David
• Member for 8 years, 1 month
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• India

$\clubsuit$ Try to view every thing to one dimension or two dimension where we can see geometrically.. $\quad$ It is quite often helps to realize the things and to generalize it. $\clubsuit$ Aim ...

From the Hint given by @cameron Williams, I did the following. $$\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}}\left(\frac{\sqrt{1}-\sqrt{3}}{-2}+\frac{\sqrt{3}-\sqrt{5}}{-2}+\ldots+\frac{\sqrt{2n-1}-\... View answer 5 votes If you know the Sylvester inequality then it's a two lines proof. Sylvester Inequality: Consider A_{m\times r} and B_{r\times n}. Then$$r(AB)\geq r(A)+r(B)-r \tag{1}where r(\cdot) ... View answer Accepted answer 4 votes (i) is correct. And for the second one you just see from the co-domain to domain.. that is, \begin{align} (P_1\circ P_2)^{-1}&=\begin{pmatrix}1&2&3&4&5&6\\ 2&4&5&... View answer 3 votes \begin{align} \sqrt{5x+19}&=\sqrt{x+7}+2\sqrt{x-5}\\ 5x+19&=x+7+4(x-5)+4\sqrt{(x+7)(x-5)}\\ 32&=4\sqrt{(x+7)(x-5)}\\ 8&=\sqrt{(x+7)(x-5)}\\ 64&=x^2+2x-35\\ 0&=x^2+2x-99 \end{... View answer 3 votes The diagram deals with the upper sum of Riemann integral.. The criteria for Riemann integrablity,U(P,f)-L(P,f)<\epsilonis hard to realise. But with the diagram its highly easy. View answer 3 votes \begin{array}{c|cccccc|c} Case:1\to &A&\underline{4\;ways}&\underline{3\;ways}&\underline{2\;way}&\underline{1\;way}& A&=24 ways\\~\\ Case:2\to &A&\underline{3\;... View answer Accepted answer 3 votes \begin{align} |a_{n}-a_m|&\leq |a_n-a_{n+1}|+|a_{n+1}-a_{n+2}|+\ldots+|a_{m-1}-a_m|\\ &\leq \frac{n^2}{2^n}+\frac{(n+1)^2}{2^{n+1}}+\ldots+\frac{m^2}{2^m}\\ &=\frac{n^2}{2^n}\left\{1+\frac{... View answer 3 votes The eigen values are 1,3  clearly. So it is diagonalizabe(distinct eigen values). And so, there exits P such thatA=P\left(\begin{array}{cc}1& 0\\ 0& 3\end{array}\right)P^{-1}.$$Now ... View answer 3 votes Suppose every element z\in\mathbb Z can be written as$$z=\sum_{i=1}^k \alpha_i z_i, k<n$$Every element of \mathbb Q^n also can be writen as$$ \frac p q=\sum_{i=1}^k\alpha _i \frac {p_i}{...

Remember One can do in this way also, I am going to use these two aspects. $1.$ If $f$ is contiuous and $A$ is open, then $f^{-1} (A)$ is also open. $2.$ Finite intersection of open sets is again ...

Hint: The equation you have quoted is pair of straight lines. The blue and green colour represents the required pair of straight lines and the intersection points have also been quoted...

\begin{align} \int_0^1 f(x)dx&=\int_0^\frac 1 2 f(x) dx +\int_\frac 1 2 ^ 1 f(x) dx\\ &\geq \int_0^\frac 1 2 f(x) dx+\int_0^\frac 1 2 f(x) dx \\ & = 2\int_0^\frac 1 2 f(x) dx \end{align}...

