# Thread: Derivative of e^x from first principles

1. ## Re: Derivative of e^x from first principles

Originally Posted by HallsofIvy

That is important for the following reason: If X is any positive real number, then $\displaystyle ln(2^{2X})= 2X ln(2)\ge (2X)(1/2)= X$. That is, ln(x) is not bounded above.
Could you please explain this line? How does the inequality imply that $\displaystyle ln(x)$ is not bounded above?

Anyway, thanks very much for your detailed response.

The fact that there are so many equivalent definitions kind of bothers me. Its like everything becomes kind of circular. For example when I was learning Analysis(I am still learning it, in fact), one book would define a closed set as the complement of an open set and it would ask me to prove that a closed set contains all of its limit points. Another book would instead define a closed set as a set which contains all of its limit points and ask me to prove that its complement is open. I was like, "What's the point to all this?". I had fun proving it, but it seemed pointless.

2. ## Re: Derivative of e^x from first principles

Originally Posted by sakodo

Looking at posts #3 and #4 by Prove it and chisigma respectively, which one came first? The definition of $\displaystyle e=\lim_{n\to 0} (1+\frac{1}{n})^{n}$ or the defintion of $\displaystyle e^x$ as a function whose derivative is itself? Since if you use either one definition, you can always prove the other.

I think this is what caused my confusion.
In the XVIII° century Leonhard Euler extablished the following definitions for the exponential and logarithm functions of a real variable x ...

$\displaystyle e^{x} = \lim_{n \rightarrow \infty} (1+\frac{x}{n})^{n}$ (1)

... and defined the function $\displaystyle \ln x$ as the inverse of the exponential. The alternative to define...

$\displaystyle \ln x = \int_{1}^{x} \frac{dt}{t}$ (2)

... and the exponential the inverse of logarithm has been proposed in the succesive century so that (1) did come first. Apparently each definition approach is reasonably and there are no reason to prefer one of them... but that is not true if we extend the functions to a complex variable z. Ever Euler discovered the famous formula with his name...

$\displaystyle e^{z}= e^{x + i y} = e^{x} (\cos y + i \sin y)$ (3)

... which permits to define the exponential as a single value function for any value of z. Define the logarithm in the complex field as the inverse of exponential is of course possible but if we do that the function $\displaystyle \ln z$ defined from (3) is a multivalued function , i.e. is defined unless a constant $\displaystyle 2 \pi i$ and that is true also if we use the definition...

$\displaystyle \ln z = \int_{1}^{z} \frac{d s}{s}$ (4)

... because the (4) depends from the path connecting the points $\displaystyle s=1$ and $\displaystyle s=z$...

The conclusion, in my opinion, is the following: taking the Euler's way You never will have surprises... taking other ways I don't know ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. ## Re: Derivative of e^x from first principles

Originally Posted by sakodo
How does the inequality imply that $\displaystyle \ln(x)$ is not bounded above?
Here is the idea. For any $\displaystyle a>0$ we want to show that for some $\displaystyle b>0$ we have $\displaystyle a<\ln(b)$.
That is, there is no upper bound for $\displaystyle \ln(x)$.

We know that $\displaystyle \ln(3)>1$ and that implies $\displaystyle a<a\ln(3)=\ln(3^a)$.
There we have it: $\displaystyle b=3^a$.

4. ## Re: Derivative of e^x from first principles

Originally Posted by chisigma
In the XVIII° century Leonhard Euler extablished the following definitions for the exponential and logarithm functions of a real variable x ...

$\displaystyle e^{x} = \lim_{n \rightarrow \infty} (1+\frac{x}{n})^{n}$ (1)

... and defined the function $\displaystyle \ln x$ as the inverse of the exponential. The alternative to define...

$\displaystyle \ln x = \int_{1}^{x} \frac{dt}{t}$ (2)

... and the exponential the inverse of logarithm has been proposed in the succesive century so that (1) did come first. Apparently each definition approach is reasonably and there are no reason to prefer one of them... but that is not true if we extend the functions to a complex variable z. Ever Euler discovered the famous formula with his name...

$\displaystyle e^{z}= e^{x + i y} = e^{x} (\cos y + i \sin y)$ (3)

... which permits to define the exponential as a single value function for any value of z. Define the logarithm in the complex field as the inverse of exponential is of course possible but if we do that the function $\displaystyle \ln z$ defined from (3) is a multivalued function , i.e. is defined unless a constant $\displaystyle 2 \pi i$ and that is true also if we use the definition...

$\displaystyle \ln z = \int_{1}^{z} \frac{d s}{s}$ (4)

... because the (4) depends from the path connecting the points $\displaystyle s=1$ and $\displaystyle s=z$...

