How are they different? Cube root vs the exponent of 1/3
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- Опубліковано 28 вер 2024
- 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
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Complex Variables and Applications, James Ward Brown, Ruel V. Churchill, 7th edition, Chapter 1 Section 8, Roots of Complex Numbers
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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
frr its so annoying 😭😭
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 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
@@Artaxo They are nothing compared to void-floors. High risk of disappearing into nothingness & experiencing spaghettification and potential disintegration. Thankfully, in the best case scenario, you'll be warped to the other side of space
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.
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.
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).
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
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.
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
Wish I had this video in April for my linear algebra final. Good explanation for finding imaginary roots!
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) .
I got lost when he rotated the parabola to the right 🥴
I appreciate your definition of the principal root - it‘s simply the zeroth solution of the corresponding equation (dividing the circle). 🤓
this is literally the same. if you do complex evaluations for both you'll get the same result
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
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.
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.
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.
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.
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 )
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 ...
I think it is not about the difference between the cube root of -1 and taking -1 to the exponent 1/3. One can be right by saying they both are just the same thing. The main question is in what field you are looking for solutions. If it is the field of real numbers then both concepts represent real-valued functions which must have one output for each x. And if it is the field of complex numbers then they represent complex-valued functions which can be multivalued with several outputs (branches). I believe thinking this way is much simpler and certain. And Math is about simplicity, simplification, and certainty. Otherwise it would be just a mess.
I did not understand all of it, but even the part that I could understand surprised me and intrigued me.
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 is simply a definition problem. In fact if you pay attention to the video you will see that wolframalpha shows the "input" it thinks you are asking at:
0:45
(-1)^(1/3) is exactly the SAME as ∛-1, but "cuberoot(-1)" is parsed as something different there, with a special down make after the "cube root" symbol.
Correct Way of Talking on this Follows:
First of all we talk of any-order root of a positive number and notice that for every positive number its any-order root is a positive number at the least. This is called as our Root-Operation. Examples: Take 2, its 2nd, 3rd, 4th, 5th, etc roots are all again some distinct positive numbers; Take 3.456, its 2nd, 3rd, 4th, 5th, etc roots are all again some distinct positive numbers.
Now, we take the negative numbers for which we define the Root-Operation only when the Root is of odd order. Examples: Take -2, its 3rd, 5th, 7th, etc roots are all again some distinct negative numbers; Take -3.456, its 3rd, 5th, 7th, etc roots are all again some distinct negative numbers.
Now, we move to consider the case wherein a number is to be found out whose some integer-power is an arbitrarily given real number. This is equivalent to solving the Algebraic Equation, namely x^n = p. This shows there are n such numbers, real or complex. The fact of the matter is while writing any such solution we may need the numbers which are root-operation, like square-root of 3, cube-root of 11 etc.
One way of solving x^n = p type equation is the use of Generalized Complex Number Representation of p and use of Euler’s Theorem.
Finally, let’s remember that Finding Zeros of a Mathematical Expression and defining it so as to Make it a Function are two different concepts. They are nothing to do here. Here is the subject-matter of basic root-operation and solving an algebraic equation.
Love this, what a great explanation !
So basically it's almost the same reason why, if x > 0, sqrt(x) > 0, but the roots of "f(x) = x²" are equal to ±sqrt(x)? It's not inherently "wrong", just depends on the context and which solutions should be considered.
sqrt(x) is a function, and hence can have only one unambiguous output.
x^2 = 9 is not a function, it's an equation, and because it is a second-order polynomial equation it has two different solutions (or "roots"). Just as, for example, x^2 - 9x + 14 = 0 has two different roots (namely, x=2 and x=7). Does that mean that suddenly we should define an expression like "sqrt(x^2 - 9x +14)" to give two different outputs at the same time? No, of course not.
(Also note that there is a conceptual difference between "sqrt(x^2 - 9)" and "x = sqrt(9)" ; the first one is an expression, the second one is a statement, in particular an equation.)
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.
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
Are there uses for an imaginary principal root? When would this principal root function be more useful than the real root?
Thanks, it was the best math lesson I attended in almost 20 years !
of course they are the same thing, it is just Wolfram makes some assumptions what you want to see
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". ^^
This was in my A-level Pure Maths text book, but after 5 minutes here I actually understand it.
