Tenth Square Root

hasan4444 · Riemann Hypothesis Offline Joined: 28 Nov 2008 Posts: 486

Find $\sqrt [10]{256}$

**Posted:** Fri Nov 06, 2009 3:45 am

_________________
New "Inequalities Marathon" join it now Welcome

isabella2296

Hmm, isn't it just $\sqrt[10]{256} \implies \sqrt[10]{2^8} \implies \sqrt[5]{16}$ ? Or am I missing something?

**Posted:** Fri Nov 06, 2009 4:28 am

_________________
Czech out $\mathfrak {Encyclopedia } \mathfrak { Mathematica}$
Latest entry: Morley's Theorem (Proof)

hasan4444 · Riemann Hypothesis Offline Joined: 28 Nov 2008 Posts: 486

I think you are right. I don't know if it can be more simplified but I posted a strange solution that I'm sure it's wrong. It goes like this
$\sqrt[10]{2^8} \implies 0.8^2$ Neutral

**Posted:** Fri Nov 06, 2009 4:39 am

_________________
New "Inequalities Marathon" join it now Welcome

Arrange your tan · Riemann Hypothesis Offline Joined: 08 Nov 2007 Posts: 383

**Posted:** Fri Nov 06, 2009 8:32 am

grn_trtle

You could use the Binomial Theorem if you really wanted to...

$(1+1)^{\frac{4}{5}}=\sum_{k=0}^{\infty}\binom{4/5}{k}=1+\frac{4/5}{1}+\frac{4/5(-1/5)}{2}+\frac{4/5(-1/5)(-6/5)}{6}+\cdots$

$\approx 1.75$

Which is pretty close to the real value, which is slightly above $1.74$

**Posted:** Fri Nov 06, 2009 8:50 am

_________________
$\arctan\alpha + \arctan\beta = \text{arg}\{(1+\alpha i)(1+\beta i)\} = \arctan\left( \frac{\alpha+\beta}{1-\alpha\beta} \righ...$

fishythefish

It is most definitely not equal to $(0.8)^2=0.64$ . In fact, there is no way that it's rational. grn_turtle has a good approximation. (Hassan, I believe you meant to say $2^{0.8}$ Wink

Here's what I would have done.

$\sqrt[10]{256}=\sqrt[10]{2^8}=2^{\frac{8}{10}}=2^{\frac{4}{5}}=\sqrt[5]{16}$ , which is what isabella has.

**Posted:** Sat Nov 07, 2009 2:29 pm

_________________
I am NOBODY. NOBODY is PERFECT. Therefore, I am PERFECT. Ninja

There are 3 types of people in the world. Those who can count and those who can't. maybe

ATTENTION ALL CALCOHOLICS!!! Know your limits. Don't drink and derive. Rotfl

hasan4444 · Riemann Hypothesis Offline Joined: 28 Nov 2008 Posts: 486

I know it's wrong but I state it just to let you know about the answer I saw for this problem.

Thanks a lot Smile

**Posted:** Sun Nov 08, 2009 4:33 am

_________________
New "Inequalities Marathon" join it now Welcome

Dr Sonnhard Graubner

hello, we get $\sqrt[10]{256}=\sqrt[5]{2^4}=\sqrt[5]{\frac{2^5}{2}}=\frac{2}{\sqrt[5]{2}}$ .
Sonnhard.

**Posted:** Sun Nov 08, 2009 6:38 am

ernie

But this isn't simplified, because there's a radical in the denominator.

**Posted:** Sun Nov 08, 2009 7:49 am

_________________
If pro is the opposite of con, then what's the opposite of progress? Razz

grn_trtle

Uh, I believe the answer here is irrational. You're not going to get any nice answer here. The best you can do, I think, is manipulate the square roots, or use an approximation like I did.

**Posted:** Sun Nov 08, 2009 10:36 am

_________________
$\arctan\alpha + \arctan\beta = \text{arg}\{(1+\alpha i)(1+\beta i)\} = \arctan\left( \frac{\alpha+\beta}{1-\alpha\beta} \righ...$

boohsunil · New Member Offline Joined: 11 Jul 2009 Posts: 11

**Posted:** Mon Nov 09, 2009 4:19 am

10000th User

**Posted:** Mon Nov 09, 2009 7:09 am

_________________
$\mathrm{Hom}(\mathbb{Z},\mathbb{Z})\cong\mathbb{Z}$
Disclaimer: I'll revive old threads if: 1) something ambiguous in the solution or wrong argument, 2) no solution, 3) I got new insight to input.