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Math Help - Notation again again

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Notation again again

    I do not know how to search this but does

    C^{\infty} mean that is differentiable any number of times, or something?

    Any help appreciated.
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    Quote Originally Posted by Mathstud28 View Post
    I do not know how to search this but does

    C^{\infty} mean that is differentiable any number of times, or something?

    Any help appreciated.
    Let I be an open interval. The notation \mathcal{C}^k(I) means a function differenciable k times on I such that f^{(k)} is a continous function on I. However, if the function is infinitely differenciable i.e. it is an element of \mathcal{C}^{k} for any k\geq 1 then we say it \mathcal{C}^{\infty}(I). For example, \exp ,\sin , \cos \in \mathcal{C}^{\infty}(\mathbb{R}). While \log \in \mathcal{C}^{\infty}(0,\infty).
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Let I be an open interval. The notation \mathcal{C}^k(I) means a function differenciable k times on I such that f^{(k)} is a continous function on I. However, if the function is infinitely differenciable i.e. it is an element of \mathcal{C}^{k} for any k\geq 1 then we say it \mathcal{C}^{\infty}(I). For example, \exp ,\sin , \cos \in \mathcal{C}^{\infty}(\mathbb{R}). While \log \in \mathcal{C}^{\infty}(0,\infty).
    Thanks TPH! Always answering my notation questions .

    One question though, I understand that e^x,\cos(x),\text{etc.} are C^{\infty}\left(\mathbb{R}\right)

    But since C^{\text{whatever}} seems to only be talking about the at least first derivative wouldnt that mean that the I specified for \ln(x) should be \mathbb{R}/\left\{0\right\} since \ln^{\left(n\in\mathbb{Z^+}\right)}(x) is continous for that interval?


    Oh wait, I looked back and the interval I is specified in relation to the original function, not the derivatives. Ok so I get it.

    Thank you.


    EDIT: Actually two more things. The first is obviously that

    If p(x) is a non-descript polynomial then it is C^{\infty} Right?

    Also how would you say that some function is this on some interval, in other words how would you put this notation down so it makes sense.

    Like this?

    " f(x) is C^{\infty}(a,b)"?
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    Quote Originally Posted by Mathstud28 View Post
    But since C^{\text{whatever}} seems to only be talking about the at least first derivative wouldnt that mean that the I specified for \ln(x) should be \mathbb{R}/\left\{0\right\} since \ln^{\left(n\in\mathbb{Z^+}\right)}(x) is continous for that interval?
    It does not need to be an open interval it can be an open set. But since open sets are concepts from topology I did not want to mention them. An open set is a generalization of an open interval. For example, \mathbb{R} - \{ 0 \} = (-\infty,0)\cup (0,\infty) is an open set. Basically an open set is such a set that has no boundary. More formally S is open iff for any x\in S there is \epsilon > 0 such that (x-\epsilon,x+\epsilon) \subset S. With open sets we can say that \ln |x| is \mathcal{C}^{\infty}(\mathbb{R} - \{ 0\}).





    If p(x) is a non-descript polynomial then it is C^{\infty} Right?
    A polynomial is infinitely differenciable so it is \mathcal{C}^{\infty}.

    Also how would you say that some function is this on some interval, in other words how would you put this notation down so it makes sense.

    Like this?

    " f(x) is C^{\infty}(a,b)"?
    That is exactly how we say it.

    As an illustration of this notation we can state a stronger version of the Fundamental Theorem of Calculus. Let f be continous on [a,b] which is \mathcal{C}^k(a,b). Define F(x) = \smallint_a^x f. Then F is \mathcal{C}^{k+1}(a,b).

    Analysts like to say that integration "smoothens out a function". The theorem above is what this is all about.
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    It does not need to be an open interval it can be an open set. But since open sets are concepts from topology I did not want to mention them. An open set is a generalization of an open interval. For example, \mathbb{R} - \{ 0 \} = (-\infty,0)\cup (0,\infty) is an open set. Basically an open set is such a set that has no boundary. More formally S is open iff for any x\in S there is \epsilon > 0 such that (x-\epsilon,x+\epsilon) \subset S. With open sets we can say that \ln |x| is \mathcal{C}^{\infty}(\mathbb{R} - \{ 0\}).






    A polynomial is infinitely differenciable so it is \mathcal{C}^{\infty}.


    That is exactly how we say it.

    As an illustration of this notation we can state a stronger version of the Fundamental Theorem of Calculus. Let f be continous on [a,b] which is \mathcal{C}^k(a,b). Define F(x) = \smallint_a^x f. Then F is \mathcal{C}^{k+1}(a,b).

    Analysts like to say that integration "smoothens out a function". The theorem about is what this is all about.
    Ok, amazing, I completely understand that.

    Like the last part since F'(x)=\frac{d}{dx}\int_a^{x}f(t)dt=f(x) by the fundamental theorem of calculus and we have stated that f(x)\quad\text{is}\quad{C^{k}}

    F(x) takes on all the k amount of differentiabilities of f(x) and adds more since we know that if it is integrable it is differentiable back to f(x)!

    Thanks TPH
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    Quote Originally Posted by Mathstud28 View Post

    Also how would you say that some function is this on some interval, in other words how would you put this notation down so it makes sense.

    Like this?

    " f(x) is C^{\infty}(a,b)"?
    Yes, but also:

    f \in C^{\infty}(a,b)

    RonL
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Yes, but also:

    f \in C^{\infty}(a,b)

    RonL
    Thanks captain black, I want to be able to write stuff with as few words as possible (no sarcasm).

