1
$\begingroup$

Power Series

I was told that I could simplify the power series into a a^k geometric series by setting a= (6*e^t)/(11^n). I remember from Calculus 2 that I could calculate the sum of a geometric series in this form using the formula 1/1-r.

I don't remember how this is applicable to a situation in which I have multiple variables or how it would help me find the function as asked in the question.

$\endgroup$
0

2 Answers 2

Reset to default
1
$\begingroup$

The choice of $a$ you mention is not correct. The series is a function of $t$ and you are told this explicitly: "$M(t)$ is the power series." Therefore, the choice of $a$ must be such that it is independent of the index variable $n$; e.g., $$a = \frac{6 e^t}{11},$$ and then noting that $$\frac{5}{6} a^n = \frac{5}{6} \cdot \frac{6^n e^{nt}}{11^n} = \frac{5 \cdot 6^{n-1} e^{nt}}{11^n}.$$ Therefore, your series can be written $$M(t) = \frac{5}{6} \sum_{n=1}^\infty a^n = \frac{5}{6} \cdot \frac{a}{1-a}, \quad |a| < 1,$$ where $a$ is given above. Note that because the index variable begins at $n = 1$, the series has sum $a/(1-a)$, not $1/(1-a)$. The radius of convergence of this sum is $1$, hence the requirement $|a| < 1$, which leads to the criterion on $t$ for the original series to converge.

$\endgroup$
4
  • $\begingroup$ That makes sense, the multivariable bit was tripping me up some. So Writing our series in terms of a, as you did, is essentially our explicit function? Or should I solve the series with a plugged in? $\endgroup$
    – CcS
    Mar 17, 2015 at 20:23
  • $\begingroup$ As it is defined in the first equation, $a$ is a function of $t$; so then $M$ is also a function of $t$, and your solution should express $M$ in terms of $t$ rather than the intermediate/auxiliary variable $a$, which you introduced only for the convenience of calculation. $\endgroup$
    – heropup
    Mar 17, 2015 at 20:50
  • $\begingroup$ M(t)= 5/6 * [((6e^t)/11) / (1-((6e^t)/11)]? So I replaced the auxiliary variable with its definition and simplify and I have an explicit function? $\endgroup$
    – CcS
    Mar 17, 2015 at 20:52
  • 1
    $\begingroup$ Yes that is correct. Try the other parts of your question now that you have $M(t)$. $\endgroup$
    – heropup
    Mar 17, 2015 at 21:11
0
$\begingroup$

Hint: $\frac{5\cdot 6^{n-1}e^{nt}}{11^n}=\frac{5}{6}\cdot\left(\frac{6e^t}{11}\right)^n$

$\endgroup$
1
  • $\begingroup$ This was the simplification I was referring to. How does one calculate the summation of that? How does said summation lead to an explicit function? $\endgroup$
    – CcS
    Mar 17, 2015 at 20:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.