Sunday, August 7, 2016

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

Indian Statistical Institute B.Math & B.Stat Solved Problems, Vinod Singh ~ Kolkata 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 \( P = \frac{a^2}{(a-c)(a-b)} +\frac{b^2}{(b-a)(b-c)}+\frac{c^2}{(c-a)(c-b)} = \frac{1}{a-b} \big( \frac{a^2}{a-c} - \frac{b^2}{b-c} \big)+\frac{c^2}{(c-a)(c-b)}\) $$$$ \( \implies P = \frac{ab-ca-cb}{(c-a)(c-b)}+ \frac{c^2}{(c-a)(c-b)} = \frac{ab-ca-cb+c^2}{(c-a)(c-b)} = \frac{(c-a)(c-b)}{(c-a)(c-b)} = 1\)$$$$

Tuesday, July 26, 2016

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

Indian Statistical Institute B.Math & B.Stat Solved Problems, Vinod Singh ~ Kolkata 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, \( A = \frac{1}{\pi +2} + \frac{1}{2} - 2 \int_{0}^{\frac{\pi}{2}} \frac{\sin t \cos t}{(1+t)}dt \)$$$$ \( \implies \int_{0}^{\frac{\pi}{2}} \frac{\sin t \cos t}{(1+t)}dt = \frac{1}{2} \bigg( \frac{1}{\pi +2} + \frac{1}{2} -A \bigg) \) $$$$ \( i.e., \) \( \int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{(1+x)}dx = \frac{1}{2} \bigg( \frac{1}{\pi +2} + \frac{1}{2} -A \bigg) \) $$$$

Monday, July 18, 2016

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

Indian Statistical Institute B.Math & B.Stat Solved Problems, Vinod Singh ~ Kolkata Evaluate \( lim_{n \rightarrow \infty} \bigg(\prod_{r=1}^{r=n} (1+\frac{2r-1}{2n}) \bigg)^{\frac{1}{2n}} \) $$$$ Let \( A = \bigg(\prod_{r=1}^{r=n} (1+\frac{2r-1}{2n}) \bigg)^{\frac{1}{2n}} \) $$$$ \( \implies A = \frac{\bigg(\prod_{r=1}^{r=2n} (1+\frac{r}{2n}) \bigg)^{\frac{1}{2n}}}{\bigg(\prod_{r=1}^{r=n} (1+\frac{2r}{2n}) \bigg)^{\frac{1}{2n}}} \) $$$$ \( \implies ln A = \frac{1}{2n} \sum_{r=1}^{r=2n} ln ( 1+\frac{r}{2n})- \frac{1}{2n} \sum_{r=1}^{r=n} ln ( 1+\frac{r}{n})\) $$$$ \( \implies lim_{n \rightarrow \infty} ln A = \frac{1}{2} lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{r=2n} ln ( 1+\frac{r}{2n}) - \frac{1}{2} lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{r=n} ln ( 1+\frac{r}{n}) \) $$$$ \( \implies lim_{n \rightarrow \infty} ln A = \frac{1}{2} \int_{0}^{2} ln (1+ \frac{x}{2}) dx - \frac{1}{2} \int_{0}^{1} ln (1+ x ) dx \) $$$$ \( \implies lim_{n \rightarrow \infty} ln A = ln 4 - 1 - ( ln 2 - \frac{1}{2}) = ln 2 - \frac{1}{2}\)$$$$ \( \implies lim_{n \rightarrow \infty} A = e^{ln 2 - \frac{1}{2}} = \frac{2}{\sqrt{e}} \) $$$$ \( \implies lim_{n \rightarrow \infty} \bigg(\prod_{r=1}^{r=n} (1+\frac{2r-1}{2n}) \bigg)^{\frac{1}{2n}} = \frac{2}{\sqrt{e}} \) $$$$

Friday, May 6, 2016

Thursday, April 14, 2016

Friday, March 18, 2016

Binomial Series

BINOMIAL SERIES, Vinod Singh ~ Kolkata 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, $$$$ \( = \sum_{k=1}^{m} \frac{1}{2^{nk}} = \frac{1}{2^n} \frac{\bigg(1- \big(\frac{1}{2^n}\big)^m \bigg)}{1 - \frac{1}{2^n} } \) $$$$ \( = \frac{1}{2^n} \frac{2^n(2^{nm}-1)}{2^{nm}(2^n-1)} = \frac{2^{nm}-1}{2^{nm}(2^n-1)} \)

Friday, March 11, 2016

Integration

INTEGRATION, Vinod Singh ~ Kolkata Evaluate \(\int \frac{dx}{e^x \big(1+e^{2008x}\big)^{\frac{2007}{2008}}} \) $$$$ \(\int \frac{dx}{e^x \big(1+e^{2008x}\big)^{\frac{2007}{2008}}} = \int \frac{dx}{e^x \big(e^{2008x}(1+e^{-2008x})\big)^{\frac{2007}{2008}}}\) $$$$ \(=\int \frac{dx}{e^x e^{2007x}\big(1+e^{-2008x}\big)^{\frac{2007}{2008}}} = \int \frac{e^{-2008x}dx}{\big(1+e^{-2008x}\big)^{\frac{2007}{2008}}} \) \(= \frac{-1}{2008}\int \frac{d(1+e^{-2008x})}{\big(1+e^{-2008x}\big)^{\frac{2007}{2008}}} = \frac{\frac{-1}{2008}\big(1+e^{-2008x}\big)^{-\frac{2007}{2008}+1}}{-\frac{2007}{2008}+1} + c \)
google.com, pub-6701104685381436, DIRECT, f08c47fec0942fa0