I've come up with this equation at work, and can't
solve the durn
thing. I've tried using logarithms, but I can't
figure out how to
separate the variables. I'm sure this is simple for
one of you guys.

Feel free to just post a link to some how-to if you
want.

(x^(n+1) - 1)/(x - 1)=n^2

I've come up with this equation at work, and can't
solve the durn
thing. I've tried using logarithms, but I can't
figure out how to
separate the variables. I'm sure this is simple for
one of you guys.

Feel free to just post a link to some how-to if you
want.

(x^(n+1) - 1)/(x - 1)=n^2

You can start with

(x^(n+1) - 1)/(x - 1) = 1 + x + x^2 + ... + x^n.

Try solutions in x for n = 1, 2, 3, 4 and see if there's a pattern. (There are closed-form solutions for cubics and quartics.)

You could also try to solve it numerically for different n.

Jack

On Jul 9, 12:58 pm, Jack Tomsky
jtom... at (no spam) ix.netcom.com> wrote:
I've come up with this equation at work, and
can't
solve the durn
thing. I've tried using logarithms, but I can't
figure out how to
separate the variables. I'm sure this is simple
for
one of you guys.

Feel free to just post a link to some how-to if
you
want.

(x^(n+1) - 1)/(x - 1)=n^2

You can start with

(x^(n+1) - 1)/(x - 1) = 1 + x + x^2 + ... + x^n.

Try solutions in x for n = 1, 2, 3, 4 and see if
there's a pattern. (There are closed-form solutions
for cubics and quartics.)

You could also try to solve it numerically for
different n.

Jack


Thanks for that. There's no algebraic solution?
--Buck

On Jul 9, 12:58 pm, Jack Tomsky
jtom... at (no spam) ix.netcom.com> wrote:
I've come up with this equation at work, and
can't
solve the durn
thing. I've tried using logarithms, but I can't
figure out how to
separate the variables. I'm sure this is simple
for
one of you guys.

Feel free to just post a link to some how-to if
you
want.

(x^(n+1) - 1)/(x - 1)=n^2

You can start with

(x^(n+1) - 1)/(x - 1) = 1 + x + x^2 + ... + x^n.

Try solutions in x for n = 1, 2, 3, 4 and see if
there's a pattern. (There are closed-form solutions
for cubics and quartics.)

You could also try to solve it numerically for
different n.

Jack

Thanks for that. There's no algebraic solution?
--Buck

Numerically, I get for the first 10 n's the following x's.

0.000, 1.303, 1.578, 1.607, 1.578, 1.538, 1.499, 1.462, 1.430, 1.402.

These are not the only solutions. Some additional ones may be negative or complex numbers. I don't know if someone can find some clever way to get a closed-form solution.

Jack

On Jul 9, 1:58 pm, Jack Tomsky
jtom... at (no spam) ix.netcom.com> wrote:
On Jul 9, 12:58 pm, Jack Tomsky
jtom... at (no spam) ix.netcom.com> wrote:
I've come up with this equation at work, and
can't
solve the durn
thing. I've tried using logarithms, but I
can't
figure out how to
separate the variables. I'm sure this is
simple
for
one of you guys.

Feel free to just post a link to some how-to
if
you
want.

(x^(n+1) - 1)/(x - 1)=n^2

You can start with

(x^(n+1) - 1)/(x - 1) = 1 + x + x^2 + ... +
x^n.

Try solutions in x for n = 1, 2, 3, 4 and see
if
there's a pattern. (There are closed-form
solutions
for cubics and quartics.)

You could also try to solve it numerically for
different n.

Jack

Thanks for that. There's no algebraic solution?
--Buck

Numerically, I get for the first 10 n's the
following x's.

0.000, 1.303, 1.578, 1.607, 1.578, 1.538, 1.499,
1.462, 1.430, 1.402.

These are not the only solutions. Some additional
ones may be negative or complex numbers. I don't
know if someone can find some clever way to get a
closed-form solution.

Jack


Thanks, but my n can be between 0 and 3000. Maybe if
I get enough
numbers, I can do a regression of some sort...

"bukzor" <workithar... at (no spam) gmail.com> wrote in message

news:e2e2a470-21a1-40eb-b5b3-61c2b9f29dea at (no spam) 34g2000hsh.googlegroups.com...

Thanks, but my n can be between 0 and 3000. Maybe if I get enough
numbers, I can do a regression of some sort...

Assume n >= 2.

Let f(x) = 1 + x + x^2 + ... x^n - n^2

f(x) is strictly increasing for x>0 so there is only one real positive
solution.

f(1) = n+1 - n^2 < 0 for all n >= 2.

f(n^(2/n)) > 0.

So the root always lies between 1 and n^(2/n). A binary search will find it
quickly.

Graham