Proving Continuity of e^x - A Delta Epsilon Proof

[drama2]

i'm battling unsuccessfully to find a delta epsilon proof for continuity of exponential function.
this is what I've tried so far but its failed either because I've gone down a blind alley or got stuck on the right path I'm not sure which one:

find
lim e^x
x->a

therefore (epsilon) or error = e^a - e^a/e^(a-x)


e^a(1- e^(x-a)) = epsilon/error

as you can see i don't know what's going on. this is frustrating for me and i would really appreciate a convincing demonstration of why e^x is continuous.

 

[HallsofIvy]

In order to prove that any function has a given property, you have to use the definition of that function. The fundamental question is "How are you defining e^x?"

I know several definitions for e^x and in all of them continuity comes directly from the definition:

1) If n is a positive integer, then e^n is defined as "e multiplied by itself n times". If n= 0, e^0= 1. If n is a negative integer, e^n= 1/e^{-n} and -n is a positive integer so e^{-n} is already defined. If x= m/n, a rational number, then e^x= nth root of e^m. Finally, if x is any real number, then there exist a sequence of rational numbers, {r_n}, converging to x, we define e^x to be the limit e^{r_n}. If you you use that definition, the only thing left to do is to prove that the sequence {e^{r_n}} is a Cauchy sequence so that it does converge.

2) Define e^x as the inverse function to ln(x) where ln(x) is itself defined as the integral, from 1 to x, of (1/t)dt. It follows form that definition that ln(x) is defined for all positive x and that its derivative is 1/x which is never 0 for x positive and that ln(x) maps the positive numbers one-to-one onto the real numbers. That means that e^x is well-defined as a function from the real numbers to the positive real numbers and, since ln(x) is differentiable for all positive x, it is continuous for all x so its inverse, e^x is continuous for all x.

3) Define e^x as the function, y, that satisfies y'= y for all x, y(0)= 1. Since y satisfies that differential equation for all x, it is differentiable and so continuous for all x.

4) Define e^x as the power series 1+ x+ (1/2)x^2+ ...+ (1/n!)x^n+ ... It is easy to show that this series converges for all x. For closed and bounded sub-interval, then, it is uniformly convergent and so continuous for any point inside that sub-interval. Since any x is in some closed and bounded subinterval of the real numbers, e^x is continuous for all x.

 

[lavinia]

what definition of the exponential are you using?

 

[micromass]

1) If n is a positive integer, then e^n is defined as "e multiplied by itself n times". If n= 0, e^0= 1. If n is a negative integer, e^n= 1/e^{-n} and -n is a positive integer so e^{-n} is already defined. If x= m/n, a rational number, then e^x= nth root of e^m. Finally, if x is any real number, then there exist a sequence of rational numbers, {r_n}, converging to x, we define e^x to be the limit e^{r_n}. If you you use that definition, the only thing left to do is to prove that the sequence {e^{r_n}} is a Cauchy sequence so that it does converge.

And you also need to show that the limit e^{r_n} does not depend on the sequence r_n. That is, if s_n is another sequence of rational numbers, then the limit of e^{s_n} equals the limit of e^{r_n}...

 

[lavinia]

I think showing that the exponential is continuous at a single point implies that it is continuous everywhere. This follows easily because it is a homomorphism from the reals under addition to the reals under multiplication.

If one agrees that exp(x) > 1 for x>0 in a neighborhood of zero, then exp is strictly increasing in that neighborhood and so must have a point of continuity - I think.
For this one just need that the radius of convergence of its power series in a neighborhood of x=0 is not zero.

 

[jostpuur]

How do you prove that the mapping

<br /> \mathbb{Q}\to\mathbb{R},\quad x\mapsto e^x<br />

is continuous at x=0, when the definition is e^{m/n}=\sqrt[n]{e^m}?

--------------

I figured this out now. Hopefully this is interesting to drama2 too!

Here is continuity from right only. But continuity from left can be deduced later from e^{-x}=1/e^{x}.

If one first succeeds proving that the exponential mapping (on \mathbb{Q}) is monotonically increasing, then it will be enough to prove this intermediate result:

<br /> \forall\; \epsilon &gt; 0\;(\epsilon\in\mathbb{R})\quad \exists\; \delta &gt; 0\;(\delta\in\mathbb{Q})\quad\textrm{s.t.}\quad e^{\delta} &lt; 1 + \epsilon<br />

This follows from an intermediate result:

<br /> \forall\; \epsilon &gt; 0\; (\epsilon\in\mathbb{R})\quad \exists\; n\in\mathbb{N}\quad\textrm{s.t.}\quad e &lt; (1 + \epsilon)^n<br />

Which follows from the result

<br /> \forall\; \epsilon &gt; 0\; (\epsilon\in\mathbb{R})\quad\quad \lim_{n\to\infty} (1 + n\epsilon) =\infty.<br />

 

[dimitri151]

You want to show that e^{x} is continuous at x . Let \delta=\ln(\epsilon\cdot e^{-x}+1) .
Let x^{\prime}&gt;x and x^{\prime}-x&lt;\delta=\ln(\epsilon\cdot e^{-x}+1) .
Then e^{x^{\prime}-x}&lt;\epsilon\cdot e^{-x}+1\Rightarrow e^{x^{\prime}}-e^{x}&lt;\epsilon.
(Sorry have to follow through.) Therefore for every epsilon there is a delta such that bla-bla-bla.

