Wikipedia:Reference desk/Archives/Mathematics/2008 May 6

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May 6

"Continuity" of integrals of L¹ functions

I'm working on a problem, and I've almost got it, except this one little lemma which seems to me ought to be true:

Let $f\in L^{1}$. For any $\epsilon >0$ there exists some $\delta >0$ such that if $\mu (E)<\delta$ then $\int _{E}|f|<\epsilon$.

Is this true or false? If it's false, does it become true if we assume we're on a finite measure space? —Bkell (talk) 04:33, 6 May 2008 (UTC)

Funny how sometimes just taking the time to write out what it is you want to prove makes something click. The lemma is true. Let $\phi =\sum a_{j}\chi _{E_{j}}$, with the $E_{j}$ disjoint, be an integrable simple function such that $\int |f-\phi |<\epsilon /2$. Let $a=\max\{|a_{j}|\}$. Take $\delta =\epsilon /2a$. Then if $\mu (E)<\delta$ we have

$\int _{E}|f|\leq \int _{E}|\phi |+\int _{E}|f-\phi |\leq a\mu (E)+\int |f-\phi |<a{\frac {\epsilon }{2a}}+{\frac {\epsilon }{2}}=\epsilon$.

—Bkell (talk) 04:46, 6 May 2008 (UTC)

Intimate with integrals

I'm nearing a certain understanding of integrals. Looking at http://en.wikipedia.org/wiki/Image:Integral_approximations.svg, I thought time was ripe for a question. More of a curiousity on my part, really; the picture shows the difference between 5 and 12 approximations for sqrt(x). Is there anything special about approximations with numbers that, when drawn on such a graph, has the line cross the middle of the approximation rectangles - and doing so for all other rectangles? 213.161.190.228 (talk) 10:41, 6 May 2008 (UTC) (edited afterwards, typo)

In order to have a sum that converges to the integral, it actually doesn’t matter where the function crosses the rectangles – the points could actually be chosen randomly, so long as the width of the rectangles goes to zero.

For a numerical approximation though, crossing at the middle of the rectangle is in general pretty good, this is called midpoint rule. See Riemann sum for that. I say it’s “pretty good” because given some simple conditions I forget at the moment, it has convergence to the integral at the same rate as the trapezoidal rule, up to a constant scalar (and rectangles are easier to compute than trapezoids).

Simpson's rule, however has faster convergence. The significant difference between these sums being not where the rectangles (or other shapes) and function meet, but by how many points they meet. The midpoint rule meets at a single point, and so is flat on top. The trapezoidal meets at two points so makes a line. Simpson’s rule meets at three points so makes a parabola. You could keep going if you wanted to.GromXXVII (talk) 11:12, 6 May 2008 (UTC)

In the illustration the difference between the 5-rectangle and the 12-rectangle approximation is more subtle than just the fact that one has more and thinner rectangles than the other. The top of each yellow rectangle in the 5-rectangle approximation lies above the curve - the height of each yellow rectangle is the maximum value of the function within the width of the rectangle (because the illustrated function is monotonic, this maximum point is always at right hand corner of the rectangle, but that need not be true in the general case). This approximation is an example of an upper Darboux sum.

In the 12-rectangle approximation, on the other hand, the top of each green rectangle lies below the curve, because the height of each green rectangle is the minimum value of the function within the width of the rectangle. This is an example of a lower Darboux sum. We could calculate an upper Darboux sum for the 12-rectangle case by making the green rectangles taller so that their tops were above the curve - the 12-rectangle upper Darboux sum would be greater than the 12-rectangle lower Darboux sum, but less than the 5-rectangle upper Darboux sum.

If the function is Darboux integrable (which well-behaved functions are) then the upper Darboux sum and the lower Darboux sum will get closer and closer together as we make the rectangles thinner and thinner, and both sums will converge to a single value, which we call the integral of the function (if we were being very precise, we would say the Darboux integral, but the distinction between different definitions of integral and integrability is not important in elementary calculus).

Incidentally, a Riemann sum is calculated by setting the height of each rectangle to be the value of the function at an arbitrary point within the width of each rectangle - it is not necessarily the mid-point. Gandalf61 (talk) 11:28, 6 May 2008 (UTC)

Similar matrices

I have a conceptual doubt on n by n matrices: Consider the collection of all equivalence classes of similar matrices and the collection of all equivalence classes of similar linear operators. Suppose I choose any 1 equivalence class say of a matrix A. Suppose A is the matrix representation of some linear operator T in any basis (Every matrix is bound to be the rep of some linear operator for any basis). Now is every matrix similar to A the matrix representation of every operator similar to T? In other words can we say that there is no actual difference between the members of the class of T and A. Can we make a 1-1 function between the collection of equivalence classes in this fashion? If not how can we relate the similar operators and the the matrices? Much appreciated.--Shahab (talk) 17:39, 6 May 2008 (UTC)

The answer is yes. The proof can vary depending on how exactly you have defined everything, but should be pretty straightforward either way.

It can help to think of $\mathbb {R} ^{n\times n}$ (the set of all $n\times n$ matrices over $\mathbb {R}$; you can choose any other field instead) and of $\mathrm {Hom} (\mathbb {R} ^{n})$ (the set of linear transformations from $\mathbb {R} ^{n}$ to itself) as rings (or even just semigroups). These two rings are isomorphic to one another. Equivalence classes with respect to similarity can be expressed using nothing but the ring operations, thus they have the same structure in both. -- Meni Rosenfeld (talk) 18:21, 6 May 2008 (UTC)

Thank you.--Shahab (talk) 18:32, 6 May 2008 (UTC)

Equation

How do you solve the following equation?

