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

  • Thread starter mesa
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  • #26
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.
 
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  • #27
646
18
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?
 
  • #28
221
0
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## .
 
  • #29
646
18
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!
 
  • #30
221
0
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.
 
  • #31
646
18
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.
 
  • #32
221
0
Plug and Chug...danger, danger
 
  • #33
221
0
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.
 
  • #34
221
0
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....
 
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  • #35
646
18
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 :)
 
  • #36
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?
 
  • #37
22,097
3,278
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:

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

So I get the equation

[tex]\frac{2}{2- i} = \frac{3}{8}k(2 + i)[/tex]

Solving for ##k##, we get

[tex]k = \frac{16}{3(2-i)(2+i)} = \frac{16}{15}[/tex]

like expected.

Solving for ##i## is not possible.
 
  • #38
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.
 
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  • #39
646
18
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!
 
  • #40
529
28
##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
 
  • #41
646
18
##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 text books 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.
 
  • #42
529
28
That is pretty awesome, I think it is high time I open up some of my calc text books 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.
 

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