How are they different? Cube root vs the exponent of 1/3
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- เผยแพร่เมื่อ 2 ส.ค. 2023
- WolframAlpha gives different results if you try the cube root of -1 versus taking -1 to the exponent 1/3. Why is this happening? There is a good reason for the difference, which involves complex numbers and roots of equations. Special thanks this month to: Kyle, Lee Redden, Mike Robertson, Daniel Lewis. Thanks to all supporters on Patreon! / mindyourdecisions
References
www.wolframalpha.com/input?i=...
www.wolframalpha.com/input?i=...
www.wolframalpha.com/input?i=...
en.wikipedia.org/wiki/Square_...
en.wikipedia.org/wiki/Functio...)
en.wikipedia.org/wiki/Princip...
en.wikipedia.org/wiki/Princip...
en.wikipedia.org/wiki/Nth_root
math.stackexchange.com/questi...
blog.wolframalpha.com/2013/04...
Complex Variables and Applications, James Ward Brown, Ruel V. Churchill, 7th edition, Chapter 1 Section 8, Roots of Complex Numbers
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When Graham Bell invented the mobile phone,he had 2 missed calls from the director of Wolfram.
Or, the principal square root of four calls - but not minus the principal cube root of minus eight!
@@rogerkearns8094😂
Heehee yX-D 😂🏆🥇❤️
I appreciate Wolfram's contributions and for my personal use I have found mathematica a better language and tool than MATLAB but I do know where MATLAB can be significantly better and I appreciate it reading his book a New kind of Science but fundamentally I have to disagree with his wanting to basically take a whole new approach to make math more of a science I still view it as more a useful and but powerful tool than a paradigm shift
Nah, not from Wolfram. This thing is always from Chuck Norris!
This (roots equispaced around a circular locus in the complex plane) was literally the *one* single moment in A-level maths lessons where I perked up and thought "ooh, that's interesting". Of course, the maths teacher immediately followed that with "but we don't go into that on this course".
A level further maths
@@deltavalley4020 yep. I believe it was covered in that course,, which I wasn't doing. It all came up in my engineering degree later anyhow
As soon as I saw the intro, I knew there were 3 roots to the cube root of -1 on the complex plain. The first root of course is -1 at 180 degrees on the unit circle. Then you know the other two roots are at +/- 60 degrees on the unit circle. You can work out what these have to be by simple trigonometry of the 60 / 30 / 90 triangle.
@@deltavalley4020Yeahh I loved doing it in Further Maths I’m surprised I managed to figure it out before my teacher taught me it
The take away is that math is a language, and like all languages, context matters. So many forget that math is just a way we are trying to describe a concept.
Math uses a language. It's not true that it "is" a language. You may think the distinction doesn't matter, but pedantic attention to detail is required in mathematics.
@@rickdesper I guess technically math is the science of the relationship of sets. But the numbers, the symbols, the rest of it is all a language to represent that and is what the OP was referring to when saying that the meaning of the two equations isn't the same. I just think it is important to remember that all those numbers and symbols are a convention to describe a concept, they aren't some sort of universal law. PEMDA, BODMAS, GEMA are all just conventions; for example, if you didn't write the equation using those conventions you wouldn't solve it using them either.
Absolutely
@@christopherkopperman8108 you should use brackets- round (), square [], curly {}, line and angle to specify which operation to solve first.
Forget about conventions like BODMAS or PEMDAS. They are all nonsense.
"This angle theta will be theta" Great wording right there.
Ah, yes, the floor here is made out of floor
@@h20dynamoisdawae37only the best floors are made of floor. Floors made of non-floor can be quite dangerous.
Non-floors commonly have the property of either being made out of wood or not being made out of wood
So basically, they ARE the same. You are just picking a different principal value.
Yes, I was waiting for Presh to say something like "wolfram alpha is just putting the roots in a different order".
Yea there are no convention. The correct way is to specify if the solution is to be real or complex. Cube root gives you -1 because when you specify it that way the Real solutions is probably what you want (think of a simple calculator).
no, he is using the real root for the first one (commonly used) and the principal root for the 2nd one (also commonly used)
Indeed, and for every nth root, there are n values, even some professors will wrongly insist that it will be take into consideration only one value. For practical purpose I agree that the positive value (in case of square toot) is useful (e.g. in engineering and in practice in general, you can't obtain negative lengths or time), in purely math, there are 2, 3, n values for square root, cube root and n-th root. Ignoring them, that doesn't mean, they are not "there".
@@radupopescu9977 There are n values that satisfy an n'th degree polynomial equation if the domain in consideration is that of the complex numbers. But the number of values depends on the considered domain. For example, in the domain of quaternions, an n'th degree polynomial equation will have _more than_ n values as solutions.