I feel like sometimes on this channel there's a whole video that could be summed up by "because it depends on the convention chosen" AoPS resolves this "mystery" in one sentence. 7 of the 8 minutes and 19 seconds of this video lead nowhere. If someone wasn't familiar with functions, complex numbers, and fractional exponents, they'd be worse off after watching this. Why do you keep doing this in your videos?
AoPS?
@@ewthmatthThe art of problem solving. It's a basic level competition maths book and it's very good.
The real question is: Why does WolframAlpha give these different answers?
In our school we got the definition that the real power functions and the real root functions are only defined on non-negative arguments. This is to make the mappings actual functions (that is, unique mappings) in a consistent way. That is, there is no cubic root of -1, but of course the cubic equation x³ = -1 has three solutions, one of which is real.
\sqrt[3]{-1} is the same as (-1)^{1/3}. this is THE same notation. there is one root in a case of real numbers and there are 3 roots in a case of complex numbers
TI n-spire calculator yields same for each expression
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.
Does anybody else hear "Fresh Toewalker" at the beginning of every video rather than Presh Talwalkar?
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".
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.
Well done. I think you kind of skipped a little quickly over the y=x^n has n roots (fundamental thm). But otherwise nicely explained and visualized.
Can have a maximum of n roots. Not necessarily n roots.
Same in Octave: (-1)^(1/3) is a complex number and nthroot(-1, 3) is -1.
No sense for me, as the two expressions are strictly equivalent.
For a student who is using W.A., your explanation is then pretty useless
@@mathisnotforthefaintofheart don't understand your point.
I am by no means a mathematician or in any way close to math, but from what I understood based on the video, the nth root of x is defined as a function(because for every x from X there is only one y from Y in R^2), while the x^(1/n) is not a function in R^2, because for every x there are different values of y. Thus, the “root sign” function can’t give more than 1 real root, while x^(1/n) can.
@@user-hi6nr5je6d for me, no difference between those 2 fonctions as long as you work in the same set R. No différence again in the same set C. The difference comes from working in R or C ( in R one root, in C 3 roots) but the 2 ways of writing the cube root are strictly equivalent IMHO.
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.
0.19 "Assuming 'cuberoot' is the real-valued root" and 0.36 "Assuming the principal root".....Just press the other links and you will get the right answers...
You describe the process of taking multiple roots as "not being a function". I don't think this is correct. It's not injective, but a non-injective function can still be called a function. I would also argue that the most correct solution to taking roots (with either notation) is to list all roots, not just the principal root. i.e.
√4=±2 and 4^(1/2)=±2. If you want to take only the principal root, you can use either notation to do so but should describe your choice to do so if it is not already obvious from the context (e.g. if you are working with physical objects, it may already be obvious that the solution space is constrained to the reals or to the non-negativite reals).
sqrt(2) times sqrt(2) is 2, but 1.41421356237 times 1.41421356237 is approx. 2 (1.9999....) so why when we draw a square with a side a = 1, it can be shown on paper, but its diagonal is = a*sqrt(2), so it is irrational. The same situation with a circle, how can we draw some circle "X", if Pi (circle or disc circumference) is irrational too: 3.141592....
Hello. Wonderful video demonstration about our maths are "conventions", so, a „math truth" is just a Convention.
And on/over this "truth type", a convention a belief all our our science was and is build.
Regards.
Next exp(z) not the same e^z . e^z := exp ( z*Ln(e))= exp(z*(ln(e)+ i*2pi*k))= exp(z( 1 +i*2pi*k)
Mean while back at the farm........
the battle of conventions rages on!
I think one thing that is consistent is that the symbol √ has only one value. It denotes a function.
Why can't we also just say that there are three cube roots of -1: one real, and two complex? That's what we would expect anyway. Then the only difference between the two expressions is whether we state the real root first, or the "principal" root first. Personally, I think that if a real root exists, but it is NOT identified as the principal root, then the notation system that produces such a result is unhelpful. But either way, both expressions would still yield the same three roots in the end.
Thanks!
You forgot the parentheses around pi + 2pik at 5:29
The polar form of e^i*pi+2*pi*k should have parenthesis, e^i(pi+2*pi*k)
To all Mathematicians:
Who has ever asked you to make sqrt(x) a function?
A multi-function is fine and Much Much More Accurate.
Your Convenience has been causing a lot of inconvenience.
Just define a new function like Psqrt(x) for principal root or something like that.
True.
