- #1

- 5

- 0

^{x}is itself.

I have a feeling this might require some definition of a

^{b}.

Surely this isn't a complicated problem, but for some reason I can't work it out or even find a good answer online.

Thanks for the help.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter DylanG
- Start date

- #1

- 5

- 0

I have a feeling this might require some definition of a

Surely this isn't a complicated problem, but for some reason I can't work it out or even find a good answer online.

Thanks for the help.

- #2

I like Serena

Homework Helper

- 6,577

- 176

Actually, the property that the derivative of e

From wikipedia:

wikipedia said:In mathematics, the exponential function is the function e^{x}, where e is the number (approximately 2.718281828) such that the function e^{x}is its own derivative.[1][2]

and also from wikipedia:

wikipedia said:From any of these definitions it can be shown that the exponential function obeys the basic exponentiation identity,

exp(x+y) = exp(x) exp(y),which is why it can be written as e^{x}.

- #3

- 5,439

- 9

[tex]{e^p} = 1 + p + \frac{{{p^2}}}{{2!}} + \frac{{{p^3}}}{{3!}} + \frac{{{p^4}}}{{4!}} + ...[/tex]

differentiate with respect to p

[tex]\frac{{d\left( {{e^p}} \right)}}{{dp}} = 0 + 1 + \frac{{2p}}{{2*1}} + \frac{{3{p^2}}}{{3*2*1}} + \frac{{4{p^3}}}{{4*3*2*1}} + ...[/tex]

=[tex]1 + p + \frac{{{p^2}}}{{2!}} + \frac{{{p^3}}}{{3!}} + ... = {e^p}[/tex]

does this help?

- #4

lurflurf

Homework Helper

- 2,440

- 138

Such as f(x)=a

f(x+y)=f(x)f(y)

and

f'(0)=1/f

then we have

f(1)=a

f'(x)=f'(0)f(x)

so we take a=e<->f'(0)=1 as the standard case of such functions

- #5

- 1,768

- 126

That just tells you there is such a function. Doesn't tell you it's a^x for some a. But you can can argue that it satisfies the property exp(x + y) = exp(x) exp (y), which implies what you want.

There's also a geometric construction of the exponential function, but I can't draw pictures on here, so it's hard to explain. Basically, you start out at the point (0,1) and you do a discrete approximation and take the limit. It leads to the definition of e as a limit. I came up with it in high school when I was studying calculus, and later I found the same argument appears in Needham's Visual Complex Analysis, somewhere in the book, as a warm-up for the complex number version.

- #6

- 5,439

- 9

It all depends upon the order you define things when you build up your theory of analysis.

Another way is to state

y = exp(x) if and only if x = log(y)

@ homeomorphic

The following properties may be proved directly from the series definition:

exp(0) = 1

exp(-x) = 1/exp(x)

exp(x) > 0 ie it never vanishes

We may also use series comparison techniques to establish the derivative from first principles by rearranging and expanding

[tex]\frac{{\exp (x + h) - \exp (x)}}{h}[/tex]

But my original offering is simpler.

go well

exp(x+h)

- #7

AlephZero

Science Advisor

Homework Helper

- 6,994

- 293

When x > 0, [tex]\ln x = \int_1^x \frac{dt}{t}[/tex]

It is then easy to see that[tex]\frac{d}{dx}\ln x = \frac 1 x[/tex] and ln(x) is a continuous and monotonic function.

From the definition ln(1) = 0, and by approximating the "area under the graph" by three rectangles, ln(4) > 1/2 + 1/3 + 1/4 > 1.

So there is a number e such that 1 < e < 4 and ln(e) = 1.

It is straightforward to show properties like ln(xy) = ln(x) + ln(y) direct from the definition, using techniqes from elementary calculus.

And finally, define [itex]e^x[/itex] as the inverse function of ln(x).

The advantage of this, IMO, is that it doesn't require anything more complcated than "calculus", which is much more accessible to "beginners" than a proper justification of operations on an infinite series that was "pulled out of a hat" as the definition of [itex]e^x[/itex].

And a few years later, if and when you have taken a course on analytic functions, you can use the power series to define [itex]e^z[/itex] when z is

- #8

- 1,768

- 126

And in this context (flow along a vector field on the line or slope field), I might add that the property exp(x+y) = exp(x)exp(y) has the very elegant interpretation that time evolution by x + y is the same thing as time evolution by x followed by time evolution by y.

Share: