Wikipedia:Reference desk/Archives/Mathematics/2010 March 18

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March 18

Integrating differential terms independently of each other in a single equaiton

Hi, I'm having trouble understanding a derivation of Bernouilli's equation given by my lecturer.

The derivation basically gets to the point of:

dp⁄ρ + d(v²)⁄2 + gdz = 0

And the next line simply integrates the d(v² and dz terms independently to become:

∫dp⁄ρ + v²⁄2 + gz = constant

Is this a legitimate operation? To the best of my knowledge you can't integrate one term in an equation independently (with respect to a different variable) to the others. In a triple integral for example you integrate the entire equation with respect to each of the three variables.

I'd appreciate any clarifications. --118.139.13.221 (talk) 00:25, 18 March 2010 (UTC)

Integration is additive, so $\int (\mathrm {d} x+\mathrm {d} y)=\int \mathrm {d} x+\int \mathrm {d} y$. --Tango (talk) 01:06, 18 March 2010 (UTC)

Note that there are already differentials present in the original equation. You can differentiate both sides just by adding an integral sign alone. What we're doing is basically summing up the values of the differentials over some range. If you've got an equation without differentials that you want to integrate, you would need to introduce some differentials. In that case you can't introduce whatever differentials to different terms; it has to be consistent. So for example if x + y = 0 you can't say that ∫xdx + ∫ydy = 0. You would have to do something like ∫xdx + ∫ydx = 0. On the other hand, if you have dx + dy = 0, then it's fine to say that ∫(dx + dy) = ∫0, which is ∫dx + ∫dy = 0. Rckrone (talk) 04:52, 18 March 2010 (UTC)

Thanks heaps for that - very useful thing to find out! --118.139.13.221 (talk) 23:13, 18 March 2010 (UTC)

Forget them differentials - they ain't nothin' but trouble. The derivation of the equation of motion for the fluid should lead you to

${\frac {1}{\rho }}{\frac {dp}{dt}}+{\frac {1}{2}}{\frac {dv^{2}}{dt}}+g{\frac {dz}{dt}}=0$

then you can integrate between times t₁ and t₂ to get:

$\int _{t_{1}}^{t_{2}}\left[{\frac {1}{\rho }}{\frac {dp}{dt}}+{\frac {1}{2}}{\frac {dv^{2}}{dt}}+g{\frac {dz}{dt}}\right]dt=0$

$\Rightarrow \int _{t_{1}}^{t_{2}}\left[{\frac {1}{\rho }}{\frac {dp}{dt}}\right]dt+\int _{t_{1}}^{t_{2}}\left[{\frac {1}{2}}{\frac {dv^{2}}{dt}}\right]dt+\int _{t_{1}}^{t_{2}}\left[g{\frac {dz}{dt}}\right]dt=0$

$\Rightarrow \int _{p_{1}}^{p_{2}}{\frac {1}{\rho }}dp+\int _{v_{1}^{2}}^{v_{2}^{2}}{\frac {1}{2}}dv^{2}+\int _{z_{1}}^{z_{2}}gdz=0$

$\Rightarrow {\frac {1}{\rho }}(p_{2}-p_{1})+{\frac {1}{2}}(v_{2}^{2}-v_{1}^{2})+g(z_{2}-z_{1})=0$

where p₁, p₂ are pressures at times t₁, t₂ etc. As this is true for any two times t₁, t₂, then we must have

${\frac {p}{\rho }}+{\frac {v^{2}}{2}}+gz={\text{constant}}$

Gandalf61 (talk) 10:13, 18 March 2010 (UTC)

Thanks heaps for that derivation, was a much 'nicer' way to think about it :)

formulas which induce functions

If $\varphi$ is a formula defined by: $\varphi (x,y)$ ::= x=y², then $\varphi$ induces a function from the set X of positive numbers x to the set Y of negative numbers y. However, $\varphi$ does not strongly induce a function from X to Y, i.e. I can only make sure that every positive x has a negative y such that $\varphi (x,y)$, and that there isn't any other negative y satisfying $\varphi (x,y)$, but I can't make sure that there isn't any other arbitrary y satisfying $\varphi (x,y)$. Here is a counter example - for strongly-inducing a function: The formula $\varphi$, does strongly induce a function from the set Y of negative numbers y to the set X of positive numbers x, namely: not only can I make sure that every negative y has a positive x such that $\varphi (x,y)$, and that there isn't any other positive x satisfying $\varphi (x,y)$, but I can also make sure that there isn't any other arbitrary x satisfying $\varphi (x,y)$.