Sir, then how √-1 is i
You said that the root will give only the real number
Because i is not a real number. It's called imaginary number
In Mathematica, (-1)^(1/3) == Power[-1, (3)^-1] yields True, even SameQ[(-1)^(1/3) , Power[-1, (3)^-1]] does. This is due to their FullForm are identical: Power[-1,Rational[1,3]].
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?
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 ?
The most simple way to understand, assuming n>0 and otherwise nice values: What's the nth root of a? Only b, the principal root. What can be raised to power n to get a? There are n values (including b).
Yes, and those n values can be easily found by multiplying b with the n complex vectors e^(i*2πk/n) = (cos(2πk/n) + i*sin(2πk/n)) , for k = 0, 1, 2, ..., (n-1) ; regardless of which of them was chosen/defined/standardized as the principal root b.
Note: those n complex vectors are the n solutions to the equation z^n = +1 in the complex plane.
nice to see Euler's identity just popup at the end.
So i is -sqrt(3). We've solved it.
_"Call the whole thing off..."_ 😃 nice reference.
Why does this man's shorts have stuff like the most basic fraction addition ever, but then his long form vids don't explain what's happening? Like, where'd the circle came from, and why is there some angle there?
The fundamental theorem of algebra, also known as d'Alembert's theorem,[1] or the d'Alembert-Gauss theorem,[2] states that every non-constant single-variable polynomial with complex coefficients has at least one complex root.
x^1/3 must have three results. In physics it must have context when you can discard results with complex numbers.
You wrote it wrong. It's x^(1/3).
Okay, that was excellent. But I definitely have to watch it a second time.
Can you report this to Microsoft since their calculator gives -1 as the answer to the 1/3 root of (-1)...
"This angle Theta would be Theta "
-Presh Talwalkar 2023
I don't think there is really a mathematical difference between them. It is more about what has been defined by the programmer and what a programmer might have overseen or included.
loved the video - nice work
Thks & similar to the inverse of a sinusoidal function
Completely identical. This will be interesting.
the inverse of y = x^2 is not y = sqrt x, it is y = ± sqrt x
Fake news. Both are the same even on Wolfram Alpha. You need to scroll further down to see the other solutions. Theyre just not in the same order on these pages.
Yes, noticed the same thing. I wouldn't inmediately say he is wrong because there may be some other sources that use the convention that he is saying, but in the sites that I have consulted, both expressions are interchangeable.
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
I don't like the ambiguity in Math. I have a problem with the Sqr of -1 times Sqr of -1. I find it to be +1, if we follow the rule that says Sqr of x times Sqr of x is x squared. Then if you insert -1 and -1 to the two xs it is positive 1. Why is it that Sqr of -1 times Sqr of -1 equal -1? I don't understand. Please explain it to me. Thanks.
Sqrt x * sqrt x = x, not x^2
I think you got a little confused there lil bro. Sqrt -1 = i.
i^2 = -1
Kudos to everyone, who lost in the middle of the video but still continued to watch until the end.
thx
Great video
What is the reasoning for functions to have only one output for each input? Other than convention, I've never really understood it from a mathematical perspective. Speaking specifically to the following two graphs:
y² = x; this is not a function as the graph will have two _y_ values for every _x_ > 0
y = √x; this is a function because we consider only the principal root
Yet mathematically, I can take the function y = √x (half a parabola), square both sides of the equation, and I'm left with y² = x (a full parabola). I have a pretty decent understanding of math as it's a big part of my job, but that has never really made sense to me when I can derive either equation from the other.
There's a difference between equations and functions. Functions specifically have separate inputs and outputs that are not treated the same. There can be many different inputs or input combinations that flgive the same output, but a specified input combination can't result in multiple outputs. This is to make the function able to be differentiated, integrated, approximated with series, and all sorts of other things that we do with functions. If the output isn't uniquely defined none of that is possible and you get all sorts of ambiguity.
@@bananatree2527 I appreciate the response, and I do understand in terms of functions why we prefer to have only one y output for every x input. So we define the square root as the primary root when functions are involved. I guess what I don't understand is why we don't generally consider both roots in everyday circumstances when we can show mathematically that the secondary root is also a solution, especially when we have precedent outside of the realm of functions to do so. For example, if I were to give you the following problem:
√y = x; find x when y = 4
The solution to this problem is x = ±2, and indeed if we were to graph the equation √y = x, we'd see that the line y = 4 intercepts that graph at x = 2 and x = -2. Subbing those values back into the original equation, we see that:
√4 = ±2
Yet if I were to ask this problem instead:
y = √x; find y when x = 4
We would see that y = 2 ≠ ±2. And if we plug those values into the original problem, we see:
√4 = +2
Both original problems are formatted exactly the same, both use the square root operator, yet in one case the solution involves both roots while the other involves only the primary root. To me, it seems that the ±2 solution is more accurate from a purely mathematical standpoint than, "Well, it's just positive 2 because that's how we define functions."
