Wikipedia:Reference desk/Archives/Mathematics/2011 May 26

= May 26 =

chord and arc leangth
If the radius and arc leangth are known, how can you calculate the length of chord? On the otherhand, if the radius and length of chord are known how can you calculate the arc length? thank you.124.43.233.225 (talk) 02:13, 26 May 2011 (UTC)


 * For circles of unit radius, the length of the chord is twice the sine of half the arc. So if the radius is r, then the chord is r times twice the sine of half the measure &theta; of the arc in radians.  The length of the arc is ℓ = r&theta;.  Hence the chord is
 * $$ \text{chord} = 2r \sin\left(\frac\ell r\right). $$
 * Consequently
 * $$ \ell = r \arcsin\left(\frac{\text{chord}}{2r}\right). $$
 * See also Ptolemy's table of chords. Michael Hardy (talk) 02:56, 26 May 2011 (UTC)

Backwards induction optimisation problem (Bellman eqn)
Hello everyone,

I'm trying to solve an optimisation problem using backwards induction but it's not quite the same as those I've encountered before so I was hoping you could let me know if I've gone wrong!

The problem is as follows: let U be the function $$ U(x)=\frac{x^{1-R}}{1-R}, x \geq 0,\, R \in [0,\,1) $$. An investor, with wealth $$w_0 > 0$$ at time 0 wishes to invest it in such a way as to maximise $$\mathbb{E}(U(w_N))$$, where $$w_N   $$ is the wealth at the start of day N. Let $$\alpha \in [0,\,1]   $$ fixed. Then on each day n, he chooses a proportion $$\theta \in [\alpha,\,1]$$ of his wealth to invest in a single risky asset, so the wealth at the start of day n+1 is $$w_{n+1} = w_n(\theta X_n + (1 - \theta)(1+r))   $$, where r is the per-period (riskless) interest rate and $$X_n$$ is a family of IID positive random variables. If $$V_n(w) = sup \mathbb{E}(U(w_N)|w_n = w)$$ denotes the value of the optimisation problem, show that $$V_n(w_n) = a_n U(w_n)$$, giving a formula for $$a_n$$.

Now my problem is that when I've worked with these value functions before, I haven't been working with functions where the 'input' comes into play as part of a conditional expectation, it is usually more "directly" involved. My thinking was as follows: apply backwards induction and suppose the formula holds for $$V_{n+1}$$. Then $$V_n(w) = sup_{\theta} \,\mathbb{E}(U(w_N)|w_n = w) = sup_{\theta} \, \mathbb{E}[U(w_N)|w_{n+1} = w(\theta X_n + (1 - \theta)(1+r))]$$. Let "$$\Gamma w$$" be the value of $$w_{n+1}$$ at which this supremum is obtained, $$\Gamma$$ some constant (I think this is okay since we're working with $$\theta  $$ on a compact interval): so then $$V_n(w) = V_{n+1}(\Gamma w) = a_{n+1}U(\Gamma w) $$ by the induction hypothesis so $$= a_{n+1}\Gamma^{1-R} U(w) = a_{n}U(w)$$.

I would be extremely grateful if anyone has any comments: I'm not sure whether or not the approach is right, since I'm a bit concerned about when we pass from the conditional expectation in $$w_n $$ to $$w_{n+1} $$, and also about when we involve the $$\Gamma $$. Could anyone advise me on whether I have it right or what I've done wrong? Thank you, wonderful helpful people! :) Sdnahzzaj (talk) 10:18, 26 May 2011 (UTC)

Equation help, please
Hi guys. I'm trying to work out how to do an equation but really struggling. Any help at all would be very much appreciated. The equation is:

A = 242 x 10-2 mol2 dm-6 divided by 5.85 x 10-3 mol dm-3

My problem with working out an answer has to do with all these different powers that are going on. The value for the first part of the equation is times by 10-2 whereas in the second part it's 10-3. Also, the units on the first part are mol2 dm-6 whereas in the second part it's mol dm -3. All of this is really confusing me. Does anybody have any advice at all?

Many thanks.

Pantscat (talk) 11:46, 26 May 2011 (UTC)


 * ab divided by ac is equal to ab &minus; c. This manipulation works whether a is a number (like your 10) or a unit (like your mol and dm). 86.160.209.202 (talk) 12:36, 26 May 2011 (UTC)
 * PS, I finally found the section where is explained, at Exponentiation. This and surrounding sections also explain a whole bunch of related stuff. 86.160.209.202 (talk) 12:52, 26 May 2011 (UTC)


 * Another way to do it is remove the negative powers. You have $$\frac {24^2 mol^2} {10^2 dm^6}$$ divided by $$\frac {5.85 mol} {10^3 dm^3}$$. Remember, dividing by a fraction is the same as multiplying by the inverse of the fraction. So you are multiplying $$\frac {24^2 mol^2} {10^2 dm^6} \times \frac {10^3 dm^3} {5.85 mol}$$. Now, you can see that the powers will start cancelling themselves out making a simple problem. -- k a i n a w &trade; 12:57, 26 May 2011 (UTC)

"Equation" isn't the right word here. You're evaluating an expression, not solving an equation.