$-x^{2}*\ln(x-1)+3x+x*\ln(x-1)+2=0\,$

I want to calculate where $y=4-kx\,$ touches $y=1/(x-1)+1\,$ (an indifference curve). Jacob Lundberg (talk) 18:47, 6 May 2008 (UTC)

The equation cannot be solved symbolically. You can solve it numerically using Newton's method or something similar. The solution is 4.09131... .

I'm not sure what it has to do with the second question, though. For $k=1$, the curves touch at $x=2$. -- Meni Rosenfeld (talk) 19:01, 6 May 2008 (UTC)

I want y = y and y' = y' for the equations above.

I then got this:

y = 4 - kx
y' = -k

y = 1/(x - 1) + 1
y' = ln (x - 1)

I somehow got the first equation, but I have probably made a mistake somewhere. Jacob Lundberg (talk) 19:06, 6 May 2008 (UTC)

ln is the integral of 1/x, not its derivative. The derivative of 1/x is ${\frac {-1}{x^{2}}}$. -- Meni Rosenfeld (talk) 19:35, 6 May 2008 (UTC)

For the second one why not just set the two equations equal to each other, they’re both solved for y. GromXXVII (talk) 23:16, 6 May 2008 (UTC)

I have to concur - why not just solve for when $4-kx=1/(x-1)+1$, which is just a quadratic equation. -mattbuck (Talk) 01:16, 7 May 2008 (UTC)

Just my 2¢: Since Jacob spoke of indifference curves, I wonder if, instead of just one indifference curve, we may speak of a family of contour sets $y={\frac {a}{x-1}}+1$. The points (x,y) that solve simultaneously both y1=y2 and y'1=y'2 would build what is normally known as an offer curve (as the one depicted in this section), and represent the solution to the utility maximization problem
$\max(x-1)(y-1)$
subject to the budget constrain
$p_{x}x+p_{y}y\leq income$, where income=1=p_y, and p_x is the parameter k. As the problem is stated, for each k we have two equations and two variables: consumption of y and utility level a. Pallida  Mors 04:45, 7 May 2008 (UTC)

Another possible interpretation of the question: k is an unknown parameter, and the problem is to determine the value of k for which the line is tangent to the curve. One way of solving this is to require that both the y-values and the y'-values coincide. Another approach, slightly simpler in this case, is solving the equation that equates the y-values for x and to require that two of the roots (and in this case there are at most two) coincide. For this case you get (after simplification) a quadratic equation, and then this means you have to find the value of k for which the discriminant of the quadratic form vanishes. This happens in fact for two values of k, both of which are integers, so there are two tangent lines through the point (4,0).  --Lambiam 15:12, 7 May 2008 (UTC)

Was there another interpretation for the question? The OP was very specific about wanting the curves to touch, and the derivatives to be equal at the point of contact. -- Meni Rosenfeld (talk) 16:00, 7 May 2008 (UTC)

I think what Lambian meant was that, instead of adding a variable a and finding for each value of the parameter k which indifference curve touches and is tangent to the corresponding line, why just not take the indifference curve the poster mentioned and find for which value of k the line touches and is tangent to the fixed indifference curve? Both approaches make use of the equations y1=y2 and y'1=y'2, but Lambian's approach is pertinent to the data given by Jacob. Pallida  Mors 16:49, 7 May 2008 (UTC)

Kuritosis

In my statistics class, we were told that in hypothesis testing, if n was too small, we should make a box plot to show that the distribution is approximately normal and the skewness and kurtosis aren't too large. First off, how can you even tell kurtosis with a box plot to any reliable degree? Second, why not just use a modified normality tests that shows how normal the distribution is, and thus if it's close enough to normal? — DanielLC 23:59, 6 May 2008 (UTC)

Indeed -- you can't reliably estimate the kurtosis from the box plot. A quick look at the degrees of freedom gives us an intuition for your thoughts: the box plot gives us five numbers (min, lower quartile, median, upper quartile, max); if we ignore the minimum and maximum, then the box plot provides three constraints. The three most "obvious" degrees of freedom for our unknown distribution are mean, variance, and skew, so we don't have enough information to constrain kurtosis. The minimum and maximum are of course very sensitive to outliers, but barring that, a distribution with a larger kurtosis will have observed minimum and maximum more distant from the median relative to how distant the lower/upper quartiles are from the median. So if you had equal-sized samples from two distributions with different mean, variance, skew, and kurtosis, then their respective box plots might help indicate the relative values of these four characteristics, including kurtosis: scale the box plots to have the same value for (upper quartile - lower quartile), then the box plot with the larger range (i.e., maximum - minimum) may correspond to the distribution with larger kurtosis, if you're lucky.

But if you wish to test your distribution for normality, it would of course be much more useful (though rather more work) to use one of the normality tests, as you suggested, than to stare and muse at box plots. With a small number of samples, some of these tests become quite erratic, but still more useful than a box plot. Eric 144.32.89.104 (talk) 17:07, 7 May 2008 (UTC)