2015 Harvard-MIT Math Tournament #25
ฝัง
- เผยแพร่เมื่อ 26 พ.ค. 2024
- This question required Vieta's formula, but before applying the formula, there was a lot of work done to find the relevant polynomial by factoring and using Limits and intermediate value theorem.
![[UNCUT] ฝันรักห้วงนิทรา | EP.8 (4/4)](http://i.ytimg.com/vi/RDYJYiDwsj0/mqdefault.jpg)
Once you have established the quartic polynomial with 4 real roots, you can also use newton's sum to obtain the sum of the squares of the roots.
Thank you for sharing. I never knew it existed.
Note that when you have
x⁴ + 4x² + x + 1 = 0
you can rewrite this as
x⁴ = −(4x² + x + 1)
x⁴ = −((2x + ¹⁄₄)² − ¹⁄₁₆ + 1)
x⁴ = −((2x + ¹⁄₄)² + ¹⁵⁄₁₆)
The left hand side is nonnegative for any real x, and the right hand side is negative for any real x, so there can be no real solutions for this quartic equation.
That's smart 👌
At 11:00 you can rewrite x^4+4x^2+x+1 as x^4 + (x+2)^2/4 + 15x^2/4 which is obviously always positive
Your voice and speed make things easier. The way you explained and those steps you arranged to digest the problem, that is what ideal teacher did. I am medical personnel yet I can easily follow what you are doing, well with some self research and basic calculus, which I got C, in university time.
Anyway I think many teachers should learn how to explain just like you do.😊
the vieta's formula is fire . do a proof of it pleaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaase.
Any polynomial is divisible by x-r. Write p(x)=a(n)*x^n+a(n-1)*x^(n-1)+...+a2*x²+a1*x+a0=an(x-r1)(x-r2)...(x-rn). Distribute the parentheses on the right side and Bob's your mother's brother, it works for any power.
Amazing video bro, in the beginning I was thinking about solving the real quartic head on but then realized those roots are going to be irrational since the rational root theorem failed and then said there must be a easier way and then remembered Vietas formula. If the roots were rational then squaring them and adding them up would have been too easy. Also I want to point out that the intermediate value theorem only applies to continuous functions for those who don’t know.
I would have glossed over the word “real”, written 14² - 2, and wondered why I had so much time left.
The answer would be 0, not 14^2 - 2, even if glossing over the word 'real,' via Vieta's formulae.
You made this problem look so easy! I would have to do many examples like this one to develop the skills needed.
All videos are outstanding
Hello there, from Russian Olympiad Community! Love you videos very much❤
Very good! I would have never gotten past the first aha! moment where you broke up the 14 term as 16 and 2. Brilliant!
This is mind blowing
Amazing teacher
Awesome problem explained in awesome way on how to solve it.
Love your videos! I was curious, as I couldn't really find anything particularly useful googling, is it at all possible to differentiate a tetrated function, where the base is the constant and the variable is the superexponent?
Thank you Sir
great video!
Very nice and intelligent question
Thanks!
Thank you!
Bravo
Just wonder for clear explanation how to find solution, trying follow the lessons in past and go for future. Thanks a lot for your time, Sir.
Sir kindly upload videos on advance analysis
Nice video
Great 😊
Wow!
Is there any info to be gotten if you take the derivative of the quartic that provides the real roots and set it to zero to find the max and min points?
Sure, but the derivative P'(x) = 4x³ − 8x − 1 is a cubic which has three real roots and which can therefore only be solved trigonometrically (or numerically). Not really worth it.
But, I agree that the explanation in the video about the number of real zeros of the polynomial
P(x) = x⁴ − 4x² − x + 1
may not have been entirely clear if you haven't seen something like this before. The purpose of rewriting the polynomial as
P(x) = x⁴(1 − 4/x² − 1/x³ + 1/x⁴)
is to note that | −4/x² − 1/x³ + 1/x⁴ | < 1 for any sufficiently large |x| so
−1 < −4/x² − 1/x³ + 1/x⁴ < 1
and therefore
0 < 1 − 4/x² − 1/x³ + 1/x⁴ < 2
for any sufficiently large |x| so P(x) will be _positive_ for any sufficiently large |x|. And since P(−1) = −1, P(0) = 1, P(1) = −3 this means there will be at least one zero on each of the four intervals
(−∞, −1), (−1, 0), (0, 1), (1, ∞)
And since there can be no more than four zeros, the conclusion follows that P(x) has _exactly one zero_ on each of these four intervals.
