MHB Why Use Half Range Fourier Series for Functions Like x and x^2?

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Using half-range Fourier series for functions like x and x^2 defined between 0 and pi allows for the extension of these functions into periodic signals, such as sawtooth or triangle waves. This extension is not merely a mathematical exercise; it has practical applications, particularly in signal processing. For example, such Fourier series representations are utilized in CRT televisions for pixel tracing. The process helps in analyzing and reconstructing signals in various engineering fields. Overall, the use of half-range Fourier series provides valuable insights into periodic function behavior and real-world applications.
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If we have a standard function like x or x^2 defined between 0 and pi. Then why should we be interested in extending this function to give a Fourier series which resembles this function between 0 and pi? What is the whole purpose of this process? Does it have any real life application or is it just a mathematical exercise?
 
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matqkks said:
If we have a standard function like x or x^2 defined between 0 and pi. Then why should we be interested in extending this function to give a Fourier series which resembles this function between 0 and pi? What is the whole purpose of this process? Does it have any real life application or is it just a mathematical exercise?

Hi matqkks!

If we have $x$ between $0$ and $\pi$ and extend it to give a Fourier series, we have a sawtooth or triangle signal.

This type of signal is used for instance in a (CRT) television for tracing the pixels on the screen.
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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