infinite series

May 2010
4
0
Dear All, could you help me to solve following?

I need to know, how much i is and how to calculate it.




Thanks milion times
 
Oct 2009
4,261
1,836
Dear All, could you help me to solve following?

I need to know, how much i is and how to calculate it.




Thanks milion times

Sum of geometric sequence: \(\displaystyle \sum^n_{k=1}a^k=\frac{a(a^{n+1}-1)}{a-1}\) , and here:

\(\displaystyle \sum^{60}_{n=1}\left(\frac{1}{1+i}\right)^n=\sum^{60}_{n=1}\left(\frac{1-i}{2}\right)^n=\frac{\frac{1-i}{2}\left[\left(\frac{1-i}{2}\right)^{61}-1\right]}{\frac{1-i}{2}-1}\) \(\displaystyle =\frac{\frac{(1-i)^{62}}{2^{61}}-(1-i)}{-1-i}=\)...--- Since \(\displaystyle (1-i)^2=-2i\) and \(\displaystyle i^{4m+j}=i^j\,,\,\,m=0,1,2,3\) , we get ---...:

\(\displaystyle =-\frac{\frac{(-2i)^{31}}{2^{61}}-1+i}{1+i}=\) \(\displaystyle -\frac{-\frac{1}{2^{30}}(-i)-1+i}{1+i}=\) ...

By the way, check your indexes...

Tonio
 
May 2010
4
0
Hello Tonio,

thank you very much for your help. But to be honest, I know nothing from math, I need to calculate this for loan purposes. Is there any change you continue in it and get me something like: i=..... so I can input it into excel and have the number?

Or let me know how to calculate this in excel?

Thank you in advance.

Eske
 
Oct 2009
4,261
1,836
Hello Tonio,

thank you very much for your help. But to be honest, I know nothing from math, I need to calculate this for loan purposes. Is there any change you continue in it and get me something like: i=..... so I can input it into excel and have the number?

Or let me know how to calculate this in excel?

Thank you in advance.

Eske


Too bad you didn't say the first time you don't know anything about maths: I would have saved myself a lot of trouble and time...(Punch)

Anyway, if you want this for loan purposes then we will need more information: how come that complex unit i got there, say....or perhaps it is another thing altogether since I know of any use of complex numbers to calculate loans interests and stuff.

Tonio
 
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HallsofIvy

MHF Helper
Apr 2005
20,249
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I suspect the "i" is an interest rate. Although I am still confused as to how he has an index of "n" on the sum and "k" inside. I believe he meant
\(\displaystyle \sum_{k=1}^{60} (1+i)^{-k}\).

That is, as Tonio says, as geometric sequence and its sum is
\(\displaystyle \frac{1- (1+i)^{-61}}{1-\frac{1}{1+i}}\)\(\displaystyle = \frac{1-(1+i)^{-61}}{\frac{i}{1+i}}\)\(\displaystyle = (1- (1+i)^{-61})\frac{1+i}{i}\).

You still have that \(\displaystyle (1- (1+ i)^{-61}\) but I suspect that you don't need dozens of decimal place and can take \(\displaystyle (1+ i)^{-61}\) to be approximately 1- 61(1+ i) so that (1- (1+i)^{-61}) is approximately 61(1+ i) and the whole thing is \(\displaystyle 61\frac{(1+i)^2}{i}= 39.55\). that will give you \(\displaystyle 61i^2+ 122i+ 61= 39.55i\) or \(\displaystyle 61i^2+ 122i+ 21.45= 0\), a quadratic equation for i. Solve it using the quadratic formula.

And, by the way, this was titled "infinite series". 60 is hardly infinite!
 
Dec 2009
3,120
1,342
Dear All, could you help me to solve following?

I need to know, how much i is and how to calculate it.




Thanks milion times
\(\displaystyle (1+i)^{-60}\) 60 times is just \(\displaystyle 60(1+i)^{-60}=39.55\ or\ =3,955\)

so, assuming the sum should be from k=1 to 60
and for convenience letting x=1+i, solving for x,
when finally i=x-1

\(\displaystyle \frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+....+\frac{1}{x^{60}}\)

\(\displaystyle =\frac{1}{x}\left(1+\frac{1}{x}+\frac{1}{x^2}+.....+\frac{1}{x^{59}}\right)\)

is a geometric series and equals \(\displaystyle \frac{1}{x}\left(\frac{\frac{1}{x^{60}}-1}{\frac{1}{x}-1}\right)=\frac{\frac{1}{x^{60}}-1}{x(1-x)}\)

Solving this graphically, using root-finding software, we get

\(\displaystyle \frac{\frac{1}{x^{60}}-1}{x(1-x)}-3,955=0\)

for x=0.907783 approximately.

Slight variations in x yield quite different results.

This gives i=x-1=0.907783-1=-0.092217

If the sum of the series is 39.55 rather than 3,955,
this yields x=1.0143, so i=0.0143 approximately.