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naiinip_ako
guest
May-07-02, 01:52 PM (EST)
 
"infinite series"
 
   (1/2) + (1/4) + (1/8) + (1/16) + ...
Does this equal 1 or approach 1?


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  Subject     Author     Message Date     ID  
infinite series naiinip_ako May-07-02 TOP
  RE: infinite series alexb May-07-02 1
  RE: infinite series eipi May-08-02 2
  RE: infinite series jman_red Jul-03-02 3
     RE: infinite series mathmaster Jul-04-02 4
         RE: infinite series Chris Tynan Jul-11-02 5
             RE: infinite series pradeep Jul-12-02 6
                 RE: infinite series Michael Klipper Aug-19-02 12
  RE: infinite series jman_red Jul-14-02 7
     RE: infinite series alexb Jul-14-02 8
         RE: infinite series jman_red Jul-17-02 9
             RE: infinite series stapel Jul-18-02 10
  RE: infinite series chris tynan Jul-18-02 11
  RE: infinite series Paul Chen A.K.A Aug-20-02 13
     RE: infinite series alexb Aug-21-02 14
  RE: infinite series Bo Jacoby Aug-22-02 15
     RE: infinite series alexb Aug-22-02 16
         RE: infinite series Bo Jacoby Aug-28-02 17
             RE: infinite series alexb Aug-30-02 18
                 RE: infinite series Bo Jacoby Sep-02-02 19
                     RE: infinite series alexb Sep-02-02 20
                         RE: infinite series Bo Jacoby Sep-03-02 21

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alexb
Charter Member
802 posts
May-07-02, 02:01 PM (EST)
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1. "RE: infinite series"
In response to message #0
 
   >(1/2) + (1/4) + (1/8) + (1/16) + ...
>Does this equal 1 or approach 1?

There's a common definition of a series. An expression like the above represents a number, if convergent. It represents nothing, if divergent. In any event, it does not move anywhere to be able to approach anything. Therefore, according to the common definition, to say that 1/2 + 1/4 + 1/8 + 1/16 + ... approaches 1 would be entirely wrong. On the other hand, the sequence of partial sums

1/2, 1/2 + 1/4, 1/2 + 1/4 + 1/8, 1/2 + 1/4 + 1/8 + 1/16, ...

can be said to approach 1, the number to which 1/2 + 1/4 + 1/8 + 1/16 + ... is equal.


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eipi
guest
May-08-02, 07:15 AM (EST)
 
2. "RE: infinite series"
In response to message #0
 
   It equals 1 - but only by a devious definition of "equals".

By normal arithmatic, addition is a binary operation. It only has a meaning when applied to two numbers at a time. The sum of a finite series, say

1 + 2 + 3 + 4 + 5 + 6 + 7

is actually (((((1 + 2) + 3) + 4) + 5) + 6) + 7

This doesn't work with infinite series, as we would have an infinite row of opening brackets.

So.... addition for infinite series is defined as the limit of the sequence formed by adding first one, then two, then three, then four etc.etc. terms together. This sequence never equals 1, it tends to 1, but the series does equal 1, by the definition outlined above.


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jman_red
Member since Jul-2-02
Jul-03-02, 09:11 PM (EST)
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3. "RE: infinite series"
In response to message #0
 
   It does not equal one because no matter how infinitely you add, your fraction will still be x-1/x, instead of the x/x that is one.

Ex. 1/2, 1/2 + 1/4 = 3/4, 1/2 + 1/4 + 1/8 = 7/8... etc.

jman_red


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mathmaster
guest
Jul-04-02, 10:48 PM (EST)
 
4. "RE: infinite series"
In response to message #3
 
   ok u can never say that an infinite series is equal to something. there is no exact value for it. what we can do is find a finite part of the infinite series and find what the sum is and say what the series is approaching. that's what limits are for.


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Chris Tynan
guest
Jul-11-02, 09:10 PM (EST)
 
5. "RE: infinite series"
In response to message #4
 
   It may be said that the sequence:

1/2 + 1/4 + 1/8 ... + 1/2^n tends to one as x tends to infinity.

Your problem can be compared another famous problem.


Lets say that we build a lamp that stays on for 30 seconds, then goes off for 15 seconds, then comes back on for 7.5 seconds, and so on.

After a minute, will the lamp be on or off?



