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#1 Re: Help Me ! » The Last Few Terms in an Infinite Series » 2019-11-02 07:42:46

Richard Feynman said: "If you think you understand quantum mechanics, you don't understand quantum mechanics."

I have heard that. If nobody understands it, how can they work with it? Thanks for your input. I will go on to enjoying your site.

#2 Re: Help Me ! » The Last Few Terms in an Infinite Series » 2019-11-02 05:01:48

Hi Bob,

Thanks for your reply. It gives some something to think about and work on.

bob bundy wrote:

So, when you subtract one series from another the subtraction process goes on for ever.  So there is no last term left over.  The parts of the two series that go on for ever 'cancel' each other out completely and so you're left with some algebra you can deal manage without worrying about infinity.

I understand and agree. This is a neat trick and this is what excited me. However, if you look at just one infinite series, there is a lot of hand waving going on.

bob bundy wrote:

I doubt that anyone can 'understand' infinity.

How can people discuss this with each other if they cannot understand it? Also, when someone proposed it first, how could the others agree that something well-defined was being proposed?

bob bundy wrote:

Your investigative result is correct (well done!)  It is a special case of the above sum formula with a = 1.

Thanks. Doing stuff like this is a part of the fun for me of being in sites like this. Thanks for your link. I will look at it.
VC1

#3 Re: Help Me ! » The Last Few Terms in an Infinite Series » 2019-11-01 04:58:02

I also realized this.

The sum of a geometric infinite series with a ratio < 1 can also be calculated as follows:

1. Multiply by the denominator of the ratio to get another infinite series.

2. Subtract the original series from this.

For S = 1/2 + 1/4 + 1/8 + 1/16 +......

2S = 1+  1/2 + 1/4 + 1/8 + 1/16 +......

2S-S = 1
S=1

The more general solution is this (after multiplying by n and subtracting the original series from the new series; 1/n is the ratio; n > 1).

S(n-1) = 1
S = 1/(n-1)

#4 Re: Help Me ! » The Last Few Terms in an Infinite Series » 2019-10-31 10:04:48

Summing geometric infinite series with ratios of 1/2, 1/3, 1/4, etc.

1. Divide by the denominator to get a new infinite series.

2. Subtract the second series from the first series. This will get rid of all the infinity mess and will get results like

S/2 = 1/2  for a ratio of 1/2. S = 1

2/3 S = 1/3  for a ratio of 1/3. S = 1/2

3/4 S = 1/4  for a ratio of 1/4. S = 1/3

4/5 S = 1/5  for a ratio of 1/5. S = 1/4

5/6 S = 1/6  for a ratio of 1/6. S = 1/5

6/7 S = 1/7  for a ratio of 1/7. S = 1/6

7/8 S = 1/8  for a ratio of 1/8. S = 1/7

One can generalize:

(n-1)/n S = 1/n for a ratio of 1/n. S = 1/(n-1)

Any typos aside, am I right? I realized all these only after seeing the proof in MathIsFun. I am not showing off. This may be of interest to others too. However, if you want me to stop posting posts like this, let me know and I will stop.

Whether I post or not, this is the kind of activity that your site evokes in me. Posting motivates me to complete this process.

Thanks.

#5 Help Me ! » The Last Few Terms in an Infinite Series » 2019-10-31 07:51:37

veeceeone
Replies: 7

I don't understand "infinity" at all but can live with it. [ What choice do I have? :-) ]

I do understand that the last few terms have extraordinarily small values in a series like:

S = 1/2 + 1/4 + 1/8 + 1/16 + ...

The following is from the proof for the sum of this series, which I think is outstanding. My question is a conceptual one.

***
First, we will call the whole sum "S":   S = 1/2 + 1/4 + 1/8 + 1/16 + ...
Next, divide S by 2:S/2 = 1/4 + 1/8 + 1/16 + 1/32 + ...
***

Let's say that S has n terms. (I understand that n is  ∞.)
S/2 may be thought of the terms in S shifted to the left by one position throwing out the first term. This series also will have n terms. What is the value of the nth term? This term will not have a corresponding term in S.

Am I just wasting my time asking questions like this? The proof is great by the way. I can now find the sum of geometric series like

1/8 + 1/64 + 1/512 etc very easily. [I didn't do the math. I am guessing that it is 1/7.]

#6 Re: Introductions » veeceeone an Introduction » 2019-10-31 07:41:42

I will post my questions in the Help me section. Thanks.

#7 Re: Introductions » veeceeone an Introduction » 2019-10-31 05:35:41

ganesh wrote:

Hi veeceeone,

Welcome the the forum!

The first line in Tamil is for Ganesh :-)

---------------------------------
யான் பெற்ற இன்பம் பெறுக இவ்வையகம்.

Thanks!

The Tamil quote may be roughly translated as:

"Let the whole world enjoy the joy I was fortunate to receive."

I wanted other folks in a certain Tamil speaking Facebook group also to enjoy Math Is Fun that I discovered and that's why I started my post there with this Tamil line.

I have some questions on the last few terms in an infinite series. The question is very low priority for me. I won't die if I don't know the answer but would love to. Where (i..e, which group here) can I ask this?

Thanks.

#8 Introductions » veeceeone an Introduction » 2019-10-30 05:13:01

veeceeone
Replies: 4

I am retired. I see the site and the forum as a place to relax and improve myself. My education is in engineering.

Unfortunately, I completed my various courses without really understanding large parts of the subjects I studied. Now that I have time, I flit from subject to subject, this time really trying to understand. At this time, I have no hesitation declaring that the emperor has no clothes [ even if I happen to be the one who is the emperor :-) ]

Here is what I wrote about Math Is Fun yesterday on a Facebook group. The first line in Tamil is for Ganesh :-)

---------------------------------
யான் பெற்ற இன்பம் பெறுக இவ்வையகம்.

I stumbled into this site while doing a search (for infinite series). This site seems to be exceptionally well done. (This seems like an understatement, really.)

https://www.mathsisfun.com/

This proof blew me away:
First, we will call the whole sum "S":   S = 1/2 + 1/4 + 1/8 + 1/16 + ...
Next, divide S by 2:S/2 = 1/4 + 1/8 + 1/16 + 1/32 + ...
Now subtract S/2 from S
All the terms from 1/4 onwards cancel out.
And we get:S − S/2 = 1/2
Simplify: S/2 = 1/2
And so:S  = 1

You can start at whatever level your are in and have some fun.
---------------------------

I certainly hope to have fun here.

VC1
PS: I did have a bit of a problem registering.

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