There is mathematical symmetry about the line y = x when the x and y variables are reversed in equations, yet if one of the mirrored graphs violates the vertical line rule for functions, we simply ignore half the solutions. Seems rather incomplete and arbitrary to me.
And it's not isolated to square roots. We see the same thing in circles. The graph of a circle with center (h, k) and radius r is (x - h)² + (y - k)² = r². If we plot that on a graph, we see a full circle. Yet if we reformat that equation in terms of y:
y = √((r + (h - k)(r + (h - k)) + k
We no longer see a full circle. We see a semi-circle because we've now written it as a function of y. It's aggravating and often confusing.
@@cdmcfall I don't know why you are thinking that depending on where you put the √ symbol (in the "x" or in the "y") you can get two different solutions or only one.
Provided that √a refers to the principal square root of "a", that is, a non negative value in real numbers, then in the problem:
1) √y = x; find x when y = 4
The answer is only x=2, not x=±2, because you are evaluating
√4 = x
2 = x
The second example is equivalent, only that the unknown value is "y" instead of "x":
2) y = √x; find y when x = 4
y = √4
y = 2
@@RonaldABG The graph of √y = x looks exactly like the graph of y = x², a complete parabola, because that's how we would reformat it as a function of x. It has a positive and negative x-value for every y ≥ 0. In the instance of y = 4, x is either 2 or -2, which is allowed because it doesn't violate the vertical line test for functions. You can plot it yourself to verify. If we were to consider only the principal square root in the case of √y = x, then that graph would look like √y = |x|, which is only half a parabola. So again, in one instance, we consider both roots. In the other, we consider only the principal root, which by giving it the qualifying term "principal," acknowledges that there is at one other root that isn't the "principal" that we are choosing to ignore, not for mathematical reasons, but for simplicity. So that's the part I don't understand. I someone asks, "What is the square root of 4?" I should answer 2. But if someone asks, "What number multiplied by itself is equal to 4?" Then I would be correct by answering both positive and negative 2. Yet both questions are asking the same thing. To me, it's silly in cases where functions aren't involved, just simple expressions, to ignore half the solutions.
2:50 I lost you on √9 = ±3 because IT IS NOT, √9 =3 AND -√9=-3.
yeah, you can cross check
Since when is root only a principle value?
PLEASE quote authoritative sources, books and send details fir your statement.
As discussion,
1. Why we have two notations x^(1/2) and √x if they are same ...??
If they are same why not have more notation, why only two? Each thing is done with a purpose, isn't it ..
2. If x^1/3 refers to positive 3rd root of x , x real, then why people write,
a^3=1, raise to power 1/3 on both sides, a=1^1/3 !!!
if former ( a^3 =1) refers, in a way, to collection of roots, put in the form of equation then, why not latter ?
3. If (-1)^1/3 or for complex no. z, if z^1/3 refers to collection of 3 roots and complex is extension of reals then, does concept change on expansion of number system or there should be harmonious extension of concept ??
OR IS IT THAT,
WE should write positive 4 th root of unity as 4√1, with 4 inside radical sign while 1^(1/4) refers to collection of 4 roots !!!
LET US HAVE A LOGICAL EXPLANATION.
Is it that it has become the practice to use 2^3/5 to denote only real just due to compulsions of laws of exponents...
Eg. In context if progressions, the centuries old books of hall and knight and euler etc state, common difference non zero etc...terms should keep changing while these days, people allow common difference to be zero as per convenience. Abuse of definition isn't justifiable as extension of definition.
,
Even books have flaws: S.L. loney says division by 0 is infinity while it is not defined ..He quotes, division by infinity is 0 while infinity is not a number.. so please tell sources of authority where definitions were penned first. These days extensions and abuse, misuse of terms is rampant and internet is a place where one is taken for ride by all authors.
Arguments like :
3^2=9=(-3)^2 so can we raise all sides to power 1/2 ?? 3=9^1/2=-3 !!!
