Finding sum of infinite series

In summary: I decided to throw in ##n = 10## just to give some "wiggle room" but it wasn't really necessary.In summary, the series $$3-3^3/3!+3^5/5!-3^7/7!$$ is a Maclaurin series for ##sin(x)## evaluated at ##x=3##. By approximating the series using the ##n^{th}## partial sum, we can estimate the sum to a certain precision. For example, to estimate the sum accurate to 5 decimal places, we can use the 7th or 10th partial sum, as the terms after that become negligible.
  • #1
Shinaolord
92
4

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?
 
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  • #2
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?
 
  • #3
Cosine is an even function and so has only even powers in its power series. Sine is an odd function and so ...
 
  • #4
Joffan said:
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 $$a$$?
 
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  • #5
Shinaolord said:
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.
 
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  • #7
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.
 
  • #8
Thank you. That makes sense, I was just overthinking the problem.
 
  • #9
Zondrina said:
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##?
 
  • #10
Joffan said:
##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.
 

FAQ: Finding sum of infinite series

1. What is an infinite series?

An infinite series is a sequence of numbers that goes on forever, with each term in the sequence being added to the previous term. It can be written in the form of ∑n=1an, where n is the index and an is the nth term of the series.

2. How do you find the sum of an infinite series?

To find the sum of an infinite series, you can use various methods such as the geometric series test, the telescoping series test, and the integral test. These methods involve evaluating the limit of the series as n approaches infinity.

3. What is the importance of finding the sum of infinite series?

Finding the sum of infinite series is important in many areas of mathematics and science, including calculus, physics, and engineering. It allows us to approximate and analyze complex functions and phenomena, and can help us understand the behavior of systems over time.

4. Can all infinite series be summed?

No, not all infinite series can be summed. There are some series that do not have a finite sum, such as the harmonic series (1 + 1/2 + 1/3 + 1/4 + ...), which diverges. It is important to use convergence tests to determine if an infinite series can be summed.

5. What are some real-life applications of infinite series?

Infinite series have many real-life applications, such as in finance, where they are used to calculate compound interest and loan amortization. They are also used in physics to model natural phenomena, and in computer science to analyze algorithms and data structures.

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