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pradeep
guest
Jul-12-02, 07:14 AM (EST)
 
6. "RE: infinite series"
In response to message #5
 
   let

x = 1/2 + 1/4 + 1/8 .........inf


hence this satisfies the equation

x/2 + 1/2 = x

so

x= 1



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Michael Klipper
guest
Aug-19-02, 06:42 AM (EST)
 
12. "RE: infinite series"
In response to message #6
 
   I really like your method of explaining the sum neatly and succintly. This method is actually very practical when dealing with recurrence relations and infinite series. The key is that you're taking the series as a whole and applying manipulations to all the terms at once.


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jman_red
Member since Jul-2-02
Jul-14-02, 12:22 PM (EST)
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7. "RE: infinite series"
In response to message #0
 
   The infinite series 1/2 + 1/4 + 1/8 + 1/16 ... is not equal to 1.

The series will always follow an x-1/x form

1/2
1/2 + 1/4
3/4
3/4 + 1/8
7/8
7/8 + 1/16
15/16
...
and so on.

It will remain in the form x-1/x, instead of the x/x that is one.

jman_red


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alexb
Charter Member
802 posts
Jul-14-02, 12:30 PM (EST)
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8. "RE: infinite series"
In response to message #7
 
   >The infinite series 1/2 + 1/4 + 1/8 + 1/16 ... is not equal
>to 1.

Before you say that you should probably look up a definition of a series. (In at least a couple of messages in this thread you can find it mentioned.)

The infinite series 1/2 + 1/4 + 1/8 + 1/16 ... is equal to 1! Period.
Please do not confuse the above series with the sequence

>1/2
>1/2 + 1/4
>3/4
>3/4 + 1/8
>7/8
>7/8 + 1/16
>15/16
>...
>and so on.

First, you have to ask yourself, is whatever you have in mind while writing

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

is a number or not. I suspect you think it is a number, although you have been missing its definition. As any other number, it is fixed. It has a value. What is it? You say it is less than 1. Fine. How much less? Think of that.



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jman_red
Member since Jul-2-02
Jul-17-02, 09:26 PM (EST)
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9. "RE: infinite series"
In response to message #8
 
   (1/2)+ 1/4 + 1/8 + 1/16
(1/2 + 1/4)+ 1/8 + 1/16
3/4 + 1/8 + 1/16
(1/2 + 1/4 + 1/8)+ 1/16
7/8 + 1/16
(1/2 + 1/4 + 1/8 + 1/16)
15/16

My series is the same as the one in question

"How much less?"

Read the last post on "site name?" to see how much less.

Although the sum of the series is smaller only by the most infinetly small value, it'should be said to approach one instead of equaling one.


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stapel
Member since Mar-5-02
Jul-18-02, 07:17 AM (EST)
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10. "RE: infinite series"
In response to message #9
 
   You have not shown a series, but a sequence of partial sums. These are two quite-different things. The values in the sequence of partial sums do "approach" the value "1". And this sequence, along with it's limit value, is used to prove, according to the definition, that the associated (but different) series has a value of 1 -- not that the series "approaches" 1, because series do not "approach" anything, but that the value of the infinite sum is "1".

Please take some time to study the differences between sequences and series, to learn the definition of a limit, and to learn what a limiting procedure is. Once you understand the terms and processes that are under discussion, you will likely have a better chance at understanding the answers that have been given.

Thank you.


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chris tynan
guest
Jul-18-02, 01:14 PM (EST)
 
11. "RE: infinite series"
In response to message #0
 
   Let's put an end to this.

1/2 + 1/4 +1/8 .. is a decaying geometric series with first term 1/2 and common ratio 1/2.

Using the formula for the sum of a decaying geometric series:

S = first term / <1 - r>

= 1/2 / <1- 1/2>

= 1/2 / 1/2


= 1

There we go - the series equals 1


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Paul Chen A.K.A
guest
Aug-20-02, 05:48 AM (EST)
 
13. "RE: infinite series"
In response to message #0
 
   Let me ask you all a question:
Let say there's a snail who is crawling on a stick. The length of the stick is 1 meters. Everytime the snail moves 1cm, the stick grows an extra 1m. So this is what it looks like....
=--------- <-- 1m -->
-=------------------ <-- 1m -->
--=---------------------------
---=------------------------------------
= the snail
= the stick

Question: Will the snail make it to the other end of the stick.


