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visits member for 1 year, 10 months
seen May 26 '13 at 18:18

Feb
20
awarded  Yearling
May
20
awarded  Caucus
Mar
28
revised Curious $\sum _{n=1}^{\infty} \frac{1}{n^2 - x^2}$ identity
\bigg paranthesis :)
Mar
28
suggested approved edit on Curious $\sum _{n=1}^{\infty} \frac{1}{n^2 - x^2}$ identity
Mar
27
awarded  Commentator
Mar
27
comment What's wrong in this equation?
@StefanHansen, your answer is completely correct. Inceptio we know nothing about $x$. It is not mentioned that $x$ is an integer, or anything. Stefan makes a very valid point about that, by demonstrating the case of 2.5. This is the problem in the proof, we cannot say '$\text{x times}$'.
Mar
27
answered What's wrong in this equation?
Mar
25
comment Mathematical way of determining whether a number is an integer
Yes, floor is a purely mathematical function.
Mar
24
revised $x$ is rational, $\frac{x}{2}$ is rational, and $3x-1$ is rational are equivalent
Improved definition
Mar
24
suggested approved edit on $x$ is rational, $\frac{x}{2}$ is rational, and $3x-1$ is rational are equivalent
Mar
24
comment $x$ is rational, $\frac{x}{2}$ is rational, and $3x-1$ is rational are equivalent
What is your definition of rational? If you can answer that, the result is immediate.
Mar
24
accepted Twice a triangle is triangle
Mar
23
comment Twice a triangle is triangle
@AndréNicolas, the Pell's equation I have is $x^2 - 2y^2 = -1$, and not $x^2 - 2y^2 = 1$. I know how to do the latter, squaring and re-arranging gives the recurrence relations. Its the case of $-1$ which I can't deal with. Squaring gives extraneous incorrect solutions, and cubing gives unhelpful result.
Mar
22
awarded  Editor
Mar
22
comment Twice a triangle is triangle
@Alvaro No, not closed form, in terms of the previous solutions.
Mar
22
revised Twice a triangle is triangle
corrected link
Mar
22
comment Twice a triangle is triangle
I already read this and similar questions, but they don't tell how to solve for the recurrence.
Mar
22
asked Twice a triangle is triangle
Mar
22
answered inscribed angles on circle
Mar
18
comment Establishing formula from recurrence
With the help of your excellent response, I managed to write a general form of $f(n)$, ie $f(n) = (y + \frac{b}{a-1}) \cdot a^{n-x} - \frac{b}{a-1}$ given $f(x) = y$ (base condition) and $f(n) = a\cdot f(n-1) + b$. In the case of the given problem, $f(n) = (1 + \frac{3}{-4-1}) \cdot (-4)^{n-1} - \frac{3}{-4-1}$ which simplifies to $f(n) = \frac{2}{5} \cdot (-4)^{n-1} + \frac{3}{5}$. Thanks a lot for your help.