In the domain of the real numbers, an n'th degree polynomial may have less than n solutions.
Moreover, there can be only _one_ n'th root - that's why it's called " _the_ n'th root". Which value it is, depends on your definition of "n'th root". Your professors are not wrong about this. But there is a difference between "n'th root" on one hand, and "the roots of a polynomial" (or "the solutions of a polynomial equation") on the other hand.
In math, we _need_ the notion of a unique "n'th root" in order to be able to describe the _complete_ solution (i.e. all values) of the polynomial equation. For example, we need the square root function to be defined in a unique way, so that we can say that the solution to the equation x^2 = 5 is
x = sqrt(5) OR x = -sqrt(5) .
If we don't have that definition of the square root function, then we can't communicate clearly and unambiguously what the two solutions are.
The Principal Root equation is taught at your University's graduate level Complex Analysis class. However, the fractional exponent notation is a concept more often utilized in an Abstract Algebra Course where your Commutative Ring must be specified. Your choice of Commutative Ring is the Complex Field. However, in some settings, say the Ring of Integers Modulo 5, 3^(1/3)=2 since 2 is that element whose cube is 5 in this setting.
I think I learned it in an undergrad course... except I completely forgot in 15 years fast forward
Your comment just has me wondering if you'd be able to help me with a question I've had regarding non-UFDs and how to find how many possible factorizations something might have in that context
The language in which you presented your view made me rethink about liking maths
This is really a question of social convention, rather than a principle of mathematics. Aliens would have the same mathematics, but they would also have completely different definitions for all these terms.
Aliens may not have the same mathematics, if they live a life in a very different context and are oriented toward very different life goals. Whereas our maths are built around goals and purposes that focus primarily on addition, multiplication, powers and exponentiation (with the definition of a positive sign directed towards "higher" and "more"), aliens may have for example developed a math centered on balancing and sharing, where one plus one equals one, and one divided by many still equals one.
@@yurenchu Math is the abstract truth everywhere, while physics is the truth about our universe (the part we know).
@@howareyou4400 I disagree. Both math and physics are not "the truth", they are just (useful) _models of_ the truth. And they only apply as far as the models are designed to apply.
Moreover, math is shaped by our (= humans') physical experiences. So we humans are only inventing the math that is/seems relevant to our human experiences, and neglect any potential of math that doesn't seem relevant to our experiences.
@@yurenchuNah, math is not the model. You use math to describe your model. Or we could say, math is part of your model, while the other part is the connection between the math and the problem in real world.
Your model might be outdated for the problem, so you updated your model, using different math and making different connections. But the math itself is still correct.
@@yurenchuAddition is addition though. Unless these aliens are microscopic where the effects of quantum physics affect them or they live in a different universe.
This really is dependent on whether you're talking about the real or complex power function here. Even then we sometimes fudge this line since we'll define square roots of negative numbers to be imaginary numbers. (-1)^(1/3) is indeed -1 if interpreted as the real power function, but if interpreted as the complex power function, the definition I use most defines that as the set of all complex cube roots of -1 rather than just the principal root.
Why use a definition that outputs multiple values?
When you write down an expression, e.g. an answer to a problem, you want it to be a singular value, exactly and unambiguously
@@AuroraNora3Note really when dealing with comolex roots or logarithms you should specify if you consider the principal value. If not you are cosidering them as polyfunctions, i. e. functuons whise output is a set and not a number
"When you write down an expression, e.g. an answer to a problem, you want it to be a singular value, exactly and unambiguously"
@@AuroraNora3 Nope. You want that. I want every possible solution.
@@devhermit Fair. Tbh aything is fair in math as long as you communicate beforehand how you define expressions. I would just argue that it should be *standard* (keyword) to have just the principal value, exactly like we do with square roots.
"The length of the pendulum is 3^(⅓) m"
x³ = 3 has three solutions, so should the statement above refer to all of them? No imo
This is a very bad idea, you don't want to use polyfunctions unless you know exactly what you are doing and write it explicitely. First, you definitely want to keep the property that x = y => f(x) = f(y). This is how you solve equations and polyfunctions don't allow that. Second, this has many less properties than regular functions. I mean, you could not even compute its derivative or a path integral, why would you use it for? There are probably many other arguments against the use of polyfunctions.
I really don't see a point where you would actually require polyfunctions. The typical way to go is to define x^z = exp(z*ln(x)), exp(x + yi) = exp(x)(cos(y) + isin(y)) and ln(z) = ln(|z|) + i Arg(z) where Arg(z) is the principle argument of z (defined in the interval you want, but single-valued). This allows you to find all solutions to equations such as x^z = y, while still having d/dz x^z = ln(x) x^z, d/dx x^z = z*x^{z-1} and d/dz ln(z) = 1/z for z where ln(z) is holomorphic (i.e. differentiable on the complex plane). Fun note, the logarithm behaves so badly on the complex plane that, in fact, it is not continuous (and thus not holomorphic) for negative real numbers. You can still work with it and that's why it's important.
