Special thanks to everyone who watched the video before it went public! I shared this video early in the "community" tab (see ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-f3indrh6GEI.html for the original) and you guys gave some great feedback! The result was this new video with 2 very, very minor changes. (It did take me 5 hours to make these edits; that's how insanely hard it is to make these math animations!). But I felt it was worth making the edits to be as accurate as possible. The changes were: a) At 5:12 I changed the subscripts to 1=j1
He said that at the beginning. You can solve it using clever tricks and simple algebra, however he introduced us to a systematic approach, which will solve every problem of this type.
From the thumbnail, I thought that the question was asking for the value of xⁿ+yⁿ+zⁿ ; so I was looking for a closed-form solution for that expression. (At one point I even hesitated and wondered if this was a trick question, but the video was listed as over 12 minutes long, which made me believe there must be some very advanced legitimate solution.) When I had spent much time trying to figure out how to turn the recursive formula that I found into a simple closed-form solution, and I finally watched the video, it turns out that it only asks for x⁵+y⁵+z⁵. Anyway, the closed-form solution is xⁿ + yⁿ + zⁿ = cⁿ + (a+ib)ⁿ + (a-ib)ⁿ , where a,b,c are real constants, and where c is the real root and {(a+ib), (a-ib)} are the two complex roots of the cubic equation q³ = q² + (1/2)q + (1/6) which is the characteristic equation of the recursive formula p[n+3] = p[n+2] + (1/2)*p[n+1] + (1/6)*p[n] Due to the recursive formula (with rational coefficients and rational start values), the value of xⁿ + yⁿ + zⁿ for integer n must be rational. Since the absolute values of (a+ib) and (a-ib) turn out to be less than 0.5 , the (a+ib)ⁿ and (a-ib)ⁿ terms vanish for increasing values of n . This and the fact that the value for integer n is rational, means that we may find an alternative exact closed-form solution for integer n that uses a rounding function instead of these complex roots (a+ib) and (a-ib). This exact closed-form solution (only for integer n) is: xⁿ + yⁿ + zⁿ = F(n) = round( h(n)*c^n )/h(n) , where c = [ 2 + ³√(44+6√26) + ³√(44-6√26) ]/6 = 1.43084957... is the real root of q³ = q² + (1/2)q + (1/6), round(X) is the rounding function which rounds X to the nearest integer, h(n) is a factor that depends on the value of m = (n mod 6) and k = floor(n/6) : n = 0 + 6k ==> h = (1/6) * 72^k n = 1 + 6k ==> h = (1/2) * 72^k n = 2 + 6k ==> h = 1 * 72^k n = 3 + 6k ==> h = 1 * 72^k n = 4 + 6k ==> h = 6 * 72^k n = 5 + 6k ==> h = 6 * 72^k This formula works for integers n > 3 . So, some examples to see this formula in action: For n = 5, m = (n mod 6) = 5, k = floor(5/6) = 0 , and thus x⁵+y⁵+z⁵ = F(5) = = round( h(5) * c⁵ ) / h(5) = round( 6*72⁰ * (1.43084957...)⁵ ) / [6*72⁰] = round( 6 * 5.99749489... ) / [6] = round( 35.9849693... ) / 6 = 36/6 = 6 For n = 6, m = (n mod 6) = 0, k = floor(6/6) = 1 , and thus x⁶+y⁶+z⁶ = F(6) = = round( h(6) * c⁶ ) / h(6) = round( (1/6)*72¹ * (1.43084957...)⁶ ) / [(1/6)*72¹] = round( 12 * 8.58151299... )/ [12] = round( 102.978156... ) / 12 = 103/12 = 8.5833333... For n = 13, m = (n mod 6) = 1, k = floor(13/6) = 2 , and thus x¹³+y¹³+z¹³ = F(13) = = round( h(13) * c¹³ ) / h(13) = round( (1/2)*72² * (1.43084957...)¹³ ) / [(1/2)*72²] = round( 2592 * 105.371146... )/ [2592] = round( 273122.01... ) / 2592 = 273122/2592 = 136561/1296 = 105.371142... (Note: these answers are _exact_ , not approximations.) Anyway, thanks for another challenging math puzzle! [ EDIT: Edited to fix an error in the recursive formula (I had left out the factor p[n] in the last term). ]
@ImPresSiveXD , Which formula or expression do you want to see the derivation of? There are several formulas and expressions in my comment, and giving the derivation of each of them would be too much for just one reply (and would possibly also not be what you want). To see why the final formula is exact, even though it uses a rounding function, I'll illustrate the principle behind it with a simpler case: Fibonacci numbers. A reminder of what Fibonacci numbers are: you start with F[0] = 0 and F[1] = 1, and every number after that is the sum of its two preceding numbers: F[n+2] = F[n+1] + F[n]. So the sequence of Fibonacci numbers starts 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,... The n'th Fibonacci number F[n] can be determined by the formula F[n] = round( (1/√5)*φⁿ ) where φ = (1+√5)/2 = 1.61803399... is the so-called _golden ratio_ , and round(X) is the rounding function that rounds X to the nearest integer. This formula for F[n] is _exact,_ even though it uses a rounding function. The reason why it works, is this: The exact closed-form formula of F[n] (without rounding function) is given by F[n] = (1/√5)*φⁿ - (1/√5)*(-1/φ)ⁿ for any integer n ≥ 0 (you can probably easily look this formula up on Wikipedia or something like that). Now consider the second term, which is (1/√5)*(-1/φ)ⁿ = (0.447213595...)*(-0.618033989...)ⁿ . For n=0 its value is (0.447213595...), and for increasing n this value is repeatedly multiplied by (-0.618033989...). So the absolute value of this term will always be less than 0.5 . So that means that the first term, which is (1/√5)*φⁿ , is always less than a distance of 0.5 removed from F[n]. But we also know that F[n] must be an integer (because that's obviously the nature of Fibonacci numbers; if you start with two integers, then each subsequent "sum of two preceding numbers" will also be an integer). So what the addition of the second term (1/√5)*(-1/φ)ⁿ essentially does, is bring the value of the first term (1/√5)*φⁿ to the nearest integer. In other words: we may replace the addition of the second term with the operation of rounding the first term to the nearest integer, since they are the same thing. And that's how we obtain the Fibonacci formula that uses the rounding function. The same principle also applies in my formula for p[n] (with p[n] being short-hand for xⁿ + yⁿ + zⁿ) , except (since p[n] is rational instead of just an integer) we're now applying the rounding function to the numerator of p[n] (because we know that that numerator must be an integer). So suppose that we'd have expressions g(n) and h(n) such that p[n] = g(n)/h(n) , and g(n) and h(n) are integers (and not necessarily reduced), then we can write h(n)*p[n] = g(n) ... substitute p[n] = cⁿ + (a+ib)ⁿ + (a-ib)ⁿ ... h(n)*cⁿ + h(n)*(a+ib)ⁿ + h(n)*(a-ib)ⁿ = g(n) We defined g(n) to be integer. So if we can show that |h(n)*(a±ib)ⁿ| < 0.25 (which is indeed the case with the proposed h(n) in my comment), then the addition of the second and third term in the lefthandside is equivalent to rounding the first term to the nearest integer: round( h(n)*cⁿ ) = g(n) And then we'll have p[n] = g(n)/h(n) = round( h(n)*cⁿ ) / h(n) I hope that helps.
