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Indian Statistical Institute ( ISI ) B.Math & B.Stat : Algebra

Let $a, b$ and $c$ be such that $a+b+c=0$ and \( P =\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}\) is defined. What is the value of $P$? $$$$ $Solution: $ \( 2a^2+bc = a^2 + a^2 -(a+c)c\), since $b=-(a+c)$. $$$$ \( \implies 2a^2+bc = a^2 + a^2 -ac-c^2 = (a-c)(a+c)+a(a-c) = (a-c)(2a+c)=(a-c)(2a-a-b)=(a-c)(a-b)\) $$$$ Therefore \( \frac{a^2}{2a^2+bc} = \frac{a^2}{(a-c)(a-b)}\) $$$$ Similarly, \( \frac{b^2}{2b^2+ac} = \frac{b^2}{(b-a)(b-c)} \) and \( \frac{c^2}{2c^2+ab} = \frac{c^2}{(c-a)(c-b)} \) $$$$ Therefore \( P = \frac{a^2}{(a-c)(a-b)} +\frac{b^2}{(b-a)(b-c)}+\frac{c^2}{(c-a)(c-b)} = \frac{1}{a-b} \big( \frac{a^2}{a-c} - \frac{b^2}{b-c} \big)+\frac{c^2}{(c-a)(c-b)}\) $$$$ \( \implies P = \frac{ab-ca-cb}{(c-a)(c-b)}+ \frac{c^2}{(c-a)(c-b)} = \frac{ab-ca-cb+c^2}{(c-a)(c-b)} = \frac{(c-a)(c-b)}{(c-a)(c-b)} = 1\)$$$$

Indian Statistical Institute ( ISI ) B.Math & B.Stat : Calculus

If \( A = \int_{0}^{\pi} \frac{cos x}{(x+2)^2}dx\), then show that \( \int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{(x+1)}dx = \frac{1}{2} \big( \frac{1}{2} +\frac{1}{\pi+2} - A\big)\). $$$$ Solution: \( A = \cos x \int_{0}^{\pi} \frac{dx}{(x+2)^2} - \int_{0}^{\pi} \bigg( \frac{d}{dx} \cos x \int \frac{dx}{(x+2)^2} \bigg) dx \) $$$$ \( \implies A = \frac{-\cos x}{x+2}\mid_{0}^{\pi} - \int_{0}^{\pi} \frac{\sin x}{(x+2)} dx \) $$$$ \( \implies A = \frac{1}{\pi +2} + \frac{1}{2} - \int_{0}^{\pi} \frac{2 \sin \frac{x}{2} \cos \frac{x}{2} }{2(1+\frac{x}{2})} dx \) $$$$ Putting $\frac{x}{2}=t$ in the last integral, we have $dx=2dt$ and $t$ varies from $0$ to $\frac{\pi}{2}$ $$$$ Therefore, \( A = \frac{1}{\pi +2} + \frac{1}{2} - 2 \int_{0}^{\frac{\pi}{2}} \frac{\sin t \cos t}{(1+t)}dt \)$$$$ \( \implies \int_{0}^{\frac{\pi}{2}} \frac{\sin t \cos t}{(1+t)}dt = \frac{1}{2} \bigg( \frac{1}{\pi +2} + \frac{1}{2} -A \bigg) \) $$$$ \( i.e., \) \( \int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{(1+x)}dx = \frac{1}{2} \bigg( \frac{1}{\pi +2} + \frac{1}{2} -A \bigg) \) $$$$

Indian Statistical Institute ( ISI ) B.Math & B.Stat : Calculus

Evaluate \( lim_{n \rightarrow \infty} \bigg(\prod_{r=1}^{r=n} (1+\frac{2r-1}{2n}) \bigg)^{\frac{1}{2n}} \) $$$$ Let \( A = \bigg(\prod_{r=1}^{r=n} (1+\frac{2r-1}{2n}) \bigg)^{\frac{1}{2n}} \) $$$$ \( \implies A = \frac{\bigg(\prod_{r=1}^{r=2n} (1+\frac{r}{2n}) \bigg)^{\frac{1}{2n}}}{\bigg(\prod_{r=1}^{r=n} (1+\frac{2r}{2n}) \bigg)^{\frac{1}{2n}}} \) $$$$ \( \implies ln A = \frac{1}{2n} \sum_{r=1}^{r=2n} ln ( 1+\frac{r}{2n})- \frac{1}{2n} \sum_{r=1}^{r=n} ln ( 1+\frac{r}{n})\) $$$$ \( \implies lim_{n \rightarrow \infty} ln A = \frac{1}{2} lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{r=2n} ln ( 1+\frac{r}{2n}) - \frac{1}{2} lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{r=n} ln ( 1+\frac{r}{n}) \) $$$$ \( \implies lim_{n \rightarrow \infty} ln A = \frac{1}{2} \int_{0}^{2} ln (1+ \frac{x}{2}) dx - \frac{1}{2} \int_{0}^{1} ln (1+ x ) dx \) $$$$ \( \implies lim_{n \rightarrow \infty} ln A = ln 4 - 1 - ( ln 2 - \frac{1}{2}) = ln 2 - \frac{1}{2}\)$$$$ \( \implies lim_{n \rightarrow \infty} A = e^{ln 2 - \frac{1}{2}} = \frac{2}{\sqrt{e}} \) $$$$ \( \implies lim_{n \rightarrow \infty} \bigg(\prod_{r=1}^{r=n} (1+\frac{2r-1}{2n}) \bigg)^{\frac{1}{2n}} = \frac{2}{\sqrt{e}} \) $$$$

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Binomial Series

Find the sum of the series \( \sum_{r=0}^{n} (-1)^r \binom{n}{r} \big( \frac{1}{2^r}+\frac{3^r}{2^{2r}}+\frac{7^r}{2^{3r}}+ \dots \) to m terms \( \big) \) $$$$ Given series is equal to \( \sum_{r=0}^{n} (-1)^r \binom{n}{r} \sum_{k=1}^{m} \frac{(2^k -1)^r}{2^{kr}} \) $$$$ \( = \sum_{r=0}^{n} \sum_{k=1}^{m} (-1)^r \binom{n}{r} \frac{(2^k -1)^r}{2^{kr}} \) $$$$ \( = \sum_{r=0}^{n} \sum_{k=1}^{m} (-1)^r \binom{n}{r} \lambda^r \) where $\lambda = \frac{(2^k -1)}{2^{k}}$ $$$$ \( = \sum_{k=1}^{m} \sum_{r=0}^{n} (-1)^r \binom{n}{r} \lambda^r \) Interchanging the summation $$$$ \( = \sum_{k=1}^{m} (1- \lambda)^n \), Substituting the value of $\lambda$ we have, $$$$ \( = \sum_{k=1}^{m} \frac{1}{2^{nk}} = \frac{1}{2^n} \frac{\bigg(1- \big(\frac{1}{2^n}\big)^m \bigg)}{1 - \frac{1}{2^n} } \) $$$$ \( = \frac{1}{2^n} \frac{2^n(2^{nm}-1)}{2^{nm}(2^n-1)} = \frac{2^{nm}-1}{2^{nm}(2^n-1)} \)

