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Functional taylor series

  1. Dec 30, 2009 #1
    Is there any place I can find the equation for the Taylor expansion of a functional around a function ??

    Particularly, I want something like:

    f[x(t)] = f[\hat{x}(t)] + (f[\hat{x}(t)] - f[x(t)] \frac{\delta f}{\delta x(t)}|_{x(t)=\hat{x}(t)} + \frac{(f[\hat{x}(t)] - f[x(t)])^2}{2!}\frac{\delta ^2f}{\delta ^2x(t)}|_{x(t)=\hat{x}(t)} \ldots

    Particularly I want to expand the functional:

    f[\Psi] [/itex] around the function [itex]\Psi \Psi^*[/itex] where [itex]\Psi^*[/itex] is the compex conjugate
  2. jcsd
  3. Jan 13, 2010 #2
    Is this problem completely ill-posed ??
    Is there no such thing as a funcitonal taylor series ??

    Is there a book where I can find the formula ??

    THank you.
  4. Jan 13, 2010 #3
    What is [tex] \hat{x}[/tex] ? If [tex] f,x \in C^{\infty} ( \mathbb{R}) [/tex] then
    [itex]f[x(t)] = f[ x( t_0 )] + \left( f[x(t)] - f[x(t_0)] \right) \frac{\delta f}{\delta x(t)}|_{ t = t_0 } + \frac{ \left( f[x(t)] - f[x(t_0 )] \right) ^2}{2!}\frac{\delta ^2f}{\delta ^2x(t)}|_{t = t_0} } + \ldots [/itex]​
    forall [tex] t \in ( t_0 - \epsilon , t_0 + \epsilon) [/tex], where [tex] \epsilon [/tex] small enough.
  5. Jan 13, 2010 #4
    [itex]\hat{x}(t)[/itex] was supposed to be the funciton about which the series is being expanded.

    What you typed out looks like just a regular taylor series (expanded about a variable), I want a functional taylor series (expanded about a FUNCTION). I realized just now that the second term should probably be negative.
  6. Jan 14, 2010 #5


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    I'm a little skeptical of being able to make sense of expanding [itex]f[\Psi][/itex] about [itex]\Psi \Psi^*[/itex], because [itex]\Psi[/itex] and [itex]\Psi^*[/itex] are typically analytically independent of each other, so I don't expect there to be a sensible notion of "closeness".

    OTOH, you do have [itex]f[\Psi] = f[\Psi \Psi^* + (\Psi - \Psi \Psi^*)][/itex], so *shrug*.
  7. Jan 14, 2010 #6
    Okay, I may be missing something that you typed ...
    which functional space are everything occurring to ?

    For example, when you said that [tex]f = g [/tex] in [tex]C(a,b)[/tex] then [tex]f (t) = g (t)[/tex] for all [tex]t \in [a,b][/tex], or, when you said that [tex]f = g [/tex] in [tex]L^1(a,b)[/tex] then
    [tex]\int_{a}^b |f(t) - g(t)| dt = 0[/tex] ​
    etc ... So, what is [itex] f[x(t)] [/itex] ? function, number, distribution ? You write [tex] f[x(t)] = .... [/tex], what did it mean ?

    pardon, which series did you mean ?
  8. Jan 14, 2010 #7

    [itex]f[x(t)][/itex] is a functional of the function [itex]x(t)[/itex].

    Now, just as I can expand x(t) in a series about some point [itex]\hat{t}[/itex], which will give me an approximation of x(t) near [itex]\hat{t}[/itex],

    I want to expand [itex]f[x(t)][/itex] in a series about the function [itex]\hat{x}(t)[/itex], which will give me an approximation of [itex]f[x(t)][/itex] near [itex]\hat{x}(t)[/itex].

    For example, let's say [itex]f[x(t)][/itex]=sin[x(t)].

    I think (although I'm not sure), [itex]f[x(t)][/itex] can be written as:

    [itex]f[x(t)] = x(t) - \frac{(x(t))^3}{3!} + \frac{(x(t))^5}{5!} + \ldots[/itex]

    Now let's say we want to see what this functional looks like for functions close to [itex]\hat{x}(t)[/itex]=cos(t).

    Again, I'm not sure, but I think the functional could be written something like:

    [itex]f[x(t)] \approx cos(t) - \frac{(x(t)-cos(t))^3}{3!} + \frac{(x(t)-cos(t))^5}{5!} + \ldots[/itex]

    so if x(t) is some function very similar to cos(t) ... let's say it's cos(t) with a few wiggles, then the above formula would give an approximation of [itex]f[x(t)] [/itex].

    I hope that wasn't too confusing (I haven't studied calculus of variations very much). Please let me know if what I said doens't make sense.
  9. Jan 14, 2010 #8
    okay, now i understand what you mean. You can do that in some special cases.

