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Posted on: 03/27/2012 at 23:03:14     Posted by maticivan

Sir,
For which values of $$n\in\mathbb{N}_0, f_n$$ is a perfect square?
Where $$f_n=f_{n-1}+f_{n-2},n\ge 2$$ and $$f_0=1, f_1=1$$.

Posted on: 06/05/2012 at 07:06:05     Posted by Dladem

Try this link: Square Fibonacci Numbers.

Posted on: 06/05/2012 at 10:06:54     Posted by maticivan

A question of the generating functions in combinatorial problems2,
is it correct if it would be better when we (can everyone or just the mods from this site?) edit to $$B_{r+1}-A_r$$ and so, isn’t it?

Posted on: 06/18/2012 at 15:06:36     Posted by Stijn C

Also, how about the negative binomialcoefficient:
$$\binom{-2}{n}=(n+1)(-1)^n$$?

Posted on: 06/18/2012 at 15:06:54     Posted by Stijn C

( sth as editing would maybe also be good as hint for a really good site)
"Find the number of ways in which $$10$$ apples can be distributed to $$7$$ students such that the first two students get an odd number of apples each, and everybody else gets an even number of apples.
"
After setting $$a_1=2b_1+1, a_3=2b_3$$ and so, we have to find the ways so we have sum $$4$$ with $$7$$ places.
If we take one number $$2$$ somewhere, we have $$7$$ places for it, then we place twice a one, so $$15$$ ways for that and rest keeps being zero, so we have $$105$$ only for this.
Where do I made a mistake or how with the $$-5$$ over...?

Posted on: 06/18/2012 at 16:06:11     Posted by Stijn C

Hi Stijn - you are right for both things. Thanks for pointing out.

First, there is a mistake in the problem, and I corrected it.

Your other comment was related to editing this site:

It is possible and we are in a desire for people who would like to join. Since you asked - you are in. If you log in, you can edit many things now.

If anyone wants to get moderator privileges for this site - please let me know ("matic" "at" "math.duke.edu") and we’ll discuss. I’d be really happy to give permissions to anyone who is nice, knows some latex, and likes math.

At the moment I don’t want to automatically allow editing for all users since this site is still in the development and is not built on some robust content management system. There was no system that could be easily modified to fit the needs, so the php programs were written from scratch. This means that there may be (many) bugs, and the only way to survive is if I keep things under control until the major bugs are fixed.

The message for now: Please know some latex, create an account, write me an email claiming so, and ask to be a moderator. You’ll see the whole new site with all the instructions on how to create and edit texts, and you will probably start doing so soon.

Posted on: 06/18/2012 at 20:06:02     Posted by maticivan

If $$x,y,z \in \mathbb{C}$$ and $$x+y+z=2$$,$$x^2+y^2+z^2=3$$, $$xyz=4$$ then calculate
$$\frac{1}{xy+z-1}+\frac{1}{yz+x-1}+\frac{1}{xz+y-1}$$

Posted on: 08/16/2012 at 17:08:43     Posted by Kasstriot

Since $$z=2-x-y$$, we have $xy+z-1=xy+2-x-y-1=xy-x-y+1=(x-1)(y-1).$ This motivates the introduction of the substitution: $$a=x-1$$, $$b=y-1$$, $$c=z-1$$. I do not want to ruin the fun you may have with this problem, so I leave you with this hint. After some calculation you will get the final solution $$-\frac29$$, but let me know if you get something different or if you run into some other sort of trouble.

Posted on: 08/22/2012 at 11:08:19     Posted by maticivan

Hi!

Let P(x) polynomial.

I know that P(x)*P(x+1)=P(x^2+x+1).

Marfak

Posted on: 09/24/2012 at 13:09:09     Posted by marfak

opps!!
I’m sorry...
P(x) is the 22th degree.
Marfak

Posted on: 09/24/2012 at 13:09:29     Posted by marfak

The answer to the problem is $$P(1)=2^{11}$$ and we will prove that $$P(x)=(x^2+1)^{11}$$.

Let us first introduce the notation $$R(x)=P(x)P(x+1)$$. Then we have $$R(x)=P(x^2+x+1)$$ hence $R(-x-1)=P\left((-x-1)^2+(-x-1)+1\right)=P(x^2+x+1)=R(x).$ This implies that $$P(x)P(x+1)=P(-x)P(-x-1)$$. By comparing coefficients it is now possible to prove that $$P(x)=P(-x)$$.