The notion of a formula which strongly induces a function, sounds very intuitive to me. Unfortunately, the very expression "strongly induces a function" - is not familiar (I invented it). Do you think that there is any familiar brief expression for indicating this intuitive notion?

HOOTmag (talk) 08:29, 18 March 2010 (UTC)

See Implicit and explicit functions. Bo Jacoby (talk) 00:45, 19 March 2010 (UTC).

I didn't find there anything that may help me to find out any familiar brief expression for indicating the intuitive notion of a formula which strongly induces a function. HOOTmag (talk) 11:07, 19 March 2010 (UTC)

Why does $\varphi$ not "strongly induce" a function $f:X\to Y$? Sure, $(4,2)$ is a point on the curve $x=y^{2}$, but $2\notin Y$, so why does the existence of that point matter? If you're specifying the codomain of the implicit function ahead of time, then "strongly induce" seems to mean the same as "induce a function." If the specified codomain is actually the reals rather than the negative reals, then it seems that what you are doing is defining a principal branch of a multivalued function, which is usually considered to be an arbitrary choice; if there is no such choice, then the set of points implicitly defines a function in the ordinary sense. What I am saying is that I don't understand the "intuitive notion" of "strongly induces." Can you provide a precise definition for what you mean? —Bkell (talk) 13:29, 19 March 2010 (UTC)

Perhaps what you mean is that $\varphi$ implicitly defines a relation from $\mathbb {R} ^{+}$ to $\mathbb {R}$, but does not define a function unless the codomain is restricted to $\mathbb {R} ^{-}$? —Bkell (talk) 13:59, 19 March 2010 (UTC)

Exactly!

To put it more formally: If $\varphi$ is a formula having two free variables only (over the universal domain of discourse), then: to state that $\varphi$ induces a function from R to S, means that:

• Every r in R has an s in S such that every v in S satisfies: $\varphi (r,v)$ iff v=s.

That's a simple definition of a function (from a given set to a given set) induced by a given formula. But what happens if we let v be totally arbitrary, i.e. let it belong to the universal domain of discourse, unnecessarily to a given set? Then we can say that $\varphi$ strongly-induces a function from R to S, namely:

• Every r in R has an s in S such that every v satisfies: $\varphi (r,v)$ iff v=s.

Can you think of any idea about how to express briefly the fact that $\varphi$ strongly-induces a function from R to S, without using the term "strongly", and without having to get into too many details (like those I've indicated above when I defined the notion of "strongly-inducing a function")?

HOOTmag (talk) 14:14, 19 March 2010 (UTC)

Hmm. Your earlier example of "strongly induces" was somewhat tautological, then—the reason $x=y^{2}$ strongly induces a function from Y to X is that x is in fact explicitly defined by the formula. (So my current working hypothesis is that "strongly induces" means "explicitly defines.") Do you have an example of a formula that strongly induces a function without explicitly defining it? —Bkell (talk) 14:29, 19 March 2010 (UTC)

I agree with you that every formula, explicitly defining a given function from X to Y, strongly induces the function, but it's not necessarily the other way around. For example, check: x=y*2. HOOTmag (talk) 18:24, 20 March 2010 (UTC)

You didn't specify the values of R or S, so I'm going to assume you intend $R=S=\mathbb {R}$. Now, what is the operation in "y*2"? Is that real multiplication? If so, how does your definition for "strongly induces" make sense if v is not a real number? —Bkell (talk) 20:02, 20 March 2010 (UTC)

What I'm saying is that the formula x = y² doesn't strongly-induce a function from the set X of positive numbers x to the set Y of negative numbers y, while the formula x= -2y does strongly-induce a function from the set X of positive numbers x to the set Y of negative numbers y.

