pass the vertical line test) and that value is always among non-negative real numbers. As about graphs, a complex-valued function has 4 dimensions (2 for the variable z = x+yi and 2 for the function f(z) = u+vi) and has its graph in 4-dim space xyuv, which is impossible for us, humans, to do. To solve this problem people plot projections of the 4-dim graph of f(z). It can be done in several ways. For example, the plot in the xyu gives the graph of real part u of f(z) as a surface. f(z) can be also plotted in the 2-dim xu-space as a real part u of f in terms of real variable x. The given function f(z) = 1^z (in the video it was 1^x) was drawn as a straight horizontal line in the xu-space (although they used y for u). If you could graph this function and the function g(z) = -1 in the xyuv-space you would see they intersect there at points z = 1/2 (x = 1/2, y = 0), z = 3/2 (x = 3/2, y = 0), and many others.

solve by comparison: 1^(x) = -1 and e^(πi) = -1 thus: 1^(x) = e^(πi) e^[ln(1)(x)] = = e^(πi) ln(1)(x) = e^(πi) x=e^(πi)/ln(1) but ln(1) = 0 hence: x=e^(πi) ÷ 0 (division by 0) therefore, x is undefined.

@garougarou1369

5 ай бұрын

ln(1)=ln(e^2inπ)=2inπ

Very interesting indeed!

for negative numbers, the exponentiation operation, with a rational degree, is not defined. (-1)^(1/2) = (-1)^(2/4) = 1

@jayconway1205

Жыл бұрын

Yes there is a definition for exponentiation of negative numbers and this problem can be solved using its definition. (-1)^(1/2) =(-i) the imaginary number. Similarly you can then easily show that i^2 = (-1) yes a real number and i^3 = (-i) back to an imaginary and i^4 = 1 back to a real We can use these identities to solve 1^x = -1 by substituting for 1 and -1 (i^4)^x = i^2 So 4x = 2, and x=1/2 and 1^(1/2) = -1. QED

Any even fractional power will do it (x=0.5, 1.5, 2.5 . . . !), since -1^2, ^4, ^6 . . . equals 1.

1^x = -1 log 1^x = log (-1) x log 1 = log (-1) x = log (-1) / log 1

The squareroot or quartic root is defined to be a postive number for instance sqrt(16)=4 but not -4 and qurt(16)=2 but not -2 that's why there's no solution.

It's not really a matter of using complex numbers; it's a matter of allowing multiple values for expressions and taking 1^(p/q) to be all of the solutions to x^q=1^p for integers p and q and the usual meaning of exponents as repeated multiplication or division. Going to complex numbers makes multiple values for expressions kind of important all of the time, since it brings up issues with picking a principal value that makes sense through the whole calculation, but the technique doesn't require using complex numbers per se.

@grrgrrgrr0202

Жыл бұрын

No! Exponentiation is defined by a^x = exp(ln(a)*x). The logarithm ln(a) is defined to be a solution of exp(x)=a. On reals, there is only one way to define the logarithm. But if you expand to complex numbers, there are are many ways to define the logarithm, with no definition being the best. For each nonzero complex number, all definitions of the logarithm agree up to multiples of 2πi.

@iabervon

Жыл бұрын

@@grrgrrgrr0202 Integer exponents are generally defined (in abstract algebra setting, for example) as repeated multiplication, and rational exponents are an obvious extension of that. In a real analysis setting, exponents with a continuous variable are as you describe, but just given an equation with a superscript and no information on the branch of mathematics, there are some other reasons the same solution as you'd get in complex analysis would be applicable, for similar reasons. For example, in modular arithmetic, there are two square roots of some numbers (depending on the modulus) and there isn't an obvious principle one.

@xXJ4FARGAMERXx

Жыл бұрын

@@grrgrrgrr0202 a^x = exp(xln(a)) ln(a) is the solution to exp(x) = a Applying this into our problem, we have: 1^x = -1 exp(xln(1)) = -1, where ln(1) is the solution to exp(x) = 1. There are multiple solutions to exp(x) = 1 in the complex numbers, exp(0) = 1, but also exp(2kπi) = 1 where k is an arbitrary integer. So we have exp(xln(1)) = -1, where ln(1) is 2kπi exp(x(2kπi)) = -1 exp(2xkπi) = -1 exp(2xkπi) = exp(ln(-1)), ln(-1) is the solution to exp(x) = -1. exp((π+2k₂π)i), where k₂ is an arbitrary integer satisfies the formula, so we have exp(2xk₁πi) = exp((π+2k₂π)i) I'm pretty sure if exp(a) = exp(b), then a = b? So 2xk₁πi = (π+2k₂π)i x = (1+2k₂)/2k₁ Substituting k₁=1 and k₂= 0 we get x = 1/2. Let's check if this is really a solution 1^x = -1 1^½ = -1 exp(½ln(1)) = -1, ln(1) is the solution to exp(x) = 1. The solution is 2kπi for arbitrary integer k exp(½(2kπi)) = -1 exp(kπi) = -1 But this equation is only satisfied when k is odd. So actually the solutions to exp(x) = 1 are 2kπi ONLY for odd k??? Something's off.

Any whole number k from 0 to infinite will give you a 1 as a complex number, so even k=0 must be valid. But in your solution you would then divide by zero which is not defined. So with k=0 your solution is not defined, but k=0 is a valid number. Your solution is therefore flawed but not entirely wrong. You see it with n=0 and k=1, that does give you -1 but also +1 and you can not define which is the solution. By using exp(x) you worked around the problem of ln(-1) but not around the problem of deviding by ln(1). But using complex numbers is a nice idea as it gives you at least some possible solutions, but no definite soluton. Maybe you could swap from euclidian to a non-euclidian geometry where paralells actually intersect, otherwise there is no definite solution to this problem.

This brings up a serious question: Should we agree that 1^(1/2) is supposed to mean the same thing as sqrt(1)?

@DARKOAMBER

Жыл бұрын

its the same only the definition of roots is incomplete and restrictedby rules

@Schockmetamorphose

Жыл бұрын

the thing is... the nth root of x^m = x^(m/n) it's not about agreeing

@DARKOAMBER

Жыл бұрын

@@Schockmetamorphose can you explain it in more detail. i dont get the point

@Schockmetamorphose

Жыл бұрын

​@@DARKOAMBER Remember that (x^a)^b=x^(ab) →(x^a)^(1/b)=x^(a/b) (x^1/a)^a=x^(a/a)=x^1=x ((a)th root of (x))^a=x therefore ((a)th root of (x))=x^(1/a) and ((a)th root of (x^b))=(x^b)^(1/a)=x^(b/a), with which we are at what I said earlier.

@DARKOAMBER

Жыл бұрын

@@Schockmetamorphose thanks for your effort to answer but i dont get it. maybe my first comment in this section caused a misunderstanding. sqrt(x) = x^(1/2) for me its the same by defintion the only trouble here which causes thoughts that sqrt(x) x^(1/2) are the incomplete defintion (how to interpret a non loss of infomation about the base number and the freedom in the sequence of operations) or restricted rules (which says a root only results into positive results for real numbers which isnt true cause we use the quadratic solution formula as example) i hope clarfied what i mean and dont confused even more ^^

Also x = i (pi) / 0 if you just ln both sides

x=1/2 if you choose 1^(1/2) to be -1 instead of 1

Why they intersect in the infinity, can you proof that?

Im not clear why 3/2 didnt work. If its a complex 1 then raising it to the 3/2 power should work just as well as raising it to the 1/2 power shouldnt it?

@RexxSchneider

Жыл бұрын

Of course x = 3/2 works if x=1/2 works. Since 1^(3/2) is the same as (1^(1/2))^3 and (-1)^3 = -1. it was just a mistake in the video.

@grrgrrgrr0202

Жыл бұрын

The value of 1^x depends on the choice of the logarithm. If ln(1) = 2πki with k odd, then both 1/2 and 3/2 work. If ln(1) = 2πki with k even, then neither work. (and regardless of the logarithm, (x^a)^b=x^(ab) if b is an integer)

@RexxSchneider

Жыл бұрын

@@grrgrrgrr0202 The complex logarithm is always a multivalued function. For situations where a single value is needed, then a "principal value" may be chosen. However, when solving equations, we always need to consider _all_ possible values of a function so that potential solutions are not missed. if x is of the form (2n+1)/2m, where n, m are integers, then 1^x will always contain -1 as one of its values, and that is why the general solution to the equation is of that form. When x = 1/2 or 3/2 or 123456789/12345678, one of the values of 1^x will be -1.

Incorrect notation, wolfram is absolutely right. 1^z is a number set! -1 is a specific number.

The end is the most important part, but cut off so quickly. I don't get it.

@SyberMath

Жыл бұрын

Sorry about that

Wow, I actually got it - thanks to the knowledge of polar coordinates!

This requires no complex numbers, as -1 is a square root of 1 without necessity of going beyond a number line. And the answer of x = 3/2 is incorrect, that is only one answer, it can be any odd numerator over 2.