    But by the way, this got me thinking, this notating would be nice to say things like,

    "Let f be a eight times differentiable function" So instead of that could you say

    "Let f(x)\in{C^{k\geq{8}}} "

    Also, can it apply to functions where \mathbb{R}^{n>1}\to\mathbb{R}

    Can I say

    C_x^{\infty}\left(\mathbb{R}\right)?

    Or would no one know what that means, haha, I think I am inventing notation here.
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    Quote Originally Posted by Mathstud28 View Post
    Thanks captain black, I want to be able to write stuff with as few words as possible (no sarcasm).

    But by the way, this got me thinking, this notating would be nice to say things like,

    "Let f be a eight times differentiable function" So instead of that could you say

    "Let f(x)\in{C^{k\geq{8}}} "
    Since:

     C^8 \supset C^9 \supset C^{10} \supset ...

    there is no need for this notation.

    Also C^8 and the set of all functions eight times differentiable (on (a,b) or whatever) are not the same thing since C^8 is the set of all 8 times differentiable function with continuous 8-th derivative.

    RonL
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    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    ...

    But by the way, this got me thinking, this notating would be nice to say things like,

    "Let f be a eight times differentiable function" So instead of that could you say

    "Let f(x)\in{C^{k\geq{8}}} "
    not really, because there is another condition: f^{(8)}(x) must also be continuous.

    so i think, f(x)\in{C^{k\geq{8}}} would mean f^{(k\geq8)}(x) must also be continuous. so y don't you get the biggest k such that f^k(x) is continuous with f is k times differentiable..

    Quote Originally Posted by Mathstud28 View Post
    Also, can it apply to functions where \mathbb{R}^{n>1}\to\mathbb{R}
    yes, like what TPH said, it can be defined in any open set, just define the open set in \mathbb{R}^{n}

    Quote Originally Posted by Mathstud28 View Post
    Can I say

    C_x^{\infty}\left(\mathbb{R}\right)?

    Or would no one know what that means, haha, I think I am inventing notation here.
    haven't seen this notation before..
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Since:

     C^8 \supset C^9 \supset C^{10} \supset ...

    there is no need for this notation.

    Also C^8 and the set of all functions eight times differentiable (on (a,b) or whatever) are not the same thing since C^8 is the set of all 8 times differentiable function with continuous 8-th derivative.

    RonL
    So what your saying is that since for a function to be f(x)\in{C^{k\geq{8}}} it must be f(x)\in{C^8} or in other words all functions of tenth order differentiablility are a subset of eight order or whatever.

    So your saying that f(x)\in{C^{n}} describes all functions who are differentiable at least n times, so this would include all functions that are differentiable n+1,n+2,n+3,\cdots times?
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  11. #11
    MHF Contributor kalagota's Avatar
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    if f \in C^{n}, then f, f', f'', ..., f^{(n)} exists and f^{(n)} is continuous..

    f \in C^{n+1} means f, f', f'', ..., f^{(n)}, f^{(n+1)} exists and f^{(n+1)} is continuous..

    now, let g \in C^{n+1}. therefore, g^{(n)} exists. g^{(n)} must be continuous otherwise, g^{(n)} will not be differentiable, hence g^{(n+1)} will not exist.

    therefore, g \in C^{n}, that is C^{n+1} \subset C^{n}

    so to answer your question, yes..
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by kalagota View Post
    if f \in C^{n}, then f, f', f'', ..., f^{(n)} exists and f^{(n)} is continuous..

    f \in C^{n+1} means f, f', f'', ..., f^{(n)}, f^{(n+1)} exists and f^{(n+1)} is continuous..

    now, let g \in C^{n+1}. therefore, g^{(n)} exists. g^{(n)} must be continuous otherwise, g^{(n)} will not be differentiable, hence g^{(n+1)} will not exist.

    therefore, g \in C^{n}, that is C^{n+1} \subset C^{n}

    so to answer your question, yes..
    So the operative phrasing here would be that f(x)\in{C^{n}} implies that f(x) is differentiable at least an n number of times, not at most.
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    MHF Contributor kalagota's Avatar
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    yes.
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  14. #14
    Grand Panjandrum
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    Quote Originally Posted by Mathstud28 View Post
    So what your saying is that since for a function to be f(x)\in{C^{k\geq{8}}} it must be f(x)\in{C^8} or in other words all functions of tenth order differentiablility are a subset of eight order or whatever.

    So your saying that f(x)\in{C^{n}} describes all functions who are differentiable at least n times, so this would include all functions that are differentiable n+1,n+2,n+3,\cdots times?
    Something differentiable m times with continuous m th derivative is obviously continuously differentiable n-times for all n \le m, so C^{m} \subset C^n, \ \forall n \le m.

    RonL
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  15. #15
    MHF Contributor Mathstud28's Avatar
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    I will give two examples at my attempt to use this notation and please tell me if it is correct.

    Say we are talking about L'hopital's and we are saying

    " Let \lim_{x\to{c}}\frac{f(x)}{g(x)} be indeterminate because either \lim_{x\to{c}}f(x)=\lim_{x\to{c}}g(x)=0 or \lim_{x\to{c}}f(x)=\lim_{x\to{c}}g(x)=\infty

    Furthermore let f(x)\in\mathcal{C}^1(c) and g(x)\in\mathcal{C}^1(c) and then blah blah blah"

    And in Rolle's Theroem

    "Let f(x) be continuous on [a,b] and f(x)\in\mathcal{C}^1(a,b) then ...."


    Did I use it right?
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