Determine a and n

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Күн бұрын

Пікірлер: 70

@9ybruhwt491 Ай бұрын

I've been watching your videos for two months now, and for the first time, I was able to solve everything on my own without looking at any hints. And my solution was the same as yours! I'm so happy.

@peterpiispanen6145 28 күн бұрын

Very inspirational... as a mathematics teacher myself I can truly appreciate the enthusiasm, structure, solution and pedagogics... good work! 😅 PS. Learning something new with every video! DS.

@rhc-weinkontore.k.7118 9 күн бұрын

Lots of fun listening to you. Thank you for sharing your thoughts.

@philipframpton9428 Ай бұрын

You can more quickly find the roots by inspection rather than trial and error. A positive sum means r1 is 4 or higher, since r2,r3 cannot be zero and must be at least -1,-1. An odd remainer from r1+r2+r3 with r1 being even means one of r2/r3 is odd, -1 is the only option. r1 + r3 = 2, the only 2^n pair with a difference of 2 are 4,2. So r1=4, r2=-1, r3=-2.

@cyberagua Ай бұрын

Good solution 👍 *r₁ = 4, r₂ = −2, r₃ = −1* (just checking my sophisticated Gboard with subscript digits 😊)

@topquark22 Ай бұрын

Great job. I do have a master's degree in mathematics, but you do a much better job of explaining this stuff to young people than I could ever do.

@cyberagua Ай бұрын

In the general case, if not a natural number, *_n_* can also be either a real transcendental number or a complex transcendental number (as follows from the Gelfond-Schneider theorem). Example: *r₁ = 5, r₂ = −3, r₃ = −1,* so that *2ⁿ = 15* and *n = log₂15* is a real transcendental (i.e. not algebraic) number.

@andrasferencz7948 13 күн бұрын

Beautiful! Another reason for no other combination: One of the roots must be odd, else we won't get the odd sum. The only odd power of two is one. So one root must have the magnitude 1. The positive number's magnitude must be greater than the sum of the negative ones, Therefore the odd root is -1. Once we fixed this, the other two roots' magnitude must be two apart, else the sum can't be 1. Only solution is 4 and -2. I hope it made sense.

@glorrin Ай бұрын

It is a great video But I want to have fun and since it was not written in the question, I want to consider n could be not natural. Then there are infinitely many solutions Let's give an exemple : r1 = 8, r2= -4, r3=-3 2^n = 8*(-4)*(-3) =2^(3)*2^(2)*2^(log2(3)) n = 5+Log2(3) a = -32 -24 + 12 = -44 And if we let n be complex r1 = 1, r2 =1, r3 =-1 r1 + r2 + r3 = 1 2^n = 1 * 1 * -1 =2^(0)*2^(0)*2^(i*Pi /ln(2)) {for those who forgot -1 = e^(i*Pi) = e^(i*Pi *ln2/ln2) = 2^(i*Pi/ln2)} n =i*Pi /ln(2) a = 1 - 1 - 1 = -2

@cyberagua Ай бұрын

Great observation 👍

@cyberagua Ай бұрын

Also some sources claim that *e^(𝒊π)* has infinitely many values, as a multivalued complex exponential function, given by *zʷ = exp(w·Ln(z)) = exp(w·(ln(|z|) + 𝒊·arg(z) + 2𝒊πk)), k ∈ ℤ,* while *exp(𝒊π) = −1* has only one value, as a single-valued function, defined by the corresponding Taylor series *exp(z) = ∑ zⁿ/n!*

@ChristodoulosTsilopoulos-vs7xb Ай бұрын

Excellent. Thank you for the presentation.

@Jono4174 Ай бұрын

n IS AN INTEGER (or rational, I suppose) otherwise, for example: n = ln(3)/ln(2) a= -5 roots = 3,-1,-1

@cyberagua Ай бұрын

Good point 👍 Prime never said that *_a_* and *_n_* should be integers, and that is an important restriction.

@Grecks75 Ай бұрын

@@cyberagua If all 3 roots are integers, a must be an integer, too (by Vieta's theorem). For the same reason, we know that 2^n is an integer. n could in theory be a real number, but it is reasonable and safe to assume that n is meant to be a non-negative integer. Otherwise the problem would not make much sense (there would be infinitely many solutions). Still, it would have been better to include that restriction in the problem statement.

@cyberagua Ай бұрын

@@Grecks75 Yes, *_n_* may also be a real or complex transcendental number, but not algebraic (like *√2̅),* which is itself an interesting math problem (compare with the famous Euler's conjecture regarding logarithms, for example). And yes, it is much better to include the full list of restrictions in the wording of the problem.

