Does anyone know an infinite series summation that is equal to i?

[mesa]

The title pretty much says it all, does anyone know of an infinite series summation that is equal to $$\sqrt{-1}$$?

 

[johnqwertyful]

Let ∑an be convergent sequence with limit L. Then ∑i an/ L converges to i

 

[mesa]

Let ∑an be convergent sequence with limit L. Then ∑i an/ L converges to i

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?
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?

 

[johnqwertyful]

Yes, you may replace "i" with anything.

 

[mesa]

Yes, you may replace "i" with anything.

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!

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?!

 

[Stephen Tashi]

have you ever seen a unique infinite series equal to 'i'?

What do you mean by a "unique" infinite series. For that matter, what would a non-unique infinite series be?

 

[mesa]

What do you mean by a "unique" infinite series. For that matter, what would a non-unique infinite series be?

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

 

[ClamShell]

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]

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?

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

 

[ClamShell]

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

Here is your homework assignment:

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.

 

[Stephen Tashi]

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

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]

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

And I'll give you half credit if the only ones you find are an infinite number
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 anyone of the infinite number of factors.

 

[ClamShell]

For those of you who do not like to cross threads,
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.

 

[ClamShell]

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

 

[mesa]

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.

Sounds like we are on the same page, although you guys at PF are masters of semantics so I look forward to your response :)

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

 

[ClamShell]

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.

 

[mesa]

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.

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!

 

[ClamShell]

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!

Keep in touch.

 

[ClamShell]

The title pretty much says it all, does anyone know of an infinite series summation that is equal to $$\sqrt{-1}$$?

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?

 

[mesa]

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 ?

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]

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?

I don't see it, please show your work...

 

[mesa]

I don't see it, please show your work...

Hang on, I'm throwing it in wolfram...

 

[mesa]

I don't see it, please show your work...

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

That's better, now it works

 

[ClamShell]

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

 

[mesa]

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

Apparently so!

So what else can we do?

 

[mariusdarie]

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]

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.

I'm not making the connection here, could you show some steps?

 

[ClamShell]

So what else can we do?

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 $0 = x^2 + 1$ .

 

[mesa]

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 $0 = x^2 + 1$ .

That's basically what I did building off of your suggestion of 1/(2i)^n.

Let's try to get something more concrete down, for example throw up another infinite series summation for some real number, stuff 'i' into the mix somewhere where it will be affected by each consecutive term and see if we can algebraically manipulate it to get more summations for 'i' like your last one.

Should be fun!

 

[ClamShell]

Next step might be to figure out what $i^2, i^3, i^4$ look like.

I'm at a disadvantage because I do not know how you arrived at your
identity.

 

[mesa]

Next step might be to figure out what $i^2, i^3, i^4$ look like.

0

I'm at a disadvantage because I do not know how you arrived at your
identity.

Sometimes you just see stuff and when it works you shout hooray! I use more 'intuition' than I do rigor (that usually comes after to prove the identity) but it seems it was along the lines of your suggestion.

 

[ClamShell]

Plug and Chug...danger, danger

 

[ClamShell]

Plug and Chug...danger, danger

Take a look at what it takes to prove some identities, in particular,
the identity in "A cool identity" in General math.

 

[ClamShell]

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.

What happens when you start at some other n?

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.

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

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...

 

[mesa]

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.

Sure could, we could rewrite it as,

$$i=\sum_{n=0}^{\infty} \frac{n.4(2i)^n-2^n}{2^n.4(2i)^n}$$
or,
$$i=\sum_{n=0}^{\infty} \frac{.8(2i)^n-2^n}{2^n.4(2i)^n}$$

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.

I am not always the biggest fan of computers either however their usefulness is hard to ignore.

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

I don't, I have only been doing this for a few months between school, work, and raising my kids. I haven't even brought Calculus into series summations yet (although I am about to start, this is going to be awesome!)

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...

If you think it's new and useful than you should look to see if it is known, the forum can be very helpful at making suggestions on where to start and (as I have seen suggested on PF) if it seems new try writing a paper and let the math community tear it up :)

 

[mariusdarie]

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.