@ImPresSiveXD , You're welcome. I determined h(n) as follows. The recursive relation that can be derived for p[n] = xⁿ+yⁿ+zⁿ , is p[n+3] = p[n+2] + (1/2)*p[n+1] + (1/6)*p[n] ....{eq.1} Given start values p[1] = 1, p[2] = 2, p[3] = 3, we can find p[4], p[5] etc. recursively by using this equation. Since this recursive equation has rational coefficients and since the start values are integers, we know that each subsequent p[n] will be rational (i.e not irrational). In other words, each p[n] can be written in the form p[n] = g/h , where g and h are integers (and g and h are not necessarily co-prime). Furthermore, since the recursive relation divides only by 2 and/or 6, we know that we can define denominator h to be of the form (2^α)*(3^β) , where α and β are nonnegative integers. We could define α(n+3) = max{ α(n+2), 1+α(n+1), 1+α(n) } β(n+3) = max{ β(n+2), β(n+1), 1+β(n) } for integer n > 3 . This matches the recursive description of p[n+3] : from the three terms in the righthandside of eq.1 , the term whose denominator (when factorised into prime factors) contains the most instances of prime factor 2, determines the quantity of prime factor 2 in the denominator of the righthandside sum, and hence also in the denominator of p[n+3]. And likewise with prime factor 3. Starting with α(1) = α(2) = α(3) = 0 and β(1) = β(2) = β(3) = 0 (because the three start values of p[n] are integers, and hence have h = 1), we can determine the following table n : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 etc. α(n) : 0 0 0 1 1 2 2 3 3 4 4 5 5 6 6 etc. β(n) : 0 0 0 1 1 1 2 2 2 3 3 3 4 4 4 etc. We see that after n = 4, α(n) is increased with 1 after every two steps, and β(n) is increased with 1 after every three steps. (Note to the reader: prove that this pattern continues indefinitely for increasing n.) Generally, for n > 3, α(n) = floor((n-2)/2) β(n) = floor((n-1)/3) From these expressions, we determine h(n) = 2^α(n) * 3^β(n) : n = 0 + 6k ==> h = 2^(3k-1) * 3^(2k-1) = (8^k)/2 * (9^k)/3 = (1/6) * 72^k n = 1 + 6k ==> h = 2^(3k-1) * 3^(2k-0) = (8^k)/2 * (9^k)/1 = (1/2) * 72^k n = 2 + 6k ==> h = 2^(3k+0) * 3^(2k+0) = 1*(8^k) * 1*(9^k) = 72^k n = 3 + 6k ==> h = 2^(3k+0) * 3^(2k+0) = 1*(8^k) * 1*(9^k) = 72^k n = 4 + 6k ==> h = 2^(3k+1) * 3^(2k+1) = 2*(8^k) * 3*(9^k) = 6 * 72^k n = 5 + 6k ==> h = 2^(3k+1) * 3^(2k+1) = 2*(8^k) * 3*(9^k) = 6 * 72^k So there we have it.
I have two degrees in EE. I once helped create something called a gamma ray resonance spectrometer. I get paid to optimize GPU code. I've taught several graduate & undergrad classes in math-focused areas. I still can't do a lot of the problems that come up on your channel, and each explanation is *very* useful. Anyone that complains... well, let them go be themselves, and the rest of us can find education and enjoyment in symmetry and solutions. (Long time fan, first time caller. Keep up the great work.)
Rather than solving anything, this video just felt like "if you know these very specific identities, which we will neither prove nor explain, we can derive the answer directly out our own asses."
How about I calculate both a⁴+b⁴+c⁴ and a^5+b^5+c^5 by short method using only algebra. Let if a,b and c are roots of a cubic polynomial f(x). f(x) = (x-a)(x-b)(x-c) = x³ - (a+b+c)x² + (ab+bc+ca)x - abc Now from the given equations. a+b+c = 1 Let squaring both the sides (a+b+c)²= 1² Let's expand the left side a²+b²+c²+2(ab+bc+ca) = 1 Now , we know that a²+b²+c²= 2 then 2(ab+bc+ca) = 1-2 = -1 Hence, ab+bc+ca = -1/2 And we know that a³+b³+c³ - 3abc = (a+b+c)(a²+b²+c²-ab-bc-ca) Putting all the values we have in the above equation we get, 3 - 3abc = (1) (2-(-1/2) 3 - 3abc = 5/2 3abc = 1/2 abc = 1/6 Hence our polynomial after substituting all these values will be f(x) = x³ - x² - (1/2)x - 1/6 Now 'a' is the root of f(x), Mathematically a³ - a² - (1/2)a - 1/6 = 0 By multiplying with 'a' we get a⁴ , a⁴ = a³ + (1/2)a² + (1/6)a We will get similar equations for b and c. After adding them, a⁴+b⁴+c⁴ = (a³+b³+c³) + (1/2)(a²+b²+c²) + (1/6)(a+b+c) = 3 + 2/2 + 1/6 = 25/6 Similarly, if we multiply with 'a²' we get equation for 'a^5' i.e. a^5 = a⁴ + (1/2)a³ + (1/6)a² Adding again, a^5 + b^5 + c^5 = 25/6 + 3/2 + 2/6 = 6 That's it! Quite simple Love from India❤️
This is a beautiful solution, but by simply multiplying the first equation with the others (and itself) and keeping track of the terms, one can find the correct values. It is very prone to error. I would like to point out that whenever you are faced with an equation, you need to justify your roots, as to weed out false roots. In this case, the roots are non-real, so a solution simply doesn't exist over R.
@@mr.moodle8836 No, it is not that difficult. By applying basic maths and multiplying the given equations, the answer can be found. The only thing is, you have to be careful and patient because the solution is quite lengthy.
@@apoorvvarshney3276 The problem is that you have to develop identities to solve the question. Using the elementary symmetric polynomials almost feels like a random step, there isn't much to prompt most people into doing it in the first place (unless they already know of the Newton-Girard formulas, but the assumption is that you aren't aware of them and have to derive the formula as shown in the video to apply it, which I probably should've been clearer about, but I think that's a reasonable assumption for a problem-solving question). I'm sure more people could do the question, given they were given a little nudge, but otherwise, probably not. If you consider the entire developing world, and acknowledge that most people with a math degree probably hold a degree in some form of applied mathematics (hence probably wouldn't specialise in solving a problem from "abstract algebra"), it seems reasonable to say that less than 0.01% have experience working with this kind of maths. Now, I imagine a far smaller proportion of non-pure math majors would be able to solve this than the proportion of math majors who can't (Ds get degrees yo), so under 0.01%, including the ENTIRE world population, not just developed countries, is a perfectly reasonable estimate.
