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

[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.