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Suppose a second rank tensor ##T_{ij}## is given. Can we always express it as the tensor product of two vectors, i.e., ##T_{ij}=A_{i}B_{j}## ? If so, then I have a few more questions: 1. Are those two vectors ##A_i## and ##B_j## unique? 2. How to find out ##A_i## and ##B_j## 3. As ##A_i## and...

I did not understand why (3) is the Fourier transform of -k instead of k. Look, I used the defination, $$ \int_{-\infty}^{\infty}f(x)e^{-ikx}\,dx=\widetilde{f}(k)\\ $$ Taking complex conjugate on both sides, $$...

I used the defination, $$\int_{-\infty}^{\infty}f(x)e^{-ikx}\,dx=\widetilde{f}(k)$$

[##f^*## represents complex conjugate of ##f##. ] [##\widetilde{f}(k)## represents Fourier transform of the function ##f(x)##.] $$\begin{align} \int_{-\infty}^{\infty}f^*(x)e^{ikx}\,dx&=\int_{-\infty}^{\infty}f^*(x)\left(e^{-ikx}\right)^*\,dx\\...

Using Fourier transformation, we have, Comparing with the equation, $$f(t)=\int_{-\infty}^{\infty}\delta(t-u)f(u)\,du$$ we have, Thus, $$\begin{align} \delta(t)&=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i\omega t}\,d\omega\\ &=\frac{1}{2\pi}\lim_{\Omega\rightarrow...

We know, $$\delta(x) = \begin{cases} \infty & \text{if } x = 0 \\ 0 & \text{if } x \neq 0 \end{cases}$$ And, also, $$\int_{-\infty}^{\infty}\delta(x)\,dx=1$$ Using Fourier Transformation, it can be shown that, $$\delta(x)=\lim_{\Omega \rightarrow \infty}\frac{\sin{(\Omega x)}}{\pi x}$$ Let's...

Suppose we have ##a \equiv 0 (mod~ m)##. And, ## a \equiv b (mod~ n)##. Is there any way to relate ##b## with ##m## and ##n## ?

What is the inverse function of ##x+sin x## ?

If, ##f(x)=x^x##, then, f-1(x)=?