Modifying taylor series of e^x

In summary, the conversation discussed the concept of fractional calculus and the use of integration with a parameter, denoted by J, to manipulate the Taylor series of e^x. The introduction of the half integral, J^(1/2), was also mentioned and how it relates to the series. The use of the error function was also discussed in relation to fractional calculus. It was noted that a background in multivariable calculus is not necessary to understand this concept.
  • #1
pierce15
315
2
I recently thought to myself about how a slight modification to the taylor series of e^x, which is, of course:

[tex]\sum_{n=0}^\infty \frac{x^n}{n!}[/tex]

would change the equation.

How would changing this to:

[tex]\sum_{n=0}^\infty \frac{x^{n/2}}{\Gamma(n/2+1)}[/tex]

change the equation? Would it still be convergent? How about:

[tex]\sum_{n=0}^\infty \frac{x^{n/2}(-1)^n}{\Gamma(n/2+1)}[/tex]
 
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  • #2
Do you know fractional calculus?
let J denote integration

[tex]J^a f(x)=\frac{1}{\Gamma(a)} \int_0^x {(x-t)^{a-1} \mathop{f(t)}} \mathop{dx}[/tex]
now the geometric series is

[tex]\sum_{k=0}^\infty J^k=\frac{1}{1-J}[/tex]

so we can say because J^k 1=x^k/k! the Taylor series of e^x is

[tex]e^x=\sum_{k=0}^\infty \frac{x^k}{k!}=\sum_{k=0}^\infty J^k \mathop{1}=\frac{1}{1-J} \mathop{1}[/tex]

J^(1/2) being the half integral
[tex]J^{1/2} f(x)=\frac{1}{\Gamma(1/2)} \int_0^x \frac{f(t)}{\sqrt{x-t}} \mathop{dx}[/tex]

your series is
[tex]\sum_{k=0}^\infty \frac{x^{k/2}}{\Gamma(n/2+1)}=\sum_{k=0}^\infty J^{n/2}\mathop{1}=\frac{1}{1-J^{n/2}} \mathop{1}=e^x (1+erf(\sqrt{x}))[/tex]
 
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  • #3
lurflurf said:
Do you know fractional calculus?
let J denote integration

[tex]J^a f(x)=\frac{1}{\Gamma(a)} \int_0^x {(x-t)^{a-1} \mathop{f(t)}} \mathop{dx}[/tex]
now the geometric series is

[tex]\sum_{k=0}^\infty J^k=\frac{1}{1-J}[/tex]

so we can say because J^k 1=x^k/k! the Taylor series of e^x is

[tex]e^x=\sum_{k=0}^\infty \frac{x^k}{k!}=\sum_{k=0}^\infty J^k \mathop{1}=\frac{1}{1-J} \mathop{1}[/tex]

J^(1/2) being the half integral
[tex]J^{1/2} f(x)=\frac{1}{\Gamma(1/2)} \int_0^x \frac{f(t)}{\sqrt{x-t}} \mathop{dx}[/tex]

your series is
[tex]\sum_{k=0}^\infty \frac{x^{k/2}}{\Gamma(n/2+1)}=\sum_{k=0}^\infty J^{n/2}\mathop{1}=\frac{1}{1-J^{n/2}} \mathop{1}=e^x (1+erf(\sqrt{x}))[/tex]

I don't know fractional calculus, nor do I really understand what just went on. Should I have a background in multivariable calc before I get into this (because I haven't studied that yet)? Also, where did the r come from in the last equation?
 
  • #4
piercebeatz said:
I don't know fractional calculus, nor do I really understand what just went on. Should I have a background in multivariable calc before I get into this (because I haven't studied that yet)? Also, where did the r come from in the last equation?

The last equation makes use of a function called the error function, denoted ##\mbox{erf}(x)##. The r is just a part of the name. The error function is defined as

$$\mbox{erf}(x) = \frac{2}{\sqrt{\pi}}\int_0^x dt~e^{-t^2}.$$
 
  • #5
Mute said:
The last equation makes use of a function called the error function, denoted ##\mbox{erf}(x)##. The r is just a part of the name. The error function is defined as

$$\mbox{erf}(x) = \frac{2}{\sqrt{\pi}}\int_0^x dt~e^{-t^2}.$$

Oh, ok. I know about the error function. It looked to me like the f in the equation was its own function and e was the constant and r a variable.

Do you know if fractional calc would be to advanced for me to learn at this time? I have taken AP calc, but not yet multivariable calc.
 
  • #6
Those differentials should have been dt not dx

lurflurf said:
Do you know fractional calculus?
let J denote integration

[tex]J^a f(x)=\frac{1}{\Gamma(a)} \int_0^x {(x-t)^{a-1} \mathop{f(t)}} \mathop{dt}[/tex]
now the geometric series is

[tex]\sum_{k=0}^\infty J^k=\frac{1}{1-J}[/tex]

so we can say because J^k 1=x^k/k! the Taylor series of e^x is

[tex]e^x=\sum_{k=0}^\infty \frac{x^k}{k!}=\sum_{k=0}^\infty J^k \mathop{1}=\frac{1}{1-J} \mathop{1}[/tex]

J^(1/2) being the half integral
[tex]J^{1/2} f(x)=\frac{1}{\Gamma(1/2)} \int_0^x \frac{f(t)}{\sqrt{x-t}} \mathop{dt}[/tex]

your series is
[tex]\sum_{k=0}^\infty \frac{x^{k/2}}{\Gamma(n/2+1)}=\sum_{k=0}^\infty J^{n/2}\mathop{1}=\frac{1}{1-J^{n/2}} \mathop{1}=e^x (1+\text{erf}(\sqrt{x}))[/tex]
 
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  • #7
http://en.wikipedia.org/wiki/Fractional_calculus
A natural place to do fractional calculus would be at the same time as the Laplace transform, similar funny integrals are computed, Laplace transforms can be used in fractional calculus. The only thing multivariable calculus is needed for is that the integrals have a parameter, which is avoided in single variable, but it is an unimportant thing as the parameter can be seen as a constant.