More generally, given the binary relation R which is determined by a given formula, I'm looking for a shorter way to express the idea that both the restriction X ◁ R is a function and the image of X ◁ R is Y.

HOOTmag (talk) 20:23, 20 March 2010 (UTC)

And what I'm saying is that I still don't understand your definition of "strongly induces." A binary relation requires the specification of a domain and a codomain. If you haven't specified those, you haven't specified a relation.

The definition you gave allows v to "be totally arbitrary," to come from "the universal domain of discourse, unnecessarily to a given set." But I don't understand how you plan to interpret the expression "-2v" if v is totally arbitrary. For example, suppose we consider the line with two origins, interpreted as the real line with an "extra" origin (call it @, say). Now we define multiplication on this space to be ordinary real multiplication, with the additional definition that @ × a = a × @ = 0 for all a. This is certainly a valid binary operation. But in this space, with this binary operation, there are two values v for which it is true that 0 = −2v. So then, apparently, x = −2y actually does not "strongly induce" a function from $X=\mathbb {R} ^{+}$ to $Y=\mathbb {R} ^{-}$.

I suppose you will object to this example, saying that it's contrived or that v isn't really a number or something. But your definition is what is allowing v to be totally arbitrary.

If by "strongly induces" you mean "both the restriction X ◁ R is a function and the image of X ◁ R is Y," as you say, then your original example of $x=y^{2}$ does not "strongly induce" a function from $X=\mathbb {R} ^{+}$ to $Y=\mathbb {R} ^{-}$, because the restriction of the domain to X does not define a function. (It defines a multivalued function, for which you must choose a principal branch.) So I am still confused about what you are trying to say. —Bkell (talk) 20:54, 20 March 2010 (UTC)

I thought it was clear that the universal domain of discourse was (in my example) the set of real numbers.

When did I say that the formula x=y² strongly induces a function from $X=\mathbb {R} ^{+}$ to $Y=\mathbb {R} ^{-}$ ? On the contrary: I said that it doesn't! See the first paragraph of my previous response! On the other hand, the formula x= -2y (defined on the set of real numbers) does strongly-induce a function from the set X of positive numbers x to the set Y of negative numbers y.

Again, I'm looking for a shorter way to express the idea that both the restriction X ◁ R is a function and the image of X ◁ R is Y. How can this simple question confuse you?

HOOTmag (talk) 21:12, 20 March 2010 (UTC)

Oh, oops, you're right. Sorry, I had it mixed up in my mind. So what's wrong with simply saying that the restriction of R to X is a function and R(X) = Y? That seems pretty short to me. —Bkell (talk) 21:23, 20 March 2010 (UTC)

Nothing wrong with that, I'm just asking if there is a shorter way to express that. If you think there isn't then it's ok. HOOTmag (talk) 21:26, 20 March 2010 (UTC)

Expected value of a difference

Please explain if there is something wrong with the following logic (I think there is).

Say the expected value of random variable a is a constant b, $E(a)=b$

Then the expected value of (a-b) should be $E(a-b)=E(a)-E(b)=b-b=0$

In that case the expected value of the cube of the difference should also be zero, right?

$E((a-b)^{3})=E((a-b)(a-b)^{2})=E(a-b)E((a-b)^{2})=(E(a)-E(b))(E(a-b)^{2})=0$

I know this can't be right (because this has to equal some non-zero term in the case of skewness)... but I don't know where the logic fails. —Preceding unsigned comment added by Damian Eldridge (talk • contribs) 10:18, 18 March 2010 (UTC)

$(a-b)$ and $(a-b)^{2}$ are not independent so you can't say $E((a-b)(a-b)^{2})=E(a-b)E((a-b)^{2})$ Zain Ebrahim (talk) 10:27, 18 March 2010 (UTC)