*New question…*
Suppose that we are given a cubic (degree 3) polynomial, and, when graphed, the polynomial exhibits the shape of a capital letter N. Now suppose that the bottom-left of the N crosses the x-axis a little to the right of the origin, and the very bottom of the bottom-right of the N BARELY TOUCHES the x-axis. How many roots would this type of cubic polynomial actually possess?
0? Oh right that was the sum of roots to 1st power. Then you showed sum of products of roots to derive the answer.
This is extremely complex, about 6 or 7 layers to this problem.
In fact this polynomial can be factored so it is octic solvable by radicals
(x^4 + 4x^2 + x + 1)(x^4 - 4x^2 - x + 1)
(x^2 - ax + b)(x^2 + ax + c) = x^4+4x^2+x+1
x^4 +ax^3+cx^2 - ax^3 - a^2x^2 - acx + bx^2+abx+bc = x^4+4x^2+x+1
x^4 + (b+c-a^2)x^2 + a(b-c)x + bc = x^4+4x^2+x+1
b+c-a^2 = 4
a(b-c) = 1
bc = 1
b+c = 4+a^2
b - c = 1/a
4bc = 4
2b = 4+a^2+1/a
2c = 4+a^2 - 1/a
4bc = 4
(4+a^2+1/a)(4+a^2 - 1/a) = 4
(4+a^2)^2-1/a^2 - 4 = 0
a^4+8a^2+16-4-1/a^2 = 0
a^4+8a^2+12-1/a^2=0
a^6+8a^4+12a^2-1=0
Here to find coefficients we must use cubic formula or derive it
For power sums there are also Newton - Girard formulas
so we can apply Newton - Girard formula and then Vieta formula
Too much work...
ESPECIALLY in a tournament where u only have such limited time
😮😮
A root of an equation is the root of a sulotion to the equation, hm?
I'm a little puzzled, cause I might not have had this in math.
What is the point of taking a root of an equation?
Do you also take log or tan of an equation?
A root of an equation is a number that makes the expression equal 0.
@@xinpingdonohoe3978
Thank you very much! ❤️🙏
I need to take more time with this one to comprehend what the professor is doing.
Now I get it.
We call it "Nullstellen" (the spots, where the function is 0).
I still don't understand, why it is called "roots" in English. 🤔
@@kragiharproots of tree are seen at the ground, ground is x-axis?
@@lcex1649
Ahhhh.
Now I get it.
❤️🙏
let say f(x)=x^2-5x+6 , x= 0, f(x)>0, is that mean f(x) doesnt cross x-axis? even q(x)>0 if x=0 in teh equation
No. The function you give can be written as
f(x) = (x − 2)(x − 3)
so we have f(2) = 0 and f(3) = 0. The graph of your function crosses the x-axis at x = 2 and at x = 3. The value of your function is negative for any real x between 2 and 3. You can graph your function online at the desmos website.
0?
Can't we plugin directly?
We don't have what to plug in
asnwer=1 isit
(x ➖ 3x+1)
The graphical arguments [and positive infinity and negative infinity] was awesome, but it involved a mature discussion, maybe not suited to younger members of your channel.
Genuinely, what are you talking about? Some strange patronisation to the уоungеr students is all you're providing.
@@xinpingdonohoe3978 It is not usual to relate solutions of a polynomial to concepts of calculus - it may not connect with the younger audience.
@@Arkapravo sure, if you're 6 it's not normal to do that, but if you're watching videos of Harvard-MIT tournaments you should be expecting *something* clever to occur.
@@xinpingdonohoe3978 Well put.
@@xinpingdonohoe3978 yeah, I am a fool
Sturm’s method finds real roots.