 

[disregardthat]

And you also need to show that the limit e^{r_n} does not depend on the sequence r_n. That is, if s_n is another sequence of rational numbers, then the limit of e^{s_n} equals the limit of e^{r_n}...

And for this, it will be very useful to prove that at each step from natural to rational to real exponents, e^x is order-preserving. With this approach I think it is almost necessary.

 

[disregardthat]

You want to show that e^{x} is continuous at x . Let \delta=\ln(\epsilon\cdot e^{-x}+1) .
Let x^{\prime}&gt;x and x^{\prime}-x&lt;\delta=\ln(\epsilon\cdot e^{-x}+1) .
Then e^{x^{\prime}-x}&lt;\epsilon\cdot e^{-x}+1\Rightarrow e^{x^{\prime}}-e^{x}&lt;\epsilon.

I don't think this proof holds. You are using the ln function assuming e^x is surjective on (1,\infty). Then you apply the order-preserving property of e, which indeed in combination implies continuity on (0,\infty). Conversely, with continuity on (0,\infty) and being order-preserving, you have surjectiveness on (1,\infty). Hence you are by assuming surjectiveness on (1,\infty) actually assuming the equivalent fact that it is continuous on (0,\infty). Given that it's order-preserving, of course.

Maybe the proof holds, but at the very least it is wholly unnecessary (tautological) as you are in fact assuming the equivalence of continuity (on a large chuck of the domain anyway).

 

[jostpuur]

disregardthat is right!

Suppose hypothetically that the image of exponential mapping had holes in it. Then the domain of logarithm would have holes in it too, and you can't know what you are allowed to substitute into it.

 

[HallsofIvy]

How do you prove that the mapping

<br /> \mathbb{Q}\to\mathbb{R},\quad x\mapsto e^x<br />

is continuous at x=0, when the definition is e^{m/n}=\sqrt[n]{e^m}?

What? The topology the rational numbers inherits from the real numbers is the discrete topology. Every set is open in that topology so every function from the rational numbers to the real numbers is continuous. That helps not at all in proving that a function from the real numbers to the real numbers is continuous.

--------------

I figured this out now. Hopefully this is interesting to drama2 too!

Here is continuity from right only. But continuity from left can be deduced later from e^{-x}=1/e^{x}.

If one first succeeds proving that the exponential mapping (on \mathbb{Q}) is monotonically increasing, then it will be enough to prove this intermediate result:

<br /> \forall\; \epsilon &gt; 0\;(\epsilon\in\mathbb{R})\quad \exists\; \delta &gt; 0\;(\delta\in\mathbb{Q})\quad\textrm{s.t.}\quad e^{\delta} &lt; 1 + \epsilon<br />

This follows from an intermediate result:

<br /> \forall\; \epsilon &gt; 0\; (\epsilon\in\mathbb{R})\quad \exists\; n\in\mathbb{N}\quad\textrm{s.t.}\quad e &lt; (1 + \epsilon)^n<br />

Which follows from the result

<br /> \forall\; \epsilon &gt; 0\; (\epsilon\in\mathbb{R})\quad\quad \lim_{n\to\infty} (1 + n\epsilon) =\infty.<br />

 

[jostpuur]

The set \mathbb{Q} is naturally a metric space with the metric

<br /> \mathbb{Q}\times\mathbb{Q}\to\mathbb{R},\quad (x,x&#039;)\mapsto |x-x&#039;|<br />

where the absolute value is the ordinary absolute value of the real numbers. When a function f:\mathbb{Q}\to\mathbb{R} is said to be continuous at x\in\mathbb{Q}, it should be interpreted to mean this:

<br /> \forall \epsilon &gt; 0 \quad \exists \delta &gt; 0\; \quad\textrm{s.t.}\quad \forall x&#039;\in\mathbb{Q},\quad |x-x&#039;|&lt;\delta\implies |f(x)-f(x&#039;)|&lt;\epsilon<br />

The topology the rational numbers inherits from the real numbers is the discrete topology.

I'm afraid this claim is incorrect. Because it is impossible to write a set \{x\}\subset\mathbb{Q} as \{x\}=V\cap \mathbb{Q} where V\subset\mathbb{R} would be open in \mathbb{R}.

