Radii of stacked circles inside the graph of y = |x|^1.5

[songoku]

Homework Statement

Please see below

Relevant Equations

Not sure

(a) The hint from question is to used geometrical argument. From the graph, I can see $r_1+r_2=c_2-c_1$ but I doubt it will be usefule since the limit is $\frac{r_2}{r_1} \rightarrow 1$, not in term of $c$.

I also tried to calculate the limit directly (not using geometrical argument at all).

$$\lim_{\frac{r_2}{r_1} \rightarrow 1} \left(\frac{r_1+r_2}{{r_2}^{1.5}-{r_1}^{1.5}}\right)$$
$$=\lim_{\frac{r_2}{r_1} \rightarrow 1} \left(\frac{1+\frac{r_2}{r_1}}{\frac{{r_2}^{1.5}}{r_1}-{r_1}^{0.5}}\right)$$

Then got stuck

I also tried rationalization:
$$\lim_{\frac{r_2}{r_1} \rightarrow 1} \left(\frac{r_1+r_2}{{r_2}^{1.5}-{r_1}^{1.5}}\right) \times \frac{{r_2}^{1.5}+{r_1}^{1.5}}{{r_2}^{1.5}+{r_1}^{1.5}}$$
$$=\lim_{\frac{r_2}{r_1} \rightarrow 1} \left(\frac{(r_1+r_2)({r_2}^{1.5}+{r_1}^{1.5})}{{r_2}^{3}-{r_1}^{3}}\right)$$
$$=\lim_{\frac{r_2}{r_1} \rightarrow 1} \left(\frac{(r_1+r_2)({r_2}^{1.5}+{r_1}^{1.5})}{(r_2-r_1)({r_2}^{2}+r_1 r_2+{r_1}^{2}})\right)$$

Then stuck again

Please give me hint, especially how to use geometrical argument.

Thanks

 

[anuttarasammyak]

$r_2/r_1 \rightarrow 1$ is realized when $r_1,r_2 \rightarrow +\infty$ where the curve is almost vertical and the center-center distance on y axis is
r_1+r_2 \approx r_2^{1.5}-r_1^{1.5}

 

[songoku]

$r_2/r_1 \rightarrow 1$ is realized when $r_1,r_2 \rightarrow +\infty$ where the curve is almost vertical and the center-center distance on y axis is
r_1+r_2 \approx r_2^{1.5}-r_1^{1.5}

I understand.

For (b), I again tried rationalization but stuck. Do we also use geometrical argument to solve (b)?

Thanks

 

[anuttarasammyak]

It seems to work that
r_2^{1.5}-r_1^{1.5}=\sqrt{r_2}^3-\sqrt{r_1}^3=(\sqrt{r_2}-\sqrt{r_1})(r_2+r_1+\sqrt{r_2r_1})

 

[fresh_42]

Hint:

Show that $\displaystyle{\lim_{n \to \infty}\dfrac{r_{n+1}}{r_n}}=1$ in case $(c)$ is true so that we can use the formula in $(c)$ instead.

Then show $(c) \Longrightarrow (b) \Longrightarrow (a)$ so we only have to prove $(c)$.

Do you know any methods to prove $(c)$?

Edit: I get $3$ as the limit in $(a)$.

 

[songoku]

Edit: I get $3$ as the limit in $(a)$.

But using method in post#2, the answer is 1

Hint:

Show that $\displaystyle{\lim_{n \to \infty}\dfrac{r_{n+1}}{r_n}}=1$ in case $(c)$ is true so that we can use the formula in $(c)$ instead.

Then show $(c) \Longrightarrow (b) \Longrightarrow (a)$ so we only have to prove $(c)$.

Do you know any methods to prove $(c)$?

Actually, I tried to prove (c) by using (b):

$\lim_{\frac{r_2}{r_1} \rightarrow 1}$ is the same as saying $r_2$ and $r_1$ are large so for large value of $r$, $\sqrt{r_2}-\sqrt{r_1}\approx \frac{2}{3}$

Then I changed it into:
$$\sqrt{r_{n+1}}-\sqrt{r_n}\approx \frac{2}{3}$$
$$r_{n+1}\approx \left(\frac{2}{3}+\sqrt{r_n}\right)^2$$
$$r_{n}\approx \left(\frac{2}{3}+\sqrt{r_{n-1}}\right)^2$$

Then I don't know what to do to get (c)

 

[fresh_42]

But using method in post#2, the answer is 1

This isn't a method. It's a heuristic at best. I get with the use of $(c)$ and for the sake of less typing with $x=r_n$ and $y=r_{n+1}$
\begin{align*}
\dfrac{y+x}{y\sqrt{y}-x\sqrt{x}}&=\dfrac{y+x}{(y+x)(\sqrt{y}-\sqrt{x})-x\sqrt{y}+y\sqrt{x}}\\[6pt]
&=\dfrac{1}{(\sqrt{y}-\sqrt{x})- \dfrac{x}{y+x}\sqrt{y}+\dfrac{y}{y+x}\sqrt{x}}\\[6pt]
&\stackrel{(y/x)\to 1}{\longrightarrow }\dfrac{1}{\sqrt{y}-\sqrt{x}-(1/2)\sqrt{y}+(1/2)\sqrt{x}}\\[6pt]
&\stackrel{(y/x)\to 1}{\longrightarrow }2\dfrac{1}{\sqrt{y}-\sqrt{x}}\stackrel{(y/x)\to 1}{\longrightarrow }3
\end{align*}
Not sure whether I made a mistake.

