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September 23

Iterated exponentiation

I wanted to ask a general question about mathematics. For example: x^x^x^x^x^... = 2

The way forward would be. x^y = 2 where y=x^x^x^x^...

But the value of y would be 2 as it is the premise.

So x^2=2

Here is why I got confused. There are two solutions to the above equation. x=root2 and x= - root2

However only one of the solution works for x^x^x^x^x^... = 2

Why is it that even though it satisfy x^2=2, only one of the solution works? Is satisfying x^2=2 only a requirement and not a guaranty of a solution? 220.239.51.150 (talk) 12:57, 23 September 2013 (UTC)[reply]

Let me clarify my confusion. Look at these equations and statements.

(Equation 1) x^x^x^x^x^… = 2
(Statement A) Every numeric value of x that satisfy Equation 1 is a valid solution to the problem

(Equation 2) x^y = 2 where y=x^x^x^x^x^…
(Statement B) Every numeric value of x that satisfy Equation 2 is a valid solution to the problem

OK, I'm gonna stop you right there: this statement is false. Why ? Because the implication in question is only one of an infinite number of implications which result from the initial equation. So the logical conclusion would be that the values which satisfy the original equation are to be found among those that satisfy the latter... In other words, inclusion rather than equality. — 79.113.229.58 (talk) 13:23, 24 September 2013 (UTC)[reply]

(Equation 3) x^y = 2 where y=2
(Statement C) Every numeric value of x that satisfy Equation 3 is a valid solution to the problem

(Equation 4) x^2 = 2
(Statement D) Every numeric value of x that satisfy Equation 4 is a valid solution to the problem

Statement A is CORRECT but Statement D is INCORRECT and yet every step of the way is perfect and flawless.

That is why I am confused. 220.239.51.150 (talk) 13:34, 23 September 2013 (UTC)[reply]

I don't really understand what exactly you are trying to do. Note that for Equation 1 there should be a unique solution for each number of xs (at least I think so). For example, there is a unique real solution to x^x = 2. -- Toshio Yamaguchi 14:01, 23 September 2013 (UTC)[reply]

"... there is a unique real solution to x^x = 2"

I have to think about that claim again. This seems to be equivalent to the claim that the real numbers are closed under exponentiation, but I am actually not sure whether that claim is true. Shouldn't be the case. On the other hand, if we restrict x to be positive, it might be true. -- Toshio Yamaguchi 14:39, 23 September 2013 (UTC)[reply]

See our article on tetration. Iterated exponentiation converges for values of x between e^-e and e^1/e, but since the positive square root of 2 is within this interval, this solution is okay. Note that the reasoning above shows that if x satisfies x^x^x^... = 2 then x^2 = 2; you cannot conclude from this that if x^2 = 2 then x^x^x^... = 2. Gandalf61 (talk) 15:43, 23 September 2013 (UTC)[reply]

I think the problem you're having is assuming that the "series" x^x^x^x… taken to it's infinite limit converges under any given input, and that 2 is in the (output) range where it converges. That is, you're neglecting the possibility that there doesn't actually exist any x for which x^x^x^x^x^… = 2 exists. It's a well known property of certain normal series (the conditionally convergent ones) that if you alter the way you compute what they converge to, you change what they converge to. [1] If you reorder the series, you change its value. I don't know anything about the properties of your exponentiation "series", but it could suffer from much the same problems. You can't just blithely chop off an x and assume the value stays the same. Especially not when negative numbers are involved. -- 67.40.209.200 (talk) 17:05, 23 September 2013 (UTC)[reply]

Statement B should be Every pair of numerical values of x, y that satisfy Equation 2 is a valid solution to the problem. But y = 2, x = minus root 2 does not satisfy the last part of Eq. 2 because the iterated expectation does not converge. So I think the logical mistake is that you've lost restricting information in writing the last part of Eq. 3. Duoduoduo (talk) 18:11, 23 September 2013 (UTC)[reply]

Statement B says "if x^y = 2 and y = x^x^..., then x^x^... = 2". This is true. However, statement C says "if x^y = 2 and y=2, then x^x^... = 2". This does not follow, because from the facts that x^y = 2 and y=2 you can't deduce that y = x^x^... and execute statement B. -- Meni Rosenfeld (talk) 20:05, 23 September 2013 (UTC)[reply]

Integrate (x^2+1)^10

Integrating a function can be like unscrambling eggs! Sometimes, there are 2 functions that look almost alike, but their integration technics are not the same. An example is to integrate each of:

$(x^{2}+1)^{10}$ and $((x^{2}+1)^{10})x$ .

Let's try both functions. To integrate $((x^{2}+1)^{10})x$ , you just relate this function to the derivative of $(x^{2}+1)^{11}$ by a constant multiplier, because the derivative is $22x(x^{2}+1)^{10}$ . Thus, the integral of $((x^{2}+1)^{10})x$ is $((x^{2}+1)^{11})/22+C$ .

But to integrate $(x^{2}+1)^{10}$ , we would need a variable multiplier. The integral of a product does not equal the product of the first function and the integral of the second unless the first function is a constant, so we cannot integrate it this way. Any way it can be done other than expanding the binomial power and integrating each term?? Georgia guy (talk) 14:38, 23 September 2013 (UTC)[reply]

Use integration by parts. Sławomir Biały (talk) 17:13, 23 September 2013 (UTC)[reply]

Will this help? For a general exponent it seems the expression involves a hypergeometric function, so I doubt there is anything simpler than expansion with a specific exponent. -- Meni Rosenfeld (talk) 19:53, 23 September 2013 (UTC)[reply]

The question was whether it can be done another way, not whether it should be done another way. Since binomial coefficients are "precomputed" you save effort over repeated integration by parts. Strangely, the standard way of handling the equivalent problem of integrating sec²²t would be integration by parts rather than doing a reverse trig substitution and applying the binomial theorem. --RDBury (talk) 21:25, 23 September 2013 (UTC)[reply]

It's a fair point that it doesn't really "help" per se. The fact is that the integral has many terms, and any integration method will have to compute those terms one way or another. Sławomir Biały (talk) 12:39, 24 September 2013 (UTC)[reply]

I am not great at all with complex analysis, but it sounds like this might also be the same as $\int (x+i)^{10}(x-i)^{10}dx$ which can be integrated by tabular integration after letting $u=x-i,du=dx$ . It's still a lot of integration by parts, but tabular integration is relatively effortless, at least for me.--Jasper Deng (talk) 22:28, 23 September 2013 (UTC)[reply]

You can define a generating function as follows. If we put

$I_{n}(x)=\int \left(x^{2}+1\right)^{n}dx$

then we can calculate the generating function:

$f(x,t)=\sum _{n=0}^{\infty }(-1)^{n}{\frac {I_{n}(x)t^{n}}{n!}}={\sqrt {\frac {\pi }{2t}}}\exp(-t)\operatorname {*} {Erf}\left({\sqrt {t}}x\right)$

Count Iblis (talk) 00:00, 24 September 2013 (UTC)[reply]

Use Newton's binomial, and the properties of the integral, namely that the integral of a sum is the same as the sum of integrals, ultimately arriving at the following identity:

\int (x^{2}+1)^{10}\ dx\ =\ \ldots \ =\ \sum _{k=0}^{10}C_{10}^{k}\cdot {\frac {x^{2k+1}}{2k+1}}

— 79.113.233.194 (talk) 01:57, 24 September 2013 (UTC)[reply]