# Looking for a function with derivatives of

• krausr79
. g(x) is a whole function so its taylor series is a still a whole function and so is f(x). the inverse laplace transform is given by a contour integral of f(z)exp(xz) where the contour goes from -\inf to \inf along the imaginary axis and around the poles of f(z) in the right half plane. the poles of f(z) are all of order 1 and are simple poles. the only pole on the imaginary axis is at z=0 and is a simple pole.this will give you a closed form representation of f(x) but it is an infinite sum. i dont know how to get rid of the integral. this probably wont help you but i thought
krausr79
I'm searching for a function expressed explicitly as a non-infinite series where
f(0)=1/1
f'(0)=1/3
f''(0)=1/5
...
fn(0)=1/(2n+1)
...

Does it exist? What is it?

If it's a strictly finite series then it's just an (n+1)-th degree polynomial with the coefficient of each x^k being f^(k)(0)/k!.

$$f(x) = 1 + f'(0)x + \frac{1}{2}f''(0)x^2 + \frac{1}{6}f^{(3)}(0)x^3 + \dots + \frac{1}{n!}f^{(n)}(0)x^n.$$

Are you really just looking for a finite series?

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Mute said:
If it's a strictly finite series then it's just an (n+1)-th degree polynomial with the coefficient of each x^k being f^(k)(0).

$$f(x) = 1 + f'(0)x + f''(0)x^2 + f^{(3)}(0)x^3 + \dots + f^{(n)}(0)x^n.$$

Are you really just looking for a finite series?

Slight error on your part: kth term has k! as divisor.

mathman said:
Slight error on your part: kth term has k! as divisor.

Whoops! You're right, I've corrected the post now.

Mute said:
If it's a strictly finite series then it's just an (n+1)-th degree polynomial with the coefficient of each x^k being f^(k)(0)/k!.

$$f(x) = 1 + f'(0)x + \frac{1}{2}f''(0)x^2 + \frac{1}{6}f^{(3)}(0)x^3 + \dots + \frac{1}{n!}f^{(n)}(0)x^n.$$

Are you really just looking for a finite series?

Nope, this is what I'm looking to avoid. I can make a maclaurin series out of it, but that doesn't help me. What I mean is that I'm hoping it can be expressed without an implied infinite pattern. For instance, I don't want
$$f(x) = 1 + f'(0)x + \frac{1}{2}f''(0)x^2 + \frac{1}{6}f^{(3)}(0)x^3 + \dots + \frac{1}{n!}f^{(n)}(0)x^n + \dots$$

I want
$$f(x) = e^x$$

For the function I'm thinking of, $$f^n(0)=\frac{1}{2n+1}$$ for all n from 0 to infinity, so it wouldn't terminate as a Maclaurin series.

krausr79 said:
Nope, this is what I'm looking to avoid. I can make a maclaurin series out of it, but that doesn't help me. What I mean is that I'm hoping it can be expressed without an implied infinite pattern. For instance, I don't want
$$f(x) = 1 + f'(0)x + \frac{1}{2}f''(0)x^2 + \frac{1}{6}f^{(3)}(0)x^3 + \dots + \frac{1}{n!}f^{(n)}(0)x^n + \dots$$
I don't understand what you mean by "an implied infinite pattern". Yes, $f(x)= 1+ x+ \frac{1}{2}x^2+ \dots+ \frac{1}{n!} x^n+ \dots$ is an infinite series and is equal to ex.

But $f(x)= 1+ x+ \frac{1}{2}x^2+ \dots+ \frac{1}{n!} x^n$, for fixed n, is not. It is a finite series and is a polynomial. If n= 3, that is $1+ x+ (1/2)x^2+ (1/6)x^3$.

I want
$$f(x) = e^x$$

For the function I'm thinking of, $$f^n(0)=\frac{1}{2n+1}$$ for all n from 0 to infinity, so it wouldn't terminate as a Maclaurin series.
So that is NOT a "non-infinite" series as you said in your original post. I have no idea what you are asking for. Nor do I know what you mean by "wouldn't terminate as a MacLaurin series".