A bit of TeX: How's this look? $$\frac {5.85\text{ mol}} {10^3\text{ dm}^3}$$ Michael Hardy (talk) 18:05, 27 May 2011 (UTC)

unlimited substitutability in logic notation
I want to write the following sentence in logical notation:

Labor L and capital K have unlimited substitutability. i.e. an appropriate level of L exists for any level of Y to be produced at any level of K.

do I write $$ \exists L | Y=(K,L) \forall Y,K $$

but that's just for L, how do I simultaneously express this property for L and Y.

And is there a particular mathematical term for this property? — Preceding unsigned comment added by Thorstein90 (talk • contribs) 18:40, 26 May 2011 (UTC)


 * You probably just want to write two statements. $$\forall (L,Y), \exists K \,\mathrm{st}\, f(L,K)=Y$$ and $$\forall (K,Y), \exists L \,\mathrm{st}\, f(L,K)=Y$$ where f is the function that, when given the labour and capital outputs the amount produced. --Tango (talk) 19:51, 26 May 2011 (UTC)

function equals infinity
Is it incorrect to say that a function equals infinity at a point? Should you instead only say that it tends toward infinity? Like, instead of

$$f(0)=\infty$$

you should only ever use

$$\lim_{k \to 0} f(k)=\infty$$


 * If the function approaches infinity at k=0, then it likely doesn't actually equal anything at k=0. It is undefined at k=0. So, showing the limit is proper. However, anyone will know what you mean if you claim it equals infinity. -- k a i n a w &trade; 19:53, 26 May 2011 (UTC)


 * It depends on what the codomain of the function is. If it's, for example, the real numbers then you can't say it equals infinity because infinity isn't a real number. However, if it's the extended real numbers, or some other set that includes infinity, then you can say it equals infinity. --Tango (talk) 19:55, 26 May 2011 (UTC)
 * I suppose you could have something like $$f(x) = \lim_{k \to x} \frac{1}{k^2}$$ in which case $$f(0)=\infty$$ Widener (talk) 00:53, 27 May 2011 (UTC)
 * Strictly speaking, the limit $$\lim_{k\to0}1/k^2$$ does not exist (following the standard definition of limits of real-valued functions), so the function $$f(x)=\lim_{k\to x}1/k^2$$ is undefined at $$x=0$$. We write $$\lim_{k\to0}1/k^2=\infty$$ to indicate that, as k approaches 0, the function $$1/k^2$$ increases without bound, but this is just a notational convention (probably an abuse of notation). The formal delta-epsilon definition of a limit does not hold: It is not true that for any $$\epsilon>0$$ there exists a $$\delta>0$$ such that for all k with $$0<|k-0|<\delta$$ we have $$|1/k^2-\infty|<\epsilon$$. After all, $$|1/k^2-\infty|$$ is meaningless in the real numbers, and even if we should try to assign it some meaning by adding $$\infty$$ to the number line (see extended real number line), we must surely say that $$|1/k^2-\infty|=\infty$$ for all $$k\not=0$$, and we must agree that $$\infty>\epsilon$$ for all real $$\epsilon$$ . —Bkell (talk) 03:07, 27 May 2011 (UTC)
 * Well, but now you're using the wrong topology on the extended reals. For any real number a, the interval (a,+&infin;] is an open set in the extended reals.  That gives you the right notion of what it means for the limit of a function to equal +&infin;.  --Trovatore (talk) 03:14, 27 May 2011 (UTC)
 * That's true. I'm not sure now exactly what I was trying to say by bringing up the extended reals. —Bkell (talk) 12:49, 27 May 2011 (UTC)
 * I think all I was trying to say is basically what Tango said above: If the codomain of the function includes &infin; (such as the extended reals), then you can just define $$f(0)=\infty$$ directly, so a "trick" like $$f(x)=\lim_{k\to x}1/k^2$$ is unnecessary to make the function equal &infin; at a point; if the codomain does not include &infin; (such as the ordinary real numbers), then you can't make $$f(0)=\infty$$ even if you try something like this limit trick. —Bkell (talk) 13:15, 27 May 2011 (UTC)
 * By the way, to the original questioner, $$f(0)=\infty$$ and $$\lim_{k\to0}f(k)=\infty$$ mean two different things, just as $$f(0)=3$$ and $$\lim_{k\to0}f(k)=3$$ mean two different things. —Bkell (talk) 13:20, 27 May 2011 (UTC)