even if the series is finite, it's value it's incredibly close to 1/(1-x) because it goes to 7th order.
for example, if you want that to equal 7, you need to input x=6/7. if the series is not infinite, then x is approximately equal to 6/7.
try it, calculate how much you get if you input x=6/7
and if x>1, then x has to be too large for the 7th order to not work, I wouldn't try anything under x=100000
>I wouldn't try anything under x=100000
I wouldn't try anything under x=100000 >I wouldn't try anything under x=100000
I wouldn't try anything under x=100000 >I wouldn't try anything under x=100000
I wouldn't try anything under x=100000 >I wouldn't try anything under x=100000
I wouldn't try anything under x=100000
>I wouldn't try anything under x=100000
I wouldn't try anything under x=100000 >I wouldn't try anything under x=100000
I wouldn't try anything under x=100000 >I wouldn't try anything under x=100000
I wouldn't try anything under x=100000 >I wouldn't try anything under x=100000
I wouldn't try anything under x=100000
what part of >and if x>1, then x has to be too large for the 7th order to not work, I wouldn't try anything under x=100000
do you not fucking understand?
>and if x>1, then x has to be too large for the 7th order to not work, I wouldn't try anything under x=100000
and if x>1, then x has to be too large for the 7th order to not work, I wouldn't try anything under x=100000 >and if x>1, then x has to be too large for the 7th order to not work, I wouldn't try anything under x=100000
and if x>1, then x has to be too large for the 7th order to not work, I wouldn't try anything under x=100000 >and if x>1, then x has to be too large for the 7th order to not work, I wouldn't try anything under x=100000
factors as (x+1)(x^2+1)(x^4+1), so need x+1 = p^l, so x^2+2x+1 = p^(2l), where x^2+1 = p^m, so p^m = p^(2l) - 2p^(l) + 2, if l = 0 we get x = 0 which doesn't work, so l!=0, so p divids lhs and rhs, so p divides 2, get p = 2, but if m > 1 and l > 1 clearly 4 stops equality. m>=l, so take l = 1, we get x = 1 which doesn't work so done none work.
even if the series is finite, it's value it's incredibly close to 1/(1-x) because it goes to 7th order.
for example, if you want that to equal 7, you need to input x=6/7. if the series is not infinite, then x is approximately equal to 6/7.
try it, calculate how much you get if you input x=6/7
and if x>1, then x has to be too large for the 7th order to not work, I wouldn't try anything under x=100000
It can be factored as (x^8 - 1)/(x - 1) so simply put: consider it as a geometric series, which leads to (x^8 - 1)/(x - 1) = (x^4 + 1)(x^2 + 1)(x + 1).
It can be factored as (x^8 - 1)/(x - 1) so simply put: consider it as a geometric series, which leads to (x^8 - 1)/(x - 1) = (x^4 + 1)(x^2 + 1)(x + 1).
The only value of x where [math](x^4 + 1)(x^2 + 1)(x + 1)[/math] is a power of a single prime is when x=1.
if x + 1 and x^2 + 1 are both powers of a prime p, then p divides 2x, so p divides 2 (p can't divide x) and p = 2. Then x is odd, say x = 2n + 1, so x^2 + 1 = 4n^2 + 4n + 2 = 2(2n^2 + 2n + 1), which is not a power of 2 unless n = 0 or -1, and if n = -1, x + 1 = 0. Hence, n = 0 and x = 1 can be verified by manually plugging it in.
that's the taylor expansion of 1/(1-x) so it's rather trivial
No it's not. Nowhere does it mention that [math]|x| < 1[/math] and the series is finite.
even if the series is finite, it's value it's incredibly close to 1/(1-x) because it goes to 7th order.
for example, if you want that to equal 7, you need to input x=6/7. if the series is not infinite, then x is approximately equal to 6/7.
try it, calculate how much you get if you input x=6/7
and if x>1, then x has to be too large for the 7th order to not work, I wouldn't try anything under x=100000
1/(1-7) = -1/6, 1+7+7^2+7^3+7^4+7^5+7^6+7^7 is nowhere near that lol.
>I wouldn't try anything under x=100000
I wouldn't try anything under x=100000
>I wouldn't try anything under x=100000
I wouldn't try anything under x=100000
>I wouldn't try anything under x=100000
I wouldn't try anything under x=100000
>I wouldn't try anything under x=100000
I wouldn't try anything under x=100000
you're retarded
still not a prime number, what's your point?
what part of
>and if x>1, then x has to be too large for the 7th order to not work, I wouldn't try anything under x=100000
do you not fucking understand?
>and if x>1, then x has to be too large for the 7th order to not work, I wouldn't try anything under x=100000
and if x>1, then x has to be too large for the 7th order to not work, I wouldn't try anything under x=100000
>and if x>1, then x has to be too large for the 7th order to not work, I wouldn't try anything under x=100000
and if x>1, then x has to be too large for the 7th order to not work, I wouldn't try anything under x=100000
>and if x>1, then x has to be too large for the 7th order to not work, I wouldn't try anything under x=100000
i've already given the answer, x=1 is the only value that works
8 is not a prime chud
is a prime power tho
It's the sum of a geometric series.
It is (1-x^8)/(1-x) with no approximation
factors as (x+1)(x^2+1)(x^4+1), so need x+1 = p^l, so x^2+2x+1 = p^(2l), where x^2+1 = p^m, so p^m = p^(2l) - 2p^(l) + 2, if l = 0 we get x = 0 which doesn't work, so l!=0, so p divids lhs and rhs, so p divides 2, get p = 2, but if m > 1 and l > 1 clearly 4 stops equality. m>=l, so take l = 1, we get x = 1 which doesn't work so done none work.
nvm x = 1 gets you 2^3, and i guess x = 0 = 1 is technically a prime power? rest is still valid
Incorrect.
Also incorrect.
It can be factored as (x^8 - 1)/(x - 1) so simply put: consider it as a geometric series, which leads to (x^8 - 1)/(x - 1) = (x^4 + 1)(x^2 + 1)(x + 1).
I forgot to mention that the x is greater than 0 and it's a whole number
but why x = 1 is the only answer?
Because this
The only value of x where [math](x^4 + 1)(x^2 + 1)(x + 1)[/math] is a power of a single prime is when x=1.
but where is it stated that it has to be a single prime, also how solving this equation can be a solution to this, please elaborate
> is a power of a prime
It's stated in the OP
> how solving this equation can be a solution to this
expand the expression and you get the OP equation, it's the same function.
but what's the logical reasoning behind this, how come two forms of the same function give us an answer
What? Are you retarded?
if x + 1 and x^2 + 1 are both powers of a prime p, then p divides 2x, so p divides 2 (p can't divide x) and p = 2. Then x is odd, say x = 2n + 1, so x^2 + 1 = 4n^2 + 4n + 2 = 2(2n^2 + 2n + 1), which is not a power of 2 unless n = 0 or -1, and if n = -1, x + 1 = 0. Hence, n = 0 and x = 1 can be verified by manually plugging it in.
actually, x = 0 is a solution.
1 is not a prime
illiterate retards
I believe that the retard is you anon.
Non sequitur.
1 is not a prime, so x=0 is not a solution.
it is a prime power tho
partial sum i think would work
x=3