Infinite Series- ISI B.MATH 2006 | Problem - 1

Problem

If $\sum_{n=1}^{\infty} \frac{1}{n^2} =\frac{{\pi}^2}{6}$ then $\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2}$ is equal to

(A) $\frac{{\pi}^2}{24}$ (B) $\frac{{\pi}^2}{8}$ (C) $\frac{{\pi}^2}{6}$ (D) $\frac{{\pi}^2}{3}$

Hint

Try to write the summation as sum of square of reciprocal of odd numbers and even numbers and take the advantage of the infinite sum

Solution

$\sum_{n=1}^{\infty} \frac{1}{n^2} =\frac{{\pi}^2}{6}$

$\Rightarrow \sum_{n=1}^{\infty} \frac{1}{(2n)^2} + \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2}= \frac{{\pi}^2}{6} $

$\Rightarrow \frac{1}{4}\sum_{n=1}^{\infty} \frac{1}{{n^2}} + \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \frac{{\pi}^2}{6} $

we know $\sum_{n=1}^{\infty} \frac{1}{n^2} =\frac{{\pi}^2}{6}$

So from the above equation we get

Hence $\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \frac{{\pi}^2}{6} - \frac{{\pi}^2}{6\cdot4}$

$\Rightarrow \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \frac{{\pi}^2}{8} $

So the correct answer is option B

ISI B.Math objective 2006 problem -2 Number theory (Euler phi function)

PROBLEM

Let $p$ be an odd prime.Then the number of positive integers less than $2p$ and relatively prime to $2p$ is:

(A)$p-2$ (B) $\frac{p+1}{2} $(C) $p-1$(D)$p+1$

SOLUTION

This is a number theoretic problem .We can solve this problem in 2 different methods. Let us see them both one by one

Method -1

Let us look at the prime factorization of $2p$ it is

$2p=2\cdot p $

Note that there are $2p-1$ numbers that are less than $2p$

$2p$ is an even number so it has a common divisor with each of the even numbers that are smaller than $2p$ i.e. the numbers in the following set

${2,4,6,\dots(2p-2)}$

So we can disregard these $(p-1)$ numbers

Next note that $p$ is an odd prime number so the only odd number smaller than $2p$ that can have a common divisor with $2p$ is $p$ so we have to disregard this number too

Taking all these into account we come to conclusion that the no of positive integers less than $2p$ and relatively prime to $2p$ is $(2p-1)-(p-1)-1=p-1$

Method-2

We can also use Euler totient function or phi function to solve this problem

We know that Euler phi function is multiplicative i.e $\phi(m\cdot n)=\phi(m)\cdot \phi(n)$ for any positive integers $m$ and $n$

So We can write $\phi(2 \cdot p)=\phi(2)\phi(p)$

Now $\phi(p)$ is the no of positive integers that are less than $p$ and are relatively prime to to $p$ .

As $p$ is a prime no ,so this number is equal to $p-1$

similarly $\phi(2)=1$

As $\phi$ is multiplicative so we get $\phi(2p)=1\cdot(p-1)=p-1$

ISI B.STAT PAPPER 2018 |SUBJECTIVE

Problem

Let $f$:$\mathbb{R} \rightarrow \mathbb{R}$ be a continous function such that for all$x \in \mathbb{R}$ and all $t\geq 0$

f(x)=f(ktx)
where $k>1$ is a fixed constant

Hint

Case-1


choose any 2 arbitary nos $x,y$ using the functional relationship prove that $f(x)=f(y)$

Case-2


when $x,y$ are of opposite signs then show that $$f(x)=f(\frac{x}{2})=f(\frac{x}{4})\dots$$
use continuity to show that $f(x)=f(0)$

Solution


Let us take any $2$ real nos $x$ and $y$.