We will prove that $$z=i$$ and $$z=-i$$ are the only possible solutions to $$P(z)=0$$. Let $$z_1\not\in\{i,-i\}$$ be a zero of $$P$$.

Let $$s:\mathbb C\to\mathbb C$$ be defined as $s(z)=\left\{\begin{array}{cc} z,& \mbox{ if Re }(z)\geq 0\\ -z,&\mbox{ if Re }(z)< 0.\end{array}\right.$ Consider the sequence $$(z_n)_{n=1}^{\infty}$$ defined by $z_{n+1}=s\left(z_n^2+z_n+1\right) \mbox{ for }n\geq 1.$ The sequence $$(z_n)_{n=1}^{\infty}$$ is the sequence of zeroes of $$P$$ hence it is periodic.

Let us define $$a_n=\mbox{Re }(z_n)$$ and $$b_n=\mbox{Im }(z_n)$$ to be real and imaginary parts of $$(z_n)_{n=1}^{\infty}$$. Then we have \begin{eqnarray*} a_{n+1}&=&\pm\left(a_n^2+a_n+1-b_n^2\right)\\ b_{n+1}&=&\pm b_n\cdot(2a_n+1). \end{eqnarray*} The sign $$+$$ or $$-$$ is chosen according to the value $$s(a_{n+1}+ib_{n+1})$$.

Consider the function $$f:\mathbb R_0^+\times\mathbb R\to\mathbb R$$ defined as $$f(a,b)=2a^2+b^2$$. Simple algebraic manipulations yield: \begin{eqnarray*} f(a_{n+1},b_{n+1})-f(a_n,b_n)&=& 2\left((a_n^2+a_n)+(1-b_n^2)\right)^2+4b_n^2\cdot (a_n^2+a_n)-2a_n^2 \\ &=&2\left( (a_n^2+a_n)^2+(b_n^2-1)^2+2(a_n^2+a_n)(1-b_n^2)+2b_n^2(a_n^2+a_n)-a_n^2\right)\\ &=&2\left( (a_n^2+a_n)^2+(b_n^2-1)^2+a_n\right)\geq0. \end{eqnarray*}

The equality holds if and only if $$a_n=0$$ and $$b_n\in\{1,-1\}$$.

Assume that $$z_1\in\{i,-i\}$$. Then the sequence $$(z_n)_{n=1}^{\infty}$$ is constant, hence periodic.

If, however, $$z_1\not\in\{i,-i\}$$ we can first prove by induction that $$(a_n,b_n)\not\in\{(0,1),(0,-1)\}$$ implies that $$(a_{n+1},b_{n+1})\not\in\{(0,1),(0,-1)\}$$. (Indeed, if $$|b_n|> 1$$ then $$|b_{n+1}|=|b_n|\cdot( 2a_n+1)> 1$$, hence $$|b_n|=1$$. This implies that $$a_n^2+a_n=0$$ which is possible only for $$a_n=0$$.)

Thus, the value $$f(a_n,b_n)$$ is strictly increasing and the sequence $$(a_n,b_n)_{n=1}^{\infty}$$ can’t be periodic.

Thus the only zeroes of $$P$$ are $$\pm i$$ and $$P(x)=(x^2+1)^{11}$$ since zeroes $$z$$ and $$-z$$ come in pairs.

Posted on: 09/26/2012 at 18:09:16     Posted by maticivan

Find all functions $$f:R\to R$$ such that $$f(xy)=f(x)f(y)$$ and $$f(x)^3f((x))=1$$

Posted on: 12/29/2012 at 22:12:18     Posted by spiralboy

The second condition seems wrong. It should probably be: $$f(x)^3f(f(x))=1$$. If that is the case, then this equation implies that $$f(x)$$ is never equal to $$0$$. Then we need to put $$y=0$$ in the first equation. It becomes $$f(0)=f(x)f(0)$$. We are now allowed to cancel by $$f(0)$$ and conclude that $$f(x)=1$$ for all $$x\in R$$.

Posted on: 12/31/2012 at 14:12:14     Posted by math11

I am not able to understand the answer 1 explanation from the link. Can anybody help in understandnig the explaination
http://www.imomath.com/index.php?options=238&lmm=1

q: In how many ways can we distribute 15 identical apples to 4 distinct students. Not all students have to get an apple.
a: Let S be the set of all distributions of 15 apples to 4 students. We can represent S as the collection of all ordered sequences of length 4 whose components add up to 15. For example, the sequence (7,0,4,4) means that the first student gets 7 apples, the second gets 0, the third and fourth get 4 apples each. We may now write:
S={(a,b,c,d):a,b,c,d∈N 0 ,a+b+c+d=15}.
Consider the set T of all sequences of length 18 that consist of 15 zeroes and 3 ones. For each (a,b,c,d)∈S we define f(a,b,c,d) in the following way:
f(a,b,c,d)=00…0        a 100…0        b 100…0        c 100…0        d .
Then f:S→T is a bijection and this is very easy to verify. Therefore |S|=|T| . However, it is easy to find the number of elements of T . This is the same problem as choosing 3 elements from the set of 18 zeroes and turning them into ones. This can be done in (18 3 ) ways. Therefore there are (18 3 ) ways to distribute 15 apples to 4 students

Posted on: 05/06/2013 at 06:05:17     Posted by MAJSATHISH

@MAJSATHISH

The solution you were referring has two things to accomplish: To introduce the set-theoretic notation to studying combinatorics, and to give an idea for solving the problem.

Let me first describe the idea for solving the problem, and then I’ll write more about the precise notation and language used in the solution.

Instead of distributing $$15$$ apples to $$4$$ students, we can think of distributing apples to students in another way: We first put $$15$$ apples in a row, and after that place $$3$$ dividers among these apples. Once the dividers are placed, we take all the apples before the first divider and give them to the first student. Then we take the apples between the first and second divider and give them to the second student. The apples between the second and third divider go to the third student, and the fourth student takes all the apples after the third divider.

The total number of distributions of apples to students is the same as the number of distributions of 15 apples and 3 dividers in a row, and that number is $$\binom{18}3$$.

The solution presented in the text uses the set theoretic notation. It first transforms the problem from counting distributions of apples to students into counting the elements of a set. The set in question is the set of all ordered sequences of length 4 whose elements add up to $$15$$.

You can think of solving the problem as a procedure consisting of the following steps:

• Step 1. Always start by asking the question: "What is the most basic way to solve the problem if I have infinite amount of time?"

The answer is: list down all the distributions of apples to students.

• Step 2. Let us try to list all the distributions of apples to students. Here they are:

$$(15,0,0,0)$$, $$(14,1,0,0)$$, $$(14,0,1,0)$$, $$(14,0,0,1)$$, $$(13,2,0,0)$$, $$(13,0,2,0)$$, $$(13,0,0,2)$$, $$(13,1,1,0)$$, $$(13,1,0,1)$$, $$(13,0,1,1)$$, $$\dots$$.

• Step 3. The things in the previous line are called ordered sequences, and we want to count all of them whose elements add up to $$15$$. We first want to describe them in the language of set theory. You may want to read the first two articles about Sets and Functions. Each ordered sequence of length 4 is of the form $$(a,b,c,d)$$, and the ones we are interested in form the set $$S$$ - the set of sequences whose elements add up to $$15$$. Using set theoretic notation we describe the set $$S$$ as: $S=\{(a,b,c,d):a+b+c+d=15\}.$

• Step 4. In step 3 we did not do anything smart. We just re-formulated the problem. Now we need to count the elements of $$S$$ instead of counting the distributions of apples to students. Instead of counting elements of $$S$$ we will count elements of another set $$T$$. The set $$T$$ is the set of all sequences of $$1$$’s and $$0$$’s, where $$1$$ represents divider, and $$0$$ represents an apple. The solution also goes on and explains a formal bijection between the set $$S$$ and the set $$T$$.

Posted on: 05/07/2013 at 15:05:10     Posted by maticivan

2 circles can have -
0 tangents in common
only 1 tangent in common
only 2 tangents in common
only 3 tangents in common
all the above
not able to understand when 2 tangents common

Posted on: 01/17/2014 at 14:01:52     Posted by awan bhati

If two circles intersect, then they have exactly two tangents in common.

