What is the Relationship Between Maclaurin Series and Infinite Series?

In summary, you need to find a pattern in the sequence 1, 2, 5, 8 to make the terms fit. Then you need to adjust something else so that the values of the terms don't change.
  • #1
sooyong94
173
2

Homework Statement


Use Maclaurin’s theorem to derive the first five terms of the series expansion for ##(1+x)^{r}##, where -1<x<1. Assuming the series, obtained above, continues with the same pattern, sum the following infinite series

##1 + \frac{1}{6} - \frac{(1)(2)}{(6)(12)} + \frac{(1)(2)(5)}{(6)(12)(18} - \frac{(1)(2)(5)(8}{(6)(12)(18)(24)}+...##

Homework Equations


Maclaurin series

The Attempt at a Solution



I have taken the derivative of ##(1+x)^{r}## several times and obtained the power series
##1+rx+\frac{r(r-1)}{2!} x^{2} +\frac{r(r-1)(r-2)}{3!} x^{3} + \frac{r(r-1)(r-2)(r-3}{4!} x^4+...##

Now, the problem is, how do I relate with the infinite series above?
 
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  • #2
The sequence 1, 2, 5, 8 is not obvious. The 1 doesn't seem to fit. What would you have to have instead of the 1 to fit with the remaining terms?
 
  • #3
haruspex said:
The sequence 1, 2, 5, 8 is not obvious. The 1 doesn't seem to fit. What would you have to have instead of the 1 to fit with the remaining terms?

I don't get it... Just rewrite them as 2, 5,8? :confused:
 
  • #4
sooyong94 said:
I don't get it... Just rewrite them as 2, 5,8? :confused:
You are given an infinite series, but only by example of the first four terms. The first thing you need to do is figure out what the general term looks like. The pattern of the denominators is straightforward, but it's not obvious what the pattern is in the numerators. Looks like you need to see a pattern in the sequence 1, 2, 5, 8 ... but what is it?
To answer that, try throwing away the 1. What would you put there instead of 1 to make a clear pattern?
 
  • #5
haruspex said:
You are given an infinite series, but only by example of the first four terms. The first thing you need to do is figure out what the general term looks like. The pattern of the denominators is straightforward, but it's not obvious what the pattern is in the numerators. Looks like you need to see a pattern in the sequence 1, 2, 5, 8 ... but what is it?
To answer that, try throwing away the 1. What would you put there instead of 1 to make a clear pattern?

-1? :confused:
 
  • #6
sooyong94 said:
-1? :confused:
Right. So, how can you rewrite the series (without changing the actual values of the terms) so that the sequence goes -1, 2, 3, 5, 8, ...
 
  • #7
(-3+2)?
 
  • #8
sooyong94 said:
(-3+2)?

No. Here's the original series:
##1 + \frac{1}{6} - \frac{(1)(2)}{(6)(12)} + \frac{(1)(2)(5)}{(6)(12)(18} - \frac{(1)(2)(5)(8}{(6)(12)(18)(24)}+...##

I'm asking you to rewrite that so that in the numerator you see the sequence -1, 2, 5, 8 instead of 1, 2, 5, 8.. You need to adjust something else so that the values of the terms don't change.
 
  • #9
haruspex said:
No. Here's the original series:
##1 + \frac{1}{6} - \frac{(1)(2)}{(6)(12)} + \frac{(1)(2)(5)}{(6)(12)(18)} - \frac{(1)(2)(5)(8)}{(6)(12)(18)(24)}+...##

I'm asking you to rewrite that so that in the numerator you see the sequence -1, 2, 5, 8 instead of 1, 2, 5, 8.. You need to adjust something else so that the values of the terms don't change.

I think I can rewrite the signs...

##1 - \frac{-1}{6} + \frac{(-1)(2)}{(6)(12)} - \frac{(-1)(2)(5)}{(6)(12)(18)} + \frac{(-1)(2)(5)(8)}{(6)(12)(18)(24)}+...##
 
  • #10
sooyong94 said:
I think I can rewrite the signs...

