What is the solution to the nested integral problem?

[docnet]

Homework Statement

$\int_1^{\int_1^{\cdots}2xdx}2xdx$

Relevant Equations

.

Let $u=\int_1^{u}2xdx$.
\begin{align}u=& \int_1^{u}2xdx=\big[x^2\big]_1^u\\
u=&u^2-1\end{align}
Which leads to $u=\frac{1\pm\sqrt{1+4}}{2}$

Assuming that the upper boundary of integration is greater than $1$, or less than $-1$, leads to $u=\frac{1+\sqrt{5}}{2}\approx 1.61$. the second case leads to a contradiction since the upper boundary of integration is $u$, and $u$ is $1.61$, which is greater than $-1$.

Assuming that the boundary of integration is $1$ or $-1$ also leads to contradictions, since this leads to $u=0$.

Then, assuming that the upper boundary of integration is between $-1$ and $1$, this leads to $u=\frac{1-\sqrt{5}}{2}\approx -.61$, which satisfies the condition that $-1<u<1$.

So, the integral is equal to $\frac{1-\sqrt{5}}{2}\approx -.61$.

 

[Office_Shredder]

I don't think I understood why $u=1.61$ leads to a contradiction.

 

[docnet]

I don't think I understood why $u=1.61$ leads to a contradiction.

I had a brain fart. You're right. It does not lead to a contradiction.

So both $u=1.61$ and $u=-.61$ are correct answers to the integral, since the the integral does not specify conditions on the other boundary of integration, right?

 

[benorin]

Homework Statement:: $\int_1^{\int_1^{\cdots}2xdx}2xdx$
Relevant Equations:: .

Let $u=\int_1^{u}2xdx$.
\begin{align}u=& \int_1^{u}2xdx=\big[x^2\big]_1^u\\
u=&u^2-1\end{align}##

I’m unclear as to why the left hand side of this eqn is $u$? Also the problem statement seems incomplete as it is just an expression? What exactly do the dots in the upper bound of the upper bound indicate? Is it kinda like a power tower or something?

 

[vela]

So both $u=1.61$ and $u=-.61$ are correct answers to the integral, since the the integral does not specify conditions on the other boundary of integration, right?

What other boundary of integration? The lower bound of 1? What conditions are you referring to?

Did you provide us the complete problem statement? You seem to be imposing conditions that weren't stated. It's not clear to me what your concerns are after you find the two roots of the quadratic.

 

[docnet]

I’m unclear as to why the left hand side of this eqn is $u$? Also the problem statement seems incomplete as it is just an expression? What exactly do the dots in the upper bound of the upper bound indicate? Is it kinda like a power tower or something?

Sorry, I forgot to say that the problem statement is to evaluate the nested integral.
the dots of the upper bound indicates that the upper bound is the nested integral, like an infinitely repeating pattern of the same integral.

What other boundary of integration? The lower bound of 1? What conditions are you referring to?

Did you provide us the complete problem statement? You seem to be imposing conditions that weren't stated. It's not clear to me what your concerns are after you find the two roots of the quadratic.

The other bound as in $u=\int2xdx$. This integral is a case of what we are solving for being an expression of itself, like a loop. Although the characteristic polynomial gives two solutions for u, one can imagine making an initial guess for u and checking if the integral is equal to u until you get the correct u (its kind of a pointless exercise in numerical methods I guess).

 

[Office_Shredder]

Sorry, I forgot to say that the problem statement is to evaluate the nested integral.
the dots of the upper bound indicates that the upper bound is the nested integral, like an infinitely repeating pattern of the same integral.
The other bound as in $u=\int2xdx$. This integral is a case of what we are solving for being an expression of itself, like a loop. Although the characteristic polynomial gives two solutions for u, one can imagine making an initial guess for u and checking if the integral is equal to u until you get the correct u (its kind of a pointless exercise in numerical methods I guess).

I tried this out actually just to see what happened and it's kind of interesting. The behavior if you start at a number smaller than 1.6 vs bigger than 1.62 is interesting.

 

[docnet]

I tried this out actually just to see what happened and it's kind of interesting. The behavior if you start at a number smaller than 1.6 vs bigger than 1.62 is interesting.

It could be fun to write an algorithm to numerically shoot for the correct $u$ starting with the initial guess $u=1.6$ or $u=1.62$ to see how quickly the solution converges for each case. $2x$ is strictly positive over the positive values of $x$, so one would have $$\int_1^{1.6}2xdx<\int_1^{1.61}2xdx<\int_1^{1.62}2xdx$$

 

[Office_Shredder]

https://www.wolframalpha.com/input/?i=int_1^{1.7}+2x+dx

Just try a couple of these, no need for a program.

