Summation Verification: Evaluating Series with -e^t Answer

[MathewsMD]

Hi,

I'm just trying to evaluate a series and would just appreciate if someone could either verify or correct me work.

Essentially, I have a series that I've produced:

-[(t^2)/2 + (t^5)/(2x5) + (t^8)/(2x5x8) + ...]

= - *sum from n = 0 to infinity* [(t^(3n+2))/(3n+2)!] = -e^t

Sorry for the poor syntax. I'm just in a slight rush here and can hopefully fix it up sooner rather than later. Regardless, i essentially get -e^t as my answer, and if someone could verify if this is correct, that would be very helpful. Thank you!

 

[Mark44]

Hi,

I'm just trying to evaluate a series and would just appreciate if someone could either verify or correct me work.

Essentially, I have a series that I've produced:

-[(t^2)/2 + (t^5)/(2x5) + (t^8)/(2x5x8) + ...]

= - *sum from n = 0 to infinity* [(t^(3n+2))/(3n+2)!] = -e^t

?
How do you figure that your series equals -e^t?
The Maclaurin series for -e^t is -(1 + t + t²/2! + ... + tⁿ/n! + ...)

Sorry for the poor syntax. I'm just in a slight rush here and can hopefully fix it up sooner rather than later. Regardless, i essentially get -e^t as my answer, and if someone could verify if this is correct, that would be very helpful. Thank you!

 

[MathewsMD]

?
How do you figure that your series equals -e^t?
The Maclaurin series for -e^t is -(1 + t + t²/2! + ... + tⁿ/n! + ...)

Yes, it seems very odd and certainly not equivalent. I tried simplifying the series, but I only got to -t² sum [t^³ⁿ/(3n+2)! But I can't quite further simplify it from here to put it in a form that seems expressible in terms of a function of e^rt. Any hints? Is there any way for me to possibly simplify the factorial in the denominator? If I change the summation from n = 0 to n = 1 (and still to infinity), this would allow it to become t^(3n - 3)/(3n - 1)!, but that still doesn't really help...

 

[Mark44]

What's the problem you're trying to solve?

 

[MathewsMD]

What's the problem you're trying to solve?

There's actually no specific problem...

I essentially have this series (that was found in another problem) and although there are other representations which could help me solve this, i was wondering if there was a way for me to transform this into a function of e^rt...I don't quite see any ways to do so, but would welcome any suggestions or methods. I don't quite see how having t^(3n) terms in the sum can allow for the sum to be expressed as a function of e^rt.
I was also wondering if there are other ways to manipulate the denominator factorial to help make the sum become another expression.

 

[lurflurf]

$$\sum_{k=0}^\infty \frac{x^{3k+2}}{(3k+2)!}=\tfrac{1}{3}e^{-x/2}\left[ e^{-3x/2}-\sqrt{3} \sin \left( \tfrac{1}{2} \sqrt{3} x \right)-\cos\left(\tfrac{1}{2}\sqrt{3}x\right) \right] $$
http://www.wolframalpha.com/input/?...qrt(3)+sin((sqrt(3)+x)/2)-cos((sqrt(3)+x)/2))

 

[lurflurf]

It is easy if you know about complex numbers
let
$$
A=\sum_{k=0}^\infty \frac{x^{3k}}{(3k)!} \\
B=\sum_{k=0}^\infty \frac{x^{3k+1}}{(3k+1)!} \\
C=\sum_{k=0}^\infty \frac{x^{3k+2}}{(3k+2)!} \\
t=-\tfrac{1}{2}(1+i\sqrt{3})
\text{then solve for C using}\\
e^x=A+B+C
e^{t x}=A+t B+t^2 C
\\e^{t^2 x}=A+t^2 B+t C
$$