I liked your x = 1/2 try surprise to satisfy 1^x = -1 as one if the two square roots of 1 results in +/- 1. But that comes from x^2 + 0x + 1 = 0 quadratic formula where now we pick -1 as the root because +1 doesn't solve the 1^x = -1 equation.

where do you get that blackboard

@SyberMath

Жыл бұрын

Notability

just starting the video. My first thought is to start with the fact that (-1)^2=1; so -1=1^½, and x=½.

@azhar9823

Жыл бұрын

✓x² = x is only true for positive values. For negative (also positive) values : ✓x² = |x|

@dataweaver

Жыл бұрын

@@azhar9823 Do you see “√” anywhere in my comment? No? That's because I know that √(x²)=|x|, and was deliberately avoiding its use for that exact reason.

@azhar9823

Жыл бұрын

You didn't write it, yes, but you did apply this. Taking the power ¹/² is equivalent to the squareroot

@dataweaver

Жыл бұрын

@@azhar9823 Not quite. The radical (i.e., √) only returns the principal root, in this case 1. The half power could return either root, depending on the context. In this context, raising the square to the half power leaves the base as is; so you go from (-1)^2=1 to -1=1^½. Technically, 1^½=either 1 or -1, and you need more context to know which one to use. But in this case, the context is supplied by what x is in x^2: if x is negative, then (x^2)^½ is negative; if it's positive, then (x^2)^½ is positive. That lets you generalize to (x^2)^½=x, which is what I did. That is specifically why I used the half power instead of the square root, to _avoid_ getting locked into the principal root and to instead use whichever root is more appropriate.

@olixx1213

Жыл бұрын

​@@dataweaver actually false x^m/n is defined as nthroot(x^m) (or nthroot(x)^m So 1^1/2 is 1

What if i express 1 as i^4 and -1 as i^2... In this case the common base Will lead to 4x=2 and x=1/2....how about that ?.... Using complex numbers

Find all possible solutions to 1^(1/π) = x. Kinda easy, but great for educationnal purpose.

@SyberMath

Жыл бұрын

Nice. 1^(1/π)=[e^(2πi)]^(1/π)=e^(2i)

Gerade Wurzeln haben positive und negative lösungen. Die negativen Lösungen bleiben aber wenn ich einen ungeraden Exponenten habe. Das ist die Kurzfassung ohne Berechnung

On a projective plane, they intersect at the point of infinity.

@BryndanMeyerholtTheRealDeal

Жыл бұрын

Yes, THE point at infinity. The unsigned infinity by joining the two ends together.

How do you get desmos in dark mode?

@SyberMath

Жыл бұрын

Click on the wrench on the top right and then check the 'Reverse Contrast' box

x=k/n where k is an odd integer and n is a nonzero even integer

Wrong.... the square root only gives a negative value when you initiate it to find X in an equation. The defeniotion of a predetermined square root like in this case, does not give a negative value like -1, but only the positive value: 1. THIS VIDEO IS A WASTE OF TIME. either that or my teachers teach math wrong.

Just logarithm both sides. On the left hand side you get x*ln(1), which is x*0 apparently... So 0=-1. Great. :-)

@justcommenting5117

Жыл бұрын

If you logged both sides, then 0=ln(-1) which is less absurd considering it's not defined and if we take absolute value inside of ln, it's also 0

@rickdesper

Жыл бұрын

But ln(-1) =/= -1 . ln(-1) isn't defined. (-1) isn't in the domain of the logarithm function.

@usdescartes

Жыл бұрын

We're talking about Complex numbers here, so x ln(1) is only one possibility. 1 = e^(i2πn) where n is an integer. This gives us ln 1 = ln e^(i2πn) = i2πn, and since n is an integer, it can be 0, which brings us back Reals. Since that gives us 0 ≠ -1, that tells us n≠0.

I did like this : 1 = (-1)*(-1) = i^2*i^2 = i^4 .... ok ? So 1^x = (i^4)^x = i^(4x) and we know that 1^x = -1 so we can write i^(4x) = -1 . And now we replace -1 by i^2 and we have i^(4x) = i^2 .... ok ? So by identification we can write 4x=2, so x=1/2

What blackboard sw are you using? I loved it.

@SyberMath

Жыл бұрын

Notability

Fantastic question...👌👌👌.

NO SOLUTIONS!

You missed to exclue k=/=0 because you can't divide by 0

well it al dpends in which filed of math you are working to solve thiq equation. it would be better to just define better the exercise

I overthought this one when I tried it before viewing the video. When I stopped without a solution, I thought he would use the Lambert function to solve it and come to with an imaginary solution. That’s not what he did, of course.

I did not get it why 3/2 does not work. 1^(3/2)= (1^1)*(1^(1/2)) = either 1, or -1. Thus 3/2 also works.

@zawatsky

Жыл бұрын

Согласен. Что есть корень? - Дословно это означает: сколько раз и какое число мы умножаем само на себя, чтобы получить имеющееся. -1 подходит в случае с любым чётным знаменателем (который, соответственно, является целой степенью для преобразования правой части уравнения в левую). Фактически дробь может быть хоть 1357/224 - результат будет тот же.

@RexxSchneider

Жыл бұрын

Yes, 3/2 also works.

@ChatwithSam

Жыл бұрын

I was bout to write the same ... Well yes it works.. 1^{3/2} ->(1^{1/2})^3->(±1)^3->±1 (-1) Hence 1^{3/2}=±1 with one of its values equal to -1

@250minecraft

Жыл бұрын

Square root is always positive

@RexxSchneider

Жыл бұрын

@@250minecraft Nonsense. Every number (except 0) has two distinct square roots that are the negative of each other. The square roots of 1 are +1 and -1.

1^x is always equal to 1

@SyberMath

Жыл бұрын

no

LN both sides use Euler's identity and solve for x. 😉

Why does he say ”intersect” instead of ”cross”? I thought ”cross” use when the graphs have the same points on a plane, and ”intersect” is for the same situation but in space. Can somebody explain me that? Because I’m awful at English science terminology. And if I made any mistakes in my text ,please indicate because I need to improve my English skills.

Interesting :)

@SyberMath

Жыл бұрын

Glad you think so!

the value is -i.log(-1) / 2.pi.n where n =/= 0 pls correct me if im wrong

@rgbstatic5823

Жыл бұрын

workings 1^x = - 1 x log (1) = log (-1) x = log (-1) / log (1) -> log 1 = 0, therefore we take e^2.pi.n and it gets cancelled with log => log (-1) / 2. pi. i. n => -i.log(-1) / 2. pi. n where n=/= 0

@akshatpratapsingh5476

Жыл бұрын

umm but i guess u cannot use a negative number inside a logarithmic function

I don't think so anything to the power of the base is 1 like 1^2 is always 1. your x-values if plugged in are false. I could be wrong but thats the rule of mathematics. If we do the logarithm way. like x = log(-1)/log(1) is undefined because any logarithm of a negative number is undefined

awesome video.

@SyberMath

Жыл бұрын

Thanks!

Ngl the awnser seems a bit forced, like it doesn't concluded backwards, but it is still better then what I got pi*i/ln(1) witch is just completely wrong

Lets simplify the given: 1^x =-1; 1=-1; Simplified 1^x as 1. Oh no. I have failed and came up with no solutions..

The answer is a real number regardless of n and k. One raised to any real power would always be 1. So there are no solutions for x.

@ManjulaMathew-wb3zn

2 ай бұрын

From the graph it was already determined solution if any are complex.

@ManjulaMathew-wb3zn

2 ай бұрын

(2n+1)/2k is real. How come the graph didn’t show the intersecting points?

There is no solution to this equation. I don't have the time to trace down your fallacy, but 1^(3/2) = 1. 1^x is a constant function.

@erikkonstas

Жыл бұрын

Well, you better had "time" to "trace down" his "fallacy" before spewing words out... it turns out that it's not a fallacy! If you watched the video, you'd notice that it's pretty much just a matter of avoiding the "principal branch", simply because, for integers n and m, e ^ (i * (2 * pi * n)) = e ^ (i * (2 * pi * m)) and, as such, 1 ^ (1 / 2) can be 1 or -1; also, 1 ^ (1 / 4) can be 1, i, -1 or -i. In general, all complex numbers z that solve (1 ^ (1 / n) = z) are called *n-th roots of unity* ; for k in {0, 1, ..., n - 1}, they are z := e ^ (i * (2 * pi * (k / n))) We can now show that z ^ n = (e ^ (i * (2 * pi * (k / n)))) ^ n = e ^ (i * (2 * pi * k)) = 1 * e ^ (i * 0) = 1

@grrgrrgrr0202

Жыл бұрын

It is a fallacy indeed. Several of the rules that are a second nature for real exponentiation, such as the power of a power, don't hold for complex exponentiation. Fact is: unless you define complex exponentiation using a logarithm that doesn't expand the real one, there are no solutions on the complex numbers either.

Простое решение. x=½. Или любая другая натуральная дробь с чётным знаменателем. Поскольку у единицы два квадратных корня, 1 и -1. Что же в уравнении может быть такого особенного?