With that said, I think what you wrote is sufficient to figure out how the convergence will work out.

 

[haruspex]

$2x$ is strictly positive over the positive values of $x$, so one would have $$\int_1^{1.6}2xdx<\int_1^{1.61}2xdx<\int_1^{1.62}2xdx$$

How is that interesting? Don't you need to look at whether $1.6>\int_1^{1.6}2xdx$ etc?

 

[docnet]

How is that interesting? Don't you need to look at whether $1.6>\int_1^{1.6}2xdx$ etc?

Yes. If $1.6>\int_1^{1.6}2xdx$ we would add a small number "$a$" to $1.6$.

If $1.6+a < \int_1^{1.6+a}2xdx$, we would subtract something like $(2/3)a$ from $1.6+a$.

If $1.6+a -(2/3)a < \int_1^{1.6+a-(2/3)a}2xdx$ we would subtract something like $(2/3)(2/3)a$ from the last number... and that would continue like that until the desired tolerance between the integral and the bound is reached. The integral might diverge for some beginning values of $a$.

 

[haruspex]

Yes. If $1.6>\int_1^{1.6}2xdx$ we would add a small number "$a$" to $1.6$.

If $1.6+a < \int_1^{1.6+a}2xdx$, we would subtract something like $(2/3)a$ from $1.6+a$.

If $1.6+a -(2/3)a < \int_1^{1.6+a-(2/3)a}2xdx$ we would subtract something like $(2/3)(2/3)a$ from the last number... and that would continue like that until the desired tolerance between the integral and the bound is reached. The integral might diverge for some beginning values of $a$.

My point was that $\int_1^{1.6}2xdx=1.56<1.6$, so it looks like starting anywhere under the 1.6... solution will not converge to that. Similarly starting above 1.6... seems not to converge. (But these observations would need to be shown to be generally true.)

It seems to me we cannot argue that the nested integral has a value unless the sequence converges.

What about convergence to the negative solution?

 

[docnet]

My point was that $\int_1^{1.6}2xdx=1.56<1.6$, so it looks like starting anywhere under the 1.6... solution will not converge to that. Similarly starting above 1.6... seems not to converge. (But these observations would need to be shown to be generally true.)

It seems to me we cannot argue that the nested integral has a value unless the sequence converges.

What about convergence to the negative solution?

Yes, if you are considering the true nested integral. Then, for any value other than the exact value, the nested integral will not converge.

Here is a simple code I wrote in Jupyter notebook to numerically solve this integral using a basic shooting method. $f$ is defined as the semidefinite integral minus $x$.

We can numerically estimate the integral using a simple iterative loop to avoid the issue.

Let $f=x^2-1-x$. Then the roots of $f$ are the solutions of the nested integral.

The algorithm finds the root of $f$ within a desired bracket. We define the bracket based on the supposed solutions of the nested integral. Each iteration involves a regular antiderivative of $2x$ evaluated at an estimate for the upper bound. The estimated bound improves with each iteration up to a reasonable desired tolerance. So, there is nothing to diverge to infinities.

The root-finding method is the "bisection method" because the bracket is bisected until it lands on the best estimate of the root.

 

[docnet]

A typo: the second line of the code should say "#semidefinite integral minus x"

 

[Fred Wright]

$$
I=\int_1^{I^{I^{...}}} 2x dx = (x^2)^{(x^2)^{(x^2)^{(x^2)^{...}-1}-1}-1}-1=\frac{(x^2)^{\frac{(x^2)^{\frac{(x^2)^{...}}{x^2}}}{x^2}}}{x^2}-1
$$
$$
=1^{1^{1^{1^{...}}}}-1=0
$$

 

[docnet]

The "integral" diverges everywhere except $x=\frac{\sqrt{1\pm5}}{2}$. So the "integral" is that function defined piece-wise:

$$\int_1^{\int_1^\cdots 2xdx}2xdx=\begin{cases}x & \text{for}\quad x= \frac{\sqrt{1\pm5}}{2}\\\infty& \text{otherwise}\end{cases}$$

 

[Office_Shredder]

I thought about this a little more and the mission here is wrong. Notice there's no x on the left hand side. You don't want an ascending chain of integrals, you want a descending chain.

$$f(z)=...\int_1^{\int_1^z 2x dx} 2x dx$$

And the convergence is more complicated than you made it out to be. For example see what happens if you start at $z=0$