interesting dilemma (1 Viewer)

kaz1 · Apr 28, 2012

does 1^-infinity=0 then

seanieg89 · Apr 28, 2012

mirakon said:
Just googled it

Sounds immensely confusing lol

Yep, I know very little about it haha.

seanieg89 · Apr 28, 2012

kaz1 said:
does 1^-infinity=0 then

No, same issue.

OH1995 · Apr 28, 2012

Guys, the simple answer is no. The rule only applies firstly for numbers.......infinite is not a number and even if it was it's far from natural. Therefore the law will not work.

funnytomato · Apr 28, 2012

Dunno if this qustion is related, but seems interesting :

x^x^x^x^x^x^x^x^x^... = 7
( infinitely interated )
Find x

Carrotsticks · Apr 28, 2012

Square root 7?

kaz1 · Apr 28, 2012

funnytomato said:
Dunno if this qustion is related, but seems interesting :

x^x^x^x^x^x^x^x^x^... = 7
( infinitely interated )
Find x

7^1/infinity

mirakon · Apr 28, 2012

Lol kaz

funnytomato · Apr 28, 2012

Carrotsticks said:
Square root 7?

is it ?

RealiseNothing · Apr 28, 2012

Carrotsticks said:
Square root 7?

But $\sqrt7$ > 2.

If you let 2 be a lower bounds, 2^2^2 is greater than 7 already, so it can't be right? I would think that anything greater than 1 wouldn't be right.

Or does the fact that the powers continue on infinitely change the whole situation?

Carrotsticks · Apr 28, 2012

Haha I cheated because we know for sure that sqrt2^sqrt2^... infinitely = 2.

So I just blindly said that therefore sqrt7^sqrt7^sqrt7^... = 7, which is clearly not true because the tetration x^x^x^x^... only converges for:

$x \in \left [ e^{-e}, e^{\frac{1}{e}} \right ]$

And sqrt 7 is most certainly not in that domain!

rolpsy · Apr 28, 2012

funnytomato said:
Dunno if this qustion is related, but seems interesting :

x^x^x^x^x^x^x^x^x^... = 7
( infinitely interated )
Find x

is it the seventh root of 7?

Carrotsticks · Apr 28, 2012

rolpsy said:
is it the seventh root of 7?

$\\ x^{x^{x^{x^{.}}}}... = 7 \\\\ $To find the convergent value, we essentially solve the equation:$ \\\\ y = x^y \Rightarrow x = y^{\frac{1}{y}} \\\\ $For the particular case $ y=7 $ , we have:$ \\\\ x = 7^{\frac{1}{7}} $ , which is the seventh root of 7 as you have said.$$

AAEldar · Apr 28, 2012

Carrotsticks said:
$\\ x^{x^{x^{x^{.}}}}... = 7 \\\\ $To find the convergent value, we essentially solve the equation:$ \\\\ y = x^y \Rightarrow x = y^{\frac{1}{y}} \\\\ $For the particular case $ y=7 $ , we have:$ \\\\ x = 7^{\frac{1}{7}} $ , which is the seventh root of 7 as you have said.$$

That only works for 7 powers though, but the question said it was infinitely iterated...?

RANK 1 · Apr 28, 2012

seanieg89 said:
Why wouldnt 1 to the power of a chair equal 1? Infinity is not a real number, the current definition of exponentiation does not apply to this situation and we cannot extrapolate any information from the fact that 1 to the power of a real number is 1.

In any case the OP meant something different, read my earlier post on limits.

but the idea of 1 to the power of n(where n is positive) is that the 1 is multiplied by itself n times, so wouldnt that mean that it doesnt matter what n is the answer would always be 1 because you're multiplying 1 by 1. which should means it wouldnt matter what n equaled in 1^n, it would always be 1 no matter what even to infinity

Riproot · Apr 28, 2012

math man said:
1 to the power of say n (n is some finite natural number) is 1 of course, and we all are taught 1 to the power of anything is 1.
So it would be obvious to think that 1^(infinity) is 1. Is this correct or not... What do you guys believe?

But n is a number and infinity is not a number, it's an idea.

funnytomato · Apr 28, 2012

Carrotsticks said:
$\\ x^{x^{x^{x^{.}}}}... = 7 \\\\ $To find the convergent value, we essentially solve the equation:$ \\\\ y = x^y \Rightarrow x = y^{\frac{1}{y}} \\\\ $For the particular case $ y=7 $ , we have:$ \\\\ x = 7^{\frac{1}{7}} $ , which is the seventh root of 7 as you have said.$$

could some one tell me what's going on here
[as an experiment, I know it's not exactly 'Maths ']:

(using a calculator)
store A as 7^(1/7)

ans=A

then ans=A^ans, the 10 digits displayed don't seem to change at all after 30 iterations or so
and it is 1.530140119
or 1.53014011947213 (from MATLAB)

does that have something to do with machine epsilson(or whatever the right terminology is) , or is my experimental method flawed ?

RealiseNothing · Apr 28, 2012

AAEldar said:
That only works for 7 powers though, but the question said it was infinitely iterated...?

$\\ x^{x^{x^{x^{.}}}}... = 7$

But since it goes on infinitely, the power of x must equal 7 (since the power is $\\ x^{x^{x^{x^{.}}}}...$ )

Hence $x^7 = 7$

$x = 7^{\frac{1}{7}}$

Carrotsticks · Apr 28, 2012

funnytomato said:
could some one tell me what's going on here
[as an experiment, I know it's not exactly 'Maths ']:

(using a calculator)
store A as 7^(1/7)

ans=A

then ans=A^ans, the 10 digits displayed don't seem to change at all after 30 iterations or so
and it is 1.530140119
or 1.53014011947213 (from MATLAB)

does that have something to do with machine epsilson(or whatever the right terminology is) , or is my experimental method flawed ?

It is possible that it is very slowly converging, but I'm on a train ready for a party so not really in Math mode now haha. But indeed a result like that is cause for questioning. However, you would think that with each further iteration, the power > 1 would compound upon itself and the result magnified every iteration.

I will be sure to do this properly at home if I'm not too tired.

lolcakes52 · Apr 28, 2012

But what if we define a series of terms such that a term is equal to 1 raised to the power of the previous term? The value of the first term is one

interesting dilemma (1 Viewer)

et tu

Well-Known Member

Well-Known Member

Member

Active Member

Retired

et tu

nigga

Active Member

what is that?It is Cowpea

Retired

Member

Retired

Premium Member

Active Member

Addiction Psychiatrist

Active Member

what is that?It is Cowpea

Retired

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)