Question

Lots and lots of questions. I don't know where to start... there are just so many. Gonna post all of them in this post.

Answer 1

Let \(a, b, c\) be integers, and suppose the equation \(f(x) = ax^2 + bx + c = 0\) has an irrational root \(r\). Let \(u = p/q\) be any rational number such that \(|u-r|<1\). Prove that\[\frac{1}{q^2}\le |f(u)| \le K|u-r|\]for some constant \(K\). Deduce that there is a constant \(M\) such that\[\left|r\frac{p}q\right| \ge M/q^2\](This is useful in approximating the non-rational zero of a polynomial by a rational number.)

Answer 2

Find the value of the positive integer \(n\) for which the quadratic equation
\[\sum_{k=1}^{n} (x+k-1)(x+k) = 10n\]has solutions \(\alpha\) and \(\alpha+1\) for some \(\alpha\).

Answer 3

Find all positive integers \(n\) for which the quadratic equation\[a_{n+1}x^2 - 2x \sqrt{\sum_{k=1}^{n+1}a_k^2 }+ \sum_{k=1}^n a_k=0\]has real roots for every choice of real numbers \(a_1, a_2, \cdots, a_{n+1}\)

Answer 4

Let the polynomial \(p(z) =z^2 + az + b \) be such that \(a\) and \(b\) are complex numbers and \(|p(z)|=1 \) whenever \(|z| = 1\). Prove that \(a = 0\) and \(b = 0\).

Answer 5

Find necessary and sufficient conditions on the coefficients \(a, b, w\) so that the roots of the equations\[z^2 + 2az + b = 0,~ z - w = 0\]are collinear in the complex plane.

Answer 6

Let \(ax^2 + bx + c \) be a polynomial with real coefficients such that\[|ax^2+bx+c| \le 1 \]for \(0 \le x \le 1 \).

Prove that \(|a| + |b| + |c| \le 17\).

Answer 7

Hi Parth. I'm going to see if I can help you out with these. I've got a couple of days off. How do I save this page so that I can come back to it tomorrow?

Answer 8

@alekos Thanks! You can probably bookmark this.

Answer 9

Consider the number\[\alpha = \left(\frac{n + \sqrt{n^2 - 4}}{2}\right)^m\]where \(n\ge 2\) and \(m \) are natural numbers. Prove that \[\alpha = \frac{k + \sqrt{k^2 -4 }}{2}\]for some natural number \(k\).

Answer 10

Ok. I'll start in about 3 hours

Answer 11

Let \(p(x) = ax^2 + bx + c\) be a polynomial in \(\mathbb R[x]\) such that \(p(\alpha ) \le 1 \) for \(|\alpha | \le 1\). Prove that \(|2a \alpha+b |\le 4 \) for \(|\alpha| < 1 \).

Answer 12

Where are these questions coming from?

Answer 13

From a book.

Answer 14

I'd happily avoid typing out these questions and rather send you a copy, but a PDF doesn't seem to exist anywhere on the web.

Answer 15

Do you work at a university or something?

Answer 16

Let \(p(x)\) be a monic quadratic polynomial over \(\mathbb Z\). Show that for any integer \(n\), there exists an integer \(k\) such that \(p(n) p(n+1) = p(k)\).

Answer 17

Nah, just an average high-school student trying to understand how solving olympiad problems works.

Answer 18

Are you thinking of entering?

Answer 19

I started out too late, so the best I'll do by learning this would be personal satisfaction. Nothing else.

Answer 20

Prove the division algorithm in \(\mathbb R[x]\): given polynomials \(a(x)\) and \(b(x)\) in \(\mathbb R[x]\), there are unique polynomials \(q(x)\) and \(r(x)\) such that \(a(x) = b(x)q(x) + r(x)\) where either \(r(x)\) is the zero polynomial or \(\deg r(x) < \deg b(x)\).

Answer 21

for the 4th one-
\[a=x_{1}+ib_{1}\]\[b=x_{2}+ib_{2}\]
|z| can be 1
so z can take these values- 1,-1, i

we have 3 cases-

1)z=1
\[1=|1+x_{1}+ib_{1}+x_{2}+ib_{2}|\]\[1=|(1+x_{1}+x_{2})+(y_{1}+y_{2})i|\]\[1=\sqrt{(1+x_{1}+x_{2})^2+(y_{1}+y_{2})^2}\]
from here we can easily conclude that x1,x2,y1,y2 are 0 and hence a,b are 0

2)z=-1
its similar and we will get this thing at last-
\[1=\sqrt{(1-x_{1}+x_{2})^2+(y_{2}-y_{1})^2}\]
here also u can get the same conclusion

3)z=i

\[1=|1+(x_{1}+iy_{1})i+(x_{2}+y_{2}i)|\]
\[1=\sqrt{(y_{2}+x_{1})^2 +(1+x_{2}-y_{1})^2}\]
here also u can conclude the same

Answer 22

oops there will be one more case of z= -i
u can follow the same conclusion there :)

Answer 23

There are actually infinitely many cases! I guess we'll have to work for the general case...

