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

In summary: Summary: In summary, the conversation discusses the search for an infinite series summation that is equal to the imaginary number, i. It is mentioned that this can be achieved by replacing "i" with any other value in a convergent sequence. The conversation also explores the idea of finding a unique infinite series equal to "i" and the potential use of infinite products in this search. One participant offers a challenging homework assignment to find an infinite number of infinite product identities for "i". The conversation concludes with a discussion on the benefits of hiding or exposing "i" in mathematical equations.
  • #36
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.

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?
 
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  • #37
mariusdarie said:
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
mariusdarie said:
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
##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
DrewD said:
##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.
 
  • #42
mesa said:
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.
 
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