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 $\mathbb{Z^+}$

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

Let $z$ be a non-zero complex number such that \( |z −\frac{1}{z}| = 2.\) What is the maximum value of $|z|$? $$$$ Given \( 2 = |z −\frac{1}{z}| \geq \big||z|- |\frac{1}{z}|\big| \) $$$$ Let $t = |z|$ \( \implies \big|t- \frac{1}{t}\big| \leq 2 \) $$$$ \( \implies -2 \leq t- \frac{1}{t} \leq 2 \) $$$$ \( \implies -2t \leq t^2- 1 \leq 2t \) $$$$ \( \implies t^2 +2t- 1 \geq 0 \quad and \quad t^2 -2t- 1 \leq 0 \) $$$$ The first inequality gives \( t \in ( - \infty, -1-\sqrt{2}] \cup [\sqrt{2}-1, \infty)\). Since $t \geq 0$ \(\implies t \in [\sqrt{2}-1, \infty) \). $$$$ The second inequality gives \( t \in [1-\sqrt{2} , 1+\sqrt{2}] \). Again since $t \geq 0$ \(\implies t \in [0,1+\sqrt{2}] \) $$$$ Combining the two case we see \( t \in [\sqrt{2}-1,\sqrt{2}+1] \implies |z| \in [\sqrt{2}-1,\sqrt{2}+1] \). Thus the maximum value of $|z|$ is $\sqrt{2}+1$. $$$$ $Practice-Problem$ Let $z$ be a non-zero complex number such that \( |z +\frac{1}{z}| = a, a \in \mathbb{R^+}.\) What is the maximum and minimum value of $|z|$? $$$$

Indian Statistical Institute ( ISI ) B.Math & B.Stat : Co-ordinate Geometry

Let $A$ be the set of all points $(h, k)$ such that the area of the triangle formed by \((h, k), (5, 6)\) and \((3, 2)\) is $12$ square units. What is the least possible length of a line segment joining $(0, 0)$ to a point in $A$? $$$$ Take the base of the triangle to be the line segment obtained by joining the points $(5,6)$ and $(3,2)$. Equation of the base is \( 2x-y-4 =0 \). Length of the base is \( \sqrt{(5-3)^2+(6-2)^2} = 2 \sqrt{5}. \) Let $p$ be the length of the perpendicular from the point $(h,k)$ onto the base. ( Note that the point $(h,k)$ cannot lie on the base. Why?) Since the area is given to be $12$, \( 12 = \frac{1}{2} \times p \times 2 \sqrt{5} \implies p = \frac{12}{\sqrt{5}}\). Therefore the point $(h,k)$ lies at a distance of $\frac{12}{\sqrt{5}}$ units from the base on both sides. Thus $A$ is the set of all points on the line $parallel$ to the base and at a distance $\frac{12}{\sqrt{5}}$ units away from the base. In the diagram, the lines colored green represents the set $A$. Clearly the least possible length of a line segment joining $(0, 0)$ to a point in $A$? is the distance between the point $(0,0)$ and the line drawn parallel to the base and to the left side of the base. Let $XY$ be the line segment perpendicular to the base and the line and passing through the orgin as shown in the diagram. Required distance is $OX$ and \( OX = XY - OY = \frac{12}{\sqrt{5}} - \big|\frac{-4}{\sqrt{2^2+(-1)^2}}\big| = \frac{12}{\sqrt{5}} - \frac{4}{\sqrt{5}} = \frac{8}{\sqrt{5}} \)

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

Find the ratio of the areas of the regular pentagons inscribed and circumscribed around a given circle. $$$$ Let $a$ be the side of the circumscribed pentagon and $b$ be that of the inscribed pentagon. $$$$ First note that for the circle is inscribed for the exterior pentagon and circumscribed for the interior pentagon. Therefore the $in-radius$ of the exterior polygon, say $r$ is equal to the $circum-radius$, say $R$ of the interior pentagon, i.e., $R=r$. See the figure below. Using standard formula, \[ a = 2 r \tan \frac{\pi}{5}, \quad b = 2 R \sin \frac{\pi}{5} \]. Area of a regular polygon having $n$ sides is \( n \times \frac {(side)^2}{4} \cot \frac{\pi}{n} \). $$$$ Therefore the required ratio is \( \bigg( \frac{5 \times \frac {b^2}{4} \cot \frac{\pi}{5}}{5 \times \frac {a^2}{4} \cot \frac{\pi}{5}} \bigg) = \frac{b^2}{a^2} = \frac{(2 R \sin \frac{\pi}{5})^2} {(2 r \tan \frac{\pi}{5})^2} = \cos^2 \frac{\pi}{5}\)

Indian Statistical Institute B.Math & B.Stat : Trigonometry

Let \( \theta_1 = \frac{2 \pi}{3}, \theta_2 = \frac{4 \pi}{7}, \theta_3 = \frac{7 \pi}{3} \). Then show that \( ( \sin \theta_1)^{ \sin \theta_1} < ( \sin \theta_3)^{ \sin \theta_3} < ( \sin \theta_2)^{ \sin \theta_2} \). $$$$ First note that \( \pi > \theta_1 > \theta_3 > \theta_2 > 0\) and all of them belong to the $second$ quadrant. $Sine$ function strictly decreases from $1$ to $0$ in the $second$ quadrant. Also \( \sin \theta_1 \neq \sin \theta_2 \neq \sin \theta_3 \neq 0 \) and each of them are posititve. $$$$ Using the strictly decreasing property of $Sine$ in the second quadrant we have \( \sin \theta_1 < \sin \theta_3 < \sin \theta_2 \). Now the result follows the standard inequality \( x^c < y^d \) for \( x,y,c,d > 0 \quad where \quad x < y, \quad c < d \).