Answer 24

yep, infinite cases, take any \(\theta\) on the unit circle of complex plane

Answer 25

But still, that was the correct initiation.

Answer 26

that jst slipped outta my mind sry
yeah we must consider this thing-
\[Z=\sqrt{(x_{1}^2+y_{1}^2)}e^{i \theta}\]

so
\[1=|1+(\sqrt{(x_{1}^2+y_{1}^2)}e^{i \theta})i +\sqrt{(x_{2}^2+y_{2}^2)}e^{i \theta}|\]\[1=|1+(\sqrt{(x_{1}^2+y_{1}^2)}e^{i \theta})i +\sqrt{(x_{2}^2+y_{2}^2)}e^{i \theta}|\]
now what:/

Answer 27

Find the value of the positive integer \(n\) for which the quadratic equation
\[\sum_{k=1}^{n} (x+k-1)(x+k) = 10n\]has solutions \(\alpha\) and \(\alpha+1\) for some \(\alpha\).

Solution :
Since \(\alpha\) and \(\alpha+1\) are the roots,
\[\sum_{k=1}^{n} (\alpha +k-1)(\alpha +k) = 10n\tag{1}\]
\[\sum_{k=1}^{n} (\alpha+k)(\alpha+k+1) = 10n\tag{2}\]

\((2)-(1) \implies\)
\[\sum_{k=1}^{n} (\alpha+k) = 0\tag{3}\]

\((2)+(1) \implies\)
\[\sum_{k=1}^{n} (\alpha+k)^2 = 10n\tag{4}\]

solving \((3)\) and \((4)\) gives \(n=11\)

Answer 28

Jesus, I didn't even try solving that problem. Thank you so much!

Answer 29

These problems are really nice... each seem to have a clever solution..

Answer 30

Wait, have you solved all of them?

Answer 31

nope, still working on #1

Answer 32

Could I have a hand with geometry?

Answer 33

Um..Just looking at those "equations" gave me a headache.. o.e

Answer 34

umm your very worderous how much do you wanna know your brain is gonna be sooo big dude

Answer 35

I'm gonna give a go at one of those problems.\[z = w\tag {1}\]For \(z^2 +2az+b = 0 \) to have roots that are collinear with those of the first equation, clearly the roots are scalar multiples of \(w\). Let \(\lambda w, -2a - \lambda w\) be the roots. We have to eliminate \(\lambda\). How?!

Answer 36

Or not. Collinearity does not imply that three complex numbers are scalar multiples of each other.

Answer 37

If \(z_1, z_2\) are the roots, then\[w - z_1 = \lambda(z_1 - z_2)\]\[z_1 + z_2 = -2a\]\[z_1z_2=b\]Eliminating \(z_1, z_2, \lambda\) is what we have to do. OK.

Answer 38

I wanna TL;DR this lol which ones are unanswered still?

Answer 39

All except the one ganeshie answered. Also you need the discriminant in this question:
Find all positive integers \(n\) for which the quadratic equation\[a_{n+1}x^2 - 2x \sqrt{\sum_{k=1}^{n+1}a_k^2 }+ \sum_{k=1}^n a_k=0\]has real roots for every choice of real numbers \(a_1, a_2, \cdots, a_{n+1}\)

Answer 40

\[\sum_{k=1}^{n+1}a_k^2 \ge a_{n+1}\sum_{k=1}^n a_k\]Some kinda inequality? I don't know.

Answer 41

exactly... but how do I apply that?

Answer 42

I've never really understood this, but it's supposedly pretty important https://en.wikipedia.org/wiki/Polynomial_remainder_theorem#Proof

Answer 43

yes, but proving the division algorithm itself seems like a huge deal.

Answer 44

http://www.artofproblemsolving.com/wiki/index.php/Power_Mean_Inequality

gives

\[\dfrac{1}{n}\sum_{k=1}^{n}a_k^2 \ge \left(\dfrac{1}{n}\sum_{k=1}^n a_k\right)^2\]

Answer 45

thank you sire

Answer 46

hmm, the sum indices are a little different. how do I handle all that?

Answer 47

@ganeshie8 could you post it again

Answer 48

I think we need to massage it a bit more before applying the inequality..