I don't think even 1% could solve this one. With two maths 'A'-levels and an engineering degree (a long time ago, admittedly) I found this video way over my head compared to most of your videos.
Maybe because group theory is hardly ever covered in engineering math courses...nevertheless, imho one great thing about internet is this chance we have to learn new topics (almost for free) with great math channels from all around the world like this one, bprp, mathologer, 3b1b, numberphile, etc.
I'm currently studying an engineering major and I'm at algebra 3 now but I've never heard of the method. I think this is the theoretic approach of math which hasn't much to do with application which is we usually learn.
How about I calculate both a⁴+b⁴+c⁴ and a^5+b^5+c^5 by short method using only algebra. Let if a,b and c are roots of a cubic polynomial f(x). f(x) = (x-a)(x-b)(x-c) = x³ - (a+b+c)x² + (ab+bc+ca)x - abc Now from the given equations. a+b+c = 1 Let squaring both the sides (a+b+c)²= 1² Let's expand the left side a²+b²+c²+2(ab+bc+ca) = 1 Now , we know that a²+b²+c²= 2 then 2(ab+bc+ca) = 1-2 = -1 Hence, ab+bc+ca = -1/2 And we know that a³+b³+c³ - 3abc = (a+b+c)(a²+b²+c²-ab-bc-ca) Putting all the values we have in the above equation we get, 3 - 3abc = (1) (2-(-1/2) 3 - 3abc = 5/2 3abc = 1/2 abc = 1/6 Hence our polynomial after substituting all these values will be f(x) = x³ - x² - (1/2)x - 1/6 Now 'a' is the root of f(x), Mathematically a³ - a² - (1/2)a - 1/6 = 0 By multiplying with 'a' we get a⁴ , a⁴ = a³ + (1/2)a² + (1/6)a We will get similar equations for b and c. After adding them, a⁴+b⁴+c⁴ = (a³+b³+c³) + (1/2)(a²+b²+c²) + (1/6)(a+b+c) = 3 + 2/2 + 1/6 = 25/6 Similarly, if we multiply with 'a²' we get equation for 'a^5' i.e. a^5 = a⁴ + (1/2)a³ + (1/6)a² Adding again, a^5 + b^5 + c^5 = 25/6 + 3/2 + 2/6 = 6 That's it! Quite simple Love from India❤️
@@sajitsama7764 What you wrote,my friend,is essentially the internal working we did when we used the Girard-Newton identities and I admit,I did it the same way when I solved it. :) Also from India ❤
IMPORTANT QUESTION. Dear Pradesh, the solution is very clever. However there is a point which is not obvious and without explanation. Namely, you use en=0 for n>3, which applies to our 3 variables case, but the n-th relation between p and e coefficients (which you use to find p4, p5) is derived for an n-grade polynomial (resp. 4, 5), where there are n solutions and en (which is a function of these solutions) is obviously not 0 (it is actually x1x2x3 ...xn). I understand that it works, but I think that it deserves a complete explanation. Thanks. Samuele
How about I calculate both a⁴+b⁴+c⁴ and a^5+b^5+c^5 by short method using only algebra. Let if a,b and c are roots of a cubic polynomial f(x). f(x) = (x-a)(x-b)(x-c) = x³ - (a+b+c)x² + (ab+bc+ca)x - abc Now from the given equations. a+b+c = 1 Let squaring both the sides (a+b+c)²= 1² Let's expand the left side a²+b²+c²+2(ab+bc+ca) = 1 Now , we know that a²+b²+c²= 2 then 2(ab+bc+ca) = 1-2 = -1 Hence, ab+bc+ca = -1/2 And we know that a³+b³+c³ - 3abc = (a+b+c)(a²+b²+c²-ab-bc-ca) Putting all the values we have in the above equation we get, 3 - 3abc = (1) (2-(-1/2) 3 - 3abc = 5/2 3abc = 1/2 abc = 1/6 Hence our polynomial after substituting all these values will be f(x) = x³ - x² - (1/2)x - 1/6 Now 'a' is the root of f(x), Mathematically a³ - a² - (1/2)a - 1/6 = 0 By multiplying with 'a' we get a⁴ , a⁴ = a³ + (1/2)a² + (1/6)a We will get similar equations for b and c. After adding them, a⁴+b⁴+c⁴ = (a³+b³+c³) + (1/2)(a²+b²+c²) + (1/6)(a+b+c) = 3 + 2/2 + 1/6 = 25/6 Similarly, if we multiply with 'a²' we get equation for 'a^5' i.e. a^5 = a⁴ + (1/2)a³ + (1/6)a² Adding again, a^5 + b^5 + c^5 = 25/6 + 3/2 + 2/6 = 6 That's it! Quite simple Love from India❤️
Sajit Sama How are you sure there is no other value rather than 1 in front of the factors? I mean, it could be p(x)= A(x-a)*(x-b)*(x-c) with A≠1 [sorry if it is obvious, i have just learned polynomial concepts]
@Renno Vilela , the value of your parameter A doesn't affect the roots of p(x); so eventually, this value will cancel in the analysis anyway. The question is asking for the value of x⁵+y⁵+z⁵ , not for the value of A(x⁵+y⁵+z⁵) .
I remember the n=5 problem. My approach was to consider different ways to produce polynomials with 5 as the highest order (i.e. (x^4+y^4+z^4)(x+y+z), (x^2+y^2+z^2)(x^3+y^3+z^3) and so on), and use those results to get rid of certain types of terms from the expansion of (x+y+z)^5 (i.e. xy^2z^2, x^2yz^2,x^2y^2z) one by one, (after doing the same for n=4 to get the value of x^4+y^4+z^4). I wonder if this is actually the same idea, just with more sensible notation (my calculations got very messy).
that’s exactly how I approached it too. You can group together all terms with equal powers, like x^2y^2z plus the other two permutations to keep it managaeble. You’ll get a 5x5 dim matrix equation which is straightforward to solve by hand by suitably adding rows. This only requires basic math skills; not the knowledge of particular theorems. Having said that; it was great to learn about the properties of symmetric polynomials, afterwards!
You completely explained the Girard-Newton Identity for solving that problem! This is what makes a channel exceptional. I am a school going student and could precisely understand everything you explained. Lots of love from Bangladesh! Keep going! ❤️💯🇧🇩
When I read other channels' title as this I'm like " I come in that 1%" When I read this channels title as this I'm like " who else is there in that 1% except for u"
I did 2 years of math after highschool and couldn't solve it. So... maybe people who did 3 years of math after highschool ? which is definitely less than 1% of the population.
