Wikipedia:Reference desk/Archives/Mathematics/2017 October 24

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October 24

Landau's Function: g(52)?

Does anyone have what g(52) is where g is Landau's function? I'm looking to find out what the value is for 52 since that is the count of a deck of cards. OEIS stops with the value for 47. Also, g(8) and g(15) are odd, are they the only ones?Naraht (talk) 09:42, 24 October 2017 (UTC)

180180 according to [1], got at from the links section of (sequence A000793 in the OEIS). I would guess all higher ones are divisible by 2. And by 3 and 5 and.., hmm I guess finding the first number that doesn't divide every sufficiently high value might be an interesting quest. Dmcq (talk) 09:48, 24 October 2017 (UTC)

Osculating circle formula for curvature

My last questions did not get any responses:

1. Ruslik told me that the formula for an osculating Circle at a point is

$(\mathbf {r} -\mathbf {r} _{c})^{2}=r^{2}\ ,\quad \mathbf {r} _{c}=\mathbf {r} _{0}+r\mathbf {n} \ ,\quad \mathbf {n} \perp \mathbf {r} _{0}'$

I think there is a problem with this solution, because $\mathbf {n} \perp \mathbf {r} _{0}'$ in 2 different ways.

Isn't $\mathbf {n}$ suppose to be unique?

2. In this weird page they wrote the center of curvature

$\left(a-{\frac {f'(a)}{f''(a)}}{\bigl (}1+f'(a)^{2}{\bigr )},f(a)+{\frac {1+f'(a)^{2}}{f''(a)}}\right)$

but amazingly, they forgot to write its derivation – saying 3 facts:

1. The center lies on the line $y=-{\frac {1}{f'(a)}}x+f(a)+{\frac {a}{f'(a)}}$ perpendicular to the tangent.
2. The center is distance $r$ from ${\bigl (}a,f(a){\bigr )}$ (Duh!).
3. These 2 details are enough to find it.

יהודה שמחה ולדמן (talk) 12:52, 24 October 2017 (UTC)

Yes there are two unit normal vectors, but only one of them is on the right side, and the usual definition of the symbol n is that it is computed as a certain derivative that guarantees that it is on the correct side. Also, given the formula it is easy to verify that it satisfies the two needed conditions, so what's the problem? (It's not like the algebraic manipulations are very informative.) --JBL (talk) 15:39, 24 October 2017 (UTC)

That page also tells you that the radius of curvature is given by

${\frac {1}{R}}={\frac {f''(x)}{(1+(f'(x))^{2})^{\frac {3}{2}}}}$

You can derive this from the following:

Using $x$ to parameterize the curve, a tangent vector (of no particular magnitude) is

$\mathbf {t} =\left({\frac {\partial }{\partial x}}x\right)\mathbf {\hat {x}} +\left({\frac {\partial }{\partial x}}y\right)\mathbf {\hat {y}}$

where $y=f(x)$. The unit tangent vector is

$\mathbf {\hat {t}} ={\frac {\mathbf {t} }{\sqrt {\mathbf {t} ^{2}}}}$

The unit normal to the curve is

$\mathbf {\hat {n}} =\mathbf {\hat {t}} \times \mathbf {\hat {z}}$

The radius vector (to within a sign) is $R\mathbf {\hat {n}}$. Consider this radius vector and the unit vector $\mathbf {\hat {n}}$, both based at the centre of curvature. As you travel a distance $\partial s$ along the curve, the tip of the radius vector sweeps out just that ($\partial s$), while the tip of $\mathbf {\hat {n}}$ sweeps out an arc $R$ times smaller. So the magnitude of this smaller arc gives ${\frac {1}{R}}$

${\sqrt {\left({\frac {\partial }{\partial s}}\mathbf {\hat {n}} \right)^{2}}}={\frac {1}{R}}$

The curve is parameterized by $x$, not distance $s$, so the chain rule is needed

${\frac {\partial }{\partial s}}\mathbf {\hat {n}} ={\frac {{\frac {\partial }{\partial x}}\mathbf {\hat {n}} }{{\frac {\partial }{\partial x}}s}}$

Pythagoras gives the distance travelled along the curve as $x$ is increased

${\frac {\partial }{\partial x}}s={\sqrt {\left({\frac {\partial }{\partial x}}x\right)^{2}+\left({\frac {\partial }{\partial x}}y\right)^{2}}}$

Put all these together, and you will eventually get

${\frac {1}{R}}={\frac {f''(x)}{(1+(f'(x))^{2})^{\frac {3}{2}}}}$

An intermediate result (which is also useful for finding the centre of curvature) is

$\mathbf {\hat {n}} ={\frac {-\mathbf {\hat {y}} +f'(x)\mathbf {\hat {x}} }{\sqrt {1+(f'(x))^{2}}}}$

--catslash (talk) 16:05, 24 October 2017 (UTC)

"Given the formula it is easy to verify that it satisfies the two needed conditions, so what's the problem?"

