Norm of a function ||f|| & the root mean square of a function.

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SUMMARY

The discussion focuses on the relationship between the norm of a function ||f|| and the root mean square (RMS) of the function, specifically within the context of the inner product space C[0,π]. The inner product is defined as (f,g) = ∫0π f(x)g(x) dx. For the function f(x) = sin(x), the norm is calculated as ||f|| = √(∫0π sin²(x) dx), which directly relates to the RMS of the function. The discussion emphasizes that the norm is defined as the square root of the inner product of the function with itself, highlighting the mathematical connection between these two concepts.

PREREQUISITES
  • Understanding of inner product spaces
  • Familiarity with continuous functions on the interval [0,π]
  • Knowledge of integration techniques
  • Basic concepts of norms and root mean square calculations
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  • Study the properties of inner product spaces in functional analysis
  • Explore the calculation of norms for various functions in C[0,π]
  • Learn about the applications of root mean square in signal processing
  • Investigate the relationship between norms and convergence in functional spaces
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norm of a function ||f|| & the "root mean square" of a function.

How do I explain the connection between the norm of a function ||f|| & the "root mean square" of a function. You may like to consider as an example C[0,[tex]\pi[/tex]], the inner product space of continuous functions on the interval [0,[tex]\pi[/tex]] with the inner product
(f,g) = [tex]\int_0^\pi[/tex] f(x)g(x) dx.
Let f(x) = sin(x). How do I find ||f||. Also find "root mean square" of f. What do you notice?
 
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Well, have you tried using the definitions and applying them to this particular f?
 


Morphism's point is that |f| is normally defined, in an inner product space, as the square root of the inner product of f with itself. So if
[tex](f, g)= \int_0^\pi f(x)g(x)dx[/tex]
Then
[tex]|f|= \sqrt{\int_0^\pi f^2(x) dx[/tex]

Now, what is
[tex]\sqrt{\int_0^\pi sin^2(x) dx[/tex]?
 

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