How about I calculate both a⁴+b⁴+c⁴ and a^5+b^5+c^5 by short method using only algebra. Let if a,b and c are roots of a cubic polynomial f(x). f(x) = (x-a)(x-b)(x-c) = x³ - (a+b+c)x² + (ab+bc+ca)x - abc Now from the given equations. a+b+c = 1 Let squaring both the sides (a+b+c)²= 1² Let's expand the left side a²+b²+c²+2(ab+bc+ca) = 1 Now , we know that a²+b²+c²= 2 then 2(ab+bc+ca) = 1-2 = -1 Hence, ab+bc+ca = -1/2 And we know that a³+b³+c³ - 3abc = (a+b+c)(a²+b²+c²-ab-bc-ca) Putting all the values we have in the above equation we get, 3 - 3abc = (1) (2-(-1/2) 3 - 3abc = 5/2 3abc = 1/2 abc = 1/6 Hence our polynomial after substituting all these values will be f(x) = x³ - x² - (1/2)x - 1/6 Now 'a' is the root of f(x), Mathematically a³ - a² - (1/2)a - 1/6 = 0 By multiplying with 'a' we get a⁴ , a⁴ = a³ + (1/2)a² + (1/6)a We will get similar equations for b and c. After adding them, a⁴+b⁴+c⁴ = (a³+b³+c³) + (1/2)(a²+b²+c²) + (1/6)(a+b+c) = 3 + 2/2 + 1/6 = 25/6 Similarly, if we multiply with 'a²' we get equation for 'a^5' i.e. a^5 = a⁴ + (1/2)a³ + (1/6)a² Adding again, a^5 + b^5 + c^5 = 25/6 + 3/2 + 2/6 = 6 That's it! Quite simple Love from India❤️
@@sajitsama7764 I've already seen your comment a couple times replaying other's comments. This it's really an easier to understand and more elegant approach to the problem. Please share it again as a normal comment so that others can find it easier than posting it in other comments. I don't know if less or more than 1% can solve the problem, but this solution is worth sharing. I also thought about the first step of your solution, but I didn't try it, so I don't know if I would be able to solve it or not.
without knowing these symmetric polynomials... if you square the first eq and use the value of the second, you get: (x+y+z)^2 = 1 = (x^2+y^2+z^2)+2(xy+yz+zx) = 2+2(xy+yz+zx) so xy+yz+zx = -1/2 now cube the first eq, rearrange the terms to get expressions you know, and you get: (x+y+z)^3 = 1 = (x+y)^3+3z(x+y)^2+3(x+y)z^2+z^3 = x^3+3yx^2+3xy^2+y^3+3zx^2+6xyz+3zy^2+3xz^2+3yz^2+z^3 (x^3+y^3+z^3)+3(x^2(y+z)+y^2(x+z)+z^2(x+y))+6xyz = 1 the first term is 3; y+z=1-x 3+3(x^2(1-x)+y^2(1-y)+z^2(1-z))+6xyz = 3+3(x^2+y^2+z^2)-3(x^3+y^3+z^3)+6xyz = 3+6-9+6xyz = 1 so xyz = 1/6 now let's consider p(n-1) p(n-1) = p(n-1)*1 = (x^(n-1)+y^(n-1)+z^(n-1))(x+y+z) = x^n+x^(n-1)(y+z)+symmetrical terms = x^n+x^(n-2)(xy+xz)+symmetrical terms [using xy+yz+zx = -1/2] p(n-1)= x^n+x^(n-2)(-1/2-yz)+symmetrical terms = p(n)-p(n-2)/2-x^(n-3)*xyz+y^(n-3)*xyz+z^(n-3)*xyz = p(n)-p(n-2)/2-p(n-3)/6 [using xyz = 1/6] so we get p(n) = p(n-1)+p(n-2)/2+p(n-3)/6 we already know the first 5 p's: 1, 2, 3, 25/6, 6 p(6) = p(5)+p(4)/2+p(3)/6 = 6+(25/6)/2+3/6 = 103/12 p(7) = 221/18 p(8) = 1265/72 p(9) = 905/36 p(10) = 15539/432...
For middle or high school students who participate in math competitions, it is useful to learn how to solve this problem without remembering complex theories. To this end, one should start with (x+y+z)[x^(n-1) + ...] = [x^(n) + ...] + [x^(n-2) + ...] (xy + xz + yz) - [x^(n-3) + ...](xyz). Using the given conditions, one can solve a=xy+xz+yz= -0.5 and b=xyz=(1/6). Then, one has [x^(n) + ...] = [x^(n-1)+...] + [x^(n-2)+...]/2 + [x^(n-3)+...]/6. As an interesting exercise, one may continue to find the equation to solve [x^(-1)+...], [x^(-2)+...], [x^(-3)+...], ...
This was a great topic and very clearly presented. Kudos. By the way, I’ve been a fan and supporter for a long time. I’ve learned so much from your videos over the years. I hope you never stop making them.
For those who want a longer yet a more general, 8th or 9th grade math, here's my method. First, multiply (x+y+z)(x+y+z) and get the value of xy+xz+yz, next, calculate (x2+y2+z2)(x+y+z) to get the value of xyz. Then to calculate other value you just need to multiply the equation one order below with (x+y+z), Ex. to calculate (x4+y4+z4) you will multiply (x3+y3+z3)(x+y+z), (x5+y5+z5) needs (x4+y4+z4)(x+y+z), and so on. MYD method are the simplification of these patterns. However, this way you only need one simple method, that is multiplying them, instead of remembering one more formula. (the reason you need those 2 values is because all of the multiplication here could be simplified into xyz and xy+xz+yz)
Ex. For those who already calculate you'll know xyz=1/6, xy+xz+yz=-1/2. Then x4+y4+z4= 25/6. So x5+y5+z5 can be calculated like: (x4+y4+z4)(x+y+z)=25/6*1 (x5+y5+z5)+xy4+yx4+xz4+zx4+yz4+zy4=25/6 f(5)+xy(x3+y3)+xz(x3+z3)+yz(y3+z3)=25/6 f(5)+xy(3-z3)....=25/6 f(5)+3(xy+xz+yz)-xyz(x2+y2+z2)=25/6 f(5)+3(-1/2)-(1/6)(2)=25/6 f(5)-11/6=25/6 f(5)=x5+y5+z5=36/6=6, and there u go.
Additional reply : The answer might be able to be simplified as f(5)+f(3)*(xy+xz+yz)-f(2)*xyz=f(4)*f(1) or f(n)=f(n-1)f(1)+f(n-3)*xyz-f(n-2)*(xy+xz+yz) You can simplify the x,y,z into f(x) again but i'm too tired and not worth it, as at this point MYD's way is simpler ( maybe, too tired)
I never took any college algebra courses, only calculus, so I was so lost this entire video until the last summation with the arithmatic. I have no clue how you found the formulas to get the answer, but it was fun to watch you work. It was a lot of e, i p, k, and I am accustomed to XY and Z 😂
Solved it. The idea is to express higher order polynomials in terms of lower order polynomials and cross-terms. Then, you use the given numerical values to solve for the cross terms. After seeing the video, I realized that what I did is equivalent to Girard-Newton.