Integration

Evaluate \(\int \frac{dx}{e^x \big(1+e^{2008x}\big)^{\frac{2007}{2008}}} \) $$$$ \(\int \frac{dx}{e^x \big(1+e^{2008x}\big)^{\frac{2007}{2008}}} = \int \frac{dx}{e^x \big(e^{2008x}(1+e^{-2008x})\big)^{\frac{2007}{2008}}}\) $$$$ \(=\int \frac{dx}{e^x e^{2007x}\big(1+e^{-2008x}\big)^{\frac{2007}{2008}}} = \int \frac{e^{-2008x}dx}{\big(1+e^{-2008x}\big)^{\frac{2007}{2008}}} \) \(= \frac{-1}{2008}\int \frac{d(1+e^{-2008x})}{\big(1+e^{-2008x}\big)^{\frac{2007}{2008}}} = \frac{\frac{-1}{2008}\big(1+e^{-2008x}\big)^{-\frac{2007}{2008}+1}}{-\frac{2007}{2008}+1} + c \)

Indian Statistical Institute ( ISI ) B.Math & B.Stat : Algebra

Let $a, b, c$ be real numbers greater than $1.$ Let $S$ denote the sum $S = log_{a}{bc} + log_{b}{ca} + log_{c}{ab}.$ $$$$ Find the smallest possible value of $S.$ $$$$ \( S =log_{a}{\frac{abc}{a}} + log_{b}{\frac{bca}{b}} + log_{c}{\frac{abc}{c}} = log_{a}{abc} + log_{b}{abc} + log_{c}{abc}-3 \) $$$$ \( = log_{a}{e} \times log_{e}{abc} + log_{b}{e} \times log_{e}{abc} + log_{c}{e} \times log_{e}{abc}-3 \) $$$$ \( = log_{e}{abc} \bigg( \frac{1}{log_{e}{a}} +\frac{1}{log_{e}{b}}+\frac{1}{log_{e}{c}} \bigg) -3 \) $$$$ \( = (log_{e}{a}+log_{e}{b}+log_{e}{c}) \bigg( \frac{1}{log_{e}{a}} +\frac{1}{log_{e}{b}}+\frac{1}{log_{e}{c}} \bigg) -3 \) $$$$ Now Using the inequality $A.M \times H.M \geq n^2 $ for $n$ positive real numbers, we see that $$$$ \( (log_{e}{a}+log_{e}{b}+log_{e}{c}) \bigg( \frac{1}{log_{e}{a}} +\frac{1}{log_{e}{b}}+\frac{1}{log_{e}{c}} \bigg) \geq 3^2 = 9 \) $$$$ Thus \( S \geq 9-3 = 6 \). Note \( log_{e}{a},log_{e}{b},log_{e}{c} \) are all positive since $a,b,c > 1$.

Indian Statistical Institute ( ISI ) B.Math & B.Stat : Algebra

Show that the polynomial $x^8 − x^7 + x^2 − x + 15$ has no real root. $$$$ Let $f(x)=x^8 − x^7 + x^2 − x + 15$, we will show that $f(x) >0$ for all $ x \in \mathbb{R}$. $$$$ \( f(x) = x^7(x-1)+x(x-1)+15 = (x-1)x(x^6+1)+15 \) $$$$ Now, \(\mathbb{R} = (- \infty , 0] \cup (0,1] \cup (1, \infty) \). Note that \( f(0)=f(1)=15\) $$$$ When $ x \in (1, \infty), $ \( x,x-1 \quad and \quad x^6+1 > 0 \implies f(x) > 15 \quad \forall \quad x \in (1, \infty)\) $$$$ When $ x \in (- \infty , 0), $ \( x \quad and \quad x-1 < 0 \quad thus \quad x(x-1) > 0\) since $x^6+1 > 0$ for any $x$, $f(x) > 15$ in this case too. $$$$ When $ x \in (0,1) $, $1 < x^6+1 < 2$ and $ 0 < x < 1 $. Since both of them are positive $ 0 < x(x^6+1) < 2$. Further $ -1 < x-1 < 0 $, thus $x(x-1)(x^6+1) < 0$. Again $|x-1| < 1$ this implies $ -2 < x(x-1)(x^6+1) < 0 $. Thus \(f(x) > 13 > 0\). $$$$ Combining all the cases we see that \( f(x) > 0 \quad \forall \quad x \in \mathbb{R} \) which shows $f(x)$ has no real root.