Actually, I tried to prove (c) by using (b):

$\lim_{\frac{r_2}{r_1} \rightarrow 1}$ is the same as saying $r_2$ and $r_1$ are large so for large value of $r$, $\sqrt{r_2}-\sqrt{r_1}\approx \frac{2}{3}$

Then I changed it into:
$$\sqrt{r_{n+1}}-\sqrt{r_n}\approx \frac{2}{3}$$
$$r_{n+1}\approx \left(\frac{2}{3}+\sqrt{r_n}\right)^2$$
$$r_{n}\approx \left(\frac{2}{3}+\sqrt{r_{n-1}}\right)^2$$

Then I don't know what to do to get (c)

That doesn't work that way.

$(b)$ is also easy with $(c)$:
\begin{align*}
\lim_{n \to \infty}\left(\sqrt{r_{n+1}}-\sqrt{r_n}\right)&=\lim_{n \to \infty}\left(\dfrac{2}{3}(n-1)+\dfrac{2}{3}+\sqrt{r_1}-\dfrac{2}{3}(n-1)-\sqrt{r_1}\right)=\dfrac{2}{3}
\end{align*}
So it remains to show that $\lim_{n \to \infty}\left(\sqrt{r_{n+1}}-\sqrt{r_n}\right)=\lim_{(r_2/r_1) \to 1}\left(\sqrt{r_{2}}-\sqrt{r_1}\right).$ As I understand the limit on the right, they are the same. But maybe that is my mistake. What does $r_2/r_1 \longrightarrow 1$ mean? $r_2$ is a function of $r_1$ isn't it, so they are not independent.

 

[anuttarasammyak]

This isn't a method. It's a heuristic at best. I get with the use of (c) and for the sake of less typing with x=rn and y=rn+1
y+xyy−xx=y+x(y+x)(y−x)−xy+yx=1(y−x)−xy+xy+yy+xx⟶(y/x)→11y−x−(1/2)y+(1/2)x⟶(y/x)→121y−x⟶(y/x)→13
Not sure whether I made a mistake.

I tried it a different way referring my post #4
\frac{y+x}{y^{1.5}-x^{1.5}}=(\sqrt{y/x}-1)^{-1}\sqrt{x}^{-1}(1+\frac{\sqrt{y/x}}{1+y/x})^{-1}
In the limit RHS first coefficient goes to infinity, the second goes to zero because geometry requires x $\rightarrow \infty$ to pursuit the limit, and the third goes to 2/3.

 

[anuttarasammyak]

From geometry I observe
Relation of t >0 and r, evaluating line equation and length of r,
t=-\frac{2}{9}+\sqrt{\frac{4}{81}+r^2}
Recurrence formula with t from $c_2-c_1=r_1+r_2$
c_2-r_2=c_1+r_1thus
3t_{n+1}^{3/2}-\sqrt{13}t_{n+1}+6t_{n+1}^{1/2}=3t_{n}^{3/2}+\sqrt{13}t_{n}+6t_{n}^{1/2}
if I do not make mistakes.

 

[songoku]

I tried it a different way referring my post #4
\frac{y+x}{y^{1.5}-x^{1.5}}=(\sqrt{y/x}-1)^{-1}\sqrt{x}^{-1}(1+\frac{\sqrt{y/x}}{1+y/x})^{-1}
In the limit RHS first coefficient goes to infinity, the second goes to zero because geometry requires x $\rightarrow \infty$ to pursuit the limit, and the third goes to 2/3.

I think the problem is not in post#4 but post#2 because @fresh_42 got 3 (which is also different from question). I follow his working and it also makes sense.

That doesn't work that way.

$(b)$ is also easy with $(c)$:
\begin{align*}
\lim_{n \to \infty}\left(\sqrt{r_{n+1}}-\sqrt{r_n}\right)&=\lim_{n \to \infty}\left(\dfrac{2}{3}(n-1)+\dfrac{2}{3}+\sqrt{r_1}-\dfrac{2}{3}(n-1)-\sqrt{r_1}\right)=\dfrac{2}{3}
\end{align*}
So it remains to show that $\lim_{n \to \infty}\left(\sqrt{r_{n+1}}-\sqrt{r_n}\right)=\lim_{(r_2/r_1) \to 1}\left(\sqrt{r_{2}}-\sqrt{r_1}\right).$ As I understand the limit on the right, they are the same. But maybe that is my mistake. What does $r_2/r_1 \longrightarrow 1$ mean? $r_2$ is a function of $r_1$ isn't it, so they are not independent.

I got (c) but not sure whether my method is valid. I used something like telescoping

$$\sqrt{r_n}-\sqrt{r_{n-1}}=\frac{2}{3}$$
$$\sqrt{r_{n-1}}-\sqrt{r_{n-2}}=\frac{2}{3}$$
$$\sqrt{r_{n-2}}-\sqrt{r_{n-3}}=\frac{2}{3}$$
$$.$$
$$\sqrt{r_2}-\sqrt{r_1}=\frac{2}{3}$$

Adding all the above:
$$\sqrt{r_n}-\sqrt{r_1}=\frac{2}{3}(n-1)$$
$$r_n=\left(\frac{2}{3}(n-1)+\sqrt{r_1} \right)^2$$

But the question states "for large r" while I also included small r so I am not sure

 

[anuttarasammyak]

I think the problem is not in post#4 but post#2 because @fresh_42 got 3 (which is also different from question). I follow his working and it also makes sense.

The fraction, thanks to post #7

=\sqrt{x}^{-1} \frac{1+a}{a\sqrt{a}-1}
=\sqrt{x}^{-1} \frac{1+a}{(\sqrt{a}-1)(a+\sqrt{a}+1)} or
=(\sqrt{y}-\sqrt{x})^{-1} \frac{1+a}{a+\sqrt{a}+1}
where
a=\frac{y}{x}
Say the fraction has limit c for a $\rightarrow$ 1
\sqrt{y}-\sqrt{x}\ \ \rightarrow \ \frac{2}{3c}