They are equivalent statements. WolframAlpha provided the real root for the cube root and principal root for the rational exponent. The real roots are the same for both expressions, and the principal roots are the same for both expressions.
So just to be clear, I’m not gonna be penalised in my exams for turning a fractional indice into a root??
Still pretty new to anything more than everyday maths, so this video mostly went over my head
@@jagobot1487you wouldn't be wrong to do that, but your math teacher might call it out as wrong anyway (if they're a bad math teacher, as many are).
7:34 This is important; thank you for your thoroughness. People need to be aware of this.
We learned it slightly different.
For us, both are the same.
But you always have to specify when you are going to use complex solutions or not.
So it seems that (from your video) some cultures give one version the implied complex version.
I can see that being handy, but also causing confusion for those (like me) that don't follow the same convention.
It's not even really a cultural thing, it's more like an individual taste thing.
For example, I prefer to reserve the notation a^(1/n) to refer to the *set* of all n-th roots of a number, while using the surd notation for the principal root, but I have had professors that use the two interchangeably as the principal root. I have rarely dealt with any confusion about my usage, but it is important to clarify whenever there is a chance of ambiguity (like with software output ala Wolfram Alpha).
@@whannabi
My entire point was that multiple teachers at the same school had different styles... and this is all happening within the same 'culture' of mathematics.
I understand that there are regional differences in how mathematics is taught, conceptualized, and notated--my comment doesn't detract from that.
The fact remains that there is individual variation even *within* that, and the notation for roots is one of the things I have observed to exhibit such individual variation.
Wish I had this video in April for my linear algebra final. Good explanation for finding imaginary roots!
I didn't understand much of it at all, but I trust that you know what you're talking about from having watched a boatload of your videos and having been a subscriber for a long time.
It requires a good grasp on complex numbers, polar coordinates and Euler's relation (e^(i.θ)=cos(θ)+i.sin(θ)). I understood, but had to stop to remember the content a bit. He does know what he's talking about
Basically, it means when you turn a function sideways to get its inverse, and are only considering the "first" root (k=0) as the first answer, that first root isn't necessarily going to cleanly lie on the real number line. But going the other way, you can always take the 2nd or 6th or sch-fiftyth root of an inverse of an n-root function that _just so happens_ to be a real number. So (-1)³ is equal to -1, but that's the 2nd root of the root3(-1) function (k=1) rather than the principle root (k=0). The principle root is complex because when you take x³ and flip it around its diagonal to invert it, the 1st and 3rd roots end up floating out in complex space while the 2nd root is fixed on the real line.
If anyone has any doubt about Presh they can pick up any book on the subject and study the topic on their own :D
Look for complex numbers and complex roots
One really important part (briefly mentioned) is, that the convention is one thing, but often what matters is the context for a particular problem. The other thing is, that conventions may differ, so sometimes (e.g. when writing a paper) the question arises, how to express something unambiguously without much hassle.
Anyone else get lost at the complex plane part?
No 😂
In the complex plane, the "real" axis is equivalent to the x-axis in the usual rectangular coordinate system, and the "imaginary" axis is equivalent to the y-axis. So, if you have an expression like 3 + 4i, you would (starting from the origin) move three units right and then four units up to locate a point on the plane.
Yes. So it's basically like plotting a point in Cartesian plane of coordinates (3,4) .
One of the best videos! Thanks a lot. 👍
One of the best yet.
Thanks, it was the best math lesson I attended in almost 20 years !
loved the video - nice work
I don't remember having studied this (but time has passed), it is very interesting indeed and makes sense
If your degree is not in engineering or science it's unlikely you encounter it before. Normally in High school, if you touch complex numbers is at a very basic level, not including multivalued functions or any of the complexities.
@@jaimeduncan6167 I am a mechanical engineer, that's why I was surprised. But it might be I simply forgot about this, since a lot of years has passed
@@milanfanas I'm also an engineer and like you, I don't remember ever studying this. It's been quite a few years for me. I do remember an old engineering professor (no doubt long dead now) speaking about how engineers and mathematicians view math differently. For us, complex math is just a stepping stone to get to real answers.
@@JimLambier ..no no i said thos beams must be -13 inches long...
he is wrong though so it does not make sense.
Nicely explained-thanks!
wrongly explained.
Isn’t the n-th root of x defined as x^(1/n)? The difference in Wolfram being standards of the principal value from it, like the first point rotating from 1 anti-clockwise.
I know this stuff but still watch. I like your way of explaining (plus, in 'my time', we had no youtube; wish we had).
