-5.48...
and
-15.52...

I was asking for the coon to factor it

Coon should factor x² + 7 = 0

Of course, it can be done easily, unlike certain annoying polynomials.

Uhm, couldn't you just...not factor it?
x² + 7 = 0
x² = -7
x = ±(7)^½i?
or did I = t3h fail?

BTW: alt+0178 = better than [sup]2[/sup]

I know that, but I'm talking about a different identity.

But you essentially concluded that if log(1/n) = -log(n) then log(a/n) = -a*log(n). (Which it isn't, it's log(a) - log(n)).

I concluded it works that way because log(1/n) = log(1) - log(n) = -log(n).

I'd never used that identity before, so obvs it doesn't.

Doesn't that only work because the log₁₀(1) is equal to zero?

Yes and no. In general, log(u^v) = v*log(u).

Yeah, I know that, but because it was 1/n instead of 5/n, that's why it worked. If it was 5/n, then it would have to be log(5) - log(n). Which is what you were saying before anyways. I don't even know why we're discussing this to be honest. The question has been solved!

Raccoon should instead solve (x²-9)/(x+3) = 0.

That would be x = 3, correct?
x²-9 factors to (x + 3)(x - 3) = 0, (x + 3)/(x+3) = 1, set x - 3 = 0, yeah, that works.

Your new homework assignment: Find the domain and range of this piecewise function:

ƒ(x) =
{ -x, x < 0
0, x = 0
x, x > 0

EDIT: I'm gonna replace that with a picture...the forums can't display it correctly.
EDIT 2: here.


Woops, sorry, hope you guys can see it

Post has been edited 3 time(s), last time on Aug 31 2009, 3:43 am by poison_us.

I like these better.
[(([(x^1/2)x]^1/2))x]^1/2.... and so on equals 777. What is x?

x equals why I'm not a math major

I'm not sure if I'm right or if I get your parentheses, but is this right?

By the recursive definition of the series, it can be rewritten as (x(x(x....)^1/2)^1/2)^1/2. Right?

If so,


(x(x(x....)^1/2)^1/2)^1/2 = x^1/2(x(x....)^1/2)^1/4 = _n=1Π^∞ x^(2-n)

= x^{(_n=1Σ∞ 2-n)}

= x.


So the answer is 777. Ostensibly.. I hate competition problems like this.

@poison: Your function maps (-∞,∞) to real values on an image of [0,∞).

Post has been edited 1 time(s), last time on Aug 31 2009, 7:23 am by DT_Battlekruser.

Hm, I'm not sure whether the operation order is well-defined or not:

1) Let's define the following sequence:
a_0 = x^1/2
a_n = (x * a_(n-1))^1/2

a_0 = x^1/2, a_1 = ((x^1/2)x)^1/2, a_2 = ((((x^1/2)x)^1/2)x)^1/2, ...
If a_n converges to a then:
a_n = (x * a_(n-1))^1/2 -> a = (x*a)^1/2 -> x = a
Since we want a = 777, we can take x = 777

Let's prove it converges:
a_0 = x^1/2 > 0 and if a_(n-1) > 0 then a_n = (x * a_(n-1))^1/2 > 0 so a_n > 0 for all n.
a_n - a_(n-1) = (x * a_(n-1))^1/2 - (x * a_(n-2))^1/2 = (x*a_(n-1) - x*a_(n-2))/((x * a_(n-1))^1/2 + (x * a_(n-2))^1/2) ->
-> sign (a_n - a_(n-1)) = sign (a_(n-1) - a_(n-2))
And since a_1 - a_0 = ((777^1/2)777)^1/2 - 777^1/2 > 0 -> a_n increases
a_0 = x^1/2 < 777. Let's assume a_(n-1) < 777 -> x*a_(n-1) < 777² -> a_n = (x*a_(n-1))^1/2 < 777 so a_n < 777 for all n.
Since a_n increases and is bounded -> converges, QED.