It would be dangerous .
or IS IT THAT in context of EQUATION, where we talk of value , there alone 9^1/2 refers to positive square root !! while, left alone, as expression, 9^1/2 refers to collection..
what is the notation then to refer to 2nd roots of 9 if 9^1/2 is a number, not a collection...If a number, was √9 insufficient to represent .. a discussion..
At 3.20 minutes you say x^(1/n) is pne number for x >0. At 4.30 minutes you say z^(1/n) is collection of n roots for z
At about 4:00, the axes are x and y. They should be the real axis and the imaginary axis
Cute reference to "you say tomAITO I say tomOTTO."
The symbol referred to as the square root symbol is actually the principal square root symbol. That’s why we take only the principal root when that symbol is given.
Well said! To highlight the interpretation error being made, at time index 2:49 the equation is shown as radical(9) = +/- 3. (Sorry for using "radical(9)", but my PC doesn't have the radical symbol!)
But if the radical symbol is used without negative sign in front of it, then by definition it is the principal square root being expressed; it doesn't convey both roots by itself. So the equation as shown is incorrect.
I would also add here that fractional notation is defined to be equivalent to radical symbol notation. We shouldn't be getting different results by using one form or the other.
@@jaimehunter8504 At 2:49 , the text " √9 = ±3 " momentarily appears on screen, but not because Presh (the video creator) agrees with it. The text merely appears in order to illustrate the issue that he's discussing at that moment: "So this always comes up in on-line discussions: when you have something like the square-root of nine, is it equal to negative three ánd plus three, or just plus three?"
He immediately concludes with "Typically, we want it to take just the _non-negative_ value", and only the formula "√9 = 3" remains on-screen.
Welcome to the real world math, where context matters!
"The angle theta will be theta" always sounds funny when people say it. It would be wild if you wrote it as theta but called it sigma. Id laugh atleast lol
I also want to add that (-1)^(1/3) is not the same as (-1)^(2/6). Because in the second case it’s the same as squaring -1 which gives 1 and any root of 1 is 1
However, 1/3 = 2/6
both are the same, u cant do the squaring operation first, it changes the supposed root which you should be getting. both HAVE to be equal, same number cant have 2 different values. (-1)^(2/6) isnt equal to (1)^(1/6) .
@@adityakulkarni2343 But what the result of (-1)^(2/6) then? (-1)^(1/3) is the same as root_3((-1)^1) or root_3(-1)^1 which both yields the same result. When you apply 2/6 to the same rule (numerator is exponent, denominator is root) it breaks
@@ic6406 see, simplifying 2/6 is required, because it changes the answer significantly, as [6th root of (-1)]^2 is different than 6th root of [-1]^2 . this confusion occurs only when ur number inside is negative, hence you tend to simplify exponents every time to avoid confusions. (-1)^2/6 is nothing but (-1)^1/3
@@ic6406 also u cant apply that rule when number inside is negative
@@adityakulkarni2343 Yes, that's what I wanted to say in original comment. Specific case when rational exponents can give unexpected results
he wrote something wrong
its e^[i(θ+2kπ)] , not e^(iθ+2kπ)😂
note: I'm talking about 5:56
5:26
It's simple, cube roots have 3 solutions. So, cube root of -1 will be 1 real (-1) and ; two imaginary (1+√3i)/2 and (1 - √3i)/2.
so all in all, x^ 1/n and nth root of x are the exact same thing, and both refer to the real value? however, if its written like x^3 = -1, and if we choose to find x's value, then the imaginary aspect comes in? It makes no sense tho, how are x^1/n and nth root of x different?
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.
Let's always do alot of good 🙏
wow, startet well, but I didn't understand anything after 3:46 😂
This principal root thing is annoying. All roots are roots. We choose which ones we want, but ultimately all the roots solve the problem. Some make sense in some appications (areas) while other applications need all of them (electromagnetism).
No one should be suppressing any root until you have stated what you are looking for. Again, another educator imposing bias because their awareness is limited and is ultimately detrimental to any youth following these.
You seem the one with limited awareness. Once you have one solution b to the equation x^n = c , you have pretty much already all of them. All you need to do is multiply b to each of the n complex vectors
z_k = e^(i*2πk/n) = (cos(2πk/n) + i*sin(2πk/n))
where k = 0, 1, 2, ..., (n-1) .