Mac Laurin
$$\frac{1}{1-x}= \sum_{j=0}^{\infty} {x} ^{j}$$
x=i/2
then
$$\frac{1}{1- \frac {i} {2}}= \sum_{j=0}^{\infty} {\left(\frac{i}{2}\right) ^{j}}= \sum_{j=0}^{\infty} {\left(\frac{i}{2}\right) ^{4j}}+ \sum_{j=0}^{\infty} {\left(\frac{i}{2}\right) ^{4j+1}}+ \sum_{j=0}^{\infty} {\left(\frac{i}{2}\right) ^{4j+2}}+ \sum_{j=0}^{\infty} {\left(\frac{i}{2}\right) ^{4j+3}}=k+i \cdot k-k-i \cdot k$$
where:
$$k=\sum_{j=0}^{\infty} {\left(\frac{1}{2}\right) ^{4j}}$$
we obtain an equation in "i"
$$\frac{1}{1- \frac {i} {2}}=k+i \cdot k-k-i \cdot k$$
$$i=-2 \frac{15k-4}{15k-16}$$
but we have problems when we replace k from denominator Mac Laurin reverse.
$$k=\sum_{j=0}^{\infty} {\left(\frac{1}{2}\right) ^{4j}}= \frac {1} {1- \frac {1}{2^4}}=\frac{16}{15}$$
where could be error?

 

[micromass]

Mac Laurin
$$\frac{1}{1-x}= \sum_{j=0}^{\infty} {x} ^{j}$$
x=i/2
then
$$\frac{1}{1- \frac {i} {2}}= \sum_{j=0}^{\infty} {\left(\frac{i}{2}\right) ^{j}}= \sum_{j=0}^{\infty} {\left(\frac{i}{2}\right) ^{4j}}+ \sum_{j=0}^{\infty} {\left(\frac{i}{2}\right) ^{4j+1}}+ \sum_{j=0}^{\infty} {\left(\frac{i}{2}\right) ^{4j+2}}+ \sum_{j=0}^{\infty} {\left(\frac{i}{2}\right) ^{4j+3}}=k+i \cdot k-k-i \cdot k$$

I fear this expression is wrong (since it is also easy to see that it equals zero).

I get the following expression:

k + \frac{ik}{2} - \frac{k}{4} - \frac{ik}{8} = \frac{3}{8}k(2 + i)

So I get the equation

\frac{2}{2- i} = \frac{3}{8}k(2 + i)

Solving for $k$, we get

k = \frac{16}{3(2-i)(2+i)} = \frac{16}{15}

like expected.

Solving for $i$ is not possible.

 

[mariusdarie]

Yes it is true. I was focused writing the equation in Latex.
Before 1/(1-x) formula I tried e^(i*pi) but I haven't finished it. Maybe Mac Laurin for e^(i*pi) =-1 could help.

 

[mesa]

Yes it is true. I was focused writing the equation in Latex.
Before 1/(1-x) formula I tried e^(i*pi) but I haven't finished it. Maybe Mac Laurin for e^(i*pi) =-1 could help.

Interesting approach, I am not sure if we have any infinite series for 'i' that come out of well known math but micromass is our local expert so I am interested to see!

 

[DrewD]

$i=e^{i\frac{\pi}{2}}=\sum\left(\frac{i\pi}{2}\right)^n\cdot\frac1{n!}$

but really this is trivial since we can, for just about any function*, set it equal to $i$ and solve for $x$. Then substitute that into the maclaurin/taylor series.

$i=\frac{1}{1-x}\rightarrow x=1+i$ so we have $\sum (1+i)^n=i$. In a matter of minutes I bet you can come up with 20 others for any number you choose. Some might look nice and others will be ugly.

Have you posted this before or am I having deja vu? I feel like this question was asked and answered in the same way recently.

*well, maybe a lot of functions won't work, but enough will that it doesn't matter

 

[mesa]

$i=e^{i\frac{\pi}{2}}=\sum\left(\frac{i\pi}{2}\right)^n\cdot\frac1{n!}$

but really this is trivial since we can, for just about any function*, set it equal to $i$ and solve for $x$. Then substitute that into the maclaurin/taylor series.

$i=\frac{1}{1-x}\rightarrow x=1+i$ so we have $\sum (1+i)^n=i$. In a matter of minutes I bet you can come up with 20 others for any number you choose. Some might look nice and others will be ugly.

Have you posted this before or am I having deja vu? I feel like this question was asked and answered in the same way recently.

*well, maybe a lot of functions won't work, but enough will that it doesn't matter

That is pretty awesome, I think it is high time I open up some of my calc textbooks and see how these things are done. Everything I know about summations at this point is self taught, it seems there are a wealth of well known neat tricks I could use.

 

[DrewD]

That is pretty awesome, I think it is high time I open up some of my calc textbooks and see how these things are done. Everything I know about summations at this point is self taught, it seems there are a wealth of well known neat tricks I could use.

Definitely. Taylor series is well worth your time.