The conclusion, in my opinion, is the following: taking the Euler's way You never will have surprises... taking other ways I don't know ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
this is why defining the exponential function as a power series seems to me, in the long run, the most sensible. it also makes Euler's identity (definition of $\displaystyle e^{iy}$) seem less arbitrary. from such a definition, one can quickly recover the differential equations $\displaystyle e^x$ sine and cosine satisfy, as well the values these functions and their derivatives have at 0. moreover, this approach yields an analytic definition of the real numbers $\displaystyle \pi$ and $\displaystyle e$ as certain well-defined rational cauchy sequences, even giving a practical way of calculating (approximations to) them. this in turn, justifies why the function 1/x is a derivative of $\displaystyle exp^{-1}$. although the integral definition of ln is perfectly adequate for the real numbers, it is hard to understand "why" we would do such a thing (other than it works!).

5. ## Re: Derivative of e^x from first principles

Personally, I despise definitions in terms of power series. The really do seem to be arbitrary - especially for trigonometric functions. I can see no reason why anyone would seriously be interested in investigating such things - unless they had already spotted that $${\mathrm d \over \mathrm d x}\mathrm e^x = \mathrm e^x$$

Conversely, it is extremely easy to see why people might have wanted a function for which $$f(xy) = f(x) + f(y)$$ to simplify multiplication and division. The only non-trivial and differentiable solutions of this functional equation are the logarithmic functions (up to certain minor distinctions). The particular form of the logarithmic function obtained via this method is$$\log x = \int_1^x {1 \over t} \, \mathrm d t$$
You then have to define $\mathrm e$: but doing so such that $\log \mathrm e = 1$ is certainly no more arbitrary than $$\mathrm e = \lim_{x \to \infty} \left(1+{1 \over n}\right)^n$$
or any other definition you care to name.

You then move on to define the exponential function $\mathrm e^x$ for any $x$ as the inverse of the logarithm and then and so get all exponentials $a^x = \mathrm e^{x \log a}$. This avoids all the nastiness of defining the exponential operation (traditionally done for the naturals as $a^1 = a$ and $a^{n+1} = a \cdot a^n$ and then extending this definition to the rest of the integers, the rationals and finally the reals via Dedekind cuts or some similarly messy route).

No other approach seems to me to be so comprehensively simple, natural and straightforward.

6. ## Re: Derivative of e^x from first principles

Originally Posted by Prove It
The function is DEFINED to be its own derivative. You CAN use this definition to evaluate the value of $\displaystyle \displaystyle e$ though.

\displaystyle \displaystyle \begin{align*} \lim_{h \to 0}\frac{e^{x + h} - e^x}{h} &= e^x \\ \lim_{h \to 0}\frac{e^xe^h - e^x}{h} &= e^x \\ \lim_{h \to 0}\frac{e^x(e^h - 1)}{h} &= e^x \\ e^x\lim_{h \to 0}\frac{e^h - 1}{h} &= e^x \\ \lim_{h \to 0}\frac{e^h - 1}{h} &= 1 \\ \lim_{h \to 0}(e^h - 1) &= \lim_{h \to 0}h \\ \lim_{h \to 0}e^h &= \lim_{h \to 0}(1 + h) \\ \lim_{h \to 0}(e^h)^{\frac{1}{h}} &= \lim_{h \to 0}(1 + h)^{\frac{1}{h}} \\ \lim_{h \to 0}(e) &= \lim_{h \to 0}(1 + h)^{\frac{1}{h}} \\ e &= \lim_{h \to 0}(1 + h)^{\frac{1}{h}} \end{align*}

And if you let $\displaystyle \displaystyle h = \frac{1}{n}$, then $\displaystyle \displaystyle n \to \infty$ as $\displaystyle \displaystyle h \to 0$, so we can write $\displaystyle \displaystyle e$ as its more common limit form

\displaystyle \displaystyle \begin{align*} e &= \lim_{h \to 0}(1 + h)^{\frac{1}{h}} \\ e &= \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n \end{align*}
Wait a minute. I am completely confused after reading this thread. If you can prove something then why would we define. We can prove using the definition of a derivative that (d/dx)(ln(x)) = 1/x.

Now x = ln(e^x). Taking derivative of both sides we get 1 = [(d/dx)(e^x)]/e^x. So (d/dx)(e^x) = e^x.
What is wrong with this proof? Why define (d/dx)(e^x) = e^x.

7. ## Re: Derivative of e^x from first principles

Originally Posted by chisigma
All right!... in that case I propose to all You the following 'easy little task': demonstrate on the basis of the defintion above that is...

$\displaystyle \ln (x_{1}\ x_{2}) = \ln x_{1} + \ln x_{2}$ (1)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
Of course. Though I would prefer to use "x" and "y" as variables.
I would start by proving that ln(1/x)= -ln(x).
By the definition I gave, $\displaystyle ln(1/x)= \int_1^{1/x} \frac{1}{t}dt$. Let u= xt so that $\displaystyle t= \frac{u}{x}$ and $\displaystyle dt= \frac{du}{x}$. When t= 1, u= x and when t= 1/x, u= 1. So the integral becomes $\displaystyle \int_x^1 \frac{x}{u}\frac{1}{x} du= -\int_1^x \frac{1}{u}du= -ln(x)$.

