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

• songoku
In summary, the conversation involves a discussion about finding the limit of a particular expression involving geometric arguments. The question is to use a geometrical argument to solve the problem, and the conversation explores different approaches to this problem. One method involves using rationalization, while another involves proving a separate limit and then using it to solve the original problem. Ultimately, the conversation leads to the conclusion that the limit is equal to 3, but there is some uncertainty about this result due to the inclusion of small values of r.
songoku
Homework Statement
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

##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
anuttarasammyak said:
##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

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

songoku
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)##.

Last edited:
songoku
fresh_42 said:
Edit: I get ##3## as the limit in ##(a)##.
But using method in post#2, the answer is 1

fresh_42 said:
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)

songoku said:
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.

songoku said:
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.

songoku
fresh_42 said:
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.

songoku and fresh_42
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##
$$c_2-r_2=c_1+r_1$$thus
$$3t_{n+1}^{3/2}-\sqrt{13}t_{n+1}+6t_{n+1}^{1/2}=3t_{n}^{3/2}+\sqrt{13}t_{n}+6t_{n}^{1/2}$$
if I do not make mistakes.

Last edited:
songoku
anuttarasammyak said:
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.

fresh_42 said:
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

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

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

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

songoku said:
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
$$\sqrt{y}-\sqrt{x}\ \ \rightarrow \ \frac{2}{3c}$$

songoku

• Calculus and Beyond Homework Help
Replies
4
Views
1K
• Special and General Relativity
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
17
Views
790
• Introductory Physics Homework Help
Replies
11
Views
780
• Calculus and Beyond Homework Help
Replies
8
Views
898
• Engineering and Comp Sci Homework Help
Replies
8
Views
2K
• Topology and Analysis
Replies
12
Views
2K
• Engineering and Comp Sci Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
12
Views
1K
• Special and General Relativity
Replies
40
Views
2K