Agreed that it is a lot harder. I can visualize the 1st equation (plane) intersecting with the 2nd equation (spherical shell). Their intersection is a circle. Harder to visualize is the cubic in 3D (it's looks wavy like the bottom of an egg carton). I enlisted the help of mathematica to do a contour plot of all 3 graphs superimposed. If you rotate the graph, you can see that there is no intersection of the 3 equations in 3D space. Thus, the solution would involve imaginary numbers. Once you get to the complexity where you need mathematica to solve for x,y,z, you've taken away the relevance to 90% of the audience. Now that's just my opinion. The analysis to get the exact answer is wonderful, but not elegant. By elegant, I mean within reach and can be appreciated at the college level. It doesn't do it for me.
Or you can find a 3rd order difference equation for which x^n+y^n+z^n is a solution, and use the given values 1,2 and 3 as initial conditions. Edit: I now see that this is essentially what you've done (but using a more algebraic approach)! Spolier: I believe the correct difference equation is I_n=I_{n-1}+I_{n-2}/2 + I_{n-3}/6. where |_n=x^n+y^n+z^n. The proof is by induction! So I_4 = 3+2/2 +1/6 =25/6 I_5= 25/6 + 3/2 + 2/6 =36/6=6 I_6=6+25/12+3/6=103/12 I_7=103/12+6/2+25/36=442/36=221/18 etc...
Hi Benom8; very good! Here is my post [for all]: This problem is not that hard as it seems, and can be solved in a basic way! You just need to know a few formulas including those of Newton's binomial. It leads to a 3rd degree equation in z (for example), which one can avoid solving [by Cardan's method, or by mine (visible on my draftWiki page)]; ... ... because the question just asks to determine the value of the expression E = x⁵ + y⁵ + z⁵ → which can be written as a function of the 3rd degree in z, and therefore be linked to the equation in z. There are a few calculations, but nothing insurmountable.
Interesting generalization. I discovered a similar pattern in a different polynomial expression. For this specific problem, by nature of symmetry, I got 3, third degree polynomial = 0. I then multiplied each polynomial of x, y and z by one power at a time. Since RHS=0, LHS becomes a recursive expression of known expression. Rest is simplification. Now answering the last part of your question, proving it rational: As long as we are adding/subtracting rational terms of power coefficients, intuitively it is a rational number. However, proving that irrational or complex may be necessary condition. Will try. 😊
Elementary approach using first principles as follows useful for those who have not exposed to higher Algebra. by squaring (x + y + z) and applying the known result x^2 + y^2 + z^2 = 2; we get xy + yz + zx = -1/2 Then multiplying (x + y + z)(x^2 + y^2 + z^2) and applying above results we get xyz = 1/6 By squaring (xy + yz + zx) we get (x^2)(y^2) + (y^2)(z^2) + (z^2)(x^2) = -1/12 By squaring x^2 + y^2 + z^2 we get x^4 + y^4 + z^4 = 25/6 Then multiplying (x^3 + y^3 + z^3)(x^2 + y^2 + z^2) and applying above results we get x^5 + y^5 + z^5 = 6 Squaring (x^3 + y^3 + z^3) and applying above results we get x^6 + y^6 + z^6 = 103/12
Purely based on the mathematical content, I feel like this was one of your best videos to date. It took me a few different approaches until I solved this myself using elementary algebraic manipulations. I had to solve a linear system of three equations in the end. When I solved it, I realized this was one of the more difficult problems and I was curious what your approach would be. I am glad to say that your method of solving is much more interesting than mine. Very interesting method of solving!
I also solved this by elementary algebra: 1. By squaring the first equation you can find the sum x*y+x*z+y*z 2. By multiplying the first equation with the second you can find the sum x^2*y + x^2*z + y^2*x + y^2*z + z^2*x + z^2*y 3. By cubing the first equation you can find the value of x*y*z because the sum from step 2 is already known 4. Then you can get a system of linear equations for the sums z^5 + y^5 + z^5, x^4*y + x^4*z + y^4*x + y^4*z + z^4*x + z^4*y and x^3*y^2 + x^3*z^2 + y^3*x^2 + y^3*z^2 + z^3*x^2 + z^3*y^2. Notice, that sums like x^2*y^2*z + x^2*z^2*y + y^2*z^2*x can be factored like x*y*z*(x*y+x*z+y*z) so are already known. You can get three equations for the system by: a) multiplying the second equation with the third equation b) multiplying the first equation squared with the third equation c) multiplying the first equation with the second equation squared Solve that system and you will find z^5 + y^5 + z^5.
The main utility of the Girard-Newton identities is that the roots do not need to be calculated. The roots only approximate the answer which is rational. This is covered in the chapter on Galois theory.
A lot simpler method sir Let us assume that a,b and c are zeroes of some cubic polynomial f(x). So we have f(x) = (x-a)(x-b)(x-c) = x³ - (a+b+c)x² + (ab+bc+ca)x - abc Now our task is to simply find those coefficients. a+b+c = 1 2(ab+bc+ca) = (a+b+c)² - a² - b² - c² = 1² - 2 = -1 Hence, ab + bc + ca = -1/2 Using identity: a³+b³+c³ - 3abc = (a+b+c)(a²+b²+c²-ab-bc-ca) Hence abc = 1/6 There f(x) = x³ - x² - (1/2)x - 1/6 Since 'a' is a zero of f(x), a³ - a² - (1/2)a - 1/6 = 0 After multiplying both sides of this equation by 'a' and a little bit of rearranging, we get: a⁴ = a³ + (1/2)a² + (1/6)a We will get similar equations for b and c. After adding them, a⁴+b⁴+c⁴ = (a³+b³+c³) + (1/2)(a²+b²+c²) + (1/6)(a+b+c) = 3 + 2/2 + 1/6 = 25/6 Similarly, a5= a⁴ + (1/2)a³ + (1/6)a² Adding again, a5+b5+ c5 = 25/6 + 3/2 + 2/6 = 6
Same here once i understand it i got a good grip but not paying much attention on high school, i had a ziroo idea on the video just watching it because it's interesting 😂
I remember I was able to solve a problem exactly like this while I was in high school. Expanding (x + y + z)^3 and doing some algebra there. Keep going to (x + y + z)^4 and then (x + y + z)^5. Nothing like the solution you presented here
@@jaibhagwanyadav2387 just to warn you it would seem like magic if you don't solve it on your own (x^2+y^2+z^2)(x+y+z) = (x^3+y^3+z^3) + (x^2*y+x^2*z)+(y^2*z+y^2*x)+(z^2*x+z^2*y) (x+y+z)^3 = (x^3+y^3+z^3) + 3[(x^2*y+x^2*z)+(y^2*z+y^2*x)+(z^2*x+z^2*y)] + 6xyz as you can see, there are terms that can be cancelled when 3*(x^2+y^2+z^2)(x+y+z) - (x+y+z)^3 which gives xyz = 1/6 let s(n) = (x^n+y^n+z^n) s(n)*s(1) = s(n+1) + [(x^n*y+x^n*z)+(y^n*z+y^n*x)+(z^n*x+z^n*y)]
Technically, the main "polynomial formula" for any degree would be the resolvent. Resolvent polynomials help split the original groups into smaller ones. Our goal is to only get the polynomial invariants stabilized by the identity. Example: x1+x2+x3, x1x2 + x1x3 + x2x3, and x1x2x3 are fully stabilized by S3. Take this invariant: x1x2^2 + x2x3^2 + x3x1^2 which is stabilized by A3, a subgroup of S3. We can solve for the A3 invariant by using a resolvent polynomial. In this case, the resolvent polynomial is a quadratic, (Y-(x1x2^2 + x2x3^2 + x3x1^2))(Y-(x1^2*x2+x2^2*x3+x3^2*x1)). This resolvent will give you symmetric polynomials, which can be made up of the elementary symmetric polynomials, because they are stabilized by S3.