Case-1

$x$ and $y$ are of same sign . WLG $0<x<y$

Then$\frac{y}{x}>1$
so there is a no $t\geq 0$ such that
$\frac{y}{x}=k^t$
$f(y)=f(k^tx)=f(x)$ [using$f(x)=f(k^tx)$]

case-2

$x,y$ are of opposite sign. WLG $x<0<y$
Then $f(x)=f(k^tx)$

$\Rightarrow f(k^tx)=f(k^t2\frac{1}{2}x)$


$\Rightarrow f(k^t2\frac{1}{2}x)=f(k^tk^{log_k2}\frac{x}{2})$


$\Rightarrow f(k^tk^{log_k2}\frac{x}{2})=f(k^{t+log_k2}\frac{x}{2})$

$\Rightarrow f(k^{t+log_k2}\frac{x}{2})=f(\frac{x}{2})$


Using this logic repeatedly we get


$f(x)=f(\frac{x}{2})=f(\frac{x}{4})\dots =f(\frac{x}{2^n})$


Now $\frac{x}{2^n}\rightarrow0$ and $f$ is a continous function hence $\lim_{n\to\infty}f(\frac{x}{2^n})=f(0)$.


[Because we know if $f$ is a continous function and $x_n$ is a sequence that converges to $x$ then $\lim_{n\to\infty}f(x_n)=f(x)$]


using similar logic we can show that $f(y)=f(0)$ so $f(x)=f(y)$ for any $x,y\in \mathbb{R}$


I.S.I B.STAT 2018 | SUBJECTIVE -4

PROBLEM

Let $f (0,\infty)\rightarrow \mathbb{R}$ be a continous function such that for all $x \in (0,\infty)$ $f(x)=f(3x)$ Define $g(x)= \int_{x}^{3x} \frac{f(t)}{t}dt$ for $x \in (0,\infty)$ is a constant function

HINT

Use leibniz rule for differentiation under integral sign

SOLUTION

using leibniz rule for differentiation under integral sign we get
$g'(x)=f(3x)-f(x)$

$\Rightarrow g'(x)=0$ [ Because f(3x)=f(x)]
Since the derivative of $g(x)$ is $0$ for all $x$, Hence $g(x)$ is a constant function

TESTING THE CONCEPT OF COPRIME NUMBERS | CMI 2015 PART B PROBLEM-3

PROBLEM

Show that there are exactly $2$ numbers $a$ in the set $\{1,2,3\dots9400\}$ such that $a^2-a$ is divisible by $10000$

HINT

Use Modular arithmetic and concepts of coprime numbers

SOLUTION

we know

$10000=2^4*5^4$

In order for $10000$ to divide $a^2-a$ both $2^4$ and $5^4$ must divide $ a^2-a $

Write $a^2-a=a(a-1)$

Note that $a$ and $a-1$ are coprime numbers so they can not have any common divisors

That means either

$2^4$ divides $a$ and $5^4$ divides $(a-1)$

or

$2^4$ divides $a-1$ and $5^4$ divides $a$

Case-1

$2^4$ divides $a$ and $5^4$ divides $(a-1)$

AS $5^4=625$ divides $a-1$ we can write

$a-1=625k$ for some integer $k$

$\Rightarrow a=625k+1$

Now $2^4=16$ divides $a=625k+1$

Note that $16$ divides $624$ hence $16$ will also divide $624k$

We can write $625k+1=624k+k+1$

As $16$ divides $625k+1=624k+k+1$ so it must divide $k+1$

$\Rightarrow k$ is among the numbers $T=\{15,31,47\dots\}$

For $k=15$ we get $a=625*15+1=9376$

For $k=31$ the value of $a$ is $19376>9400$

So from $31$ onward none of the elements of $T$ is acceptable

So in this case the only possible value of $a$ is $9376$

Case-2

$2^4$ divides $a-1$ and $5^4$ divides $a$

As $5^4=625$ divides $a$ so $a=625k$ for some integer $k$

$\Rightarrow a-1=625k-1$

Now $2^4=16$ divides $a-1$

Write $625k-1=624k+k-1$

As in the last case we can argue that $16$ divides $624$ so it must divide $k-1$

So $K$ must be from the set $U=\{1,17,33\dots\}$

For $k=17 , a=10625>9400$ so any value of $k$ greater or equal $17$ does not count

So the only possible value that $k$ can take in this case is $k=1$

So $a=625$

So $625$ & $9376$ these are the only $2$ values that $a$ can take