Derivatives of Exponentials (why e?)

  • #1
278
0
Why is it that the constant e is defined as being the unique case where the limit as h goes to 0 of (e^h - 1)/h = 1? I mean every exponential function like a^x equals 1 when x equals 0, right? So would it be fair to say that (a^h) approaches 1 as h approaches zero? And that (a^h - 1) approaches zero as h approaches zero? And finally that the limit as h goes to 0 of (a^h - 1)/h also equals one?

I am confused as to why is it that only e is such that d/dx(e^x)=e^x
 

Answers and Replies

  • #2
pwsnafu
Science Advisor
1,080
85
Why is it that the constant e is defined as being the unique case where the limit as h goes to 0 of (e^h - 1)/h = 1?
We can give you mathematical proofs, but it seems like in your case, you should see what's going on. Get out a calculator and calculate the left hand side by taking smaller and smaller h. Then repeat with something not e. It's one thing to know the theory, its another to see the raw data infront of you.

I mean every exponential function like a^x equals 1 when x equals 0, right?
a=0

So would it be fair to say that (a^h) approaches 1 as h approaches zero? And that (a^h - 1) approaches zero as h approaches zero? And finally that the limit as h goes to 0 of (a^h - 1)/h also equals one?
But do they approach at the same rate?
 
  • #3
881
40
I mean every exponential function like a^x equals 1 when x equals 0, right?
a=0
ax, when x=0 is taken to be 1, even for a=0. Read under exponentiation : https://www.physicsforums.com/showthread.php?t=530207 [Broken].


To the OP, try taking y = ax as a general function and find its derivative, by taking logarithm on both sides. See where this leads...Also, as pwsnafu said, raw data will help you get an initial 'feel' of it.
 
Last edited by a moderator:
  • #4
278
0
Do you mean to say that (e^h - 1) approaches zero at the same rate as h does, but (a^h - 1) doesn't ? Why that happens exclusively to e?
Also, is rate the only thing that matters? So if I had two things approaching zero at the same rate, but one of them would reach zero earlier because it starts lower, would the ratio of those two things still be equal to one?
And if that is not the case, does that mean that (e^h - 1) = h as h approaches zero?
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,833
956
Yes, it is true that [itex]a^h- 1[/itex] approaches 0, as h goes to 0, for different a at different rates. It should be no surprise to you that different functions can converge to the same number at different rates- that's the whole point of the "slope" or derivative. To verify that, for example, look at graphs of [itex]y= a^x- 1[/itex] for different a. And it happens that the rate is "1" for a= e, essentially because e is defined to be that number.
For example, if we take h= .1, .01, .001, successively, [itex]2^h- 1[/itex] becomes
0.07177, 0.00695, and .00069. Each of those, divided by h, is 0.7177, 0.695, and 0.690, respectively. Each of those is less than 1 and it can be shown that they converge to a number less than 1.

But [itex]3^h- 1[/itex] becomes 0.11612, 0.01104, and 0.00109. Again dividing by h they give 1.1612, 1.104, and 1.09. Each of those is larger than 1 and converges to a number larger than 1.

We could do the same with, say, [itex](2.7^h- 1)/h[/itex] and [itex](2.75^h- 1)/h[/itex] showing that the first converges to a number slightly less than 1, the second to a number just larger than 1. There exist a between 2 and 3, between 2.7 and 2.75, such that [itex](a^h- 1)/h[/itex] converges to 1. We call that value, "e".
 

Related Threads on Derivatives of Exponentials (why e?)

  • Last Post
Replies
4
Views
5K
  • Last Post
Replies
5
Views
3K
Replies
3
Views
2K
Replies
13
Views
3K
Replies
4
Views
3K
Replies
3
Views
2K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
3
Views
548
  • Last Post
Replies
4
Views
541
  • Last Post
Replies
7
Views
8K
Top