i didnt undestand, but i solved it the "simple algebra" way. didnt understand this method, but only watched video once. watchign it again to understand
How about I calculate both a⁴+b⁴+c⁴ and a^5+b^5+c^5 by short method using only algebra. Let if a,b and c are roots of a cubic polynomial f(x). f(x) = (x-a)(x-b)(x-c) = x³ - (a+b+c)x² + (ab+bc+ca)x - abc Now from the given equations. a+b+c = 1 Let squaring both the sides (a+b+c)²= 1² Let's expand the left side a²+b²+c²+2(ab+bc+ca) = 1 Now , we know that a²+b²+c²= 2 then 2(ab+bc+ca) = 1-2 = -1 Hence, ab+bc+ca = -1/2 And we know that a³+b³+c³ - 3abc = (a+b+c)(a²+b²+c²-ab-bc-ca) Putting all the values we have in the above equation we get, 3 - 3abc = (1) (2-(-1/2) 3 - 3abc = 5/2 3abc = 1/2 abc = 1/6 Hence our polynomial after substituting all these values will be f(x) = x³ - x² - (1/2)x - 1/6 Now 'a' is the root of f(x), Mathematically a³ - a² - (1/2)a - 1/6 = 0 By multiplying with 'a' we get a⁴ , a⁴ = a³ + (1/2)a² + (1/6)a We will get similar equations for b and c. After adding them, a⁴+b⁴+c⁴ = (a³+b³+c³) + (1/2)(a²+b²+c²) + (1/6)(a+b+c) = 3 + 2/2 + 1/6 = 25/6 Similarly, if we multiply with 'a²' we get equation for 'a^5' i.e. a^5 = a⁴ + (1/2)a³ + (1/6)a² Adding again, a^5 + b^5 + c^5 = 25/6 + 3/2 + 2/6 = 6 That's it! Quite simple Love from India❤️
II did it this way: from first equation, we can infer x+y = 1-z and x^2+y^2 +2xy = 1-2z+z^2. From second one: x^2 + y^2 = 2-z^2. So 2z^2 - 2z - 1= 2xy. From the third one, X^3 + y^3 = 3 - z^3, and hence (x+y)(x^2+y^2-xy) = 3- z^3 and hence, (1-z)(2-z^2-1/2-z-z^2) = 3-z^3 . Thus, 6z^3 -6z^2 - 3z - 1 = 0 (a). Symmetrically, 6y^3 -6y^2 - 3y - 1 = 0 (b), and 6x^3 -6x^2 - 3x - 1 = 0 (c). (a) + (b) + (c) yields 6(x^3 + y^3 + z^3) -6(x^2 + y^2 + z^2)-3(x + y+ z) - 3 = 0 Now, supposing as x, y, and z 0 (one can prove from (a) that z0, from (b) that y0, and (c) that x0), we calculate (a)*z + (b)*y + (c)*x), and get: 6(x^4 + y^4 + z^4) -6(x^3 + y^3 + z^3)-3(x^2 + y^2 + z^2) - (x + y+ z) = 0 So 6 (x^4 + y^4 + z^4) -6*3-3*2-1 = 0 and (x^4 + y^4 + z^4) = 25/6. Applying the same reasoning with (a)*z^2 + (b)*y^2 + (c)*x^2, we get: 6 (x^5 + y^5 + z^5)-6*25/6-3*3-2 = 0 and hence: (x^5 + y^5 + z^5) = 6. Applying the same reasoning, we can find x^n + y^n + z^n, and as it will be a linear combination of ((x^(n-m) + y^(n-m) + z^(n-m)) who are all rationales, x^n + y^n + z^n will be a rational number for all n.
Here we can find: X^3=X^2+1/2X+1/6 for given X=x,y and z, and figure p(4) and p(5) out. We also can find above equation has 3 different real solutions, then x,y,z in R. PROOF of p(n) in Q: Using above equation, we write X^n by linear combination of X^2, X and const. as following: X^n = a_nX^2 + b_nX + c_n On the other hand, X^n = X^(n-1)X leads a_n = a_(n-1)+1/2a_(n-2)+1/6a_(n-3) b_n = 1/2a_(n-1)+1/6a_(n-2) c_n = 1/6a_(n-1) for natural n>3. Since obviously a_1=0, a_2=1, a_3=1, the sequence a_n in Q so b_n and c_n are in Q too. Then X^n is in Q, so p(n) is. (END)
Lesser degrees equations can be solved without this method, I have solved this problem by some methamatical operations(sqauring both side, substitution ect)
I used only some algebraic identities of class 9 and my answer was 25/6 for x^4+y^4+z^4. I don't claim that I am one of those 1% but it was very interesting sum. Thanks for the video. 👍👍👍👍
I solved it using only simple algebra. Square the first equation and substitute the second equation and you have xy+xz+yz=-1/2 (eq. a). Multiply the first equation by the second and substitute the third and you have xy²+x²y+xz²+x²z+yz²+yz²=-1. Grouping you have xy(x+y)+xz(x+z)+yz(y+z)=-1, and if you substitute x+y, x+z; y+z using the first equation you have xy(1-z)+xz(1-y)+yz(1-x)=-1. Continuing xy-xyz+xz-xyz+yz-xyz=-1, and using the (eq. a), you have xyz=1/6 (eq. b). Squaring (eq. a) you have x²y²+x²z²+y²z²+2x²yz+2xy²z+2xyz²=1/4. Continuing x²y²+x²z²+y²z²+2xyz(x+y+z)=1/4. Substituting (eq. a) and (eq. b), we have x²y²+x²z²+y²z²=-1/12 (eq. c). Multiplying the second by the third equation give us x^5+y^5+z^5+x²y³+x³y²+x³z²+x²z³+y³z²+y²z³=6 Grouping give us x^5+y^5+z^5+x²y²(x+y)+x²z²(x+z)+y²z²(y+z)=6, and substituting x+y, x+z; y+z using the first equation you have x^5+y^5+z^5+x²y²(1-z)+x²z²(1-y)+y²z²(1-x)=6. Then x^5+y^5+z^5+x²y²+x²z²+y²z²-xyz(xy+xz+yz)=6. Finally substituting (eq. a), (eq. b) and (eq. c), we have x^5+y^5+z^5+(-1/12)-(1/6)(-1/2)=6, which is x^5+y^5+z^5=6.
