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  1. #1 dumb question about differentials... 
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    Hi there,

    This no doubt will sound very dumb but excuse me as I've had very little exposure to multivariable calculus, so maybe there is something I missed there... or maybe it's not related to multivariable calculus...

    I'm reading a book on manifolds, and so he defines the differential of a map as a linear map between the tangent spaces.

    More precisely, if F:N-->M is a C-oo map between two manifolds, the differential F_*:TpN-->TF(p)M is given by F_*(Xp)(f)=Xp(foF).

    Ok, I understand everything in this section, the only thing I don't understand is that he says that this generalizes the derivative in euclidean spaces. What troubles me is that the differential F_* is a linear map, whereas the derivative of say x^3:R-->R is 2x^2, a nonlinear map.

    Thanks a lot!


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  3. #2 Re: dumb question about differentials... 
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    Quote Originally Posted by Stranger
    Hi there,

    This no doubt will sound very dumb but excuse me as I've had very little exposure to multivariable calculus, so maybe there is something I missed there... or maybe it's not related to multivariable calculus...

    I'm reading a book on manifolds, and so he defines the differential of a map as a linear map between the tangent spaces.

    More precisely, if F:N-->M is a C-oo map between two manifolds, the differential F_*:TpN-->TF(p)M is given by F_*(Xp)(f)=Xp(foF).

    Ok, I understand everything in this section, the only thing I don't understand is that he says that this generalizes the derivative in euclidean spaces. What troubles me is that the differential F_* is a linear map, whereas the derivative of say x^3:R-->R is 2x^2, a nonlinear map.

    Thanks a lot!
    The derivative, at a point, is a linear map. In one dimension the derivative at a point is usually discussed as a number, but linear maps in one dimension are identified with numbers, or 1 x 1 matrices if you like.

    So the idea of a derivative in multimple dimensions as a linear map between tangent spaces is a generalization of the notion from ordinary calculus, but you have to look at it just right to see what is going on. Remember that the tengent space is defined at a point. Do not confuse the linearity of the map on the tangent space at a point, with map that is induced on the tangent bundle (at all points), which is what you are doing when you think about the derivative of X^3 as 2x^2. For any single value of x, 2x^2 is just a number.

    What book are you reading ?

    It is generally a good idea to learn and understand multivariable calculus before you start a serioius study of manifolds, although there are good introductory books that handle multivariable calculus and introductory concepts of manifolds simultaneously. Spivak's Calculus on Manifolds is one such book.


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  4. #3  
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    Oh, I think I understand now.

    Indeed, if for example f(x) = x^3, then f'(x)=3x^2, which we can write as (3x^2), a 1x1 matrix.

    For f:R^n-->R^m, this notion is then replaced with the jacobian. Instead of a (1x1) matrix, we get a (mxn) matrix.

    Aha, that's why the differential indeed generalizes that: when he computes the matrix associated with the differential in the special case of f:R^n-->R^m, he gets the jacobian.

    Hum, so the question now is why is the jacobian the good generalization of the derivative. For example, if we take the function
    f(x,y) = (x^2,y^3,x+y)
    The jacobian matrix will be
    J(x,y) =
    (2x 0
    0 3y^2
    1 1)
    So why does that relate to the derivative? I saw in the article at wikipedia that we should have
    f(x,y) = f(x1,y1) + (J(x1,x2))(x-x1,y-y1) + o(||(x-x1,y-y1)||)
    In that case it would indeed generalize the 1D case.
    But I tried to work that out with that example and I couldn't.

    I've actually had a course on multivariable calculus, but I fogot everything now )

    The book I'm reading is called "An introduction to manifolds" by Loring Tu. It's really manifolds for babies ) yeah, very elementary, I really liked it. I already read the first 22 chapters )

    Thanks again!
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    (T.B)
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  5. #4  
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    (duplicated post due to bad connection)
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  6. #5  
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    The derivative of a function at a point is essentially the best approximation of that function by a linear map.

    For example, fix a point x_0, and consider the function f(x) for x "near" x_0. Let h be the vector of displacement from x_0 to x.

    Then for small h, we'd like to be able to write

    f(x_0+h) = f(x_0) + Lh + (small error), where L is a linear map acting on h. If we can do this in such a way that the error term is small compared to ||h||, then we say that L is the derivative of f at x_0.

    In the one variable case, L is simply the map given by multiplication by f'(x_0). That is,

    f(x_0+h) = f(x_0) + f'(x_0)*h + error

    The idea is that the functions we call "differentiable" act "locally" like linear maps. This is a slightly different way of thinking about differentiability than you may be used to, but it's really how most mathematicians think of the derivative (as opposed to thinking about slopes of tangent lines, etc.)
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  7. #6  
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    Quote Originally Posted by Stranger
    Oh, I think I understand now.

    Indeed, if for example f(x) = x^3, then f'(x)=3x^2, which we can write as (3x^2), a 1x1 matrix.

    For f:R^n-->R^m, this notion is then replaced with the jacobian. Instead of a (1x1) matrix, we get a (mxn) matrix.

    Aha, that's why the differential indeed generalizes that: when he computes the matrix associated with the differential in the special case of f:R^n-->R^m, he gets the jacobian.

    Hum, so the question now is why is the jacobian the good generalization of the derivative. For example, if we take the function
    f(x,y) = (x^2,y^3,x+y)
    The jacobian matrix will be
    J(x,y) =
    (2x 0
    0 3y^2
    1 1)
    So why does that relate to the derivative? I saw in the article at wikipedia that we should have
    f(x,y) = f(x1,y1) + (J(x1,x2))(x-x1,y-y1) + o(||(x-x1,y-y1)||)
    In that case it would indeed generalize the 1D case.
    But I tried to work that out with that example and I couldn't.