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alexb
Charter Member
802 posts
Aug-21-02, 05:52 AM (EST)
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14. "RE: infinite series"
In response to message #13
 
   The question is only interesting if you assume that both the snail and the stick are in possession of eternal life, right?

Please do not reply to that thread, but start a new one instead.


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Bo Jacoby
guest
Aug-22-02, 07:20 AM (EST)
 
15. "RE: infinite series"
In response to message #0
 
   I like your distinction between 'equal' and 'approach'.
These two words do not necessarily refer to the same concept.

YES, the series x=1/2+1/4+1/8+1/16+... EQUALS 1
because it'satisfies the equation 2x=x+1
and the only solution of this equation is x=1.
So if you want the series to have any value at all
it had better be 1.

YES, the series 1/2+1/4+1/8+1/16+... APPROACHES 1
because the SEQUENCE
(1/2, 1/2+1/4, 1/2+1/4+1/8, 1/2+1/4+1/8+1/16,...)
= (1/2, 3/4, 7/8, 15/16, ...)
approaches 1.

Now consider another series: x=1+2+4+8+16+...

This series EQUALS -1
because it satisfies the equation 2x=x-1
and the only solution to this equation is x=-1.
So if you want the series to have any value at all
it had better be -1.

But the series does not APPROACH -1
because the sequence
(1, 1+2, 1+2+4, 1+2+4+8, ...)
=(1, 3, 7, 15,...)
does not approach -1.


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alexb
Charter Member
802 posts
Aug-22-02, 07:37 AM (EST)
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16. "RE: infinite series"
In response to message #15
 
   >I like your distinction between 'equal' and 'approach'.
>These two words do not necessarily refer to the same
>concept.
>
>YES, the series x=1/2+1/4+1/8+1/16+... EQUALS 1
>because it'satisfies the equation 2x=x+1
>and the only solution of this equation is x=1.

The series 1/2+1/4+1/8+1/16+... equals 1 because it is convergent and satisfies 2x=x+1.

>So if you want the series to have any value at all
>it had better be 1.

Being convergent, the series does have a sum. It's not a matter of wanting or not wanting it.
>
>YES, the series 1/2+1/4+1/8+1/16+... APPROACHES 1
>because the SEQUENCE

The series does not approach 1, because from the above, it is just a number. The sequence below does approach 1.

>(1/2, 1/2+1/4, 1/2+1/4+1/8, 1/2+1/4+1/8+1/16,...)
>= (1/2, 3/4, 7/8, 15/16, ...)
>approaches 1.

Series (possibly) equal, sequences (possibly) approach.

>Now consider another series: x=1+2+4+8+16+...
>
>This series EQUALS -1

No, it does not. The series is divergent and can't be assigned a value in any meaningful sense.

>because it satisfies the equation 2x=x-1

Consider y = 1 - 1 + 1 - 1 + 1 - ...

Then y = (1 - 1) + (1 - 1) + ... = 0,
but also y = 1 - (1 - 1) - (1 - 1) - ... = 1,
and also y = 1 - y, so that t = 1/2.

It's more difficult to manipulate 1+2+4+8+16+..., but the principle is the same. The series is divergent and could not be assigned any value.
>and the only solution to this equation is x=-1.
>So if you want the series to have any value at all
>it had better be -1.

>But the series does not APPROACH -1
>because the sequence
>(1, 1+2, 1+2+4, 1+2+4+8, ...)
>=(1, 3, 7, 15,...)
>does not approach -1.


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Bo Jacoby
guest
Aug-28-02, 10:26 PM (EST)
 
17. "RE: infinite series"
In response to message #16
 
   Anything satisfying the equation 2x=x+1 equals 1, 'convergent' or not.
Anything satisfying the equation 2x=x-1 equals -1.
Anything satisfying the equation -x=x-1 equals 1/2.

Therefore 1/2+1/4+1/8+1/16+... = 1
and 1+2+4+8+... = -1
and 1-1+1-1+... = 1/2
and, generally, (1-q)(1+q+qq+q^3+...)=1
This gives meaning to all expressions x=1+q+qq+q^3+...
except for q=1, where 0x=1 shows us that 1+1+1+1+... cannot be assigned any value.