The recursion is: Sum of the nth powers = the sum of the (n-1)st powers + 1/2*(sum of the (n-2)nd powers) + 1/6(sum of the (n-3)rd powers), so they'll all be rational and the sum of the 6th powers is 6 + 1/2 * 25/6 + (1/6)*3 = 103/12. Sheesh, this is so easy. I hear they give this problem to 3rd graders in Singapore.
A bit late to the table, but here is a recurrence relation. it is {1/6, 1/2, 1} on {1, 2, 3}, which gives (1*1/6) + (2*1/2) + (3*1) = 25/6. The next iteration is (2*1/6) + (3*1/2) + (25/6*1) = 6 and then (3*1/6) + (25/6*1/2) + (6*1) = 103/12 etc. The first 10 values are :- {1, 2, 3, 25/6, 6, 103/12, 221/18, 1265/72, 905/36, 15539/432, 11117/216, 21209/288, 136561/1296, 260531/1728, 559171/2592}.
It is quite easy to derive that S(n)=S(n-1)+S(n-2)/2+S(n-3)/6 , where S(n)=x^n+y^n+z^n. We are given S(1)=1, S(2)=2 and S(3)=3 hence we can calculate S for any n.
@@umbragon2814 Vieta's theorem links a system of sums and products of N numbers to the coefficients of an N- degree polynomial (in this case a polynomial of deg 3)
@Adam Romanov Yes, it does. all you have to do is some substitutions to end up with x+y+z=1, xy+xz+yz=1/2, and xyz=1/6 which leads you (using Vieta's formula) to the equation X^3+X^2-1/2*X+1/6=0. All what's left is solving the 3rd deg eq with Cardaro method. Besides, I precised that this is "hardcore method", it is not optimal but, for instance, has other uses like proving that the system S(x,y,z) has no solutions in R^3
@@khaledshell507 6x^3 - 6x^2 -3x -1 = 0 There is no need to actually solve this equation. It gives you a linear recursive formula for all higher powers of x: x^n=(6x^(n-1)+3x^(n-2)+x^(n-3))/6, and the same for y and z. Now simply add the three recursion formulas and you get p_n=(6p_(n-1)+3p_(n-2)+p_(n-3))/6, where p_n is x^n+y^n+z^n. Plug in the initial values 1,2,3 and you're done.
I was well on my way to not understanding another amazing video explaining things I gave up on trying to understand way back in high school in Mr Backer’s algebra class, when I spent a week’s worth of after school tutoring in quadratic equations because I didn’t want to give up on it without a fight. At the end of that week Mr Baker politely informed me that I’d never understand them. The reason: there’s only so much stuff you can memorize without understanding at least some of it. What sealed the deal in this video was the commercial in the middle of it. It wiped away any entertainment value for me, and I only finished watching it out of the desire to be polite.
How about I calculate both a⁴+b⁴+c⁴ and a^5+b^5+c^5 by short method using only algebra. Let if a,b and c are roots of a cubic polynomial f(x). f(x) = (x-a)(x-b)(x-c) = x³ - (a+b+c)x² + (ab+bc+ca)x - abc Now from the given equations. a+b+c = 1 Let squaring both the sides (a+b+c)²= 1² Let's expand the left side a²+b²+c²+2(ab+bc+ca) = 1 Now , we know that a²+b²+c²= 2 then 2(ab+bc+ca) = 1-2 = -1 Hence, ab+bc+ca = -1/2 And we know that a³+b³+c³ - 3abc = (a+b+c)(a²+b²+c²-ab-bc-ca) Putting all the values we have in the above equation we get, 3 - 3abc = (1) (2-(-1/2) 3 - 3abc = 5/2 3abc = 1/2 abc = 1/6 Hence our polynomial after substituting all these values will be f(x) = x³ - x² - (1/2)x - 1/6 Now 'a' is the root of f(x), Mathematically a³ - a² - (1/2)a - 1/6 = 0 By multiplying with 'a' we get a⁴ , a⁴ = a³ + (1/2)a² + (1/6)a We will get similar equations for b and c. After adding them, a⁴+b⁴+c⁴ = (a³+b³+c³) + (1/2)(a²+b²+c²) + (1/6)(a+b+c) = 3 + 2/2 + 1/6 = 25/6 Similarly, if we multiply with 'a²' we get equation for 'a^5' i.e. a^5 = a⁴ + (1/2)a³ + (1/6)a² Adding again, a^5 + b^5 + c^5 = 25/6 + 3/2 + 2/6 = 6 That's it! Quite simple Love from India❤️
Another solution: let a = (x + y), b = xy From x + y + z = 1, we get z = 1 - (x + y) From x^2 + y^2 + (1 - (x + y))^2 = 2, we get b = a^2 - a - 1/2 From x^3 + y^3 + (1 - (x + y))^3 = 3, we get a^3 - 2a^2 + (1/2)a = -2/3 x^5 + y^5 + (1 - (x+y))^5 = (-15/2) (a^3 - 2a^2 + (1/2)a) + 1 = 5 + 1 = 6
Remember, it's not the "Mind Your Decisions guy that solved this", Girard and Newton solved this, and even them took at least a few days or even weeks to prove their findings. The key to this problem is at the Girard Newton Identities, so it's not 1% of the world, it's like whoever studied the Girard Newton Identities, which is very rare.