Great, the Rogers-Astaire reference at the end; thanks! 😂❤
thanks for nice explanation and gardening tutorial
great video. Coincidentally I got to know that such thing as complex plane exists a few days ago, so the concept was easier to understand.
I believe it is up to the individual product to treat the two root notation and the power notation the same or differently. To state that an odd root of a negative real number and its reciprocal power are treated as the real and principal such roots, respectively, seems to me to just happen to be how Wolfram does it. They could have just as easily defined the opposite.
I think Wolfram defines the principal root as the root closest to the positive real axis, which would explain why they give a complex number for the power form, but it would not explain why they don’t also give that same answer for the root form.
yeah, I don't see a problem ising reciprocal to write root down
An excellent presentation.
Maybe it is worth mentioning that the whole issue comes from the fact that you cannot define a _continous_ complex logarithm funktion on C, or on C\{0}; you have to also exclude e.g. the negative real numbers whoch gives us the standard branch of complex log C\{z: z∈R and z ≤0 } -> C; z = r*exp(i*t) --> log(r) + i*t
in other cases it might be useful to exclude the positive real numbers, e.g. if you want to take roots of negative numbers, or add some multiple of 2π, but that depends on context.
No sense for me, as the two expressions are strictly equivalent.
In lower classes we actually do teach at school that √x is only to be used for non-negative x and open up the definition later in A-level classes. It really depends a lot , so I dare say ³√-1 can also be "not defined" for good reasons.
No reason to. Cubic root is a function across all the real numbers
@@sakesaurus1706 Yes, the nth root is well-defined for n odd, for all real numbers.
@@sakesaurus1706 It's better to keep these things separate: real solutions, polynomials with real coefficients, complex solutions, and polynomials with complex coefficients. Roots (denoted √) should only be used for calculating the modulus of a complex number. The " ³√-1 " expression should be referred to as the roots of x³+1=0.
@@vivvpprof Then you'd have to change a lot of maths books. Students usually learn about roots before they learn about complex numbers
What are you talking about? Mathematicians define the square root to be a function, period. Granted, you have to make choices, like a suitable (simply connected) domain and a branch of the inverse of the power (or a branch of the complex logarithm...).
But (almost) nobody defines roots as inverse _relations_
Nicely done - Both cube root and exponents are human constructs suspectable to paradox - that can only be resolved by exploring their constructs - and you did that "expertly"
So in other words both answers are correct answers to both equations, and it really just depends on what sort of answer you want. Sometimes you want all the possible answers, like often when solving the quadratic formula.
"Both answers are correct answers to both equations". Well....a properly stated question has only one answer.
@@rickdesper
I disagree, because in any math beginning algebra and higher you often find situations where there are multiple answers.
Take the equation x^2 + x - 6 = 0
There are two answers if asked to find x: 2 and -3. Giving only one wouldn't be a complete solution, and would only be worth partial credit.
Going into higher mathematics, when we get to integrals there are often problems where we only want the approximate integral because the full integral is difficult. There are lots of ways of doing that though, and the best answer depends on how much time you are willing to put into better precision, how much precision you want, what method is most precise for the particular situation, etc. But often you don't need the best answer, or have time to figure out what method achieves it, you need a good enough one. So different people would give slightly different close enough answers depending on the method and precision they use, but they would all work as a solution.
@@rickdesperso anything such as x^2 = 9 is an improper? Seems wrong to me.
@@rickdesperWere you even watching the video? There's literally an infinite number of math questions that have multiple answers. What a silly claim to make lol
I did not understand all of it, but even the part that I could understand surprised me and intrigued me.
Thanks for your video
It's weird. I've ran into that issue once. When I stumbled upon the apparent contradiction that:
sqrt[(-2)²] = sqrt(4) = 2
But
[(-2)²]^1/2 = (-2)^(2 * 1/2) = (-2)^1 = -2
You went from basic functions to complex numbers real quick 😂😂
I read once somewhere that the root symbol with a long overscore is one type (all n roots?), while the short overscore is another, perhaps the principal root. Though I can't find anything about that on the internet or Wikipedia...
This is a very good and clear explanation of the problem.
Multi-valued functions is maybe a amusement for some theory gurus, but if you need to make calculation, you should simply ignore them and consider either the principal value, which is well defined, or the real value if exist, as explained here.
The root symbol is basically used only for the n-th root with n natural (that means exponent is 1/n). We can extend easily to any rational exponent.
But you can always write exponentiation operation where exponent is irrational ... if you apply multi-valued operators to (-1)^pi for example, then you're in big trouble, because there are an infinity of values. The whole unit circle is solution ...
Great video
this is literally the same. if you do complex evaluations for both you'll get the same result
But in general if we want the set of all solutions, (Nth root of X) and (X ^ 1/n) should have the same N solutions, right?
