# Modifying taylor series of e^x

by piercebeatz
Tags: taylor series
 P: 192 I recently thought to myself about how a slight modification to the taylor series of e^x, which is, of course: $$\sum_{n=0}^\infty \frac{x^n}{n!}$$ would change the equation. How would changing this to: $$\sum_{n=0}^\infty \frac{x^{n/2}}{\Gamma(n/2+1)}$$ change the equation? Would it still be convergent? How about: $$\sum_{n=0}^\infty \frac{x^{n/2}(-1)^n}{\Gamma(n/2+1)}$$
 HW Helper P: 2,080 Do you know fractional calculus? let J denote integration $$J^a f(x)=\frac{1}{\Gamma(a)} \int_0^x {(x-t)^{a-1} \mathop{f(t)}} \mathop{dx}$$ now the geometric series is $$\sum_{k=0}^\infty J^k=\frac{1}{1-J}$$ so we can say because J^k 1=x^k/k! the Taylor series of e^x is $$e^x=\sum_{k=0}^\infty \frac{x^k}{k!}=\sum_{k=0}^\infty J^k \mathop{1}=\frac{1}{1-J} \mathop{1}$$ J^(1/2) being the half integral $$J^{1/2} f(x)=\frac{1}{\Gamma(1/2)} \int_0^x \frac{f(t)}{\sqrt{x-t}} \mathop{dx}$$ your series is $$\sum_{k=0}^\infty \frac{x^{k/2}}{\Gamma(n/2+1)}=\sum_{k=0}^\infty J^{n/2}\mathop{1}=\frac{1}{1-J^{n/2}} \mathop{1}=e^x (1+erf(\sqrt{x}))$$
P: 192
 Quote by lurflurf Do you know fractional calculus? let J denote integration $$J^a f(x)=\frac{1}{\Gamma(a)} \int_0^x {(x-t)^{a-1} \mathop{f(t)}} \mathop{dx}$$ now the geometric series is $$\sum_{k=0}^\infty J^k=\frac{1}{1-J}$$ so we can say because J^k 1=x^k/k! the Taylor series of e^x is $$e^x=\sum_{k=0}^\infty \frac{x^k}{k!}=\sum_{k=0}^\infty J^k \mathop{1}=\frac{1}{1-J} \mathop{1}$$ J^(1/2) being the half integral $$J^{1/2} f(x)=\frac{1}{\Gamma(1/2)} \int_0^x \frac{f(t)}{\sqrt{x-t}} \mathop{dx}$$ your series is $$\sum_{k=0}^\infty \frac{x^{k/2}}{\Gamma(n/2+1)}=\sum_{k=0}^\infty J^{n/2}\mathop{1}=\frac{1}{1-J^{n/2}} \mathop{1}=e^x (1+erf(\sqrt{x}))$$
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?

HW Helper
P: 1,391

## Modifying taylor series of e^x

 Quote by piercebeatz 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

$$\mbox{erf}(x) = \frac{2}{\sqrt{\pi}}\int_0^x dt~e^{-t^2}.$$
P: 192
 Quote by Mute 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.
HW Helper
P: 2,080
Those differentials should have been dt not dx

 Quote by lurflurf Do you know fractional calculus? let J denote integration $$J^a f(x)=\frac{1}{\Gamma(a)} \int_0^x {(x-t)^{a-1} \mathop{f(t)}} \mathop{dt}$$ now the geometric series is $$\sum_{k=0}^\infty J^k=\frac{1}{1-J}$$ so we can say because J^k 1=x^k/k! the Taylor series of e^x is $$e^x=\sum_{k=0}^\infty \frac{x^k}{k!}=\sum_{k=0}^\infty J^k \mathop{1}=\frac{1}{1-J} \mathop{1}$$ J^(1/2) being the half integral $$J^{1/2} f(x)=\frac{1}{\Gamma(1/2)} \int_0^x \frac{f(t)}{\sqrt{x-t}} \mathop{dt}$$ your series is $$\sum_{k=0}^\infty \frac{x^{k/2}}{\Gamma(n/2+1)}=\sum_{k=0}^\infty J^{n/2}\mathop{1}=\frac{1}{1-J^{n/2}} \mathop{1}=e^x (1+\text{erf}(\sqrt{x}))$$
 HW Helper P: 2,080 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 $$J^{1/2} f(x)=\frac{1}{\Gamma(1/2)} \int_0^x \frac{f(t)}{\sqrt{x-t}} \mathop{dt}=\frac{1}{\sqrt{\pi}} \int_0^x \frac{f(t)}{\sqrt{x-t}} \mathop{dt}$$ 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 $$J^\alpha\left(\frac{t^k}{k!}\right)= \dfrac{t^{\alpha+k}}{\Gamma(\alpha+k+1)}$$ 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 $$J^{1/2}\left(\frac{t^k}{k!}\right)= \dfrac{t^{1/2+k}}{\Gamma(1/2+k+1)}$$ So your series is seen as $$e^x+J^{1/2}e^x$$ $$J^{1/2}e^x=\frac{1}{\Gamma(1/2)} \int_0^x \frac{e^t}{\sqrt{x-t}} \mathop{dt}=e^x \text{erf}(\sqrt{x})$$ hint: The change of variable u=sqrt(x-t) will help expressing the integral in terms of erf. Your other series is just $$e^x-J^{1/2}e^x$$
P: 192
 Quote by lurflurf 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 $$J^{1/2} f(x)=\frac{1}{\Gamma(1/2)} \int_0^x \frac{f(t)}{\sqrt{x-t}} \mathop{dt}=\frac{1}{\sqrt{\pi}} \int_0^x \frac{f(t)}{\sqrt{x-t}} \mathop{dt}$$ 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 $$J^\alpha\left(\frac{t^k}{k!}\right)= \dfrac{t^{\alpha+k}}{\Gamma(\alpha+k+1)}$$ 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 $$J^{1/2}\left(\frac{t^k}{k!}\right)= \dfrac{t^{1/2+k}}{\Gamma(1/2+k+1)}$$ So your series is seen as $$e^x+J^{1/2}e^x$$ $$J^{1/2}e^x=\frac{1}{\Gamma(1/2)} \int_0^x \frac{e^t}{\sqrt{x-t}} \mathop{dt}=e^x \text{erf}(\sqrt{x})$$ Your other series is just $$e^x-J^{1/2}e^x$$
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?
 HW Helper P: 2,080 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.
 P: 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

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