Ciro Santilli @cirosantilli 34

 Incoming links: Taylor series

Identity theorem  Updated 2024-12-15  +Created 1970-01-01

Essentially, defining an holomorphic function on any open subset, no matter how small, also uniquely defines it everywhere.

This is basically why it makes sense to talk about analytic continuation at all.

One way to think about this is because the Taylor series matches the exact value of an holomorphic function no matter how large the difference from the starting point.

Therefore a holomorphic function basically only contains as much information as a countable sequence of numbers.

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Product definition of the exponential function  Updated 2024-12-15  +Created 1970-01-01

e^{x} = lim_{n \to \infty} (1 + \frac{x}{n})^{n}

The basic intuition for this is to start from the origin and make small changes to the function based on its known derivative at the origin.

More precisely, we know that for any base b, exponentiation satisfies:

$b^{x + y} = b^{x} b^{y}$ .
$b^{0} = 1$ .

And we also know that for

b = e

in particular that we satisfy the exponential function differential equation and so:

\frac{d e ^{x}}{d x} (0) = 1

One interesting fact is that the only thing we use from the exponential function differential equation is the value around

x = 0

, which is quite little information! This idea is basically what is behind the importance of the ralationship between Lie group-Lie algebra correspondence via the exponential map. In the more general settings of groups and manifolds, restricting ourselves to be near the origin is a huge advantage.

Now suppose that we want to calculate

e^{1}

. The idea is to start from

e^{0}

and then then to use the first order of the Taylor series to extend the known value of

e^{0}

to

e^{1}

.

E.g., if we split into 2 parts, we know that:

e^{1} = e^{1 / 2} e^{1 / 2}

or in three parts:

e^{1} = e^{1 / 3} e^{1 / 3} e^{1 / 3}

so we can just use arbitrarily many parts

e^{1 / n}

that are arbitrarily close to

x = 0

:

e^{1} = (e^{1 / n})^{n}

and more generally for any

x

we have:

e^{x} = (e^{x / n})^{n}

Let's see what happens with the Taylor series. We have near

y = 0

in little-o notation:

e^{y} = 1 + y + o (y)

Therefore, for

y = x / n

, which is near

y = 0

for any fixed

x

:

e^{x / n} = 1 + x / n + o (1 / n)

and therefore:

e^{x} = (e^{x / n})^{n} = (1 + x / n + o (1 / n))^{n}

which is basically the formula tha we wanted. We just have to convince ourselves that at

lim_{n \to \infty}

, the

o (1 / n)

disappears, i.e.:

(1 + x / n + o (1 / n))^{n} = (1 + x / n)^{n}

To do that, let's multiply

e^{y}

by itself once:

e^{y} e^{y} = (1 + y + o (y)) (1 + y + o (y)) = 1 + 2 y + o (y)

and multiplying a third time:

e^{y} e^{y} e^{y} = (1 + 2 y + o (y)) (1 + y + o (y)) = 1 + 3 y + o (y)

TODO conclude.

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Taylor expansion definition of the exponential function  Updated 2024-12-15  +Created 1970-01-01

The Taylor series expansion is the most direct definition of the expontial as it obviously satisfies the exponential function differential equation:

the first constant term dies
each other term gets converted to the one before
because we have infinite many terms, we get what we started with!

e^{x} = \sum_{n = 0}^{\infty} \frac{x ^{n}}{n !} = 1 + \frac{x}{1} + \frac{x ^{2}}{2} + \frac{x ^{3}}{2 \times 3} + \frac{x ^{4}}{2 \times 3 \times 4} + \dots

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The derivative is the generator of the translation group  Updated 2024-12-15  +Created 1970-01-01

Take the group of all Translation in

R^{1}

.

Let's see how the generator of this group is the derivative operator:

\frac{\partial}{\partial x}

The way to think about this is:

the translation group operates on the argument of a function $f (x)$
the generator is an operator that operates on $f$ itself

So let's take the exponential map:

e^{x_{0} \frac{\partial}{\partial x}} f (x) = (1 + x_{0} \frac{\partial}{\partial x} + x_{0}^{2} \frac{\partial ^{2}}{\partial x ^{2}} + \dots) f (x)

and we notice that this is exactly the Taylor series of

f (x)

around the identity element of the translation group, which is 0! Therefore, if

f (x)

behaves nicely enough, within some radius of convergence around the origin we have for finite

x_{0}

:

e^{x_{0} \frac{\partial}{\partial x}} f (x) = f (x + x_{0})

This example shows clearly how the exponential map applied to a (differential) operator can generate finite (non-infinitesimal) Translation!

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