In our example, (-1)^(1/3) = cuberoot(-1) = (X_1 , X_2 , X_3) three solutions, the three we can see at 6:20 so:
X_1 = (-1 +0i) = -1
X_2 = (1/2 + i*sqrt(3)/2 )
X_3 = (1/2 - i*sqrt(3)/2 )
Excellent video. If we are looking to calculate the real positive odd root of a negative number in wolfram using fractional powers, the way to do it is raise the positive number to the fractional power and take the negative. So to get the real 5th root of -5, we can do -(5^(1/5)).
Hey man u have to precise in which set u have to calculate the value of an expression or to solve an equation.
This isn't standard notation. The √ symbol with a positive real number refers to the positive square root, but other than that scenario there is no standard convention that is refers to the principle root. It is a multi-valued function unless stated otherwise (we can write ±√ if we want the multi-valued square root). That is why Wolfram Alpha includes that note at the top - you have to explicitly say you are taking the principle root since it isn't standard.
Still not clear to me. By definition these terms are identical. But you are saying they’re not. Why?
In practice, it's simply a conflict between conventions. If you're sticking to the reals, then there's only one cube root of negative one. If you move to the complex numbers, there are three and you have to choose one as the definitive answer. It's explained in the video why that particular one was chosen, and it's explained that sources differ on which one is taken as the principle root. It's worth mentioning that in the complex world it's not just roots that have multiple solutions--some other common functions (like logarithms) do as well. The choice of k=0 as the principle value maintains consistency even if it might seem like the real value, if one exists, would be a nicer choice.
@@bobbun9630 that part I understand. the video was explaining why that computer program gave a different answer. But I'm just talking about definition. a root vs a fractional power mean the same thing. Therefore the answer (or mulitple answers) should be the same.
There was another video regarding -1^2 vs (-1)^2. in my mind they are both the same. But to avoid confusion they are treated differently.
If we're not clear on the rules for these things than people are going to get different answers to things.
@@bb55555555 "If we're not clear on the rules for these things than people are going to get different answers to things."
Indeed. But these inconsistencies exist. One of my favorites that always has to be taken from context is the term "natural numbers". Sometimes the set of natural numbers includes zero, sometimes it doesn't. It depends on how that author is using the term. I always preferred to use the terms "positive integers" or "non-negative integers" just to be more clear, but you see "natural numbers" and the "N" symbol used everywhere in mathematics texts. It's usually clear which is meant, but the need to derive from context to be sure is always a potential source of confusion.
@@bobbun9630 authors calling zero a natural number? that I was not aware of. it's these crazy inconsistencies that produce these crazy answers. In a discipline like mathematics that requires this insane level of precision, I'm surprised we don't all make more of an effort to root out these so-called inconsistencies.
Surely this is a matter of convention and notation and not mathematics itself, no?
I appreciate your definition of the principal root - it‘s simply the zeroth solution of the corresponding equation (dividing the circle). 🤓
Thanks! This has always mystified me. Good to know that we have to look at the context. Apparently, sadly, mathematics does leave room for ambiguity in the case of roots.
You can have a function of different variables than x. y=+/-sqrt(x) is just the function x=y^2, which is a function. All a function does is create a one to one correspondence between two variables.
What you defined here is a bijection not just a function.
Is there a way to move both functions so that they only touch in one point, that is, not at (0,0) and (1;1)? If so, how would you do that?
You can treat them differently and yes, their expressions lend themselves to different approaches to resolve/solve/simplify. But mathematically they are the same.
We know that the sqrt(x) is a single value and not + or -. For example, when we write the quadratic formula, we say -b +/- sqrt(...). Why +/-? Because the sqrt function is single valued. By convention we use the positive value (when I was in algebra we learned that the sqrt is the absolute value of y such that y^2 = x). Therefore, we specify +/- sqrt to get both values.
Here's an odd thing. If you use the math-entry cube root button with -1, rather than type out "cuberoot", it gives the principal root, not the real-valued root, at the top.
This question highlights a question that I have had regarding both irrational and complex exponents and the existence of solutions that are not solutions to the exponential function
What's your question?
@@GoldenAgeMath well really I have more than one and perhaps I should not have included irrationals although I consider the number i an irrational I know that's not appropriate in all context but in that context I was referring to real rationals which aren't really as much of a part of my question there is a sense in which we could say there are infinitely many solutions there I forgotten the proof of it though It does have to do with chords on a circle so similar but in regards to when you have a complex exponent ideally I would want to know how to find any and all solutions The only solutions I know how to find of course being those that are an extension of the exponential function I'm including going up to like including its Riemann surface but I would initially be satisfied of knowing a proof of simply existence of solutions or a disproof of solutions that are not part of the extension of that function
My scientific calculator isn't wolframalpha so both gave the same value
Most calculators don’t even have complex numbers anyway
@@Ninja20704 mine shows the complex numbers only when solving quadratic and cubic equations, not in regular calculation. I'm using the casio fx-991ES Plus
@@Ninja20704 mine does and gave the same answer. Also, wolfram alpha gets things wrong too
5:29 Great video! Small typo: should be e^(i*(π+2πk))
If I understand correctly, the principal root is being referred to as the one with \theta = acute, as for others \theta are obtuse. This why it is 1/2 + i* \sqrt(3)/2 i.e. the one with \theta =60deg (others being 180deg & 300deg respectively)
Nope, it's just the case of k=0, no matter whether theta is acute or not.