Thanks! —Preceding unsigned comment added by Damian Eldridge (talk • contribs) 10:43, 18 March 2010 (UTC)

(ec) Nope, $E(X)=0$ does not imply $E(X^{3})=0$. Consider $X$ which gets one of three values $\{-2,\ -1,\ 3\}$ with equal probability of 1⁄3 — then $X^{3}$ gets values $\{-8,\ -1,\ 27\}$ and its expected value is certainly not zero. --CiaPan (talk) 10:45, 18 March 2010 (UTC)

Distance along a parabola

I have a parabola y=x^2, in the range -x0<=x<=x0 (x0 is guaranteed finite) and the arc length d between (-x0, (-x0)^2) to the (x1, (x1^2)) (x1<=x0). How do I find x1?

I looked through wolfram, and the "parabola" and "arc length" pages, but I really couldn't find an answer that made sense to me. This is not at all a homework question, by the way.

Thank you. 78.245.228.100 (talk) 11:59, 18 March 2010 (UTC)

The arc length of the function $y=f(x)$ graph over $a<x<b$ is $L=\int \limits _{a}^{b}{\sqrt {1+\left[f'(x)\right]^{2}}}\,dx$. For function $f(x)=x^{2}$ you get $f'(x)=2x$, so $L=\int \limits _{a}^{b}{\sqrt {1+4x^{2}}}\,dx$. This can be expressed in an algebraic form with $a$ and $b$ as parameters (see List of integrals of irrational functions). Then substitute your given length and starting point −x₀ as L and a respectively, and try to solve for b. --CiaPan (talk) 12:13, 18 March 2010 (UTC)

Thank you, that all makes perfect sense to me: I'll end up with d=f(x0, x') which I can rearrange to solve for x'. I consulted List of integrals of irrational functions in the hope of finding the algebraic representation of the integral, but couldn't find it there: whilst calculus brings back fond memories of my university math classes, I fear that twenty five years of disuse has relagated me to the neophyte level. 78.245.228.100 (talk) 12:29, 18 March 2010 (UTC)

Hint: ${\sqrt {1+4x^{2}}}=2{\sqrt {x^{2}+\left({\tfrac {1}{2}}\right)^{2}}}$. -- Meni Rosenfeld (talk) 12:34, 18 March 2010 (UTC)

Got it (for the moment, I think). The integral I was looking for was in fact the very first line $\int r\;dx={\frac {1}{2}}\left(xr+a^{2}\,\ln \left(x+r\right)\right)$ where $r=2{\sqrt {x^{2}+a^{2}}}$ and $a={\tfrac {1}{2}}$. So I expand all of that out, and call it $f$, then try to rearrange $d=f(x')-f(x0)$ to get an expression for x' in terms of f(x0) and d.

Thank you both for making me think about it (a tiny bit) rather than just giving me the solution. I'm in engineering where we typically delight in spitting out the answer, why is it that mathematicians almost universally prefer to make the questionner work a little? 78.245.228.100 (talk) 13:00, 18 March 2010 (UTC)

Because we want the world to appreciate how beautiful it all is. I nearly lost my to be wife by explaining some maths to her :) Dmcq (talk) 13:12, 18 March 2010 (UTC)

• I think mathematicians know that almost nothing in maths is clear and obvious for everybody. So there is always place to make mistakes. If they give you some general directions, you'll be able to better understand your problem and possible ways to solve it (as well as similar problems in future). You will also be able to check your solution if it seems wrong. On the other hand if you're given a final answer, you could do nothing except using it. If it is wrong, or you misunderstand it, you fail to solve your problem.
• Another possible reason is that asking the questioner to show some of his/her own work lets them recognize the questioner's skills level, and express their advice in appropriate level of formalism.
• Asking for some work allows them also to filter those lazy pupils who want to get their homework done by someone else. At least I think so. CiaPan (talk) 13:27, 18 March 2010 (UTC)