Any "power series" that sums to a given analytic function is the Taylor's series of that function about some point.

Let us set up the relevant power series here:
$$F(x)=\sum_{n=0}^{\infty}\frac{1}{n!(2n+1)}x^{n}$$

Now, assume x>=0, and set $$x=y^{2}$$
and we have:
$$F(x)=G(y)=\sum_{n=0}^{\infty}\frac{1}{n!(2n+1)}y^{2n}$$
Then, we have:
$$yG(y)=\sum_{n=0}^{\infty}\frac{1}{n!(2n+1)}y^{2n+1}$$, and:
$$\frac{d}{dy}(yG(y))=\sum_{n=0}^{\infty}\frac{1}{n!}(y^{2})^{n}=e^{y^{2}}$$

We therefore have:
$$yG(y)=\int_{0}^{y}e^{t^{2}}dt\to{F}(x)=\frac{1}{\sqrt{x}}\int_{0}^{\sqrt{x}}e^{t^{2}}dt$$
Thus, your desired function can be described in terms of a definite integral.

Hmm..something has occurred at x=0 that I don't like, it seems F(0) equals zero.

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To HallsofIvy:
Sorry for the confusion.. I was trying to express that I wanted a function that is finitely expressible. It's true that I wasn't interested in a series.

A maclaurin series expansion will terminate if the nth derivative of the function is always zero after some n.

To arildno:
I can assure you your work is correct, because this is the inverse of what I'm trying to do. I was looking for a solution to $$\int e^{(x^2)}dx$$, and figured that if I could find a nice closed form for f(x), then $$\int e^{(x^2)}dx=xf(x^2)$$.
At this point I'm going to assume what I'm looking for doesn't exist or it would have been found already.

Don't worry about f(0)=0; the +c can take care of that. Actually, would x=0 even work with the 1/sqrt(x) at the start?

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It's quite well known that integral has no elementary solution, and it can be proven there is no solution, it's not just because no one has found it yet. Doing so however requires the Risch Algorithm, something I have only a vague idea how it works.

Correction:

F(0) can be set at 1, and still be a continuous function.

krausr79 said:
I'm searching for a function expressed explicitly as a non-infinite series where
f(0)=1/1
f'(0)=1/3
f''(0)=1/5
...
fn(0)=1/(2n+1)
...

Does it exist? What is it?

for this case i do not know, but for the case $$\frac{1}{(2n+1)2n!}$$

there is a generating function $$2x^{-1}\int_{0}^{x}dtsin(t)/t =g(x)$$

taking the inverse laplace transfomr of this function you will get a function f(x) so its taylor series includes 1/(2n+1) in its derivatives

Last edited by a moderator:

## What is a function with derivatives?

A function with derivatives is a mathematical function that has a mathematical relationship with its derivatives. Derivatives are measures of how a function changes at a specific point, and they can be used to analyze and model various phenomena in science and engineering.

## Why is it important to find a function with derivatives?

Finding a function with derivatives is important because it allows us to understand and predict the behavior of a system or phenomenon. Derivatives can provide insights into the rate of change, maximum and minimum values, and other important characteristics of a function.

## What are the different methods for finding a function with derivatives?

Some common methods for finding a function with derivatives include using the power rule, product rule, quotient rule, chain rule, and implicit differentiation. These methods involve manipulating the given function to find its derivative, which is a new function that describes the rate of change of the original function.

## Can a function have multiple derivatives?

Yes, a function can have multiple derivatives. This is because the derivative of a function is itself a function, and it can be differentiated again to find higher-order derivatives. These higher-order derivatives can provide more detailed information about the behavior of the original function.

## How are functions with derivatives used in real-world applications?

Functions with derivatives are used in various real-world applications, such as physics, economics, and engineering. For example, they can be used to model the motion of objects, predict stock market trends, or optimize the design of structures. They are also used in scientific research to analyze and understand complex systems.

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