 

[dimitri151]

disregardthat is wrong. The proof is correct. I do begin assuming that e^x is defined, injective and surjective on positive reals, but then there's nothing wrong with that. I also used a<b+c implies a-c<b, and a^b*a^c =a^(b+c). Those things are easy to show, and the question is for an epsilon delta proof of continuity not for a proof of the foundations of mathematics.

 

[jostpuur]

disregardthat is wrong. The proof is correct. I do begin assuming that e^x is defined, injective and surjective on positive reals, but then there's nothing wrong with that.

How do you prove that \mathbb{R}\to ]0,\infty[,\quad x\mapsto e^x is surjective?

If somebody suspects that the image could contain holes like this:

<br /> \exp ([0,\infty[ ) = [1,x_1]\;\cup\; [x_1+\delta, x_2]\;\cup\; [x_2+\delta, x_3]\;\cup\;\cdots<br />

with some very small \delta&gt;0 and small 1&lt;x_1&lt;x_2&lt;\cdots?

How do you convince the suspecting person, that these holes don't exist in the image?

Remember, in order for you proof to work, you must prove that the holes don't exist, before you prove that the exponential mapping is continuous! How you do accomplish that without falling victim to a circular reasoning?

 

[lavinia]

I think showing that the exponential is continuous at a single point implies that it is continuous everywhere. This follows easily because it is a homomorphism from the reals under addition to the reals under multiplication.

If one agrees that exp(x) > 1 for x>0 in a neighborhood of zero, then exp is strictly increasing in that neighborhood and so must have a point of continuity - I think.
For this one just need that the radius of convergence of its power series in a neighborhood of x=0 is not zero.

this can be done by calculating the power series for exp(x) for a single positive number, say 1

 

[HallsofIvy]

this can be done by calculating the power series for exp(x) for a single positive number, say 1

But that requires assuming that exp(x) is analytic, at least in a neighborhood of 1, which is far more strict than "continuous" which is what you are attempting to prove.

 

[dimitri151]

To show surjectivity of e^x you first show that it is monotone increasing. The you know that the only discontinuity possible is a jump discontinuity. Say the jump discontinuity is at the point c. Supposing (by way of contradiction) that there is a jump discontinuity at c, you evaluate e^{c+\epsilon}-e^{c-\epsilon}. You can then go two ways that I can see, you can take the limit of e^{\epsilon}-e^{-\epsilon} as \epsilon goes to infinity which is 1-1=0, or you can expand e^{c+\epsilon}-e^{c-\epsilon} in its series (say you've defined e^x as a certain series). Then you see all the terms have epsilon in them or powers of epsilon, so the difference e^{c+\epsilon}-e^{c-\epsilon} may not be finite as required for there to be a discontinuity but vanishes by virtue of the form of the expansion. Therefore there are no discontinuities. Hmm, maybe I should have said that in the first place...

P.S. how do you do laytex in sentence?

 

[lavinia]

But that requires assuming that exp(x) is analytic, at least in a neighborhood of 1, which is far more strict than "continuous" which is what you are attempting to prove.

no. analytic is not assumed. The general theorem is that any power series has a radius of convergence for which it converges for all values within the radius and diverges for all values without.
This can be a real or complex series and makes no assumptions about differentiability. So for the exponential function on the real line, proof of convergence at 0 and 1 proves convergence for all numbers between zero and 1 so for all of those numbers the exponential must be strictly increasing and hence have a point of continuity.

 

[Nebuchadnezza]

use itex and /itex to allign latex in sentences.

 

[dimitri151]

I mean as \epsilon goes to zero.
Thanks Neb.

 

[disregardthat]

dimitri, you are aware of that your proof of surjectivity (just skimmed it, I assume it was a proof of this), actually implies continuity given order-preservation? You have really just expanded my point, namely that your assumptions are in fact sufficient for continuity, rendering the epsilon-delta proof unnecessary.

 

[dimitri151]

disregardthat, was your point that a monotone increasing, surjective function is continuous?

 

[disregardthat]

disregardthat, was your point that a monotone increasing, surjective function is continuous?

Yes, these were two facts you assumed prior to proving continuity.

 

[dimitri151]

The monotonicity and surjectivity does prove the continuity, but we're not wondering how to prove continuity. We're trying to show a epsilon-delta proof of continuity like the asker requested.
You can also use epsilon-delta on the definition of e^x directly if you're severely allergic to using the inverse function. Say e^{x}=\underset{n\rightarrow\infty}{\lim}(1+\frac{x}{n})^n. Then for x_1=x+\delta,
e^{x+\delta}-e^x=\underset{n\rightarrow\infty}{\lim}((1+\frac{x+\delta}{n})^n-(1+\frac{x}{n})^n)= a bunch of terms all with \delta in them and all the n's disappear. You can also prove it like that and skirt the whole using-an-inverse-without-proving-surjectivity controversy. Similarly with the infinite series expansion, all the terms will have \delta in them.