    If the function [tex] f[/tex] in [tex] C^\infty ( \mathbb{R}) [/tex], ie, the [tex]n [/tex]th derivative [tex]\frac{d^n f}{ dx^n} [/tex] always exists and it is a continuos function on [tex] \mathbb{R}[/tex], for all [tex]n \in \mathbb{N} [/tex]. Then you have
    [tex] f(u) = f(v) + \sum_{n=1}^{\infty} \frac{1}{n!} \cdot \frac{d^n f}{ dx^n} (u) \, \cdot \, (u-v)^n \;\; , \; \forall u,v \in \mathbb{R} [/tex]​
    where [tex]\frac{d^n f}{ dx^n} (u)[/tex] is the [tex]n [/tex]th derivative [tex]\frac{d^n f}{ dx^n} [/tex] evaluated at the point [tex] u [/tex].

    Now, for all [tex] t \in \mathbb{R}[/tex], we put [tex] u = x(t) [/tex] and [tex] v = \hat{x}(t) [/tex] so that we can write
    [tex] f(u) = f(v) + \sum_{n=1}^{\infty} \frac{1}{n!} \cdot \frac{d^n f}{ dx^n} (u) \,\cdot \, (u-v)^n[/tex]​
    ie, for all [tex] t \in \mathbb{R}[/tex]
    [tex] f(x(t)) = f \left( \hat{x}(t) \right) + \sum_{n=1}^{\infty} \frac{1}{n!} \frac{d^n f}{ dx ^n} \left( \hat{x}(t) \right) \, \cdot \, \left[ x(t)- \hat{x} (t) \right]^n[/tex]​
    if [tex] f \in C^\infty ( \mathbb{R}) [/tex]

    For example, if [tex]f \left( x(t) \right)=\sin x(t) [/tex] then [tex] f \in C^\infty ( \mathbb{R}) [/tex] and we have
    [itex]f[x(t)] = x(t) - \frac{(x(t))^3}{3!} + \frac{(x(t))^5}{5!} + \ldots[/itex]​
    where [tex] \hat{x} (t) = 0 [/tex] for all [tex]t \in \mathbb{R} [/tex], and
    [itex]f(x(t)) = \sin \left( \cos(t) \right) + \sum_{k = 1}^{\infty} \frac{(-1)^k}{(2k) !} \sin \left( \cos t\right) \, \left( x(t) - \cos t \right) ^{2k} + \sum_{k = 0}^{\infty} \frac{(-1)^k}{(2k + 1) !} \cos \left( \cos t \right) \, \left( x(t) - \cos t \right) ^{2k +1} [/itex]​
    where [tex] \hat{x} (t) = \cos t [/tex] for all [tex]t \in \mathbb{R} [/tex].
  10. Jan 15, 2010 #9
    Thank you Brahman,

    That would be the intuitive way of thinking about it (defining the functional Taylor series pointwise where at each point the series is equal to the Taylor series at that point) ,

    but the reason why I was asking all this is because I thought the expression was much more complicated.

    On page 234 of this pdf file (It will say pg 234 in acrobat reader, but pg 226 on the actual page since the first few pages are not numbered):

    The funcitonal Taylor series is defined under the word "Proof" , but only for the very special case of x(t)=0.

    Still, this expression is much more complicated than the one you typed out (and the ones that I was suggesting). It involves n'th degree funcitonal differentiation and multiple integrations.

    I'm looking for the general expression for this formula, that is not specific to the case x(t)=0.

    I hope there is a book or reference that has this formula
    Last edited by a moderator: Apr 24, 2017
  11. Jan 16, 2010 #10
    oh, the formula doesn't look familiar with me, it was more complicated than what we typed. Because i have no background of Stochastics prossess, so hope that some one else will help you with the general formular ^-^ . Anyway, the book you sent is worthy of note. Cheers!

    PS: edit what I typed

    f(u) = f(v) + \sum_{n=1}^{\infty} \frac{1}{n!} \cdot \frac{d^n f}{ dx^n} (v) \, \cdot \, (u-v)^n \;\; , \; \forall u,v \in \mathbb{R}
  12. Jan 16, 2010 #11


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    Can't you just translate?

    e.g. in the single real variable case, you can get the Taylor series for f(x) about x=a by finding the Taylor series for f(y+a) about y=0.
  13. Jan 16, 2010 #12
    I don't think it's that simple, you see how each derivative is evaluated at gamma=0 ??
    If we are expanding around an arbitrary funciton, then we are evaluating these derivatives at those "functions" (not just variables) ... and that's confusing me since this book doesn't explain what [itex]\gamma_1, \gamma_2[/itex], etc. are

    That's why I'd like to see this formula in a real book, not just to see the formula for the general form, but to also understand why the formula looks the way it is.

    And Brahman, this functional taylor series has nothing to do with stochastic processes.. it's just used to prove something related to them. It's actually a topic of functional calculus (or 'calculus of variations')
  14. Jul 15, 2011 #13
    looks like this topic is dead now for a while. Still, in case anybody comes across the same question: I believe what Juliette is looking for is what is called a "Volterra Series" expansion.
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