Tutor profile: Rajat K.

Define Linear Independent & Linear Dependent in vector space.

Linear Independent :-The vectors in a subset S=\left \{ v_{1},v_{2},v_{3},v_{4}.........v_{n} \right \} of a vector space V over a field F are linearly independent over the field F if \sum_{j=1}^{n}a_{j}v_{j}=0,\: a_{j}\epsilon F\: \: holds \: when\: a_{j}=0\: for\: each\: 1\leq j\leq n So the vectors in S are linearly independent over the field F if the unique way the 0-vector can be expressed as a linear combination of the vectors v_{j} is to have all scalar coefficient equal to 0. Linear Dependent : If the vectors are linearly dependent over the field F, then there exists a_{j}\epsilon F for j=1,2,3......n such that \sum_{j=1}^{n}a_{j}v_{j}=0,\: a_{j}\epsilon F\: \: where \:not \: all \: a_{j}=0.

Every sequence has unique limit.

Proof:-Let {Sn} be a sequence of real number & we have to show that, the limit of sequence {Sn} is unique. Now we consider it’s opposite that is L & L’ are two limit of the sequence {Sn}. We know that if L be the limit of the sequence of {Sn}.Then \lim_{n\rightarrow \infty }S_{n}=L\: (for\: every\: \epsilon > 0) There exist a positive integer m such that \left | S_{n} -\: L\right |< \frac{\epsilon }{2}\: \: \forall n\geq m_{1}.......(1) \lim_{n\rightarrow \infty }S_{n}=L^{'}\: (for\: every\: \epsilon > 0) There exist a positive integer m such that \left | S_{n} -\: L^{'}\right |< \frac{\epsilon }{2}\: \: \forall n\geq m_{2}.......(2) If we consider a positive integer m such that m\leq m_{1}m_{2} From eq. (1) & eq.(2) \left | S_{n} -\: L\right |< \frac{\epsilon }{2}\: \: \forall n\geq m_{1}.......(1) \left | S_{n} -\: L^{'}\right |< \frac{\epsilon }{2}\: \: \forall n\geq m_{2}.......(2) If we take m=max\left ( m_{1},m_{2}^{} \right ) \left | S_{n} -\: L\right |< \frac{\epsilon }{2}\: \: \forall n\geq m.......(3) \left | S_{n} -\: L^{'}\right |< \frac{\epsilon }{2}\: \: \forall n\geq m.......(4) \left | L-L^{'} \right |=\left | L-S_{n}+S_{n}-L^{'} \right | \leq \left | L-S_{n} \right |+\left | S_{n}-L^{'} \right | < \frac{\epsilon }{2}+\frac{\epsilon }{2}\: \forall n\geq m \left | L-L^{'} \right |< \epsilon\: \: \forall \: n\geq m Which is contradiction hence every sequence has unique limit. It is only possible when L=L’.

Solve the initial value problem \frac{\mathrm{d}y }{\mathrm{d} x}=\frac{x^{2}}{y} , y(1)=3

Solution \frac{\mathrm{d}y }{\mathrm{d} x}=\frac{x^{2}}{y} Step 1:- We saw that this is differential equation with variable separable, so put y variable with dy & x variable with dx. Then y.dy=x^{2}.dx Step 2:- Now integrate on both side. \int y.dy=\int x^{2}.dx \frac{y^{2}}{2}=\frac{x^{3}}{3}+C............(1) Step 3:- Use the given initial value to omit the constant, so put x=1 , y = 3 \frac{3^{2}}{2}=\frac{1^{3}}{3}+C \frac{9}{2}=\frac{1}{3}+C \frac{9}{2}-\frac{1}{3}=C \frac{25}{6}=C Step 4:- Value of C put in the equation (1) \frac{y^{2}}{2}=\frac{x^{3}}{3}+\frac{25}{6} Answer