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I'm guessing what you mean is: Is $f(\overline z) = \overline {f(z)}$ for every function $f$ and every complex number $z.$ That is false. For example, suppose \begin{align} f(z)=i|z|. \\[8pt] \text{Then } f(i)= i|i| = i\cdot 1 & = i \\[4pt] \text{and } f(-i) = i\left|-i\right| = i\cdot1 & =i \ne -i. \end{align} However if $f$ is differentiable on ...

2

Let $f(z) = i z$. You're proposing writing $$g(z) = (-i) z$$ and hoping that's the complex conjugate of $f$. Let's see it in parts. We have $$f(a + bi) = i(a + bi) = ai -b = -b + ai\\ g(a + bi) = -ai + b = b - ai$$ But $\overline{-b + ai}$ is not $b - ai$, but is actually $-b - ai$. So no, your proposed approach does not work, even for this very simple ...

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Take any sequence of functions that form a pointwise converging series (such as an entire series), and divide every element by the limit of the series. E.g $$f_n(x)=\frac{x^n}{n!e^x}.$$ There are million other solutions.

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This is similar to writing a full fourier series in complex form. Follow this example: $f(x) = \frac{a_0}{2}+\sum_{n=-\infty}^\infty (a_n cos(nx) b_n sin(nx)$ $= \frac{a_0}{2} + \sum_{n=-\infty}^\infty a_n \frac{e^{inx}+e^{-inx}}{2} + b_n \frac{e^{inx} -e^{-inx}}{2i}$ $=\frac{a_0}{2} + \sum_{n=-\infty}^\infty \frac{a_n-ib_n}{2} e^{inx} + \sum_{n=-\infty}^\... 2 Note first that the hypothesis is just a silly way of writing $$\sum_{j=1}^\infty\theta_j^2j^{2\beta}<\infty.$$ Proof by Looking it Up The hypothesis says precisely that$f$is in the Sobolev space often denoted$H^\beta$; since$\beta>1/2$the Sobolev Embedding Theorem shows that$f$is continuous. Actual Proof $$\sum|\theta_j|=\sum|\theta_j|j^\beta j^... 0 You can do this directly in Fourier-Physical space too, in which case you don't need to transform u\partial_x u to 0.5 \partial_xu^2. Sometimes this is useful, for example, you can compare with the solution by Taozi and notice much less oscilations as a function of time. (Much of the code is taken from Taozi's post). The matrices for first and second ... 0 Hint: You can use :Fourier expansion of f(x)=|x| which is:$$|x|=\frac{\pi}2-\frac{4}{\pi}\big(\frac{ cos x}{1^2}+\frac{ cos 3x}{3^2}+ . . . \frac{ cos (2n-1)x}{(2n+1)^2}+ . . .\big)$$; (-\pi≤x≤\pi) f(x)=|x|=\frac{\pi^2}8 for x=±\pi or x=0 And that of function f(x)=x^2 which is:$$x^2=\frac{\pi^2}{3}-4\big[\frac{cos x}{1^2}-\frac{cos 2x}{2^2}... 0 This is a standard application of Fourier series. See https://en.wikipedia.org/wiki/Fourier_series for the basics. To compute the Fourier series of a given function, you first need a periodic function. For the Fourier series to involve only cosine terms, you need the function to also be even. Note that the period need not be$2\pi$. Let$f$be a$\pi$-... 0 These expressions are$a_n$and$b_n$in the Fourier Series (times L). If these limits did not equal 0 then the sum in Bessel's Inequality would diverge by the divergence test and the equality would be false. 1$\{ \sin(n\pi x) \}_{n=1}^{\infty}$is a complete orthogonal basis of$L^2[0,1]$. And$\{ e^{2\pi i nx} \}_{n=-\infty}^{\infty}$is a complete orthonormal basis of$L^2[0,1]$. You can expand$\sin(\pi x)$in$L^2[0,1]$using a series$\sum_{n=-\infty}^{\infty}a_n e^{2\pi inx}$. That might seem a little strange, but no more strange than being able to expand$\...

2

Refer to this fantastic animation from Wikipedia (here we have interchanged $x_1(t)$ in your notation with $f(t)$): Here you can see that a simple function $f$ is being decomposed into a sum of simple sinusoidal waves, some more important than others (as measured by their amplitude). (For the purposes of this answer, you can think of the simple sinusoidal ...

