Power Series Question

[Ghost3k]

Define a function of t, M, using the power series ∑ n=1 to ∞, (5*6^(n-1)e^(nt))/11^n
i.e., M(t) is the power series.

Determine

(a) the domain of the function M , i.e., the values of t for which the function is defined;
(b) the explicit function that is represented by the power series;
(c) the value of the first derivative of M at t = 0;
(d) the value of the second derivative of M at t = 0.




Could someone please help me out with this one problem. I've done the rest of my hw problems but I don't know how to approach this one. It would be incredibly helpful if anyone could at least guide me through the steps! I'm not really sure what they mean by explicit function either. Any help is appreciated!

 

[pka]

Define a function of t, M, using the power series ∑ n=1 to ∞, (5*6^(n-1)e^(nt))/11^n
i.e., M(t) is the power series.
Determine
(a) the domain of the function M , i.e., the values of t for which the function is defined;
(b) the explicit function that is represented by the power series;
(c) the value of the first derivative of M at t = 0;
(d) the value of the second derivative of M at t = 0.

Is it true that \(\displaystyle {\displaystyle\sum\limits_{n = 1}^\infty {\dfrac{{5 \cdot {6^{n - 1}} \cdot {e^{nt}}}}{{{{11}^n}}}} = \dfrac{5}{6}\sum\limits_{n = 1}^\infty {{{\left( {\dfrac{{6{e^t}}}{{11}}} \right)}^n}}}~? \)

 

[Ghost3k]

Is it true that \(\displaystyle {\displaystyle\sum\limits_{n = 1}^\infty {\dfrac{{5 \cdot {6^{n - 1}} \cdot {e^{nt}}}}{{{{11}^n}}}} = \dfrac{5}{6}\sum\limits_{n = 1}^\infty {{{\left( {\dfrac{{6{e^t}}}{{11}}} \right)}^n}}}~? \)

Yes, that is true. You separated the 6^(n-1) to 6^n * 6 ^-1 so you could bring out the the 6^-1 out with the 5 to be 5/6. And then you raised all to n. How would I proceed from this point? I'm sorry if I sound dumb. Would I have to find the interval of convergence to find the domain?

 

[pka]

Yes, that is true. You separated the 6^(n-1) to 6^n * 6 ^-1 so you could bring out the the 6^-1 out with the 5 to be 5/6. And then you raised all to n. How would I proceed from this point? I'm sorry if I sound dumb. Would I have to find the interval of convergence to find the domain?

What are the conditions under which a geometric series converges?

 

[Ghost3k]

What are the conditions under which a geometric series converges?

Okay so the common ratio r, needs to be -1 < r < 1
r = 6e^t/11
So I'm assuming I need to solve for t to find what value satisfies this?


So
-1 < 6e^t/11 < 1


-11/6 < e^t < 11/6


ln (-11/6) < t < ln (11/6)


But you can't calculate ln (-11/6) so it should just be t < ln (11/6)

Am I on the right track? Thank you for the help thus far!

 

[pka]

so it should just be t < ln (11/6)
Am I on the right track?

Yes you are. You know that \(\displaystyle \forall x,~e^x>0\). So what is the final answer?

 

[Ghost3k]

Yes you are. You know that \(\displaystyle \forall x,~e^x>0\). So what is the final answer?

I'm not really sure how to approach it. I know you are saying that for all x, e^x>0, but I'm not sure where do I go from there. Any other hints?

 

[pka]

I'm not really sure how to approach it. I know you are saying that for all x, e^x>0, but I'm not sure where do I go from there.

Let me be blunt without being mean. I am done with this thread.
If you still don't understand why, then pay for a tutor.
The answer is: \(\displaystyle t<\log(11)-\log(6)\). Good luck!