Hint: Use completing the square for $-8yz+z^2+y^2$ which gives $$-8yz+z^2+y^2=y^2+z^2-2yz-6yz=(y-z)^2-6yz$$ Now, \begin{align} 1-8yz+z^2+y^2+9y^2z^2&= 1+(y-z)^2-6yz+9y^2z^2\\ &=(y-z)^2+(...

\begin{align} f^{-1}([0,1])&=\{x\in X: f(x)\in [0,1]\} \\ &=\{x\in X: x^2\in [0,1]\}\\ %&=\{x\in A: x=\pm\sqrt y\}\\ &=[-1,\,1] \end{align} So, it is not one-one and so you can not ...

there is another way of proving this... Suppose there are only finitely many rationals in between $a$ and $a+\epsilon$, say $x_1,x_2\ldots,x_n$ such that $$a<x_1<x_2<\ldots<x_n<a+\... View answer Accepted answer 2 votes all the terms are positive, So, a_n=\frac 1 {\sqrt n}=\frac 1 p where p^2=n. So, it contains a sub-series \sum \frac 1 n which is diverges... so the original series must be diverge. View answer 1 votes Holomorphic is just talking differentiability in the complex plane. For example, in Real analysis, we look at the left and right limits only. whereas here we are looking in all possible directions. ... View answer 1 votes Topics in Algebra by I.N. Herstein is good book to read. My professors also suggesting the same. View answer Accepted answer 1 votes Let us first calculate A^H b where H is the hermitian operator. That is,$$ (A^H)_{ij} = A_{ji}^*where * represents the complex conjugate. \begin{align} A^H b b^H A & = (A^H b) (b^H A) \... View answer 1 votes Using the defn: Let 0\leq s\leq f where s is any simple function.\int f d\mu= sup \int s d\mu = sup \sum c_i\mu(E_i)\leq b sup \sum \mu (E_i)=b\mu(E)$$For the lower bound, consider the ... View answer 1 votes Its not true always... infact. Its entirely depends on which set you are working. One more notable thing is x^{-1} is just a notation to represent a inverse of an element x(mostly for ... View answer Accepted answer 1 votes$$x^{13}+1=0x^{13}=-1x=(-1)^{\frac 1{13}} x= (\cos \pi+i\sin \pi )^{\frac 1{13}}$$Then by De Moivre's formula, add 2k\pi with the arguments, and then you can use as$$ x= (\cos (...

Clearly $(1)$ and $(3)$ are in the linear span of $f$ and $g$, since $\cos 2x=\cos^2 x-\sin^2 x$ $\cos^2x+\sin^2x=1$ For the other options, Checking for $\sin x$, if $a\cos^2 x+b\sin^2 x=\sin x$ ...

Yes..its homogenious.. $$y' =\frac{y^2 + x \sqrt{4x^2 + y^2}}{xy}$$ For this, their is well known method, running by substituting, $$y=vx\implies y'=v+xv'$$ So, it will became as, \begin{align} ...

$f$ is a linear polynomial as quoted by you. So, $$f(z)=a_0+a_1z$$ and $|f(z)|\leq K|z|\implies |f(0)|\leq K|0|\implies f(0)=0$ Hence, $a_0=0$ and $f(1)=i\implies a_1=i$ therefore, $f(i)=a_0+a_1.... View answer Accepted answer 1 votes Divide the numerator by denominator first.. $$x^3-1=\frac 1 4 (4x^3-x)-\frac 1 4 (x-4)\tag{division algorithm}$$ $$\frac{x^3-1}{4x^3-x}=\frac 1 4 \frac{4x^3-x}{4x^3-x}-\frac 1 4\frac{x-4}{4x^3-x}$$ ... View answer 1 votes Observe that$f:X\to Y$is bounded linear map and bijective. Claim:$f^{-1}:Y\to X$also continuous... let$G$open in$X\$ then $$f^{-1}(G)=\{y\in Y: f^{-1}(x)=y,x\in G\}$$ $$=\{y\in Y: f(x)=y,x\in ... View answer 1 votes its given that y=f(x)=x^{x^{x^{x^\cdots}}} I can write this function as,$$y=x^{y}$$Taking log_e on both sides, we have,$$\log y=y\log x$$Now, differentiate w.r.to x on both sides,$$\...