Did it a bit more manually, knowing there was some hidden relations with elementary symmetric polynomials. If you label the equations like this : (1) : x+y+z =1 (2) : x^2+y^2+z^2 = 2 (3) : x^3+y^3+z^3 = 3 you compute (1)^2 to get (4) : xy+yz+zx = -1/2, using (2) you compute (3)^3-3(1)(2) to get (5) : xyz = 1/6, using (3) you compute (4)^2 to get (6) : x^2y^2+y^2z^2+z^2x^2 = -1/12, using (5) and (1) you compute (2)(3)-(6)(1)+(4)(5) to get the end result
Brilliant, what a great piece of teaching. I knew enough about symmetric polynomials to blunder out a semi-numerical solution, involving finding the roots of a quadratic and then a cubic, and got the answer a lot closer than wolfram. However, Newton-Girard goes into my mental card-file
My solution: Call the equations a,b and c. Square a to get x*y+x*z+y*z=-1/2. Call this d. Multiply a and b to get x*y*z=1/6. Square d to get x^2*y^2+x^2*z^2+y^2*z^2=-1/12. Now take x^2*c+y^2*c+z^2*c to get x^5+y^5+z^5=6-A, where A=x^2*(y^3+z^3)+y^2*(x^3+z^3)+z^2*(x^3+y^3). Now A, by the help of the expressions above, can be calculated to be zero, thus, the answer is 6.
This was absolutely delightful. I encountered another similar problem in a book full of mathematical puzzles, but they are not the p polynomials but sums of e polynomials. I took a crack at it and eventually gave up. Little did I know there's so much stuff behind it. Thanks and now I can sleep in peace.
SOLVE, a, b, c, then replace into a^5+b^5+c^5 =... Numeric result : {a = -.21543-.26472*i, b = 1.4308, c = -.21544+.26472*i}, or {a = -.21543+.26472*i, b = 1.4308, c = -.21544-.26472*i} Then a+b+c = 1, a^2+b^2+c^2 = 2, a^3+b^3+c^3 = 3 So : a^4+b^4+c^4 = 25/6 a^5+b^5+c^5 = 6 a^6+b^6+c^6 = 103/12 a^7+b^7+c^7 = 221/18 For people who don't like numeric resolution here exact solution : {a = -(1/6)*(44+6*sqrt(26))^(1/3)-5/(3*(44+6*sqrt(26))^(1/3))+2/3+(1/6)*(-2*(44+6*sqrt(26))^(2/3)-I*sqrt(-10*(44+6*sqrt(26))^(4/3)+75*(44+6*sqrt(26))^(2/3)-30*(44+6*sqrt(26))^(1/3)*sqrt(26)-220*(44+6*sqrt(26))^(1/3)+396*sqrt(26)+2154)+3*sqrt(26)+5*(44+6*sqrt(26))^(1/3)+22)/(44+6*sqrt(26))^(2/3), b = (1/6)*(44+6*sqrt(26))^(1/3)+5/(3*(44+6*sqrt(26))^(1/3))+1/3, c = -(1/6)*(-2*(44+6*sqrt(26))^(2/3)-I*sqrt(-10*(44+6*sqrt(26))^(4/3)+75*(44+6*sqrt(26))^(2/3)-30*(44+6*sqrt(26))^(1/3)*sqrt(26)-220*(44+6*sqrt(26))^(1/3)+396*sqrt(26)+2154)+3*sqrt(26)+5*(44+6*sqrt(26))^(1/3)+22)/(44+6*sqrt(26))^(2/3)}, or {a = -(1/6)*(44+6*sqrt(26))^(1/3)-5/(3*(44+6*sqrt(26))^(1/3))+2/3+(1/6)*(-2*(44+6*sqrt(26))^(2/3)+I*sqrt(-10*(44+6*sqrt(26))^(4/3)+75*(44+6*sqrt(26))^(2/3)-30*(44+6*sqrt(26))^(1/3)*sqrt(26)-220*(44+6*sqrt(26))^(1/3)+396*sqrt(26)+2154)+3*sqrt(26)+5*(44+6*sqrt(26))^(1/3)+22)/(44+6*sqrt(26))^(2/3), b = (1/6)*(44+6*sqrt(26))^(1/3)+5/(3*(44+6*sqrt(26))^(1/3))+1/3, c = -(1/6)*(-2*(44+6*sqrt(26))^(2/3)+I*sqrt(-10*(44+6*sqrt(26))^(4/3)+75*(44+6*sqrt(26))^(2/3)-30*(44+6*sqrt(26))^(1/3)*sqrt(26)-220*(44+6*sqrt(26))^(1/3)+396*sqrt(26)+2154)+3*sqrt(26)+5*(44+6*sqrt(26))^(1/3)+22)/(44+6*sqrt(26))^(2/3)} because : {a = -RootOf(6*_Z^3-6*_Z^2-3*_Z-1)-RootOf(2*_Z^2+(-2+2*RootOf(6*_Z^3-6*_Z^2-3*_Z-1))*_Z+2*RootOf(6*_Z^3-6*_Z^2-3*_Z-1)^2-1-2*RootOf(6*_Z^3-6*_Z^2-3*_Z-1))+1, b = RootOf(6*_Z^3-6*_Z^2-3*_Z-1), c = RootOf(2*_Z^2+(-2+2*RootOf(6*_Z^3-6*_Z^2-3*_Z-1))*_Z+2*RootOf(6*_Z^3-6*_Z^2-3*_Z-1)^2-1-2*RootOf(6*_Z^3-6*_Z^2-3*_Z-1))}
It cannot be done by normal algebra these are imaginary no.s you cannot handle themwithout permutation send me the solution or just tell how you solved it
Excelente video, en mi intento de resolver ese problema pensé que salía 5 :'/ Ya luego aplique la extensión de Horner con las potencias de las raíces y me salió también Ahora se un poquito más, Muchas Gracias ^-^ y excelente video
All you have two do is multiply x^2 multiplied by x^3 and so on We get x^5 and so on If we multiply 2•3 then we get 6 There fore the answer is 6 TELL ME IM WRONG Thanks btw
I think we can also use basic high school algebra to solve this. It's very simple actually try this way name them, eqns 1,2,3 now u want eqn5 try multplying 2,3 ur RHS is 6 and now try to simplify LHS based on terms we have like find common terms take pairs x3y2,x2y3 write it as x2y2(x+y) and u can use eqn 1 x+y=1-z. Now u'll need to do 2 more things try to obtain eqn2 from eqn1*eqn1 u'll have some set of terms and it's value similarly try obtaining eqn3 from eqn1*eqn2 u'll get some set of terms and it's value but u'll need similar solving technique like I said before using 1-z=x+y they make ur life simple. With these u can obtain eqn5 easily. They actually cancels out.
Thank you for teaching us about the Girard-Newton polynomial equations. I'd learnt in school that in a monic nth-degree polynomial, e_k, the sum of the products of the roots taken k at a time, is (-1)^k a_{n-k}, but not about G-N. I feared that the solution for x^5+y^5+z^5 would be some arbitrary degree-5 combo of e_0,...,e_5 which nobody could be fairly expected to find under exam conditions, but you showed us a systematic way. Thank you. Explanation starts at 4:20. 6:21 Nobody going to note Presh saying "negative" for "minus"? Sounds like military or air-pilot jargon. 10:05 Respecting mathematics more than engineering, I'd prefer "-1/2" to "-0.5".