Indian Statistical Institute B.Math & B.Stat : Trigonometry

Find the value of the sum \( \cos \frac{2 \pi}{1000} + \cos \frac{4 \pi}{1000} + \dots + \cos \frac{1998 \pi}{1000} \). $$$$ Let \( z = \cos \frac{\pi}{1000} + i \sin \frac{\pi}{1000} = \cos \theta + i \sin \theta \) where $ \theta = \frac{\pi}{1000}$ . It is easy to see that $ z \neq 1,-1$. $$$$ Consider the sum \( 1 +z^2+z^4+ \dots + z^{1998} \), $ z \neq 1,-1$. Putting $w = z^2$ the sum reduces to \( 1 +w+w^2+ \dots + w^{999} \), $ w \neq 1 $. $$$$ Now, \( 1 +w+w^2+ \dots + w^{999} = \frac{w^{1000}-1}{w-1}\) $$$$ Substituting back $w$ we have the following identity \( 1 +z^2+z^4+ \dots + z^{1998} = \frac{z^{2000}-1}{z^2-1}\),$ z \neq 1,-1$. $$$$ Using $De-Moivre's$ theorem we have \( z^n = \cos n \theta + i \sin n \theta \) for \( n \in \mathbb{N} \). $$$$ Substituting back in the above identity we have, \( \big(1+ \cos 2 \theta + \cos 4 \theta + \dots + \cos 1998 \theta \big) + i \big(1+ \sin 2 \theta + \sin 4 \theta + \dots + \sin 1998 \theta \big) = \frac{\cos 2000 \theta + i \sin 2000 \theta -1}{\cos 2 \theta + i \sin 2 \theta -1} \) $$$$ Equating the real part from both side we have. \( 1+ \cos 2 \theta + \cos 4 \theta + \dots + \cos 1998 \theta = Re \bigg( \frac{\cos 2000 \theta + i \sin 2000 \theta -1}{\cos 2 \theta + i \sin 2 \theta -1} \bigg) = Re \bigg( \frac{\cos 2 \pi + i \sin 2 \pi -1 }{\cos 2 \theta + i \sin 2 \theta -1}\bigg) \), since $ \theta = \frac{\pi}{1000}$. $$$$ Therefore \( 1+ \cos 2 \theta + \cos 4 \theta + \dots + \cos 1998 \theta = Re (0) = 0 \implies \cos 2 \theta + \cos 4 \theta + \dots + \cos 1998 \theta = -1\). $$$$ \( \implies \cos \frac{2 \pi}{1000} + \cos \frac{4 \pi}{1000} + \dots + \cos \frac{1998 \pi}{1000} = -1 \)

Indian Statistical Institute B.Math & B.Stat : Quadratic Equations

Consider the function \(f(x) = ax^3 + bx^2 + cx + d\), where $a, b, c$ and $d$ are real numbers with $a > 0$. If $f$ is strictly increasing, then show that the function \( g(x) =f′(x) − f′′(x) + f′′′(x) \) is positive for all \( x \in \mathbb{R} \). $$$$ First we calculate the derivatives up to the third order. \(f'(x) = 3ax^2+2bx+c, \quad f''(x) = 6ax+2b \quad and \quad f'''(x) = 6a \). $$$$ It is given that $f$ is strictly increasing which implies \( f' > 0 \) which in turn implies \( 3ax^2+2bx+c > 0\). $$$$ Let \(y = 3ax^2+2bx+c \) It is easy to see that \( y = 3a \big( x + \frac{b}{3a} \big)^2 + \frac{3ac-b^2}{3a} \). Since $y > 0$ and $a$ is given to be positive $3ac$ must be strictly greater than $b^2$. Note $\big( x + \frac{b}{3a} \big)^2$ is always non-negative. $$$$ Now \( g(x) = f′(x) − f′′(x) + f′′′(x) = 3ax^2+2bx+c -(6ax+2b) + 6a = 3ax^2 + 2x(b-3a)+(c-2b+6a) \) \( = 3a\big( x^2 + 2 x \frac{b-3a}{3a} + \frac{(b-3a)^2}{9a^2}+ \frac{(c-2b+6a)}{3a} - \frac{(b-3a)^2}{9a^2} \big) \) = \( 3a\bigg( x^2 + 2 x \frac{b-3a}{3a} + \frac{(b-3a)^2}{9a^2} \bigg) + 3a \bigg( \frac{(c-2b+6a)}{3a} - \frac{(b-3a)^2}{9a^2}\bigg) \) \( = 3a \big( x + \frac{b-3a}{3a} \big)^2 + \frac{9a^2+3ac-b^2}{3a}\) $$$$ \( 3a \big( x + \frac{b-3a}{3a} \big)^2 \geq 0 \) for all \( x \in \mathbb{R} \). (since $a$ is given to be positive) We have already shown that $3ac > b^2$ therefore \(\frac{9a^2+3ac-b^2}{3a} > 0 \). Thus \( g(x) > 0 \) for all \( x \in \mathbb{R} \).