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Prologue
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I am trying to understand the definition of the inner product of two functions on an interval. I know that the form of a scalar product in finite dimensional space is given by
[tex]\vec{\phi} \bullet \vec{\psi}= \sum_{k} \phi_{k} \psi_{k}[/tex]
and in infinite dimensional space
[tex]\langle \phi , \psi \rangle = \int_{a}^{b} \phi(x) \psi(x) dx[/tex]
If we expand out the first definition we get
[tex]\vec{ \phi} \bullet \vec{ \psi} = \phi_{1} \psi_{1} + \phi_{2} \psi_{2} + \phi_{3} \psi_{3} ...[/tex]
and for the second we get
[tex]\langle \phi , \psi \rangle = \phi(a) \psi(a) dx \: + \: \phi(a+dx) \psi(a+dx) dx \: + \: \phi(a+2dx) \psi(a+2dx) dx ...[/tex]
What is up with the factor of dx multiplying everything is this last expansion? It has to be there because that is what an integral does, it takes the function evaluated at that specific x and multiplies it by dx, but what purpose does the dx factor serve in the definition of the inner product? It is pretty obvious that apart from the factor of dx it looks identical to the finite dimensional definition. So what gives?Edit: I edited this to make my question more clear. And I'd like to add that it seems to me that
[tex]\langle \phi , \psi \rangle = \phi(a) \psi(a) dx \: + \: \phi(a+dx) \psi(a+dx) dx \: + \: \phi(a+2dx) \psi(a+2dx) dx ...= \vec{\phi} \bullet \vec{\psi}dx[/tex]
Where [tex]\vec{\phi} \bullet \vec{\psi}[/tex] is supposed to just be a recognition of the form of the initial definition.
[tex]\vec{\phi} \bullet \vec{\psi}= \sum_{k} \phi_{k} \psi_{k}[/tex]
and in infinite dimensional space
[tex]\langle \phi , \psi \rangle = \int_{a}^{b} \phi(x) \psi(x) dx[/tex]
If we expand out the first definition we get
[tex]\vec{ \phi} \bullet \vec{ \psi} = \phi_{1} \psi_{1} + \phi_{2} \psi_{2} + \phi_{3} \psi_{3} ...[/tex]
and for the second we get
[tex]\langle \phi , \psi \rangle = \phi(a) \psi(a) dx \: + \: \phi(a+dx) \psi(a+dx) dx \: + \: \phi(a+2dx) \psi(a+2dx) dx ...[/tex]
What is up with the factor of dx multiplying everything is this last expansion? It has to be there because that is what an integral does, it takes the function evaluated at that specific x and multiplies it by dx, but what purpose does the dx factor serve in the definition of the inner product? It is pretty obvious that apart from the factor of dx it looks identical to the finite dimensional definition. So what gives?Edit: I edited this to make my question more clear. And I'd like to add that it seems to me that
[tex]\langle \phi , \psi \rangle = \phi(a) \psi(a) dx \: + \: \phi(a+dx) \psi(a+dx) dx \: + \: \phi(a+2dx) \psi(a+2dx) dx ...= \vec{\phi} \bullet \vec{\psi}dx[/tex]
Where [tex]\vec{\phi} \bullet \vec{\psi}[/tex] is supposed to just be a recognition of the form of the initial definition.
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