##1 - \frac{-1}{6} + \frac{(-1)(2)}{(6)(12)} - \frac{(-1)(2)(5)}{(6)(12)(18)} + \frac{(-1)(2)(5)(8)}{(6)(12)(18)(24)}+...##
Good. (And note that the signs in front of the terms now alternate nicely, +-+-.., whereas in the OP it started ++-+..)
Next, the Maclaurin series has r(r-1)(r-2)... in the numerators, i.e. the product of a sequence decreasing by 1 at each step. You have (-1)(2)(5)(8)..., an sequence increasing in 3s. What can you do to make those match?
(If you can't think how to fix up both of those differences, just try to fix up one: i.e. either get the sequence to be decreasing, or get it to increase by 1s.)
 
  • #11
I don't get it... Do I need to rewrite the series in reverse?
 
  • #12
sooyong94 said:
I don't get it... Do I need to rewrite the series in reverse?
No. The common difference in the arithmetic sequence is +3. You need a sequence with common difference -1. What simple arithmetic operation can you apply to every term in the sequence to make that change? Remember here I'm just talking about the sequence -1, 2, 5, 8..., not the series.
 
  • #13
sooyong94 said:
I don't get it... Do I need to rewrite the series in reverse?

No, don't do that. Just take the direct approach. Equate the first terms of your two series. r*x=1/6 and r*(r-1)*x^2/2=(-1*2/(6*12)). You should be able to solve for r and x.
 
  • #14
Well, I got r=1/3 and x=1/2...
 
  • #15
sooyong94 said:
Well, I got r=1/3 and x=1/2...
That's right.
I should have thought of suggesting Dick's way. I'm sure that was much easier for you. I went the way I did because that was the easiest for me - I could see the answer straight away by making those adjustments.
 
  • #16
Alright, what should I do afterwards? :P
 
  • #17
sooyong94 said:
Alright, what should I do afterwards? :P

To get the sum of the series?? Your series is the infinite series expansion of (1+x)^(1/3) when x=1/2, right? You might want to check a few other terms to make sure. Don't make me have to say the obvious about what the sum is. Just think about it. If you are thinking for over 5 seconds then you are thinking about it wrong.
 
  • #18
Is it cube root of 3/2?
 
  • #19
sooyong94 said:
Is it cube root of 3/2?

Yes, it's (1+1/2)^(1/3).
 
  • #20
That's it? :shock:
 
  • #21
sooyong94 said:
That's it? :shock:

Apparently you've been thinking too hard about this. If you step back and look at it again, the whole thing is pretty simple.
 
  • #22
Oh wow... I don't even know about this... :/
 
  • #23
sooyong94 said:
Oh wow... I don't even know about this... :/

What? Are you doubting that it's correct?
 
  • #24
Erm.. nope. The answer is kinda unexpected... :blush:
 

Related to What is the Relationship Between Maclaurin Series and Infinite Series?

1. What is the purpose of expanding infinite series?

The purpose of expanding infinite series is to find a way to represent a function as an infinite sum of simpler functions, making it easier to analyze and manipulate. It can also help approximate the value of a function at a certain point.

2. How do you expand an infinite series?

To expand an infinite series, you can use a variety of techniques such as the geometric series, Taylor series, or Maclaurin series. These methods involve finding a pattern in the coefficients and exponents of the series and using that to write a general formula for the series.

3. Can all functions be represented as infinite series?

No, not all functions can be represented as infinite series. For example, a function that has a vertical asymptote or a discontinuity cannot be represented as an infinite series. Additionally, some functions may require an infinite number of terms to accurately represent them, making it impractical to use in calculations.

4. What is the difference between a convergent and divergent infinite series?

A convergent infinite series is one where the sum of all its terms approaches a finite value as the number of terms increases. In contrast, a divergent infinite series is one where the sum of its terms does not approach a finite value and instead either approaches infinity or oscillates between different values.

5. How is the convergence of an infinite series determined?

The convergence of an infinite series can be determined by using various convergence tests, such as the ratio test, root test, or comparison test. These tests help determine whether the series approaches a finite value, diverges, or requires further analysis.

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