Engineers place a high value on getting stuff done--they expect the questioner wants the answer with as little fuss as possible, so s/he can get onto the next thing for some practical purpose. Mathematicians are more in pursuit of understanding things and seeking mental stimulation for its own sake, so they tend to expect the questioner to also want that. Of course in any well-functioning culture, both approaches are necessary, and having everyone follow their own tendency seems to result in a pretty good mix. 66.127.52.47 (talk) 20:40, 18 March 2010 (UTC)

In my expression d=f(x')-f(-x0), I set x0 to 1 and got f(-x0) as -2.9366. Does this value have any meaning? (notice how engineers want their numbers to mean something?) 78.245.228.100 (talk) 13:34, 18 March 2010 (UTC)

Not really. As you probably know, $(f(x)+c)'=f'(x)\;\!$ so when taking integrals you can choose which constant to add. The formula in the list is for a particular choice of constant (which makes the expression the simplest). The value of f at any particular point will depend on this arbitrary constant. What matters is the differences between values, which don't depend on the constant. -- Meni Rosenfeld (talk) 14:22, 18 March 2010 (UTC)

My goodness. Now I have $2(d+2.9366)=x'r+a^{2}ln(x'+r)\;\!$. Where $r=2{\sqrt {x'^{2}+a^{2}}}$ and $a={\tfrac {1}{2}}$ as above. If I keep on trying to rearrange this will I ultimately end up with an algebraic expression for x', or is this going to end up with an iterative solution? Perhaps I'm going about what I really want to do in an overly complicated way.78.245.228.100 (talk) 08:05, 19 March 2010 (UTC)

What I have is some values in the range umin to umax which I want to map to a different coordinate system vmin to vmax such that the entire range umin to umax is visible but the ones to the center of seem bigger. My idea was to use ua-umin as the arc length and then work out the corresponding v, but this all suddenly seems like a lot of work.78.245.228.100 (talk) 08:28, 19 March 2010 (UTC)

Is there a reason to use specifically a parabola? If not, you can choose a transformation which will be much easier to work with. For example, $v=\arctan u$. Other possibilities exist, depending on the requirements. -- Meni Rosenfeld (talk) 11:50, 21 March 2010 (UTC)

Drive time commute

How long would it take to drive from Colorado to San Jose, Bolivia? --67.134.239.205 (talk) 12:45, 18 March 2010 (UTC)

This isn't really a maths question. [maps.google.com Google Maps] should be able to give you an idea (click "Get Directions" in the top left). However, you will need to be more precise - there are apparently two Colorados and five San Joses in Bolivia. --Tango (talk) 12:56, 18 March 2010 (UTC)

I tried Google maps. I'm looking for an estimate, really. Any part of Colorado, USA to any San Jose in Bolivia. My roommate and I watched an episode of South Park where the kids drove to Bolivia ("Getting Gay with Kids") and wondered if a drive like this is even possible. If so, how long would it take. --67.134.239.205 (talk) 15:14, 18 March 2010 (UTC)

Just a "back of the envelope calculation": It looks to be about 6000 miles by bus, which, at an average of 30 MPH, would take 200 hours, or about 8 days. Of course, keeping the bus moving most of the time would require driving in shifts. Note that this doesn't include customs stops at the various borders. StuRat (talk) 16:38, 18 March 2010 (UTC)

You're forgetting the (enormous) difficulties of getting a bus across the Darién Gap. Algebraist 17:37, 18 March 2010 (UTC)

Good point, that's probably why Google declined to give a time and distance estimate. StuRat (talk) 03:45, 19 March 2010 (UTC)

No, that's not why. Google Maps must only cover a certain area. I tried to get directions from Denver, Colarado to México City, México and it couldn't do it. It couldn't give directions from México City to Acapulco either. The Mexican Google Maps doesn't even have a cómo llegar (Get Directions) option. •• Fly by Night (talk) 13:59, 19 March 2010 (UTC)