Okay, that was excellent. But I definitely have to watch it a second time.
In 5:45, didn't you forget about the parentheses around the angle? In z = -1 = e...
yes, although the real timestamp where he first made that omission was 5:24. And earlier, at 4:11 he did use the parentheses correctly, to distribute the i to both terms of the addition.
PS Took me a while to understand the "In" wasn't a ln, and that it's "z = -1 = e..." prefixed by a capitalized "in". ^^
Mean while back at the farm........
the battle of conventions rages on!
The iperbola has 2 values of y for the x, the square root principle applies also on this one?
This was in my A-level Pure Maths text book, but after 5 minutes here I actually understand it.
Great video, so interesting that Math is not as pure and simple as we are all initially lead to believe but, excuse the pun, interesting complexity lies beneath, depending on your point of view!
Actually it is, because the complete set of solutions is always the same for every case.
But in different context we "prefer" one particolar solution over the others.
-1 ^ (1/3) = cuberoot (-1) = three solutions = 6:20 in the video, those 3 points are the full set of solutions.
It's the same as positive roots, if we say sqrt(9) = 3 that's only ONE particular solution we are interested in but the full set is { +3, -3 }, two solutions.
@@jimmyh2137 Those three points are the "full" set of solutions as long as you stay within the context of the complex plane. If we expand to, say, quaternions, the solution set becomes suddenly infinite!
The equation x*x = 9 has two real solutions: namely +sqrt(9) and -sqrt(9) (which can be simplified further, because 9 happens to be a perfect square; but that's beside the point). The current definition of the square-root function is sufficient and capable enough for us to help represent/communicete the solution set. There isn't really a point in making the square-root function two-valued ; or are we trying to milk _four_ solutions out of this quadratic equation?
Wolfram alpha actually has -1 as a solution, when you scroll down.
slight correction: around 5:23, the 2*pi*k part is missing an i, seeing as it's part of the angle in question
I remember learning that for something to be a function, it had to pass that vertical line test: only one output for a given input. Couldn't y^2 = x or y = √x be a function of y? x = y^2 would give one output for x for each unique input of y.
You're correct that x=y^2 is valid for x as a function of y. However, the vertical line test is a specific means of determining whether y is a function if x when considering the graph of the equation on the standard coordinate plane. If you wanted to use a geometric test on the coordinate plane for x as a function of y, you could use a horizontal line.
It can absolutely be a function of y. In formal math, a function is defined with a fixed input space (domain) and output space (codomain), whether we consider the equation y=x^2 as representing the function a -> a^2 or a -> root(a) is up to us.
@@GoldenAgeMath Nope. That arrow goes in one specific direction, and that is the function. And in order to be a function, it must map each element of its domain to exactly _one_ element of its codomain (not to multiple elements in its codomain).
Now in some cases, a function is invertible and hence we can also define an associated inverse function, in which the arrow goes the other way. But a --> a^2 and a --> sqrt(a) are not the same function (and not even their inverses).
At 5:37, z need to be corrected as e^ i( pi + 2*pi*k) on the top right of screen. I hope I understood right.
My TiNspire Calculator gave the complex root for both ways. In other words. It did not say the cube root of -1 was -1, no matter what I entered.
Can someone please explain what is the difference in the statements made from 01:24 to 02:16
So then check your work. Multiply one half plus I root 3 over two by itself 3 times and see if you get back to -1. Which should work if x to the 1/3 raised to the 3rd is just x.
All math is based on convention.
Why on earth would we define "function" in such a way that it discards some of its solutions??!
You're thinking of a multi-value function that outputs a set, but in a mathematical sense, we need functions to be well-defined and get a single, predictable value instead of a set of values. A lot breaks down when the output is not a single value.
Having a more stringent definition allows for broader generalizations and development in other areas and clarification of what you mean
U want to use functions as ways to describe phenomena in predictable manner, one input always gets one and only one output. As said in the video, with radical functions it will depend on context of what your function is describing, sometimes it will only be the principal root or the collection of all roots.
Calculus is based on this definition of a function. You set up y = F(x). Then you run differentials and integrals (anti-differentials) on the original function. As stated by other respondents, a lot of stuff breaks down unless there's a maximum of one y-value for every x-value.