Bravo, Bravo!!! What an answer!!!

"A Sleeping drug causes sleep due to its sleepy effectiveness".

Give me a break (or the derivation). יהודה שמחה ולדמן (talk) 17:31, 24 October 2017 (UTC)

As a casual observer, I've never seen a more graceless, indeed offensive response to an answer. The OP should learn some manners and a bit of humility.→31.52.213.50 (talk) 15:59, 25 October 2017 (UTC)

I myself am humiliated by circular reasoning when I ask for derivations and proof. Just take a look at that quote. יהודה שמחה ולדמן (talk) 21:29, 25 October 2017 (UTC)

OP, if you can't see what's inappropriate about demanding a further answer from volunteers who run this desk as a favour to you, you're likely to be disappointed at the subsequent non-response. And you seem to be equating humility and humiliation. Everyday definitions are "humility - a way of behaving that shows that you do not think that you are better or more important than other people"; "humiliation - the unhappy and ashamed feeling that you get when something embarrassing happens".→31.52.213.50 (talk) 15:28, 26 October 2017 (UTC)

I hope Yehuda (the OP) will not be offended by me saying this, I feel that everyone's thinking it and someone has to say it eventually. The root cause of the current argument may be that Yehuda's mathematical enthusiasm isn't matched by his mathematical skill. He frequently asks questions here about a variety of mathematical topics, some relatively advanced, but often severely lacks either the talent, knowledge or proficiency to make sense of the answers.

This is what we're seeing here. The algebraic manipulations required to derive the formula from the stated conditions are completely straightforward and trivial. Cumbersome perhaps (not anything I can do in my head), but trivial. So when he asks how is the formula derived, we have difficulty understanding what exactly it is that he is asking. It's as if he would say "This page shows that the answer is 423 * 728, but then concludes that the answer is 307944. How did they make this step?". They made this step by multiplying 423 and 728. With a calculator perhaps. Or with long multiplication with pen and paper, but nobody here is going to show him a step-by-step calculation of this long multiplication.

That was the motivation for JBL's (non-)answer. But for Yehuda, who thinks there really is a missing derivation, it seems that JBL is patronizing him and wasting his time. So I can see where his response is coming from.

On that note, I should say his 1st question here is better than the 2nd. And JBL did try to answer that. In fact, if you just plug in the 2 conditions, you get a quadratic system you can easily solve to get two solutions, one of which is the one given by the page. Finding out which is the correct one requires understanding the direction of curvature and how it relates to the sign of the 2nd derivative.

Now - it's ok to be less mathematically skilled than the rest of us. The problem is that Yehuda often asks questions that are above the level he should be asking with his current skill level. Or the other way around, he should try to improve his skills before he tackles more advanced topics. It's hard to answer advanced questions when the basics are not covered.

My suggestion - it's fine to continue asking questions, and I'm sure many people here will be happy to help. But Yehuda should also understand his special situation, and respond with more humility, appreciation and gratitude, even if he is not getting quite the answers he expects. -- Meni Rosenfeld (talk) 17:11, 26 October 2017 (UTC)

The root cause of the current argument may be that Yehuda's mathematical enthusiasm isn't matched by his mathematical skill. He frequently asks questions here about a variety of mathematical topics, some relatively advanced, but often severely lacks either the talent, knowledge or proficiency to make sense of the answers.

So true. יהודה שמחה ולדמן (talk) 19:01, 26 October 2017 (UTC)

The algebraic manipulations required to derive the formula from the stated conditions are completely straightforward and trivial.