@chaosredefined3834 Ай бұрын

Since r1, r2 and r3 are factors of 2^n, the only way they can be odd is if they are 1 or -1. And we know that r1 + r2 + r3 = 1. Since the sum of the three roots is odd, either one of them or all three of them are odd. As you argue in the vid, two of them are negative and one of them is positive. Case 1: The positive term is odd, both negative terms are even. So, r1 = 1, r2 = -2^a and r3 = -2^b. This is impossible as 1 + r2 + r3 < 1, but we need it to equal 1. Case 2: One of the negative terms is odd, the other negative term and the positive term are even. Without loss of generality, r2 is odd, r3 is even. So, r1 = 2^a, r2 = -1, and r3 = -2^b for some natural numbers a and b. So, 2^a - 1 - 2^b = 1. Or 2^a - 2^b = 2. So, 2^(a - b)(2^b - 1) = 2. Note that 2^b - 1 has to be odd. The only odd factor of 2 is 1, so 2^b - 1 = 1, and therefore, 2^b = 2, so b = 1. Furthermore, 2^(a-b) = 2, so a-b = 1, and since b = 1, that means a = 2. So, the other factors are 4 and -2. Therefore, a valid set of roots is (4, -1, -2), which gives us 2^n = 8, so n = 3, and a = (4)(-1) + (4)(-2) + (-1)(-2) = -10. Case 3: All terms are odd. Thus, r1 = 1, r2 = r3 = -1. So, r1 + r2 + r3 = 1 + (-1) + (-1) = -1. This contradicts our given info. So, no solution here. Thus, the only solution is n=3, a=-10.

@-wx-78- Ай бұрын

One of negative roots should be -1, otherwise all terms in r₁+r₂+r₃ are even and sum can't be odd. WLOG r₁=2ⁱ, r₂=-2ʲ, r₃=-1: r₁+r₂−1 = 1 r₁−2 = -r₂ 2ⁱ−2 = 2ʲ 2(2ⁱ⁻¹−1) = 2ʲ r₂ can not be -1, hence RHS is even; therefore 2ⁱ⁻¹−1 = 1 ⇒ i = 2 ⇒ r₁ = 4, r₂ = -2, r₃ = -1.

@cyberagua Ай бұрын

Hm, you too are able to type subscripts and superscripts? 🤔 How do you input them? From a smartphone or from a PC or Mac?

@Grecks75 Ай бұрын

Good observation. 👍

@-wx-78- Ай бұрын

@@cyberagua On Windows 10+ you can press Win+; and select symbols from palette, or use Alt+KeypadPlus then type Unicode codepoint in hex while holding Alt. My method is somewhat different: custom keyboard layout with true RightAlt and AutoHotkey script with RightAlt hotkeys.

@cyberagua Ай бұрын

@@-wx-78- Thank you for sharing! I'm mostly using a Gboard Dictionary with shortcuts on Android and the Character Pad App by Hussein El Feky to look for symbols: *³√−̅8̅ = 1+𝒊√3̅*

@absolutezero9874 Ай бұрын

No reply Still ignoring

@holyshit922 Ай бұрын

In Vieta formulas on LHS we have elementary symmetric polynomial and on RHS we have (-1)^{n-k}\frac{a_{n-k}}{a_{n}} Possible roots (4,-2,-1)

@itsphoenixingtime Ай бұрын

For me I recognised that the roots themselves must have absolute values of powers of 2, since the product of the roots is a power of 2. Also, by drawing a graph, there were two possibilities: 1. 3 roots are all + 2. 1 root is +. 2 are - The second case must be true as if all 3 roots are + they cannot sum up to 1 [as the sum would exceed 1 quickly setting the smallest root to 1] So we get that there are 3 roots, 1 positive, 2 negative, and the absolute values of said roots must be equal to a power of 2. For some reason, I decided to try -1,-2 and 4 for an example, and sure enough it matched. Not sure if there was a more rigorous way to do this other than by pure luck. Lol.

@yasinforughi-b1z Ай бұрын

I just love watching your videos I hope you'd make more videos on Fourier series Hello from Iran🖐

@simplicity530 17 күн бұрын

One more thing you forgot to mention about the roots. r1 + r2 + r3 = 1 1 is an odd integer so two roots must be even and one must be odd. The odd integer cannot be greater than absolute 1 otherwise the roots product will not be a multiple of 2 only.

@babitasingh8620 Ай бұрын

We can prove that there is only one solution by extending parity over negative integers. r1=2^d r2=-2^b r3=-2^c[as two roots have to be -ve] 2^d+2^b+2^c=1 If all d,b,c>1 Then LHS is even which is not possible as RHS is odd. d,b or c has to be 0 If d=0, then 1-2^b-2^c=1 -2^b=2^c which is not possible So, b or c=0 Let c=0 2^d-2^b-1=1 2^d-2^b=2 Only possible when d=2 and b=1 So,r1=2^d=4 r2=-2^b=-2 r3=-2^c=-1 Value of r2 and r3 can be interchanged by taking b=0 but it gives same value of a and n. THANK YOU

