Thanks Tiny-tim,
The actual equations I am trying to equate are pretty large, and apparently there must be a trick to calculate the variable (found in one of the two equations) - most probably it should be a closed form solution.
So far, I don't think there will be a solution and so I thank...
Thanks Tin-tim,
No, suppose I have the following two equations:
x=\sum^{N}_{t=1}\beta^{t}{C(1-g)^{t}-\beta^{t}}
y=\sum^{N}_{t=1}\beta^{t}{C(1+k)^{t}-\beta^{t}}
I want to calculate \frac{x}{y}, and in the process I need to eleminate C as I don't have its value.
Any ideas?
Thanks...
Hi again,
The last equation is not equivalent- but the problem, after factoring the (1/1-α) out, can be expressed as:
\sum^{N}_{t=1}\beta^{t}{C(1-g)^{t}-\beta^{t}}
where I need to get the constant C out.
Regards
Thanks again, but suppose I have the following which can serve as a perfect simplification of the problem:
Ʃa^t*C - a^T
your answer suggests that it becomes ƩC(a^t*C - a^T) and this is not a solution- beside, indeed the denominator can be factored out but I am interested in factoring the C...
I don't know if I got you right! because:
- I don't want to end up with c*(cƩ-Ʃ) which is nothing really
- Further, 3*2-3*1 are not the same as 2-1
Thanks anyways for the post
Just as a quick add:
I think that if I simplify to:
\sum^{N}_{t=1}\frac{\beta^{t}C(1-g)^{t}}{1-\alpha}-\frac{\beta^{t}}{1-\alpha}
I could get the C constant out, but the only problem is that I will end up with:
C\sum^{N}_{t=1}\frac{\beta^{t}(1-g)^{t}}{1-\alpha}-...
Hello All,
I have what I think an easy summation, but I haven't worked with math for very long - I don't know the term which I should search the internet for in order to solve the problem and so I would be very thankful if you help me get the constant C out of the summation...