Linked Questions

Ask Question
31 questions linked to/from Are the any non-trivial functions where $f(x)=f'(x)$ not of the form $Ae^x$
Hot Newest Score Active Unanswered
4 votes
3 answers
9k views

I've heard that $f(x) = Ae^x$ is only function (both elementary and non-elementary) that satisfies the property $f(x)=\frac{df(x)}{dx}$. Is this true (and if it's true, is there a definitive way to ...
user avatar
7 votes
4 answers
1k views

I understand that assuming an analytic solution, we look at the Taylor series and arrive at a unique solution y = exp(x). However how do we know that there are no other non-analytic solutions? (...
Danny Duberstein's user avatar
  • 94
3 votes
1 answer
122 views

This is just a curious question, but is the following true? $$f(x) = f'(x)\iff f(x) = e^x.$$ I can prove that $\dfrac{\mathrm d}{\mathrm dx}\left(e^x\right) = e^x$ from using the the formula, $e^x :=...
Mr Pie's user avatar
  • 9,774
133 votes
9 answers
29k views

I was wondering on the following and I probably know the answer already: NO. Is there another number with similar properties as $e$? So that the derivative of $ e^x$ is the same as the function itself....
Timo Willemsen's user avatar
  • 1,657
61 votes
10 answers
4k views

Motivation We all get familiar with elementary functions in high-school or college. However, as the system of learning is not that much integrated we have learned them in different ways and the ...
Hosein Rahnama's user avatar
  • 15.7k
20 votes
4 answers
5k views

How can I prove that a function that is its own derivative exists? And how can I prove that this function is of the form $a(b^x)$?
P.B.G.'s user avatar
  • 223
8 votes
4 answers
2k views

Is the property of a function being its own derivative unique to $e^x$, or are there other functions with this property? My working for $e$ is that for any $y=a^x$, $ln(y)=x\ln a$, so $\frac{dy}{dx}=\...
acernine's user avatar
  • 1,790
14 votes
7 answers
4k views

In my calculus class, to show that $\frac{d}{dx}e^x=e^x$ we did something like this: $$\lim_{h \to 0} \frac{a^{x+h} - a^x}{h} = a^x \lim_{h \to 0} \frac{a^h-1} h,$$ and then we defined $e$ to be the ...
Jacob's user avatar
  • 801
6 votes
5 answers
4k views

Suppose I knew nothing about the function $e^x$. If I wanted to find a power series that was its own derivative, I would logically start with the constant term, and first start by setting it to $1$. ...
u8y7541's user avatar
  • 758
4 votes
4 answers
693 views

Is it possible to prove that $C_1e^x$ is the unique solution to $f'(x)=f(x)$? I have tried to suppose there exists $g'=g$ and $g(x)\neq C_1e^x$. But I cannot find any contradiction by myself. Any ...
High GPA's user avatar
  • 3,950
3 votes
7 answers
778 views

I understand why $f : \mathbb{R} \to \mathbb{R}$ with $f'(x) = f(x)$ and $f(0) = 1$ must be $f (x) = e^x$, but I don't really feel it is super intuitive. Intuitively, why would you expect such a ...
katana_0's user avatar
  • 1,912
2 votes
5 answers
1k views

I occasionally see proofs where the limit definition for $e$ pops up and I don't recognize it for some reason, every time! $$e = \lim_{n \to \infty} \left( 1 + \frac{1}{n}\right)^n$$ For one thing I ...
user525966's user avatar
  • 5,985
1 vote
4 answers
360 views

Suppose I do not know that $e^x$ solves the equation $$f'(x)=f(x),\;\;\;x\in\mathbb{R}.$$ I am just given this equation and want to see if $f$ is increasing. Is there a way to prove that $f$ is ...
UserA's user avatar
  • 1,699
2 votes
5 answers
1k views

The Wikipedia page on the natural logarithm says: 'Logarithms can be defined to any positive base other than 1, not only e. However, logarithms in other bases differ only by a constant multiplier from ...
user avatar
2 votes
3 answers
2k views

So, I was wondering what is the proof of the existence of $e$. I want to define it as the number such that for which the derivative of $(b^x)'=b^x$. Now, I have to show it exists. So, I have to show ...
Sorfosh's user avatar
  • 3,688

15 30 50 per page
1
2 3