@Grecks75 Ай бұрын

Solution: I assume n is meant to be an integer, otherwise the problem wouldn't make much sense, because there would be infinitely many solutions and answers. Let x1, x2, x3 be the 3 integer roots. By Vieta's theorem, we have: 1) x1 * x2 * x3 = 2^n 2) x1 + x2 + x3 = 1 3) x1*x2 + x2*x3 + x3*x1 = a According to (1), none of the roots can be zero. Being integers they must divide 2^n, and therefore they must themselves be powers of 2 with an optional minus sign. This follows from prime factorization. They can't be all positive, otherwise their sum would be at least 3 which contradicts (2). According to (1) their product is positive; therefore one of them must be positive while the two other roots must be negative. W.l.o.g., let x1 be the positive root and |x2| = 5 which is never a power of 2. We are left with: 2^e1 = 2 + 2^e3. Here we learn that e1 >= 2 and e3 > 0. Dividing by 2 we get: 2^(e1 - 1) = 1 + 2^(e3 - 1). The only way this can be satisfied is with e3 = 1 and e1 = 2. Thus, the 3 integer roots must be x1 = 4, x2 = -1, x3 = -2 (in any order). According to (1) we get: n = e1 + e2 + e3 = 3. According to (3) we get: a = -4 + 2 - 8 = -10. In summary: a = -10, n = 3 ist the ONLY answer. The roots are x1 = 4, x2 = -1, x3 = -2 (in any order). We briefly verify that this satisfies the problem.

@alipourzand6499 Ай бұрын

Starting point: from the first equation we know that gcd(r1, r2, r3) must divide 1 and from the third we know that r1, r2, r3 must be powers of 2. So either 1 or -1 is a root.

@karamsedighi 24 күн бұрын

I LOVE SOUTH AFRICA , THANKS OF NELSON MANDELA, HE IS EVER ALIVE !!

@pijanV2 18 күн бұрын

@georgelaing2578 21 күн бұрын

Can you think of a direct way to show that 4abc - b - c (a, b and c are natural numbers) can never be a square integer?

@hervesergegbeto3352 Ай бұрын

Thank you very much

@keithrobinson2941 26 күн бұрын

Have you done a proof of Vieta's formula? I have not heard of it before today.

@williamperez-hernandez3968 16 күн бұрын

So if we are asked to factorize x^3 - x^2 - 10x - 8, we "instantly" know it is (x-4)(x+2)(x+1). Easy!

@surendrakverma555 Ай бұрын

Thanks Sir 🙏🙏🙏🙏

@ryhorabramovich1457 Ай бұрын

Because r1, r2, r3 are integer and r1*r2*r3=2^n, then each of |r1|, |r2|, |r3| must be powers of 2 (2^t). Now, from conditions r1+r2+r3=1 and "2 negatives, 1 positive", we can easily find the answer: -2, -1, 4

@0lympy Ай бұрын

16:34 I guess, it can be proved from r1 + r2 + r3 = 1 => r1 + r2 = 1 - r3; assume r1=2^n, r2=-2^m, r3=-2^k, where n, m, k >= 1 => RHS is 2^k + 1, which is always odd in these conditions; but LHS is 2^n - 2^m can't be odd!

@PauloDacosta-s1s Ай бұрын

I learnt what you call of Vieta’s formulas as Girard equations…..I found out that it was Girard that proved the general theorem and Vietas theorem was proved for positive roots only….so in this case, I think that you should call Girard equation and not Vieta’s formula….what do you think?

@mahoremujini 20 күн бұрын

On peut éliminer rapidement le cas r=1: puisque la somme des racines est 1, les deux autres racines sont opposées ; ce qui est en contradiction avec le résultat sur les signes à savoir qu’une seule racine est positive.

@Random_DBL_Player Ай бұрын

Nahh, this is total brain confusion for me, I'll learn it later when im in class 9 or 10 and me in 8 now

@Grecks75 Ай бұрын

That's reasonable. Start with simpler exercises, you can always return to these problems later when you're more experienced. It will all fall into place one day if you keep practicing. I would say his problems are for undergraduates (or for very ambitious highschool students). Not suitable for eighth grade. Btw, the reasoning at the end of this video wasn't particularly good; there are better solution developments in the comments.

@srisaishravan5512 Ай бұрын

Day 3 of asking newton sir to restart making videos on integrals

@PeymanMoradi-gv1ik Ай бұрын

perfect

@Defbarr Ай бұрын

(a,n) = (-4,2) works also

@PrimeNewtons Ай бұрын

Product of roots needs to be positive. This fails

@hervesergegbeto3352 Ай бұрын

Bonjour mon professeur

@marcclaude2065 Ай бұрын

What's done if n is not a natural number? If n= 1/2.. Vietas does not works.

@ryhorabramovich1457 Ай бұрын

r1+r2+r3=2^n As r1, r2, r3 are integer, then 2^n also must be integer

@marcclaude2065 Ай бұрын

@@ryhorabramovich1457 why r1, r2, r3 must be integer?

@yuichiro12 Ай бұрын

@@marcclaude2065 The question states that the cubic has 3 integer roots.

@DanielKniaz Ай бұрын

Let r's be 1,2,3, sum=7 which equals 2^n but is not integer. Seems it was missed in preconditions

@cyberagua Ай бұрын

Exactly. In the general case, *_n_* can also be either a real transcendental number or a complex transcendental number (as follows from the Gelfond-Schneider theorem)