Posted on: 01/22/2014 at 22:01:22     Posted by anish0202

Problem:

Let $$b=\frac{2^{3^{4^5}}}{e^n}$$, otherwise stated as "two to the power of (three to the power of (4 to the power of 5)) over e to the nth power", where n is an integer such that
$$1 < b < e$$.
Find b, with an accuracy of 10 decimal digits.

Posted on: 09/01/2014 at 17:09:14     Posted by magicworks

use these numbers only...you can use it multiple times.

if 1+3+5+7+11
then fill the blanks to match the answer

_ + _ + _ + _ + _ = 30

Posted on: 09/17/2014 at 18:09:35     Posted by scmraj

@magicworks:

The required inequality is equivalent to $$\ln 1< \ln b< \ln e$$, i.e. $$0< \ln b< 1$$. The number $$\displaystyle \ln b=\ln\frac{2^{3^{4^{5}}}}{e^n}=\ln 2^{3^{4^{5}}}-n$$ can be expressed as $\ln b=\ln 2^{3^{1024}}-n=3^{1024}\ln 2-n.$ What we need to find is the first 10 digits of the fractional part of the number $$3^{1024}\ln 2$$, or, in other words, the first ten digits of $$\left\{3^{1024}\ln 2\right\}$$. Using wolframalpha.com you get that these 10 digits are 4261674320.

Posted on: 09/19/2014 at 19:09:52     Posted by maticivan

@scmraj

If I understood your problem correctly, you are trying to pick 5 numbers from the set $$\{1,3,5,7,11\}$$ that add up to $$30$$. Since each number from the given set is odd, any sum of 5 such numbers will be odd as well and as such it can never be equal to $$30$$.

Posted on: 09/19/2014 at 19:09:52     Posted by maticivan

find lim((ln(x+1)-x)/x^2 when x → 0

Posted on: 11/17/2014 at 09:11:32     Posted by fodili

Find all functions $$f(R)$$->$$R$$ such that $$f(x+y)+xy=f(x)f(y)$$ for all real $$x,y$$.

Posted on: 11/18/2014 at 07:11:57     Posted by Neelabh Deka

@fodili

The limit $$\displaystyle \lim_{x\to 0}\frac{\ln(1+x)-x}{x^2}$$ can be evaluated by using L‘Hopital‘s theorem twice, or by using Taylor‘s expansion of the function $$f(h)=\ln(1+h)$$.

Here is the solution using L‘Hopital‘s theorem. Denote by $$A(x)=\ln(1+x)-x$$ and $$B(x)=x^2$$. The function $$A(x)$$ and $$B(x)$$ are differentiable and $$\displaystyle \lim_{x\to 0}A(x)=\lim_{x\to 0}B(x)=0$$. According to L‘Hopital‘s theorem, if the limit $$\displaystyle \lim_{x\to 0}\frac{A^{\prime}(x)}{B^{\prime}(x)}$$ exists and is equal to $$L$$ then the limit $$\displaystyle \lim_{x\to 0}\frac{A(x)}{B(x)}$$ also exists and is equal to the same $$L$$.

We will now prove using L‘Hopital‘s theorem that $$\displaystyle \lim_{x\to 0}\frac{A^{\prime}(x)}{B^{\prime}(x)}$$ exists and is equal to $$-\frac12$$. We have that $$\displaystyle A^{\prime}(x)=\frac1{1+x}-1$$ and $$\displaystyle B^{\prime}(x)=2x$$. Let us denote $$C(x)=\frac1{1+x}-1$$ and $$D(x)=2x$$. Then we have $$\displaystyle\lim_{x\to0}C(x)=\lim_{x\to0}D(x)=0$$. We also have $$\displaystyle C^{\prime}(x)=-\frac1{(1+x)^2}$$ and $$D^{\prime}(x)=2$$. Therefore $$\lim_{x\to 0}\frac{C^{\prime}(x)}{D^{\prime}(x)}=-\frac12$$ and the according to the L‘Hopital‘s theorem we have $$\displaystyle \lim_{x\to0}\frac{C(x)}{D(x)}=-\frac12$$.

Thus $$\displaystyle \lim_{x\to 0}\frac{\ln(1+x)-x}{x^2}=-\frac12$$.

Posted on: 11/25/2014 at 14:11:28     Posted by maticivan

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