Note that these z_k are the n solutions of the equation z^n = +1 .
So since b^n = c , then (b * z_k)^n = b^n * (z_k)^n = c * (+1) = c , which means each b_k = (b * z_k) is a solution to the equation x^n = c .
Demanding that the "inverse" function lists all possible solutions, is something for lazy students who don't know math (nor want to learn math). Would you also expect/demand your calculator to list all branches of the arcsine, arccosine, and arctangent function?
You've confirmed my point on this - as he stated there is no standard definition of a principal root. So, as I pointed out (and he did in the video) the user needs to look at the context (the type of problem) and consider the body of roots that are outcomes and choose accordingly. Again, The Principal Root is frustrating because it is unclear, but if you show youth how to use ALL the roots, and that some make no sense, then they learn to discriminate garbage outputs to problems. This is an education thing that so many are challenged with. Math is not about just doing the steps by rote manipulation - that is how people forget to check that they are not dividing by zero, etc.@@yurenchu
@@ApresSavant Before users can look at roots and choose the one(s) that is(/are) applicable to the problem at hand, they must first (learn how to) determine/calculate the roots. And the concept of the principal root is a helpful mathematical tool for that, which allows us to perform that task in a methodical way.
The principal root isn't unclear to users who know what they are doing. It's only unclear to lazy students who (or students of lazy teachers who) don't want to put in any effort to understand where roots come from, and what their underlying structure is. And by handing students simply all roots on a silver platter (without them having a clue what they really mean, and what to do with them) you're only encouraging their laziness and incompetence. (How would you expect them to pick the appropriate applicable root from an infinite list of _infinitely many_ roots, if they have never learned how these infinitely many roots are all related to eachother via the concept of the principal root? Do you expect them to manually verify each of these _infinitely many_ roots one by one? LOL!) Anybody can operate a Texas Instruments Graphic Calculator, or input equations into a Wolfram Alpha website and hit the "Go!" button. Learning those "computer skills" is *not* what a proper math education is about.
The principal root really isn't the problem. What's really the problem is the lacking understanding among students (and their mentors) about what roots are.
We shouldn't cater to lazy students (who later become lazy and sloppy mentors). It is appropriate that a proper and thorough math education would require students to put in the legwork to understand what roots are, how they are related to eachother, and how just knowing one of them may allow us to derive all of them; and hence why mathematicians work with the concept of a "principal root". Students who don't manage to wrap their head around those concepts are simply not fit to become math teachers.
You guys are complaining because you're having difficulty with some quadratic and/or cubic equations. Mathematicians have developed the concept of principal roots in order to tackle in a methodical way math problems that are far more complicated than mere pesky quadratic/cubic equations.
It doesn't matter that the definition of a principal root may differ between contexts. The essence is that a principal root forms a sort of "baseline". It doesn't really matter what the baseline is, as long as we do have a baseline. Abandoning any baseline is one of the most short-sighted things one can do.
Abandoning the concept of the principal root doesn't gain us more insight; it only makes the waters even more difficult (if not impossible) to navigate.
@@yurenchu Thanks for your reply, and for the most part I agree with you, until you mentioned that "Mathematicians have developed the concept of principal roots" - because they have not, which is why this argument goes on. There are all manner of principal root definitions (Michael Penn did a video where this was an aside), and thus my principal root is not your principal root. So this brings me back to my point - all roots are relevant until they are not. You mentioned figuring out the infinity of roots, which is absurd as any infinity of roots has some sort of cyclical part that can be stated as such (2*pi*n for example), so the educated student would learn to write down the relevant simplified general root and then look at whether they needed the positive ones, or the negative ones, or the just ones in the -ve complex / -ve real quadrant, etc. The point is that to be complete any proof that involves roots needs to have them all examined and either approved as valid or excluded as contradictory. If you do not do the work, then the person doing the work may get lucky with a valid answer, or they may be leaving an incomplete gap in their conclusion. So, per my original observation, because there is no agreement across time about what is a principal root is, or which version is best used when, it is still best at the elementary level of this type of lesson to fall back to first principles and look at each case - there will never be an omission if you look at them all! Perhaps it is my scientific and engineering experience bias on this, but completeness matters when lives are on the line - something I feel some mathematicians, playing with abstractions like infinity, do not always remember (the infinite sum of all integers = -1/12 comes to mind).
Obviously this notation is confusing. You need to define at first the domain whether it's ℝ or ℂ.