Now, $\displaystyle ln(xy)= \int_1^{xy} \frac{1}{t} dt$. Let u= t/y. Then du= (1/y) dt so that dt= ydu. When t= 1, u= 1/y and when t= xy, u= x. The integral becomes $\displaystyle \int_{1/y}^x \frac{1}{uy}(1/y)du= \int_{1/y}^x \frac{1}{u}du$$\displaystyle = \int_{1/y}^1 \frac{1}{u}du+ \int_1^x \frac{1}{u} du$$\displaystyle = \int_1^x \frac{1}{u}du- \int_1^{1/y} \frac{1}{u}du= ln(x)- ln(1/y)= ln(x)+ ln(y)$.

8. ## Re: Derivative of e^x from first principles

Originally Posted by JaguarXJS
Wait a minute. I am completely confused after reading this thread. If you can prove something then why would we define. We can prove using the definition of a derivative that (d/dx)(ln(x)) = 1/x.
Yes, if you first define ln(x) then you can do this. Exactly HOW do you define ln(x)? That was my point before. I would define $\displaystyle ln(x)= \int_1^x \frac{1}{t}dt$ so that $\displaystyle \frac{d ln(x)}{dx}= \frac{1}{x}$ follows immediately.

Now x = ln(e^x). Taking derivative of both sides we get 1 = [(d/dx)(e^x)]/e^x. So (d/dx)(e^x) = e^x.
What is wrong with this proof? Why define (d/dx)(e^x) = e^x.

9. ## Re: Derivative of e^x from first principles

Originally Posted by JaguarXJS
Wait a minute. I am completely confused after reading this thread. If you can prove something then why would we define.
It is very common for functions or concepts to have multiple equivalent definitions. Which one you take depends on what you are trying to prove and personal preference. It's similar to roads. Take Route 66, for example. Some people would say that it went from Chicago, Il to Santa Monica, Ca. Others would say that it went from Santa Monica to Chicago. It doesn't matter. The road passed the same waypoints whichever way you travelled.

In some cases, picking the wrong definition can destroy a proof by introducing a circular argument - this is a common problem with attempted proofs that $$\lim_{x \to 0} {\sin x \over x} = 1$$

10. ## Re: Derivative of e^x from first principles

Originally Posted by sakodo
Could you please explain this line? How does the inequality imply that $\displaystyle ln(x)$ is not bounded above?

Anyway, thanks very much for your detailed response.
The inequality came from $\displaystyle ln(2^{2x})= 2x ln(2)$ and then, since ln(2)> 1/2, $\displaystyle ln(2^{2x})> x$. The definition of "bounded above" is that a function is bounded above if and only if there exist some number, M, such that $\displaystyle f(x)\le M$ for all x. But, given any number, M, the above shows that if [tex]x= 2^{2M}[tex], then ln(x)> M so there is NO upper bound for ln(x).

The fact that there are so many equivalent definitions kind of bothers me. Its like everything becomes kind of circular. For example when I was learning Analysis(I am still learning it, in fact), one book would define a closed set as the complement of an open set and it would ask me to prove that a closed set contains all of its limit points. Another book would instead define a closed set as a set which contains all of its limit points and ask me to prove that its complement is open. I was like, "What's the point to all this?". I had fun proving it, but it seemed pointless.
So you want everyone to always do things the same way? There are always different ways of looking at the same thing, different ways of defining the same thing, and different ways of solving the same problem. Often, different definitions give an easier way of proving some property. Having more tools in your toolkit gives you more flexibility. All that is important is that the different definitions be equivalent.

11. ## Re: Derivative of e^x from first principles

Originally Posted by sakodo
The fact that there are so many equivalent definitions kind of bothers me. Its like everything becomes kind of circular. For example when I was learning Analysis(I am still learning it, in fact), one book would define a closed set as the complement of an open set and it would ask me to prove that a closed set contains all of its limit points. Another book would instead define a closed set as a set which contains all of its limit points and ask me to prove that its complement is open. I was like, "What's the point to all this?". I had fun proving it, but it seemed pointless.
Again we can use an analogy based on roads. The different definitions and exercises are journeys around the mathematical road network. What is important is not so much where you start from, but where you are aiming for. And if you have a good knowledge of the roads in that area and how they link up, you are more likely to be able to find your way to your destination.

Page 2 of 2 First 12

,

,

,

,

,

,

,

,

,

,

,

,

,

,

# differentiation of natural logarithm of x from the first principle

Click on a term to search for related topics.