Number of solutions

December 10, 2018 / Leave a comment

Find the number of solutions of the equation:

    \[x_1+x_2+\cdots + x_{100} = 2017\]

in the positive integers.

Solution

We set y_n=x_n-1 \;\;, \;\; 1\leq n \leq 100. It suffices to find the number of solutions of the equation

(1)   \begin{equation*} y_1+y_2+ \cdots + y_n =1917 \end{equation*}

in the non negative numbers. We represent each sum z_1+\cdots+z_m of non negative integers with a sequence of z_1 dots (\bullet) followed by a vertical bar (\big \mid), after z_2 dots another one vertical bar etc, till we place the last z_m dots ( without the vertical bar at the end.) For example the sum 5+2+0+1=8 can be represented as

    \[\bullet \bullet \bullet \bullet \bullet \mid \bullet \bullet \mid \mid \bullet\]

 

We note that every solution of (1) matches a sequence that has 1917 dots in total and 99 vertical bars. Conversely, every such sequence matches a solution of (1).

Thus, in total there are

    \[\binom{1917+99}{99} = \binom{2016}{99}\]

solutions.

Comment: In general the equation

    \[x_1+x_2+\cdots+x_m=n\]

has \displaystyle \binom{n-1}{m-1} solutions in the positive integers and \displaystyle \binom{n+m-1}{m-1} solutions in the non negative integers.

 

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Eulerian equality

November 26, 2018 / Leave a comment

We know that there are infinite Pythagorian triplets, that is numbers a, b, c such that

(1)   \begin{equation*} a^2 = b^2 +c^2 \end{equation*}

Let us investigate if there exist triplets such that

(2)   \begin{equation*} \varphi \left( a^2 \right) = \varphi \left( b^2 \right) + \varphi \left( c^2 \right) \end{equation*}

where \varphi denotes the Euler’s totient function.

Solution

Indeed, there are infinite triplets such that (2) is satisfied. For example noticing that

    \[\varphi \left(4^2 \right) + \varphi \left(6^2 \right) = 8 + 12 = 20 = \varphi \left(5^2 \right)\]

we deduce that for each natural N such that (N, 30) =1 we have

    \[\varphi \left((4N)^2 \right) + \varphi \left((6N)^2 \right) = 20\varphi \left(N^2\right) = \varphi \left((5N)^2 \right)\]

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A logarithmic inequality

November 22, 2018 / Leave a comment

Let x_1, x_2, \dots, x_n be n \geq 2 positive numbers other than 1 such that x_1^2+x_2^2+\cdots +x_n^2=n^3. Prove that

    \[\frac{\log_{x_1}^4 x_2}{x_1+x_2}+ \frac{\log_{x_2}^4 x_3}{x_2+x_3}+ \cdots + \frac{\log_{x_n}^4 x_1}{x_n+x_1} \geq \frac{1}{2}\]

Solution

The Engels form of the Cauchy – Schwartz inequality gives us:

    \begin{align*} \sum \frac{\log_{x_1}^4 x_2}{x_1+x_2} & \geq \frac{\left (\sum \log_{x_1}^2 x_2 \right )^2}{\sum (x_1+x_2)} \\ &= \frac{\left ( \sum \log_{x_1}^2 x_2 \right )^2}{2\sum x_1} \\ &\!\!\!\!\!\!\overset{\text{AM-GM}}{\geq } \frac{\left [ n \left (\prod \log_{x_1} x_2 \right )^{2/n} \right ]^2}{2\sum x_1} \\ &\!\!\!\!\overset{\text{C-B-S}}{\geq } \frac{n^2}{2n^2} \\ &= \frac{1}{2} \end{align*}

and the inequality is proven.

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A series limit

November 17, 2018 / Leave a comment

Evaluate the limit:

    \[\ell = \lim_{x \rightarrow +\infty}\left(\sum_{n=1}^\infty\left(\frac xn\right)^n\right)^{1/x}\]

Solution

It is quite known that

    \[\left(1+\frac{1}{n} \right)^n \leq e \leq\left(1+\frac{1}{n} \right)^{n+1}\]

Thus,

    \begin{align*} \frac{x}{(n+1)e} &\leq \frac{x}{n+1}\left(\frac{n}{n+1}\right)^n \\ &=\frac{\left(\frac x{n+1}\right)^{n+1}}{\left(\frac xn\right)^n} \\ &=\frac{x}{n}\left(\frac{n}{n+1}\right)^{n+1}\\ &\leq \frac{x}{ne} \end{align*}

Therefore, by induction , for n \geq 1,

    \[\frac{e}{n!}\left(\frac {x}{e} \right)^n \leq \left(\frac {x}{n} \right)^n \leq \frac{x}{(n-1)!}\left(\frac {x}{e} \right)^{n-1}\]

Summing the last equation we get:

    \[e\left(e^{x/e}-1\right) \leq \sum_{n=1}^\infty\left(\frac xn\right)^n \leq xe^{x/e}\]

and raising the last equation to the 1/x power , we get:

    \[\left(e\left(1-e^{-x/e}\right)\right)^{1/x}e^{1/e}\leq \left(\sum_{n=1}^\infty\left(\frac {x}{n} \right)^n\right)^{1/x}\leq x^{1/x}e^{1/e}\]

Thus, by the squeeze theorem the limit is equal to e^{1/e}.

 

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A limit on Dirichlet function

November 13, 2018 / Leave a comment

Let f:\mathbb{R} \rightarrow \mathbb{R} be the Dirichlet function;

    \[f(x) = \left\{\begin{matrix} 1 & , & x\in \mathbb{Q} \\ 0& , & x\in \mathbb{R} \setminus \mathbb{Q} \end{matrix}\right.\]

Evaluate the limit

    \[\ell = \lim_{n \rightarrow +\infty} \frac{1}{n} \left ( f \left ( 1 \right ) + f \left ( \sqrt{2} \right ) + f \left ( \sqrt{3} \right )+ \cdots + f \left ( \sqrt{n} \right ) \right )\]

We simply note that

    \[0 \leq \frac{f \left ( 1 \right ) + f \left ( \sqrt{2} \right ) + f \left ( \sqrt{3} \right )+ \cdots + f \left ( \sqrt{n} \right )}{n} = \frac{\left \lfloor n \right \rfloor}{n} \leq \frac{1}{\sqrt{n}}\]

and the limit follows to be 0. The reason why

    \[f \left ( 1 \right ) + f \left ( \sqrt{2} \right ) + f \left ( \sqrt{3} \right )+ \cdots + f \left ( \sqrt{n} \right ) = \left \lfloor n \right \rfloor\]

is because \sqrt{m} is rational if-f m is a perfect square.

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