Indian Statistical Institute (ISI) B.Math & B.Stat : Combinatorics

For $ k \geq 1$, find the value of \[ \binom{n}{0}+ \binom{n+1}{1}+ \binom{n+2}{2}+ \dots + \binom{n+k}{k}\] Using the identity \( \binom{n}{r} = \binom{n}{n-r} \), \( \binom{n}{0}+ \binom{n+1}{1}+ \binom{n+2}{2}+ \dots + \binom{n+k}{k}\) reduces to \[ \binom{n}{n}+ \binom{n+1}{n}+ \binom{n+2}{n}+ \dots + \binom{n+k}{n}\] = Coefficient of $x^n$ in $(1+x)^n$ + Coefficient of $x^n$ in $(1+x)^{n+1}$ + $\dots$ + Coefficient of $x^n$ in $(1+x)^{n+k}$ $$$$ = Coefficient of $x^n$ in \( (1+x)^n + (1+x)^{n+1} + \dots + (1+x)^{n+k} \) $$$$ = Coefficient of $x^n$ in \( (1+x)^n \frac{(1+x)^{k+1}-1}{1+x-1} = \frac{(1+x)^{n+k+1}-(1+x)^n}{x}\) $$$$ = Coefficient of $x^{n+1}$ in $(1+x)^{n+k+1}$ $= \binom{n+k+1}{n+1}$ $$$$

Indian Statistical Institute (ISI) B.Math & B.Stat : Algebra

If \(p,q,r\) are positive real numbers such that $pqr=1$, then find the value of \( \frac{1}{1+p+q^{-1}}+\frac{1}{1+q+r^{-1}}+\frac{1}{1+r+p^{-1}}\). $$$$ Throught the simplification we will use \( 1=pqr,q^{-1}=pr,r^{-1}=pq \quad and \quad p^{-1} = qr \) $$$$ Given expression is \[ \frac{1}{1+p+q^{-1}}+\frac{1}{1+q+r^{-1}}+\frac{1}{1+r+p^{-1}}\] \[= \frac{pqr}{pqr+p+pr}+\frac{pqr}{pqr+q+pq}+\frac{1}{1+r+p^{-1}} \] \[= \frac{qr}{qr+1+r}+\frac{pr}{pr+1+p}+\frac{1}{1+r+p^{-1}} \] \[= \frac{qr}{p^{-1}+1+r}+\frac{pr}{pr+pqr+p}+\frac{1}{1+r+p^{-1}} \] \[= \frac{qr}{p^{-1}+1+r}+\frac{r}{r+qr+1}+\frac{1}{1+r+p^{-1}} \] \[= \frac{qr}{p^{-1}+1+r}+\frac{r}{r+p^{-1}+1}+\frac{1}{1+r+p^{-1}} \] \[= \frac{qr+r+1}{p^{-1}+1+r} \] \[= \frac{p^{-1}+r+1}{p^{-1}+1+r} \] \[= 1\]

Indian Statistical Institute ( ISI ) B.Math & B.Stat : Number Theory

Consider the equation $x^2 + y^2 = 2007$. How many solutions $(x, y)$ exist such that $x$ and $y$ are positive integers? $$$$ \( 2007 = 2000 + 7 \equiv 0 + 3 \equiv 3 (mod \quad 4) \). Now we know that square on an integer is either divisible by $4$ or leaves a remainder $1$ when divided by $4$, said otherwise \( x \in \mathbb{Z} \implies x^2 \equiv 0 \quad or \quad 1 (mod \quad 4) \). Thus for integers $x$ and $y$, \( x^2+y^2 \equiv 0 \quad or \quad 1 \quad or \quad 2 (mod \quad 4) \). Since we have different remainders $mod \quad 4$ on the two sides, it follows there cannot be any solution in $\mathbb{Z}$ hence no solution in $\mathbb{Z^+}$