@yiutungwong315

Жыл бұрын

1/4 1/8 1/16 1/32 ...

@zawatsky

Жыл бұрын

@@yiutungwong315 числитель может быть любым. Если в дробную степень возводить в правильном порядке, при первом возведении единицы в любую целую положительную останется единица. При последующем извлечении корня любой целой чётной степени получается ±1, нас интересует только -1. Если знаменатель нечётный - то корень только +1, не выйдет.

x=m/2, m is an odd integer.

squaring on both side 1^2x=1, or x=1/2 and x=0 but since x can't be zero it is 1/2 is it correct?

@widespirit7648

Жыл бұрын

No because 1^1/2 = sqrt(1) = 1

@gtavbyrockstargames950

Жыл бұрын

@@widespirit7648 but square root of a number can be + or - cus the square of a negative number is also positive

@widespirit7648

Жыл бұрын

​@@gtavbyrockstargames950 oh wait yes you are right Sorry

@gagadaddy8713

Жыл бұрын

@@widespirit7648 I am afraid you can't simply draw a conclusion that there is only sqrt(1) as the solution, because it already entered into the complex domain. There are unlimited number of solutions .....

@gtavbyrockstargames950

Жыл бұрын

@@widespirit7648 no problemo we all make mistakes

Cool !!!

What is the point anyway ? I feel like this is not a point like is it true or not, just depends on how you define your function x->y^x honestly, which in the definition on reals with exp and Ln always says that 1^x = 1

1^1/2 is the answer because -1 is equal as i

Hey, new to your channel. I think this is an excelent video, but I think you could have gone more in depth about what the result implies. You had me very invested and the video ended too soon!

@maalikserebryakov

Жыл бұрын

You sound like ChatGPT

@yaredreinarz3244

Жыл бұрын

@@maalikserebryakov whatever, as an AI language, I was just politely suggesting that you can extend a little bit more on the conclusion to make your videos better

The result is true only if k is not equal to zero.

I simply did 1=(-1)^(1/x) 1/x=2k (-1 always become 1 when its power is multiple of 2) X=1/2k K is an integer

@RexxSchneider

Жыл бұрын

But you miss all the other possible values of x where the numerator is any odd integer. Since (-1)^(2m+1) = -1.

doesn't 1^0,5 have 2 solutions one of which is exactly -1?

If ln1=0 the problem is cancelled.

Suppose x=0.5. In that case, the square root of 1 is on the left side and -1 is one of the two solutions, the other one being 1. Other values for x also work: any integer + 0.5 will do.

Easy, all x=1/2n with n being positive whole numbers

Now analyze the integral of 1^x ;)

Alle Achtung junger Geist, Sie haben mich erbost.

I didn't get it, 1= e*(2k)×(pi)×(i), can someone explain it please

@gregorysmathchannel5357

Жыл бұрын

The exponential function is extended to complex numbers by the definition: e^(a+bi)=e^a(cos b + i sin b). This gives e^it = cos t + i sin t for any real number t. So, e^(2ki(pi)) = cos (2kpi) + i sin (2kpi)= 1 + i*0. This is because cos (2kpi) = 1 for any integer k and sin (2kpi)= 0 for any integer k.

Ich habe gut gelacht, dafür kann ich mich bedanken. Kann ich mich dafür auch bedanken, dass ich mich über solchen Unverstand anschließend geärgert habe? Diese Kleckerei wirkt, sudelt indes nicht etwa nur eine Drolligkeit mit tatsächlichem Lachwert. Fast, dass es perlt, gibt dieses Bepperl nämlich dem Barbaren augenscheinlich recht, welcher meint, dass aus der tatsächlichen Geduldigkeit des Papiers die Denkbarkeit von allem folge, wo es die Mathematik ist, die gerade im Aufprall auf die Undenkbarkeit ihre Gegenständlichkeit kristallisiert. Das ist das eine. Zum anderen spuckt dieses Bepperl auf Cantors Grab, dass der dumpfe Kronecker feixt. Immerhin ist hier ein Exempel von Blödsinn statuiert, das es in sich hat. Nicht etwa, dass sich mit diesem Beispielunfug eine mathematische Thematik ergäbe. Dieses Unfugsparadigma stellt eine epistemische Aufgabe, und zwar eine nach meinem Geschmack wenig ergiebige, nur mühselige und unerfreuliche. Toyota hat seinerzeit gescherzt. Copyright by Andreas Raab

(a^b)^c is not always a^bc with complex numbers. Sorry.

@SyberMath

9 ай бұрын

When are they equal?