Are you saying you know the derivation? I have no luck. יהודה שמחה ולדמן (talk) 21:23, 26 October 2017 (UTC)

There are two ways I can think of, one is easier to understand and one is easier to compute. I'll go with the easier to understand: The center point is some $(x,y)$ which is on the line that passes through $(a,f(a))$ and is perpendicular to the function at that point (and thus has slope ${\frac {-1}{f'(a)}}$). That is the line $y=-{\frac {1}{f'(a)}}x+f(a)+{\frac {a}{f'(a)}}$ they mentioned. Also, the distance between $(x,y)$ and $(a,f(a))$ is equal to r, meaning $(x-a)^{2}+(y-f(a))^{2}=r^{2}={\frac {(1+f'(a)^{2})^{3}}{f''(a)^{2}}}$. This is a system of 2 equations you have to solve to find x and y. Substitute y from the 1st equation into the 2nd equation and expand everything out. You get a quadratic equation in x. Solve it, and then substitute to find y.

You get two solutions - one above the point $(a,f(a))$, one below. The correct solution is the one where if $f''(a)>0$, the center point is above, and vice versa. -- Meni Rosenfeld (talk) 22:20, 26 October 2017 (UTC)

Thanks. But that is exactly what I got by trying the dot product of the tangent and radius vectors (as someone told me before):

${\begin{aligned}&(1)\ {\vec {T}}\cdot {\vec {r}}={\binom {1}{f'(a)}}\cdot {\binom {b}{c}}=b+f'(a)c=0\ ,\ b=-f'(a)c\\&(2)\ b^{2}+c^{2}={\frac {{\bigl (}1+f'(a)^{2}{\bigr )}^{3}}{f''(a)^{2}}}\\&{\bigl (}-f'(a)c{\bigr )}^{2}+c^{2}={\frac {{\bigl (}1+f'(a)^{2}{\bigr )}^{3}}{f''(a)^{2}}}\\&c^{2}{\bigl (}f'(a)^{2}+1{\bigr )}={\frac {{\bigl (}1+f'(a)^{2}{\bigr )}^{3}}{f''(a)^{2}}}\\&c^{2}={\frac {{\bigl (}1+f'(a)^{2}{\bigr )}^{2}}{f''(a)^{2}}}\ ,\ (b,c)=\left(-{\frac {f'(a)}{|f''(a)|}}{\bigl (}1+f'(a)^{2}{\bigr )},{\frac {1+f'(a)^{2}}{|f''(a)|}}\right)\\&(h,k)=\left(a-{\frac {f'(a)}{|f''(a)|}}{\bigl (}1+f'(a)^{2}{\bigr )},f(a)+{\frac {1+f'(a)^{2}}{|f''(a)|}}\right)\end{aligned}}$

Almost there, and still it does not answer how we get the complete formula at the beginning. יהודה שמחה ולדמן (talk) 23:35, 26 October 2017 (UTC)

I am taking a break from this place (though it is always nice to see Meni Rosenfeld show up to remind me that not everything here is worthless) but I thought before I go I should point out one further detail about my earlier comment: the acts of deriving and verifying are not the same, and in fact it is (at least arguably) easier to verify by hand that two points have the correct distance from each other (an application of the distance formula) and that the line through them has the correct slope (an application of the basic formula for slope) than it is to derive the location of the second point from that information. So there is no circularity in my earlier comment. Hopefully I will not return here for a good long while. --JBL (talk) 23:17, 26 October 2017 (UTC)

And by the way, $\mathbf {\hat {n}} ={\frac {f'(x)\mathbf {\hat {x}} -\mathbf {\hat {y}} }{\sqrt {1+f'(x)^{2}}}}$ is the wrong vector for finding the center of curvature after multiplying by $r$ . You could use Desmos. יהודה שמחה ולדמן (talk) 10:20, 27 October 2017 (UTC)

It looks like you are suggesting that if $c^{2}={\frac {{\bigl (}1+f'(a)^{2}{\bigr )}^{2}}{f''(a)^{2}}}$ then $c={\frac {1+f'(a)^{2}}{|f''(a)|}}$. That is not correct. The correct conclusion is that either $c={\frac {1+f'(a)^{2}}{f''(a)}}$ or $c=-{\frac {1+f'(a)^{2}}{f''(a)}}$. Out of these, the former is the correct solution, because if $f''(a)>0$ then the center point should be above the point $(a,f(a))$. -- Meni Rosenfeld (talk) 10:45, 27 October 2017 (UTC)