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.