How does x^n have n solutions? If I understand the claim it means (-2)^3 has three solutions or two more than the -8 I expected. I get how -8 ^ (1/3) has multiple/infinite solutions as explained in the video. Is this the parallel version of that? If so what are the other two imaginary/complex solutions to my specific example?
He meant y=x^n has n solutions for a chosen value of y, not x. So x^n = 1 for example has n different solutions
That's not what it means, it refers to the equations of the form x^n = y, where y is an arbitrary result. It doesn't say "any value cubed has 3 results", that would break the function, it says "for any arbitrary value y (except 0), there are exactly n numbers which will lead to it if raised to the exponent of n". When you do (-2)^3 you're already applying one of the specific solutions, so you won't get further values
To represent the case that you're talking about you would write instead x^3 = -8, which has one real solution in the form of -2 as well as 2 other imaginary ones -
x = 1 ± i sqrt(3)
Which work exactly the same way as the ones shown in the video
Isn’t this the key problem with most discussions in the comment section of so called math videos? That solving the equation x² = 4 is not the same as calculating √4.
You are incorrect. f(x) where x is an element of R is not the same function as f(x) where x is an element of C. Also note that in your own clip of Wolfram, which should not be used as a definitive authority of mathematics based upon computational results, the exponential version was converted to the root version.
Amazing 🎉
Just to clarify what I get here, so the reason the wolframalpha calculator gave a complex number was because it interpreted the exponent in (-1)^(1/3) as a principal value, it took one of the 3 solutions, which was positive?
The three solutions to the equation x^3 = -1 are basically one vector in the complex plane that is rotated repeatedly over 120 degrees, there is a standard way of enumerating them, and the one that is labeled with number "0" (or "1", depending on which start value you use for enumerating) is named the "principal root". Usually, it's the one that's closest to the positive real axis (i.e. has smallest complex argument when expressed in polar coordinates).
More concretely, the roots (solutions) of the equation x^3 = -1 are
x = e^(i(π/3 + 2πk/3)) = cos((1+2k)π/3) + i*sin((1+2k)π/3)
, for k = 0, 1, 2.
You forgot the parentheses around pi + 2pik at 5:29
Error in 5:29, the exponent should be e^(i(pi+2*pi*k))
“this angle θ will be θ”
😮
_"Call the whole thing off..."_ 😃 nice reference.
When I was studying Mathematics, one of the way we avoided this confusion was we had to say what domain is x in and what domain is the answer in and how many solutions we want.
For example, solve the equation x^2+2=1 where x is in the set of Real Number. The answer is no solution because there is no real value x that will solve this.
If I said, "Solve for x^3+2=1 where x is in the set of Real Number", now you know there is one solution and it is -1. If I change it to, "Provide all solutions of x of the following equation where x^3+2=1 and x is in the set of Complex Numbers", now you know to give me all of the solutions that this video mentioned.
This was especially important in abstract algebra and complex analysis class (class just focused on complex numbers).
It's another example of the Golden Rule... the mayhematicians ( leaving the typo on purpose) that decide the conventions are the ones with the "Gold" so they make the rules that make their work easier. Even if it conflicts with principles it is based on. If we want all of the answers, we have to specify so. If they want just one answer to make their functions work, no specificity needed.
Semper phi
Each number can be thought of as vector in the complex plane. The product of vector V and a unit vector U = exp(i * theta) is a rotated version of V. (In fact, the rotation is theta radians counterclockwise about the origin.) The vector "-1" is just exp(i*pi). So, a geometric interpretation of the solution to U^3 = exp(i*pi) is "what unit vector -- when rotated by itself twice -- results in the vector exp(i*pi)?" Clearly, a unit vector having angle pi/3 (60 degrees), when rotated by itself counterclockwise twice, results in a vector with angle 60 + 60 + 60 = 180 degrees. exp(i*pi/3) is the principle root you found. Note also that the vector (i*(-pi/3)) is also a root: if we rotate it twice using itself we also reach exp(i*(-pi)) = exp(i*pi). However, in this case the rotation is negative (clockwise).
exp(pi)=23.14069...
Smash that "edit" button and try again.
Typos fixed. Thanks.
nice to see Euler's identity just popup at the end.
I have always wondered about the fractional exponents whether or not they can be reduced.
For example (-1)^(2/6) is it equal to +-1 (6th root of (-1)^2)
or just -1 because -1^(1/3) = -1
Is it correct to reduce the fraction before doing the power arithmetic?
Yes, although -1 is probably not the best example to use since there are many unrelated values that will still give you -1
You either need to reduce the fraction, or you must do the denominator first then the numerator. So to do (-1)^(2/6) we must do [(-1)^(1/6)]^2. After accounting for the multi-valued radicals, you will get 3 answers, which are the same 3 answers as (-1)^(1/3) if you consider all the possible cube roots.