Rank-Nullity

Hi. I am currently revising a course on Vectors and Matrices and am a bit confused by part of my notes. I have been given the example of a linear map T(x,y) such that $T(x,y)=(x+y,x,y-3x,y)$ and it then goes on to say that the rank of T(x) is 2 and that the nullity is 0, which would work by the Rank-Nullity Theorem. My problem is that I can't see how the rank is 2. It actually says in the question that it is a map from R² to R⁴ so surely the rank is 4. But if so, how does this fit with the Rank Nullity? Thanks. 92.11.208.237 (talk) 20:12, 18 March 2010 (UTC)

See if you can write T as a matrix. It takes a vector in R² to a vector in R⁴, so it should have 2 columns and 4 rows. Then find the rank and nullity of that matrix. (The rank really is 2 and nullity really is 0.) Staecker (talk) 20:36, 18 March 2010 (UTC)

Think of the image in 4-space of R² under that transformation, or if it's easier to visualize, think of a comparable 3-d transformation instead of a 4-d one. What does that image look like? What is its dimensionality? Also, what does T's kernel (the set of vectors that T maps to zero) look like? Remember that the rank is just the dimension of T's image, and the nullity is just the dimension of T's kernel. 66.127.52.47 (talk) 20:47, 18 March 2010 (UTC)

A linear map to Rⁿ has rank n if and only if it is surjective. Your map isn't. --Tango (talk) 21:28, 18 March 2010 (UTC)

The rank cannot be bigger than the dimension of the domain. The fact that it maps INTO a 4-dimensional space does not mean that it maps ONTO a 4-dimensional space. And it doesn't, so the rank is not 4. Your words "surely the rank is 4" prove that you don't understand what rank is. The rank is the dimension of the image space, not the dimension of the target space. For example, consider the map that takes every point in the 2-dimensional space to (0,0,0,0) in the 4-dimensional space. Its rank is 0, not 4. Michael Hardy (talk) 23:13, 18 March 2010 (UTC)

The set of all (x,y) such that T(x,y) is the zero vector in R⁴, is precisely the kernel of T. Since T(x,y)=(0,0,0,0) if and only if (x+y,x,y−3x,y)=(0,0,0,0), it follows that only the zero vector of R² is in the kernel of T ((x+y,x,y−3x,y)=(0,0,0,0) implies that the second and fourth coordinates of (x+y,x,y−3x,y) are zero, in particular, and that therefore x=0 and y=0). From this we conclude that the kernel of T is trivial; that is, the map T is injective. By the rank-nullity theorem, r(T)+n(T)=dim(R²), where r(T) and n(T) are the rank and nullity of T respectively. Since the nullity of T is 0, as we just observed, the rank of T must be 2.

Another way to determine the dimension of the image of T (i.e. the rank of T), is to determine the images of all elements in a set of basis vectors for R² under T (any set of basis vectors will do, but for simplicity's sake, we will choose the set of standard basis vectors of R²). Note that T(1,0)=(1,1,−3,0) and that T(0,1)=(1,0,1,1). Can you see that T(1,0) and T(0,1) are linearly independent? (Suppose k₁T(1,0)+k₂T(0,1)=(0,0,0,0) for some $k_{1},k_{2}\in \mathbb {R}$. Then we have that (0,0,0,0)=k₁T(1,0)+k₂T(0,1)=k₁(1,1,-3,0)+k₂(1,0,1,1)=(k₁,k₁,−3k₁,0)+(k₂,0,k₂,k₂)=(k₁+k₂,k₁,k₂−3k₁,k₂). In particular, k₁=k₂=0 and linear independence is established. Alternatively, observe that two vectors in Euclidean space are linearly independent if and only if one is not a scalar multiple of the other, and that this is clearly the case for T(1,0) and T(0,1)) However, if T maps linearly independent vectors to linearly independent vectors, the dimension of its image should be the dimension of its domain, and as such, T has rank 2. Without necessarily applying the rank-nullity theorem, the fact that T has nullity 0 also follows from this fact (if T maps linearly independent vectors to linearly independent vectors, its kernel must be trivial, and in particular, its nullity must be 0).

Hope this helps. PST 01:10, 19 March 2010 (UTC)