If you don't understand a sentence or a formula, then it has no meaning to you. Investigation might show you that some meanings are possible and others are not. You might or might not want to assign one of the possible meanings to a formula.

You are free to miss the meaning, and I am free to get it.

>Consider y = 1 - 1 + 1 - 1 + 1 - ...
>Then y = (1 - 1) + (1 - 1) + ... = 0,
>but also y = 1 - (1 - 1) - (1 - 1) - ... = 1,
>and also y = 1 - y, so that y = 1/2.
>
This merely shows us that we are not generally allowed to insert an infinite number of parentheses.


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alexb
Charter Member
802 posts
Aug-30-02, 11:58 PM (EST)
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18. "RE: infinite series"
In response to message #17
 
   >Anything satisfying the equation 2x=x+1 equals 1,
>'convergent' or not.
>Anything satisfying the equation 2x=x-1 equals -1.
>Anything satisfying the equation -x=x-1 equals 1/2.

Yes, of course.

>Therefore 1/2+1/4+1/8+1/16+... = 1

Only because it's a convergent series you can apply arithmetic operations to show that 2x=x+1.

>and 1+2+4+8+... = -1

No, not at all. What you can claim at best is that if 1+2+4+8+... has a finite value and arithmetic operations apply in the expected manner, the value is -1.

>and, generally, (1-q)(1+q+qq+q^3+...)=1

I know how to prove that for |q| < 1. What meaning does it have for |q| > 1?

>This gives meaning to all expressions x=1+q+qq+q^3+...

In order to give it must first have ... Does it?

>except for q=1, where 0x=1 shows us that 1+1+1+1+... cannot
>be assigned any value.
>
>If you don't understand a sentence or a formula, then it has
>no meaning to you.

Yes, of course.

>Investigation might show you that some
>meanings are possible and others are not.

Yes, of course. But what investigations do you have in mind that relate to the problem at hand?


>You are free to miss the meaning, and I am free to get it.

Absolutely. Every one has the right to hug an electric pole.

>>Consider y = 1 - 1 + 1 - 1 + 1 - ...
>>Then y = (1 - 1) + (1 - 1) + ... = 0,
>>but also y = 1 - (1 - 1) - (1 - 1) - ... = 1,
>>and also y = 1 - y, so that y = 1/2.
>>
>This merely shows us that we are not generally allowed to
>insert an infinite number of parentheses.

But is not what you are doing when deriving your 1/2 or -1?


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Bo Jacoby
guest
Sep-02-02, 05:58 AM (EST)
 
19. "RE: infinite series"
In response to message #18
 
   Thank you, alexb, for your prompt reaction.

>>(1-q)(1+q+qq+q^3+...)=1
>I know how to prove that for |q| < 1. What meaning does it
>have for |q| > 1?
The condition |q|<1 does not enter into the computation:
x=1+q+qq+...=1+(q+qq+...)=1+q(1+q+qq+...)=1+qx
So the condition is unnecessary.

>what investigations do you have in mind
>that relate to the problem at hand?
That x=1+2+4+8... satisfies x=1+2x , even if the series isn't convergent. So x=-1 is a possible meaning and no other value is possible.

>Every one has the right to hug an electric pole.
What is the danger in assigning 1+2+4+8+...=-1 ?

>>This merely shows us that we are not generally allowed to
>>insert an infinite number of parentheses.
>But is not what you are doing when deriving your 1/2 or -1?
No, I only insert one pair of parentheses:
x=1+q+qq+...=1+(q+qq+...)=1+q(1+q+qq+...)=1+qx
So we need not consider conditions for infinite insertion of parentheses to be permissible.


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alexb
Charter Member
802 posts
Sep-02-02, 06:21 AM (EST)
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20. "RE: infinite series"
In response to message #19
 
   >The condition |q|<1 does not enter into the computation:
>x=1+q+qq+...=1+(q+qq+...)=1+q(1+q+qq+...)=1+qx

Yes, but what enters the computation is the assumption that x (whatever it is) is finite.

>So the condition is unnecessary.

But assume that x is infinite, which I would say is a reasonable assumption given that the terms of the infinite sum grow.

>>what investigations do you have in mind
>>that relate to the problem at hand?
>That x=1+2+4+8... satisfies x=1+2x , even if the series
>isn't convergent. So x=-1 is a possible meaning and no other
>value is possible.

