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. Power Series

I understand c & d as fairly simple first and second derivatives of a multivariable function @ t = 0. But it escapes me on whether it is acceptable to separate the functions and do the product rule.

I do not understand what I'm being asked in a & b. Isn't the domain of M defined for all values of t since e^0 is simply 1?

I have no clue what explicit function is being represented by the power series?

CcS
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2 Answers2

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Hint

a) The Ratio test should help you.

b) If you set $$\alpha=\frac{6e^t}{11}$$ your series is of the form $$\frac{5}{6}\sum_{n=1}^\infty \alpha^k$$ that you should be able to compute.

Surb
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  • That would help me computationally, but I don't see how computing the power series helps me find the answers to a through d. Should I be computing the power series first, and then computing the derivatives or trying to figure out the explicit function? – CcS Mar 17 '15 at 19:10
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    Do you know some condition for whether a geometric series converges? – GEdgar Mar 17 '15 at 19:12
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    You should know that $\sum_{k=0}^n a^k=\frac{1-a^{n+1}}{1-a}$ and that the series converge if and only if $|a|<1$. – Surb Mar 17 '15 at 19:12
  • I recall that a power series can converge for some values of n along a radius, and diverge for others. It's been a few years since I looked at series of any sort so it alludes me as to what I'm even being asked to do. – CcS Mar 17 '15 at 19:16
  • You have now all elements to solve questions a) and b) :-) – Surb Mar 17 '15 at 19:18
  • IS the ratio test applicable here? I can't find a common ratio – CcS Mar 17 '15 at 19:28
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    Of course, just compute $$\lim_{n\to\infty }\frac{\frac{6^{n+1}e^{(n+1)t}}{11^{n+1}}}{\frac{6^ne^{nt}}{11^n}}$$ and find a condition on $t$ such that the limit is smaller than $1$. (you will have some $\ln$) – Surb Mar 17 '15 at 19:30
  • I was able to simplify it all the way down to (6t/11t) * (n+1)/n). This means then that the series converges when 6t/11t < 1? – CcS Mar 17 '15 at 19:43
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    No, $$\lim_{n\to\infty }\frac{\frac{6^{n+1}e^{(n+1)t}}{11^{n+1}}}{\frac{6^ne^{nt}}{11^n}}=\frac{6}{11} e^t$$ and thus, it converge if $$\frac{6}{11}e^t<1\implies e^t<\frac{11}{6}\implies t<\ln\left(\frac{11}{6}\right).$$ – Surb Mar 17 '15 at 19:47
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For $a$, you're probably going to want to use the $root-test$ and see what happens to $\sqrt{\frac{5\cdot 6^{n-1} e^{nt}}{11^n}}$. as $n$ goes to infinity. From that you solve which $t$ lets the series converge

For $b$. Let's manipulate the series a little bit $$ \sum_{i=1}^\infty \frac{5\cdot 6^{n-1} e^{nt}}{11^n} \\ \frac{5}{6} \sum_{i=1}^\infty \left(\frac{6e^t}{11} \right)^n $$ That's just a geometric power series, and you know what the summation of that is.

maxbaroi
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