What is the Relationship Between Maclaurin Series and Infinite Series?

[sooyong94]

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?

 

[haruspex]

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?

 

[sooyong94]

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?

 

[haruspex]

I don't get it... Just rewrite them as 2, 5,8?

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?

 

[sooyong94]

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?

 

[haruspex]

-1?

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, ...

 

[sooyong94]

(-3+2)?

 

[haruspex]

(-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.

 

[sooyong94]

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...

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

 

[haruspex]

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.)

 

[sooyong94]

I don't get it... Do I need to rewrite the series in reverse?

 

[haruspex]

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.

 

[Dick]

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.

 

[sooyong94]

Well, I got r=1/3 and x=1/2...

 

[haruspex]

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.

 

[sooyong94]

Alright, what should I do afterwards? :P

 

[Dick]

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.

 

[sooyong94]

Is it cube root of 3/2?

 

[Dick]

Is it cube root of 3/2?

Yes, it's (1+1/2)^(1/3).

 

[sooyong94]

That's it? :shock:

 

[Dick]

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.

 

[sooyong94]

Oh wow... I don't even know about this... :/

 

[Dick]

Oh wow... I don't even know about this... :/

What? Are you doubting that it's correct?

 

[sooyong94]

Erm.. nope. The answer is kinda unexpected... :blush: