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- Thread starter mesa
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Let ∑an be convergent sequence with limit L. Then ∑i an/ L converges to i

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If that works then it is clever and unexpected (a wonderfully typical trait of PF:) although it will work for anything in place of 'i' correct?Let ∑an be convergent sequence with limit L. Then ∑i an/ L converges to i

Being recursive is fine but I am looking for a 'unique' solution.

I have one and have a hard time believing I am first to do so, so what else is out there?

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Yes, you may replace "i" with anything.

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Borek posted on a couple of my previous threads 'make an infinite series equal to 1 and multiply (insert the value I was attempting to get) by it.' At least this is a cleverly disguised version of the same thing, well done!Yes, you may replace "i" with anything.

Anyway, have you ever seen a unique infinite series equal to 'i'? Google searches and the such are turning up nothing. How cool would it be if this is a first?!

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Stephen Tashi

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What do you mean by a "unique" infinite series. For that matter, what would a non-unique infinite series be?have you ever seen a unique infinite series equal to 'i'?

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As in one where we don't use an infinite series equal to '1' and then multiply by 'i'. Apparently a fairly common suggestion here on PF :tongue2:What do you mean by a "unique" infinite series. For that matter, what would a non-unique infinite series be?

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Care to investigate this further? Or do you need a sum,

and can't see how a product could come in handy?

Because I have a product, ##log(I)## is a sum and is

easy to get. And powers of "I" are also easy to get

using the infinite product identity.

Would that be of any help in your project?

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Honestly I have little experience with infinite products so far but I would love to see your work!

Care to investigate this further? Or do you need a sum,

and can't see how a product could come in handy?

Because I have a product, ##log(I)## is a sum and is

easy to get. And powers of "I" are also easy to get

using the infinite product identity.

Would that be of any help in your project?

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Here is your homework assignment:Honestly I have little experience with infinite products so far but I would love to see your work!

Given the information in the new thread "A Cool Identity" in General Math,

find an infinite number of infinite product identities for "I". Partial credit

will be given if you can only find a finite number of identities.

HINT: factor the factors into complex and real conjugates.

This is not a "snipe" hunt, we are dealing with the mathematical

certainty of prohibiting polynomial cross-products. Or stated

differently, an infinite number of equations in an infinite number

of unknowns, IS solvable.

Math is very fun.

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Stephen Tashi

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Are you asking for a series that sums to i whose individual terms cannot each be divided by i ? If the individual terms can be divided by i, this gives an infinite series that sums to 1.As in one where we don't use an infinite series equal to '1' and then multiply by 'i'. Apparently a fairly common suggestion here on PF :tongue2:

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And I'll give you half credit if the only ones you find are an infinite numberHonestly I have little experience with infinite products so far but I would love to see your work!

of products equal to unity and trivially multiply by "I".

Math is very, very fun...or none of us would be here, n'est ce pas?

You don't have to be "super-brilliant", just interested in puzzles,

and then you become "super-brilliant" in some tiny, tiny niche of

reality; you learn more and more about less and less.

EDIT: another HINT: be sure to remember that "I" need only be

multiplied by any one of the infinite number of factors.

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here is the super-cool identity:

\begin{eqnarray}

\frac {1} {1-x} =

(1+x) \prod_{n=1} ^{\infty}

\left( \frac {1+x^{2^n}} {1-x^{2^n}} \right)^{2^{-n}} , {\;}

\text{for} {\;} 0\leq x<1

\nonumber

\end{eqnarray}

It's in LaTex, so feel free to move it around.

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Just food for thought, which is better? Exposing "I", or hiding it?

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Sounds like we are on the same page, although you guys at PF are masters of semantics so I look forward to your response :)Are you asking for a series that sums to i whose individual terms cannot each be divided by i ? If the individual terms can be divided by i, this gives an infinite series that sums to 1.

ClamShell, I wish I could do as you are suggesting but infinite products are not yet part of my vocabulary.

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their so easy to differentiate. Now I suspect that it

is just another fatal attraction. The answer is that

it is better to hide "I" than to expose it.

Disclaimer: But I do respect the opinion that "I"

is a pretty fun object. Adios.

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Infinite products are the next step after infinite series, I can't wait to 'dig in' to these wonderful things. When I get up to speed I look forward to accepting your challenges!

their so easy to differentiate. Now I suspect that it

is just another fatal attraction. The answer is that

it is better to hide "I" than to expose it.

Disclaimer: But I do respect the opinion that "I"

is a pretty fun object. Adios.

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Keep in touch.Infinite products are the next step after infinite series, I can't wait to 'dig in' to these wonderful things. When I get up to speed I look forward to accepting your challenges!

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Just can't keep my mouth shut...

What about 1 = 1/2 + 1/4 + 1/8 + 1/16 + ... (Zeno?)?

Each denominator is 2^n for n = 1 to infinity

or each denominator is double the previous denominator.

What happens if each denominator is (2i)^n ?

This simple sum may reveal what happens to

some class of other sums.

Or what happens when ##i = i/2 + i/4 + i/8 + ...## ?

and you need to find powers of ##i## by powering the

right hand side?

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Ah hah! With that we could do this,Just can't keep my mouth shut...

What about 1 = 1/2 + 1/4 + 1/8 + 1/16 + ... (Zeno?)?

Each denominator is 2^n for n = 1 to infinity

or each denominator is double the previous denominator.

What happens if each denominator is (2i)^n ?

$$i=.2-\sum_{n=0}^{\infty} \frac{1}{.4(2i)^n}$$

I think that will work.

Very good, thank you ClamShell!

Anyone else have anything?

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I don't see it, please show your work...Ah hah! With that we could do this,

$$i=.2-\sum_{n=0}^{\infty} \frac{1}{.4(2i)^n}$$

I think that will work.

Very good, thank you ClamShell!

Anyone else have anything?

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Hang on, I'm throwing it in wolfram...I don't see it, please show your work...

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$$i=2-\sum_{n=0}^{\infty} \frac{1}{.4(2i)^n}$$I don't see it, please show your work...

That's better, now it works

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So the sum is ##(2 - i)## ?

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