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It is’t the exponential function that’s a miracle, it’s the Maclaurin series. Think about it. The Maclaurin series says that in a very real sense, the value of

It is’t the exponential function that’s a miracle, it’s the Maclaurin series. Think about it. The Maclaurin series says that in a very real sense, the value of a function at zero, and the value of all of its derivatives at zero, contains all of the information there is to know about it. Intuitively, this should be completely unexpected. Why in the world would the derivatives of a function evaluated at a single point tell you anything whatsoever about the value of the function arbitrarily far away from that point? EDIT: as mentioned below, this is not true of all functions, even all functions that are infinitely differentiable at zero. But it is true for very large classes of functions.

obastani 7y ago
That's not actually true. There exist functions of a single real variable that have derivatives at the origin that are all zero, yet are nonzero. For an example, see the following: https://en.wikipedia.org/wiki/Taylor_series#Analytic_functions https://en.wikipedia.org/wiki/Taylor_series#Analytic_functio... Nevertheless, it is actually true that all complex differentiable functions satisfy this property, which is miraculous.
- GlenTheMachine 7y ago
  Well, sure. Not all functions are described by their Taylor expansions. But this very large class is. I agree, if the word “miraculous” applies at all in mathematics, then this surely is one.
- petters 7y ago
  > There exist functions of a single real variable that have derivatives at the origin that are all zero, yet are nonzero The function that is equal to 0 for x<1 and equal to 1 otherwise also satisfies this.
  - messe 7y ago
    The more interesting statement is: > There exist smooth functions of a single real variable that have derivatives at the origin that are all zero, yet are nonzero https://en.wikipedia.org/wiki/Non-analytic_smooth_function https://en.wikipedia.org/wiki/Non-analytic_smooth_function
  - posterboy 7y ago
    fyi, that's the heavyside function
boyobo 7y ago
This seems pretty tautological to me. this is not true of all functions... But it is true for very large classes of functions Which classes of functions? It sounds like you're saying that a function is determined by its derivatives if it's determined by its derivatives.
- GlenTheMachine 7y ago
  Not really. Sure, the technical term is an “analytic function”, the definition of which is, essentially, “any function that is described by it’s Taylor series”. The surprising thing is how many functions are analytic. All polynomials. The exponential, and sums of exponentials. The trigonometric functions. Bessel functions. All complex, differentiable functions. Combinations of the above. Nearly any function you are likely to encounter in a physics class, to the best of my knowledge, is analytic (IANAP). Another way to look at this is: “almost all” smooth functions are “nearly” polynomials. Why should trigonometric functions be “nearly” polynomials? Why should exponentials? Why should there be any connection at all between the trig functions and the polynomials? I still say this is a surprising result.
  - boyobo 7y ago
    Let's take "analytic" to mean "determined by its maclaurin series". It's not surprising that polynomials are analytic. Look at the definition of a polynomial. It's already in "power series" form, in fact there are only finitely many terms! As for the other examples, here is one explanation for why they are ubiquitous in physics. Essentially, the class of analytic functions is closed under "solving differential equations". So that's why sin/cos/exp/bessel are analytic - they are solutions to differential equations with constant/polynomial coefficients (we already know that constants and polynomials are analytic). That's why many functions found in physics are analytic - they are created from other analytic functions via a differential equation. https://math.stackexchange.com/a/190167 https://math.stackexchange.com/a/190167 P.S: Regarding your last statement "almost all smooth functions are almost polynomial", something much stronger is true: Every continuous function is almost a polynomial. https://en.wikipedia.org/wiki/Stone–Weierstrass_theorem https://en.wikipedia.org/wiki/Stone–Weierstrass_theorem
6nf 7y ago
It's true for analytical functions I think?
- boyobo 7y ago
  Analytic functions are usually defined to be those functions which can be represented by their Taylor series. This is what I meant by `tautological'.
  - devnulloverflow 7y ago
    It's one of the definitions, and perhaps not the most fundamental. In the complex plane (and all real analytic functions can be analytically extended into the complex plane), you can define the equivalent property as saying a (first!) derivative has the some value no matter how you chose to take the limit. So just by virtue of having a well defined first derivative in a particular region you get all the Maclaurin series magic.
md224 7y ago
> Intuitively, this should be completely unexpected. It actually makes intuitive sense to me. I like to think of it this way: 1) If we have the value of a function at a given point and we want to extrapolate the function's values before and after that point, we need to know how the value is changing at that point: the derivative. 2) But if that derivative isn't constant, that won't get us very far. We also need to know how the derivative is changing at the given point: the 2nd derivative. 3) But if that 2nd derivative isn't constant, that won't get us very far. We also need to know how the 2nd derivative is changing at the given point: the 3rd derivative. And so on, potentially up to infinity. But once we take into account all of the derivatives, we know how the value of the function is changing and how that change is itself changing, and as there is no additional change that comes "out of nowhere", so to speak, we have enough information to calculate the value at any other point.
- stoops 7y ago
  I love math. Thank you for the beautiful explanation.
- objektif 7y ago
  But the problem is that the definitions of those derivatives are for infonitesimal change. What about far away points?
  - AstralStorm 7y ago
    The definition works for any point. However often for higher functions you need to break the function into ranges to find closed form derivative series and get a useful result.
  - ddxxdd 7y ago
    Far away points have properties resembling infinitesimally close points, as long as the function is continuous. Now when there's a lack of continuity, then we run into a problem.
  - galaxyLogic 7y ago
    I'm not a mathematician but I assume there are functions with infinite number of derivatives. For such you cannot know all the values of the function by knowing the values of all derivatives because you would have to know infinitely many numbers. Whereas if there are a fixed number of derivatives then at some level the derivative is constant and thus you can intuitively think that based on those derivatives you can then "draw" the function.
Sharlin 7y ago
I find integrals even more magical. Just by evaluating the antiderivative of a function at a single point we get a summary of the behavior of the function everywhere between that point and minus infinity!
- TeMPOraL 7y ago
  And like proper miracles, it often takes a ton of work and some divine insight to find an integral analytically.
  - nightcracker 7y ago
    Which is funny, in the analytical world differentiation is easy and integration is hard, but in the numerical world it's the other way around.
imtringued 7y ago
It doesn't appear miraculous because knowing all derivatives for a function is quite a lot of information about a function. A polynomial of degree six can be described with six derivatives but it can also be described with six factors. There is no information advantage.
- Jeff_Brown 7y ago
  If you define a polynomial by its coefficients, and then take its derivatives at a point and forget the coefficients, it would be reasonable to wonder whether you had lost some information. The miracle is that you haven't (if you also remember that the original function is a polynomial).
Jeff_Brown 7y ago
> it is true for very large classes of functions. While it is, yes, from the point of view of a human counting them, very large, measure-theoretically it is a meager set. If you were picking functions at random from the space of all possible functions, you would never (i.e. with probability 0) pick one with a useful Maclaurin series.

obastani 7y ago

That's not actually true. There exist functions of a single real variable that have derivatives at the origin that are all zero, yet are nonzero. For an example, see the following: https://en.wikipedia.org/wiki/Taylor_series#Analytic_functions https://en.wikipedia.org/wiki/Taylor_series#Analytic_functio... Nevertheless, it is actually true that all complex differentiable functions satisfy this property, which is miraculous.

GlenTheMachine 7y ago
Well, sure. Not all functions are described by their Taylor expansions. But this very large class is. I agree, if the word “miraculous” applies at all in mathematics, then this surely is one.
petters 7y ago
> There exist functions of a single real variable that have derivatives at the origin that are all zero, yet are nonzero The function that is equal to 0 for x<1 and equal to 1 otherwise also satisfies this.
- messe 7y ago
  The more interesting statement is: > There exist smooth functions of a single real variable that have derivatives at the origin that are all zero, yet are nonzero https://en.wikipedia.org/wiki/Non-analytic_smooth_function https://en.wikipedia.org/wiki/Non-analytic_smooth_function
- posterboy 7y ago
  fyi, that's the heavyside function

boyobo 7y ago

This seems pretty tautological to me. this is not true of all functions... But it is true for very large classes of functions Which classes of functions? It sounds like you're saying that a function is determined by its derivatives if it's determined by its derivatives.

GlenTheMachine 7y ago
Not really. Sure, the technical term is an “analytic function”, the definition of which is, essentially, “any function that is described by it’s Taylor series”. The surprising thing is how many functions are analytic. All polynomials. The exponential, and sums of exponentials. The trigonometric functions. Bessel functions. All complex, differentiable functions. Combinations of the above. Nearly any function you are likely to encounter in a physics class, to the best of my knowledge, is analytic (IANAP). Another way to look at this is: “almost all” smooth functions are “nearly” polynomials. Why should trigonometric functions be “nearly” polynomials? Why should exponentials? Why should there be any connection at all between the trig functions and the polynomials? I still say this is a surprising result.
- boyobo 7y ago
  Let's take "analytic" to mean "determined by its maclaurin series". It's not surprising that polynomials are analytic. Look at the definition of a polynomial. It's already in "power series" form, in fact there are only finitely many terms! As for the other examples, here is one explanation for why they are ubiquitous in physics. Essentially, the class of analytic functions is closed under "solving differential equations". So that's why sin/cos/exp/bessel are analytic - they are solutions to differential equations with constant/polynomial coefficients (we already know that constants and polynomials are analytic). That's why many functions found in physics are analytic - they are created from other analytic functions via a differential equation. https://math.stackexchange.com/a/190167 https://math.stackexchange.com/a/190167 P.S: Regarding your last statement "almost all smooth functions are almost polynomial", something much stronger is true: Every continuous function is almost a polynomial. https://en.wikipedia.org/wiki/Stone–Weierstrass_theorem https://en.wikipedia.org/wiki/Stone–Weierstrass_theorem

6nf 7y ago

It's true for analytical functions I think?

boyobo 7y ago
Analytic functions are usually defined to be those functions which can be represented by their Taylor series. This is what I meant by `tautological'.
- devnulloverflow 7y ago
  It's one of the definitions, and perhaps not the most fundamental. In the complex plane (and all real analytic functions can be analytically extended into the complex plane), you can define the equivalent property as saying a (first!) derivative has the some value no matter how you chose to take the limit. So just by virtue of having a well defined first derivative in a particular region you get all the Maclaurin series magic.

md224 7y ago

> Intuitively, this should be completely unexpected. It actually makes intuitive sense to me. I like to think of it this way: 1) If we have the value of a function at a given point and we want to extrapolate the function's values before and after that point, we need to know how the value is changing at that point: the derivative. 2) But if that derivative isn't constant, that won't get us very far. We also need to know how the derivative is changing at the given point: the 2nd derivative. 3) But if that 2nd derivative isn't constant, that won't get us very far. We also need to know how the 2nd derivative is changing at the given point: the 3rd derivative. And so on, potentially up to infinity. But once we take into account all of the derivatives, we know how the value of the function is changing and how that change is itself changing, and as there is no additional change that comes "out of nowhere", so to speak, we have enough information to calculate the value at any other point.

stoops 7y ago
I love math. Thank you for the beautiful explanation.
objektif 7y ago
But the problem is that the definitions of those derivatives are for infonitesimal change. What about far away points?
- AstralStorm 7y ago
  The definition works for any point. However often for higher functions you need to break the function into ranges to find closed form derivative series and get a useful result.
- ddxxdd 7y ago
  Far away points have properties resembling infinitesimally close points, as long as the function is continuous. Now when there's a lack of continuity, then we run into a problem.
- galaxyLogic 7y ago
  I'm not a mathematician but I assume there are functions with infinite number of derivatives. For such you cannot know all the values of the function by knowing the values of all derivatives because you would have to know infinitely many numbers. Whereas if there are a fixed number of derivatives then at some level the derivative is constant and thus you can intuitively think that based on those derivatives you can then "draw" the function.

Sharlin 7y ago

I find integrals even more magical. Just by evaluating the antiderivative of a function at a single point we get a summary of the behavior of the function everywhere between that point and minus infinity!

TeMPOraL 7y ago
And like proper miracles, it often takes a ton of work and some divine insight to find an integral analytically.
- nightcracker 7y ago
  Which is funny, in the analytical world differentiation is easy and integration is hard, but in the numerical world it's the other way around.

imtringued 7y ago

It doesn't appear miraculous because knowing all derivatives for a function is quite a lot of information about a function. A polynomial of degree six can be described with six derivatives but it can also be described with six factors. There is no information advantage.

Jeff_Brown 7y ago
If you define a polynomial by its coefficients, and then take its derivatives at a point and forget the coefficients, it would be reasonable to wonder whether you had lost some information. The miracle is that you haven't (if you also remember that the original function is a polynomial).

Jeff_Brown 7y ago

> it is true for very large classes of functions. While it is, yes, from the point of view of a human counting them, very large, measure-theoretically it is a meager set. If you were picking functions at random from the space of all possible functions, you would never (i.e. with probability 0) pick one with a useful Maclaurin series.