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So, I think I understood how to do point (b). As suggested in the comments, first prove it for a function $f$ like $$f_m(t)= \sum_{\left| k\right|\leq m} \hat{f_k} e^{ikt}$$ Therefore, if $n \geq m$ we can write $$\int_0^{2\pi}f(t)g(nt) dt= \sum_{\left| k\right|\leq m} \hat{f_k} \int_0^{2\pi} e^{ikt}g(nt)dt = \sum_{\left| k\right|\leq m} \hat{f_k} \int_0^{2\... 3 To give you an idea of why it is true, recall Fourier series. Given a nice enough 2\pi-periodic function f : \mathbb{R} \to \mathbb{C} we can write f(x) = \sum_{n \in \mathbb{Z}} c_n e^{inx} where c_n = \frac{1}{2\pi} \int_{0}^{2\pi} f(x) \, e^{-inx} \, dx. Now take f(x) = \sum_{k \in \mathbb{Z}} 2\pi \, \delta(x-k), the 2\pi-period extension of ... 1 Let S_n f be equal to n-nth partiał sum od Fourier series od tej function f, i.e,$$S_n f (x) = \frac{a_0}{2}+\sum_{k=1}^n (a_k\cos kx + b_k\sin kx )$$. Ten thę Cęsaro partiał sum arę equał to$$c_n f(x)=\frac{S_1 f(x) +...+S_n f(x)}{n}.$$And Thę Theorem says that$$c_n f\to f$$uniformły on [-\pi ,\pi]. 1 If (S_n) is the partial sum sequence of the Fourier series of f then uniform Cesaro summability means \sigma_n \to f uniformly, where \sigma_n =\frac {S_0+S_1+...+S_n} {n+1}. This result is called Fejer's Theorem and you can find it in many books as well as Wikipedia. See, for example, 6.1.1, p. 87 of Edward's Fourier Series. (Take k=0). 3$$f''''(x)+f(x) = \cos{x} - \left( \sin{x} \right)^2$$Solve the homogeneous equation first:$$f''''(x)+f(x) = 0$$the characteristic polynomial is$$r^4+1=0$$Solve for r then the solution is$$y_h=\sum_{i=1}^4c_ie^{r_ix}$$For the particular solution$$f''''(x)+f(x) = \cos{x}-\frac{1}{2}+\frac{1}{2}\cos{2x}$$Try$$y_p=A+B\cos x +C \cos(2x)$$You find ... 1 Trying a solution of the form e^{\lambda x} for the homogeneous problem, we have the characteristic equation (or auxiliary equation) \lambda^{4}+1=0, which has the roots \lambda=e^{\frac{i\pi}{4}(1+2k)} for k=0, 1,2  and 3. For instance when k=0 we have e^{x(\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4}))}=e^{\frac{x}{\sqrt{2}}}(\cos(\frac{x}{\sqrt{2}}... 2 The only thing that differs is the domain (or the period). In the former case, f:[-L,L]\to\mathbb{C} while in the latter case, f:[-\pi,\pi]\to\mathbb{C}. They are practically equivalent. One can easily transform one case into the other by scaling f in the x-direction. 0 From @QiaochuYuan's comment: If the polynomials p_n(x) are orthogonal with respect to a weight w(x) on [a,b], which means that if we define \langle f, g \rangle = \int_a^b f(x) \overline{g(x)} w(x) \, dx then \langle p_i, p_j \rangle = 0 if i \neq j and is positive otherwise, then the coefficient of c_n in the expansion of a function f is \... 0 Because the Uniform Limit Theorem also holds if you replace continuity by uniform continuity, proving each element of the series is uniformly continuous also proves that the limit is uniformly continuous as well (assuming uniform convergence). I suppose the author assumes each element of Fourier series is uniformly continuous. 0 The partial sums of a serise of non-negative numbers is an increasing sequence. If it is bounded then it is convergent. In this case the bound is \frac 1 {\pi} \int_{-\pi}^{\pi} f^{2}(t)dt. 1 There is a typo. Equality does not hold but there is an inequality. See Wikipedia for 'Minkowski's inequality for integrals' to get \leq  in place of =. Ref: https://en.wikipedia.org/wiki/Minkowski_inequality 0 I see that both of the previous answers to part (b) of this exercise do not fully use the hint in the statement, so I'll attempt to do so. For part (c), the first few paragraphs of fonini's answer should suffice. We need to show that the series$$ \sum_{n\ne0} \left| \frac{e^{-ina}-e^{-inb}}{2\pi in} e^{inx} \right| $$converges. Rewriting the expression in ... 4 When there is a discontinuity, as you have in f(x) at x=\frac\pi2, the Fourier series converges to the average of the left and right limits: that is the sum should evaluate to \frac12, not 1. Think of what would have happened if you had defined$$ f(x)=\left\{\begin{array}{rl} 1&\text{if }x\in\left(\frac\pi2,\pi\right)\\ 0&\text{if }x\in\left[...

1

Using Fourier transform $$\mathcal{F}\{f(t)\} = \int_{-\infty}^{\infty} f(t) \, e^{-i\omega t} \, dt$$ and $$\operatorname{sinc} x = \frac{\sin x}{x}.$$ I happen to know that $$\mathcal{F}\{\chi_{[-1,1]}(t)\} = 2 \frac{\sin\omega}{\omega}$$ so to get $\operatorname{sinc}\frac{a\omega}{2}$ we should just modify $\chi_{[-1,1]}(t).$ Therefore, we ...

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