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mesa
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The title pretty much says it all, does anyone know of an infinite series summation that is equal to $$\sqrt{-1}$$?
johnqwertyful said:Let ∑an be convergent sequence with limit L. Then ∑i an/ L converges to i
johnqwertyful said:Yes, you may replace "i" with anything.
mesa said:have you ever seen a unique infinite series equal to 'i'?
Stephen Tashi said:What do you mean by a "unique" infinite series. For that matter, what would a non-unique infinite series be?
ClamShell said:I do...but it's not an infinite sum...it's an infinite product.
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?
mesa said:Honestly I have little experience with infinite products so far but I would love to see your work!
mesa said: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:
And I'll give you half credit if the only ones you find are an infinite numbermesa said:Honestly I have little experience with infinite products so far but I would love to see your work!
Stephen Tashi said: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 said:Sorry to have imposed...I used to like sums because
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.
Keep in touch.mesa said: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!
Just can't keep my mouth shut...mesa said:The title pretty much says it all, does anyone know of an infinite series summation that is equal to $$\sqrt{-1}$$?
ClamShell said: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 don't see it, please show your work...mesa said: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?
ClamShell said:I don't see it, please show your work...
ClamShell said:I don't see it, please show your work...
ClamShell said:So the sum is ##(2 - i)## ?
mariusdarie said:I like this forum!
I used Mac Laurin (1/(1-x/2) and e^(i*pi))formulas therefore "i" can be written as relation which is containing series but not just one summation series because i^0=1; i^1=i;i^2=-1;i^3=-i;i^4=1...
Now using Mac Laurin formula for 1/(1-x/2) and k=sum(j=0;oo;(1/2)^(4*j)); we have i=-2*(15*k-4)/(15*k-16).
This is not a sum of series but a relation which contains series.
mesa said:So what else can we do?
ClamShell said:Here's what I suggest, although it is very distasteful to folks
who have learned very much about ##i## .
Coming up with normalized sums and then multiplying both
sides of the equation by ##i## is a bit trivial in my book.
It's what I call "factoring it in".
I think what the many folks who try to come up with infinite
sums for ##i##, are really after, is a method for "factoring it out"
of the right-hand-side of a normalized relationship; but the only
infinite sums that seem to crop up leaves ##i## on both sides
of the equation.
An example of "factoring ##i## out" instead of "factoring ##i## in"
is ##0 = x^2 + 1## .
ClamShell said:Next step might be to figure out what ##i^2, i^3, i^4## look like.
ClamShell said:I'm at a disadvantage because I do not know how you arrived at your
identity.
Take a look at what it takes to prove some identities, in particular,ClamShell said:Plug and Chug...danger, danger
ClamShell said:Sum n/(2^n) is also neat...it converges to 2
It's the entropy of Sum 1/(2^n) if you choose
to think of it as a probability sum equal to one.
n starts at 1 in both sums.
You might use Sum n/(2^n) for the 2 in your identity,
I'm not telling you how to do this stuff, just
making suggestions.
Maybe 1/(0.4) could be rewritten as 2.5...see
where I'm going? Put both together and enclose
in just one summa.
EDIT: Sum 2/(2^n) could also be used to replace
the 2 in your identity.
ClamShell said:What does 1/(2i)^n converge to? Wolfram could
tell us that. I might try Wolfram some day.
You might ask, "where do you come up with this
stuff"...I can't even use IRS programs without
making mistakes. Go figure.
ClamShell said:I'm thinkin' that you already have enough
knowledge of sums; remember I said it was
a fatal attraction...I'm thinkin' all you need
to learn products is in:
http://en.wikipedia.org/wiki/Infinite_product
ClamShell said:And all you need to know about my infinite product
identity, is that it's not merely a polynomial, it's an
infinite product of polynomials; special polynomials;
cross-product free polynomials. And the only way to
get cross-product free polynomials is by real and
complex conjugate pairs. All the principle cross-product
free polynomials that exist. Ahem...