Finding sum of infinite series

[Shinaolord]

Homework Statement

Recognize the series $$3-3^3/3!+3^5/5!-3^7/7!$$ is a taylor series evaluated at a particular value of x. Find the sum

Homework Equations

Sum of Infinite series = ##a/1-x##

The Attempt at a Solution

So, I can't figure out what i would us as the ratio (the thing you multiply the term by each time.) I got as far as
$$ \sum\limits_{n=0}^\infty (-1)^n \frac{(x^{2n}))}{2n!} $$ is the series for $$cos(x)$$ evaluated at $$ x=0$$, and the value we're looking at is $$a=3$$.

So, I tried the following
## T(x)= \frac{3}{(\frac{(1-3^2)}{n!})}##

but I'm totally stumped on how to find ##n##.
Any hints?

 

[Joffan]

So close...

##2n!## as denominator in the cosine series is hard to fix - is there a function that has ##(2n+1)!## as the denominator in that sort of series?

 

[HallsofIvy]

Cosine is an even function and so has only even powers in its power series. Sine is an odd function and so ...

 

[Shinaolord]

So close...

##2n!## as denominator in the cosine series is hard to fix - is there a function that has ##(2n+1)!## as the denominator in that sort of series?

Yes, there is, $$ Sin(x) = \sum\limits_{i=0}^\infty (-1)^i \frac{x^{2i+1}}{(2i+1)!}$$
centering at x=0, and evaluating at 3, we get
$$\sum\limits_{i=0}^\infty (-1)^i \frac{3^{2i+1}}{(2i+1)!}$$
is that right?
I'm not entirely sure how this helps...
maybe because $$sin(x) = cos(x\pm \pi/2)$$ ?
so we'd have
$$\sum\limits_{i=0}^\infty (-1)^i \frac{(3-\pi/2)^{2i+1}}{(2i+1)!}$$
?

EDIT: I get that one is even the other is odd, I'm sorry if this is a simple example, could i use the ratio test to find $$a$$?

 

[Curious3141]

Yes, there is, $$ Sin(x) = \sum\limits_{i=0}^\infty (-1)^i \frac{x^{2i+1}}{(2i+1)!}$$
centering at x=0, and evaluating at 3, we get
$$\sum\limits_{i=0}^\infty (-1)^i \frac{3^{2i+1}}{(2i+1)!}$$
is that right?
I'm not entirely sure how this helps...
maybe because $$sin(x) = cos(x\pm \pi/2)$$ ?
so we'd have
$$\sum\limits_{i=0}^\infty (-1)^i \frac{(3-\pi/2)^{2i+1}}{(2i+1)!}$$
?

EDIT: I get that one is even the other is odd, I'm sorry if this is a simple example, could i use the ratio test to find ##a##?

I don't get why you're working so hard. The question seems to be a simple pattern recognition one. You've already recognised it as the Maclaurin series (Taylor centered at zero) for ##\sin x##. So what is ##x## here?

I doubt you have to prove convergence because it is well known that the Maclaurin series for sine is convergent for all real arguments.

 

[benorin]

your answer is sin 3.

 

[STEMucator]

evaluated at x=0, and the value we're looking at is a=3

Just a slight mix-up here. The series is evaluated at ##x = 3## and the series is centered at ##a = 0##.

The series ##\sum\limits_{n=0}^\infty (-1)^n \frac{3^{2n+1}}{(2n+1)!}## converges to ##sin(3)## because ##R = ∞##.

Why not try approximating the series by the ##n^{th}## partial sum ##s_n##? The larger ##n## is the better the approximation.

If you're looking for a certain precision in your approximation, consider placing a restriction on the precision. Suppose you wanted to estimate the sum precise to five decimal places. How large would ##n## have to be? Well:

##|error| = |R_n| = |s - s_n| < a_{n+1} ≤ 0.00001##

Thus,

$$a_{n+1} = \frac{3^{2n + 3}}{(2n + 3)!} ≤ 0.00001$$

The solution over the integers is ##n = 84##, thanks to wolfram. So you would need to approximate ##sin(3)## using ##s_{84}## to be accurate to 5 decimal places.

 

[Shinaolord]

Thank you. That makes sense, I was just overthinking the problem.

 

[Joffan]

If you're looking for a certain precision in your approximation, consider placing a restriction on the precision. Suppose you wanted to estimate the sum precise to five decimal places. How large would ##n## have to be? Well:

##|error| = |R_n| = |s - s_n| < a_{n+1} ≤ 0.00001##

Thus,

$$a_{n+1} = \frac{3^{2n + 3}}{(2n + 3)!} ≤ 0.00001$$

The solution over the integers is ##n = 84##, thanks to wolfram. So you would need to approximate ##sin(3)## using ##s_{84}## to be accurate to 5 decimal places.

##s_{10}## is accurate to 8 decimal places. As the ##|a_n|## are monotonic decreasing (by that point), and alternating in sign, the error ##|s - s_n|## is less than the absolute value of the last term, ##a_{10}≈-9.6 \times 10^{-9}##. (##s_{10}## is actually accurate to within the absolute value of ##a_{11}##, but that's gravy).

Perhaps Wolfram gave you ##8.4##, not ##84##?

 

[STEMucator]

##s_{10}## is accurate to 8 decimal places. As the ##|a_n|## are monotonic decreasing (by that point), and alternating in sign, the error ##|s - s_n|## is less than the absolute value of the last term, ##a_{10}≈-9.6 \times 10^{-9}##. (##s_{10}## is actually accurate to within the absolute value of ##a_{11}##, but that's gravy).

Perhaps Wolfram gave you ##8.4##, not ##84##?

Wolfram is not all knowing and I'm not trusting it for the sake of this discussion. ##8.4## is definitely not a possibility because ##n \in \mathbb{N}##. You shouldn't be getting negative numbers like ##-9.6 \times 10^{-9}## for a positive sequence btw.

The series converged by the A.S.T to begin with, i.e the sum ##\sum_{n=0}^{∞} (-1)^n a_n = \sum_{n=0}^{∞} (-1)^n \frac{3^{2n+1}}{(2n+1)!}##.

##a_n## is monotonically decreasing for ##n≥N=3## and bounded below (##lim(a_n) = 0##), so it converges. In fact, ##\sum |(-1)^n a_n|## converges, so the series converges absolutely (this could be seen as a consequence of ##|x-a| < R = ∞## originally).

The problem is the terms of the sequence are a bit nit picky for the first few values of ##n## apparently. I decided to plug numbers in manually and found that the sequence was in fact decreasing after ##n = 3##. At ##n=7## the terms of the series dipped below the precision required, so you could neglect them and say that ##s_7## is a good enough approximation for 5 decimal places.