    I've actually had a course on multivariable calculus, but I fogot everything now )

    The book I'm reading is called "An introduction to manifolds" by Loring Tu. It's really manifolds for babies ) yeah, very elementary, I really liked it. I already read the first 22 chapters )

    Thanks again!
    1. What salsaonline and I have told you is correct.

    2. The example that you gave from Wiki is correct and it should have worked out for you. Since you did not show what you did, I cannot tell what went wrong.

    3. You might try looking at the treatment of the derivative in higher dimensions in a good book on advanced calculus or real analysis. I am not familiar with the book of Tu. However, I can recommend Calculus on Manifolds by Spivak, Elements of Real Analysis by Bartle and Principles of Mathematical Analysis by Rudin for good treatments of the derivative in higher dimensions. Given your interest in manifolds, Spivak's book would be a good choice. It is available on the used book market (see e.g. Alibris.com) at reasonable prices.

    4. You might also try "googling" the topic "Frechet derivative". Don't worry about the application to Banach spaces if you don't know about them. Just keep in mind that ordinary n-space is a Banach space and interpret the exposition in that light.
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  8. #7  
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    Thanks for replying, both of you

    I'll check on these books then.

    As for Banach spaces, I know them well; calculus is the only thing I've really forgotten... so I'll see that Frechet derivative as I've heard of it somewhere.

    About the example, it's just that I don't like to type math here... so I hope my notation will be clear.

    We have f(x,y) = (x^2,y^3,x+y). This gives

    J(x,y) =
    (2x 0
    0 3y^2
    1 1)

    So, taking p=(x_0,y_0),

    J(p)(x-p) =
    (2x_0(x-x_0),3(y_0)^2(y-y_0),x-x_0+y-y_0)

    On the other hand,

    f(x) - f(p)
    =(x^2-(x_0)^2,y^3-(y_0)^3,x+y-x_0-y_0)

    So,

    f(x) - f(p) - J(p)(x-p)
    = (x^2-2(x)(x_0)+(x_0)^2,y^3-3(y)(y_0)^2+2(y_0)^3,0)

    So why is that o(||x-p||)?

    Thanks and have a nice day!
    Watch what thy eyes can't see... and live it.

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  9. #8  
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    Quote Originally Posted by Stranger

    So,

    f(x) - f(p) - J(p)(x-p)
    = (x^2-2(x)(x_0)+(x_0)^2,y^3-3(y)(y_0)^2+2(y_0)^3,0)

    So why is that o(||x-p||)?

    Thanks and have a nice day!
    Because all the terms in the first two coordinates are of order two or higher in x and y and the last coordinate is, as you noted, 0.
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  10. #9  
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    ahem... so could you please explain what exactly o(||x-p||) means?

    I'm not lazy, I did try to check on wikipedia and 2 books, but in wikipedia the article doesn't seem to be related, and in the books I don't find that equation (that is, F(x) = F(p) + F(p)(x-p) + o(||x-p||)). I only find the definition of the jacobian.

    The article about little o in wikipedia is just as I learned it in my first analysis course: we say that f(x) is little o of g(x) if the limit of f(x)/g(x) tends to zero. That's all. So what does it mean here?

    I understand o(||x-p||) should mean a small error, but what is the definition exactly?

    I read the article about Frechet derivatives; it's indeed a very pleasent way of viewing derivatives, and in my opinion, much easier to understand than the tangent story we usually learn... but it's useful to see both points of view of course. After that, I read the article about Banach manifolds (which naturally uses the concept :))

    Thanks again :)
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  11. #10  
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    Quote Originally Posted by Stranger

    The article about little o in wikipedia is just as I learned it in my first analysis course: we say that f(x) is little o of g(x) if the limit of f(x)/g(x) tends to zero. That's all. So what does it mean here?

    I understand o(||x-p||) should mean a small error, but what is the definition exactly?
    The definition is exactly as you stated.
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  12. #11  
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    I guess I'm very dumb then, but I still don't get it :?

    I mean, ||x-p|| is a function with values in R whereas the other side of what's supposed to be an equation has values in R^3!

    I really can't see any limits to take here...
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  13. #12  
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    Quote Originally Posted by Stranger
    I guess I'm very dumb then, but I still don't get it :?

    I mean, ||x-p|| is a function with values in R whereas the other side of what's supposed to be an equation has values in R^3!

    I really can't see any limits to take here...
    take the limit as x-p tends to the vector 0.
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  14. #13  
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    Oh... I think I finally understand

    That little o would mean that, if we divide each component of

    f(x) - f(p) - J(p)(x-p)

    by

    ||x-p||

    and take the limit as x-->p, each component would go to zero. That's why you were saying "because the orders are 2 and higher" yeah pf course, so the fraction does go to zero.

    Ok, thanks a lot

    But really, they should explain that little o notation in the article just as we did here. I don't think I'm the only one who got confused there.

    Good night!
    Watch what thy eyes can't see... and live it.

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  15. #14  
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    Quote Originally Posted by Stranger
    \

    But really, they should explain that little o notation in the article just as we did here. I don't think I'm the only one who got confused there.

    Good night!
    Probably, but "little o" notation is pretty standard in analysis and assuming the reader to be familiar with it, or able to look it up, is also pretty standard.
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