If the fraction is already reduced, then we can be happy because we can do it either way. For example, (-1)^(3/4) can be done as [(-1)^3]^(1/4) or [(-1)^(1/4)]^3 and both yield the same set of answers.
@@Ninja20704"If the fraction is already reduced..." Oh really? What about (-1)^(2/3) ?
@@yurenchu i dont see any problem ur trying to get at.
if you only consider real numbers you get only 1 whether you do [(-1)^2]^(1/3) or [(-1)^(1/3)]^2
If you consider complex numbers, then you will need to consider the multi-valued cube root. But you can verify that both of them will give you the same solution set.
[(-1)^2]^(1/3)=1^(1/3) which is the cube root of unity, and we know the 3 answers are 1 or -1/2 +/- i*sqrt(3)/2
(-1)^(1/3) = -1 or 1/2+/-i*sqrt(3)/2. Now we need to square each one individually. -1 squared is 1. 1/2 + i*sqrt(3)/2 will give -1/2 + i*sqrt(3)/2 when squared. and you can easily check that 1/2 - i*sqrt(3)/2 will give -1/2 - i*sqrt(3)/2
@@Ninja20704 Oops, you're right! Sorry, I don't remember what I was thinking when I wrote that. I probably mixed something up, and didn't notice that we'll get 1 either way, whether we multiply by 2 first, or later. Thanks for the reply!
In class I was taught that n√x has only 1 solution and x^1/n is not the same as n√x since it has n solutions. Is it correct?
The video is very interesting and draws the attention for a sometimes confusiong situation, but in "my book" cube_root (x) or x^1/3 mean exactly the same thing. You say that in some contexts they mean different things, and Wolframalpha also makes that distinction, but I don't agree. May be usually we do not think about all the complex roots of any number, but to me, it has nothing to do with the notation. The two notations have mathematically exactly the same meaning and in calculations any of them can be used at our choice. To my understanding and experience, real root or principal root have nothing to do with the notation, but only with the context.
if x
7:48 I didn't know Fred and Ginger were mathematicians!
of course they are the same thing, it is just Wolfram makes some assumptions what you want to see
Can you solve the integral of :
ln(sinx+cosx)/(cosx-sinx) dx
I haven't seen anything of the video, but I can deduce that it's because the cube root is defined as "³√(x)=k ⇔ k·k·k=x", and actually exist infinitie solutions for "³√(-1)" (such as "(-1)¹" or "[(-1)¹'⁶⁶]"), and "(-1)⅓" is just one of these solutions.
(that's de difference, or at least what I first thought).
Actually, the most common definition for the cube root simply specifies that a real value is always returned if the imaginary part of the argument is 0. If the argument is non-real-valued (i.e., non-zero imaginary part), then it returns the principal value, which is defined as the cube root with the greatest real part.
@@XJWill1 i am in 10th grade. i don't understand a single thing in the video or what you're saying. should i start learning this seriously now? i am highly interested in mathematics. but i have no idea why he brought in the pi, theta or e
@@h.smusic350 What I wrote is not complicated, but it does require an understanding of imaginary numbers. You can find plenty of information with a web search. The basic idea is that the square root of -1 is defined as the letter 'i' and then complex numbers are defined and can have a real part and an imaginary part.
z = a + i*b
where z is a complex number, and 'a' and 'b' are real-valued numbers, and i = sqrt(-1)
@@h.smusic350Def! If youre not sure where to start, Professor Leonard has good youtube videos and you can watch those until you find something else you want to watch
Why y = square root of x is not a function. The values of y still depends on x. There are 2 values of y with 1 value of x, given x > 0. And so, y can't be called a function of x ? Is that how function is defined ?
6:17 doesn't It should be quadratic root(1) and not cubic?
I'm very confused. In the middle of a Maths degree here in the UK. If square root of -1 is i [me thinking that square roots of negative numbers start using imaginary numbers] then why is cube root of -1 suddenly -1? Where's the i in all of this?
The cubed root of -1 is -1 because
-1 × -1 × -1 = -1
The other two possible answers for the cubed root of -1 are complex roots though, so both of those have i in them.
I think one thing that is consistent is that the symbol √ has only one value. It denotes a function.
thx
One runs into ambiguity when the base is negative and the exponent is fractional. The core of the issue is that a^b is defined as exp(b ln a) when b is not an integer. And ln a is not uniquely defined for anything but a > 0. Wolfram's behavior generalizes better than "take the real root, if one exists".
Funny enough, you can also do regular function stuff with the square root defined as negative instead of positive.
Hell, you can even get the e^x to behave nicely even when e^(1/2)