We can define fractional order integrals in this case the half integral

[tex]J^{1/2} f(x)=\frac{1}{\Gamma(1/2)} \int_0^x \frac{f(t)}{\sqrt{x-t}} \mathop{dt}=\frac{1}{\sqrt{\pi}} \int_0^x \frac{f(t)}{\sqrt{x-t}} \mathop{dt}[/tex]

since gamma(1/2)=sqrt(pi)

For this problem we do not especially care that the operation that appears is the half integral only that it appears.

We know the series for e^x so consider all the new terms, we can write them as half integrals of terms in the series for e^x

[tex]J^\alpha\left(\frac{t^k}{k!}\right)= \dfrac{t^{\alpha+k}}{\Gamma(\alpha+k+1)}[/tex]

This can be proven using the Riemann–Liouville fractional integral, Laplace transforms, or guessing based on the integer case among other methods. We are concerned with the 1/2 case

[tex]J^{1/2}\left(\frac{t^k}{k!}\right)= \dfrac{t^{1/2+k}}{\Gamma(1/2+k+1)}[/tex]

So your series is seen as

[tex]e^x+J^{1/2}e^x[/tex]

[tex]J^{1/2}e^x=\frac{1}{\Gamma(1/2)} \int_0^x \frac{e^t}{\sqrt{x-t}} \mathop{dt}=e^x \text{erf}(\sqrt{x})[/tex]

hint: The change of variable u=sqrt(x-t) will help expressing the integral in terms of erf.

Your other series is just
[tex]e^x-J^{1/2}e^x[/tex]
 
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  • #8
lurflurf said:
http://en.wikipedia.org/wiki/Fractional_calculus
A natural place to do fractional calculus would be at the same time as the Laplace transform, similar funny integrals are computed, Laplace transforms can be used in fractional calculus. The only thing multivariable calculus is needed for is that the integrals have a parameter, which is avoided in single variable, but it is an unimportant thing as the parameter can be seen as a constant.

We can define fractional order integrals in this case the half integral

[tex]J^{1/2} f(x)=\frac{1}{\Gamma(1/2)} \int_0^x \frac{f(t)}{\sqrt{x-t}} \mathop{dt}=\frac{1}{\sqrt{\pi}} \int_0^x \frac{f(t)}{\sqrt{x-t}} \mathop{dt}[/tex]

since gamma(1/2)=sqrt(pi)

For this problem we do not especially care that the operation that appears is the half integral only that it appears.

We know the series for e^x so consider all the new terms, we can write them as half integrals of terms in the series for e^x

[tex]J^\alpha\left(\frac{t^k}{k!}\right)= \dfrac{t^{\alpha+k}}{\Gamma(\alpha+k+1)}[/tex]

This can be proven using the Riemann–Liouville fractional integral, Laplace transforms, of guessed from the integer case among other methods. We are concerned with the 1/2 case

[tex]J^{1/2}\left(\frac{t^k}{k!}\right)= \dfrac{t^{1/2+k}}{\Gamma(1/2+k+1)}[/tex]

So your series is seen as

[tex]e^x+J^{1/2}e^x[/tex]

[tex]J^{1/2}e^x=\frac{1}{\Gamma(1/2)} \int_0^x \frac{e^t}{\sqrt{x-t}} \mathop{dt}=e^x \text{erf}(\sqrt{x})[/tex]

Your other series is just
[tex]e^x-J^{1/2}e^x[/tex]

I think I got some of that... Is this a course that you took in college? Or did you read about it? If so, can you recommend a book?
 
  • #9
Like I said some calculus or differential equations classes will do a little bit of fractional calculus when covering Laplace transforms or special functions. Whole courses are rare. The two standard books are An Introduction to the Fractional Calculus and Fractional Differential Equations by Kenneth S. Miller and Bertram Ross which might be out of print and The Fractional Calculus: Theory and Applications of Differentiation and Integration to Arbitrary Order by Keith B. Oldham an inexpensive Dover Books edition makes it a good choice for recreational reading.
 
  • #10
The proposed series can be considered as a special case of the called Mittag-Leffler function (for x>0) or of the alpha-exponential. They appear when we try to solve the equation

Dαf(t)+af(t) = δ(t)

For interested people I can send papers

mdo@fct.unl.pt
 

1. What is the Taylor series of e^x?

The Taylor series of e^x is an infinite series representation of the function e^x. It is defined as: e^x = 1 + x + (x^2 / 2!) + (x^3 / 3!) + (x^4 / 4!) + ...

2. Why would you want to modify the Taylor series of e^x?

The Taylor series of e^x can be modified to calculate the value of e^x at a specific point or to approximate the value of e^x with a finite number of terms. This can be useful in a variety of mathematical and scientific applications.

3. How do you modify the Taylor series of e^x?

The Taylor series of e^x can be modified by changing the point around which the series is centered, the number of terms used, or by adding or subtracting additional terms to the original series.

4. What is the purpose of modifying the Taylor series of e^x?

Modifying the Taylor series of e^x allows for more flexibility in calculating or approximating the value of e^x at a specific point. It can also be used to improve the accuracy of the approximation by including more terms in the series.

5. Are there any limitations to modifying the Taylor series of e^x?

Yes, there are limitations to modifying the Taylor series of e^x. As the number of terms used increases, the accuracy of the approximation may improve, but the computational complexity also increases. Additionally, the approximation may only be valid within a certain range of values, and may become less accurate outside of that range.

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