2) We could also define the following sequence:
a_0 = 1
a_n = x*(a_(n-1))^1/2

a_0 = 1, a_1 = x, a_2 = (x^1/2)x, a_3 = (((x^1/2)x)^1/2)x, a_4 = (((((x^1/2)x)^1/2)x)^1/2)x, ...
If a_n converges to a then:
a_n = x*(a_(n-1))^1/2 -> a = x*a^1/2 -> x = a^1/2
Since we want a = 777, we can take x = 777^1/2

It can be proven that converges in a similar way.

So the value of x depends on the order you do the operations...

I think DTBK nailed it. And from that stand point, theres no way you can get a 777^1/2 since it is 777^1/2+1/4+1/8+... and so on- forever. Which means the answer is 777^(1/2)/1-(1/2) = 777 o-o;\

a_1 should be x^1/2, not x- which is why I believe the answers turned out to be different.

DT_BK's method is good but has some inaccuracies. First he's not proving that that infinite operation converges. If it doesn't converge then all what he said could be wrong. And also he has moved the dots ... from outside to inside the expression, and that could also be wrong. Even with standard series the addition loses its conmutivity, there are series which are convergent if we sum its members in one order but are not if we do it in a different order.

But, rewriting my answer the way DT_BK did it:
1) Would be exactly what DT_BK did
2)

x(x(x(x....)^1/2)^1/2)^1/2 = x*x^1/2(x(x....)^1/2)^1/4 = x*_n=1Π^∞ x^(2-n)

= x*x^{(_n=1Σ∞ 2-n)}

= x².

Why not? You're not specifying whether [(([(x^1/2)x]^1/2))x]^1/2... refers to

x^1/2
[(x^1/2)x]^1/2
[(([(x^1/2)x]^1/2))x]^1/2
...

OR

(x^1/2)x
(([(x^1/2)x]^1/2))x
([(([(x^1/2)x]^1/2))x]^1/2)x
...

And the results are actually different. So what you asked is not well-defined, and thus there's no simple answer. This wouldn't happen with a standard serie since when using only addition the result doesn't vary as long as you sum the elements up in the same order (even if you sum the first and second, then third and fourth or if you sum the first, then second and third, then fourth and fifth, ...).

ok. you guys should stop dealing with easy math. solve this for me please:

_k=n+1Σ^∞((-1)^k(11/17)^(2k))/(2k)! < 10^-10 for the largest n.

Using any electronic device is prohibited only because they are finite machines (they are inaccurate); I need an exact solution.

Actually if there were a 'x' outside the semicolons, then I would have put it there. If you do the second, then of course the results are different- because you've added an extra x which is not part of the equation or the sequence.
Unless you can prove to me how (x(x(x....)^1/2)^1/2)^1/2 would equal x(x(x(x....)^1/2)^1/2)^1/2 which if divided by (x(x(x....)1/2)1/2)1/2 would be x.

Oo cheeze's problem is a pretty hard one. I'm not really sure how a pactorial would work with sigma. I shall attempt to solve... o-o;;;

They're not, that's the point. And neither the first neither the second is equal to [(([(x^1/2)x]^1/2))x]^1/2....
In fact that one doesn't even converge. What I understand from the dots is "take x, then calculate its square root, then multiply it by x, then calculate its square root, then multiply it by x, then calculate its square root, then multiply it by x, then..." to the infinity. Lets assume x = 2, then:
1) take x: 2
2) calculate its square root: 1.41421356
3) multiply it by x: 2.82842712
4) calculate its square root: 1.68179283
5) multiply it by x: 3.36358566
6) calculate its square root: 1.83400809
7) multiply it by x: 3.66801617
8) calculate its square root: 1.91520656
9) multiply it by x: 3.83041312
10) calculate its square root: 1.95714412
11) multiply it by x: 3.91428825
etc...

As you can see while odd terms approach 2, even terms actually approach 4 which is 2^2, and it continously oscillates without converging to any of the two values.
Why should I suppose you meant only using the odd terms with your formula? The dots just mean that there are infinite x)^1/2)x)^1/2)x)^1/2... there and that's no real number, because it won't converge to any real number.