Except for infinity, of course.

>>Every one has the right to hug an electric pole.
>What is the danger in assigning 1+2+4+8+...=-1 ?

Danger? There's none to anybody. Hugging an electric pole is a metaphor for a senseless, not a suicidal action. Possible impressions to the contrary, math notations are introduced for a purpose. They are seldom labels like a street address. The assignment of -1 to 1+2+... has no purpose. The effort starts and ends with that assigtnment.

>>>This merely shows us that we are not generally allowed to
>>>insert an infinite number of parentheses.
>>But is not what you are doing when deriving your 1/2 or -1?
>No, I only insert one pair of parentheses:
>x=1+q+qq+...=1+(q+qq+...)=1+q(1+q+qq+...)=1+qx
>So we need not consider conditions for infinite insertion of
>parentheses to be permissible.

Ok, then consider conditions that could justify insertion of a pair of parentheses.


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Bo Jacoby
guest
Sep-03-02, 07:46 AM (EST)
 
21. "RE: infinite series"
In response to message #20
 
   >>The condition |q|<1 does not enter into the computation:
>>x=1+q+qq+...=1+(q+qq+...)=1+q(1+q+qq+...)=1+qx
>Yes, but what enters the computation is the assumption that
>x (whatever it is) is finite.
If x has no meaning as a number then the computation is merely a formality. Still, mathematics deals with formalities. The formality tells us that x can uniquely be given meaning as a number, unless q=1.

>But assume that x is infinite, which I would say is a
>reasonable assumption given that the terms of the infinite
>sum grow.
You may prove by induction that 'a finite sum of positive terms is greater than any term', but the proof does not apply to infinite series. The assumption may be reasonable, but it is not necessary.

That 'the sum of positive numbers is positive' is proved by induction for finite sums but not for infinite series.
Counterexample: 1+2+4+8+...=-1
Even if the terms are positive, the sum is negative.

That 'the sum of integers is an integer' is proved by induction for finite sums but not for infinite series.
Counterexample: 1-1+1-1+...=1/2
Even if the terms are integers, the sum is fractional.

'a+(b+c)=(a+b)+c' allows us to insert a finite number of pairs of parentheses into a sum, but not an infinite number of parentheses.
Example: (1-1)+(1-1)+...=0 while 1-1+1-1+...=1/2
Even if the terms are the same, the sums differ.

'a+b=b+a' allows us to reshuffle a finite number of terms in a sum, but not an infinite number of terms.
Example: 1-1/2+1/3-1/4+1/5-...=log(2)
while 1+1/3-1/2+1/5+1/7-1/4+...>log(2)
Even if the terms are the same, (and even if both series are convergent!), the sums differ.

We have got two options:
1. We restrict the meaningful series to be the ones for which the above rules are true.
2. We accept that the rules are not generally true, giving meaning to more series.
I think that the latter option gives the wider horison.

>math notations are introduced for a purpose.
Mathematical ideas have often no immediate application. Some ideas get applications later in history.

>The assignment of -1 to 1+2+... has no purpose.
Actually it is used in computer design.
Computers use 8-digit and 16-digit binary numerals. '00001001' means 2^3+2^0=9. The 8-bit notation for (-1) is '11111111'. Adding (-1) to (+1) gives '00000001'+'11111111'='00000000'. The carry is thrown away and the simple arithmetics works. Extending from 8 bits to 16 bits is done by copying the leftmost bit:
'00001001'='0000000000001001'=(+9)
'11111111'='1111111111111111'=(-1)
The mathematics of this notation for negativ numbers is that
1+2+4+8+...=-1. So it is a useful idea.

In decimal notation, negative numbers without the minus sign but with '9' as the leftmost digit was used in trigonometric tables, (as log sin is negative).
'9.9999' meaning -0.0001 and '9.8265' meaning -0.1735.
This is nice for routine calculation:
0518+9377=9895 is easier than 518-623=-(623-518)=-105

So 9+90+900+...=...999=-1

>consider conditions that could justify insertion of
>a pair of parentheses.
a+b+c+... means a+(b+(c+(...)))
So if (...) is defined, then a+b+c+... can be considered a finite sum and the rules for finite sums apply.
a+b+c+...=a+b+c+(...)=((a+b)+c)+(...)
so insertion of a finite number of pairs of parentheses is generally justified.


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