Wikipedia:Reference desk/Archives/Mathematics/2008 February 23

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[edit] February 23

[edit] Finding basic limits

How does one find this limit:

\lim_{x \to 9} \frac{x-9}{\sqrt{x}-3}

When I enter it into my graphing calculator, I can trace the rational curve and the answer should be somewhere between 5.8 and 6.1. But how do I find the exact value? Thanks. Acceptable (talk) 03:12, 23 February 2008 (UTC)

Hint: (a − b)(a + b) = a2 − b2. PrimeHunter (talk) 03:19, 23 February 2008 (UTC)
\lim_{x \to 9} \frac{x-9}{\sqrt{x}-3}

=\lim_{x \to 9} \frac{(\sqrt{x}-3)(\sqrt{x}+3)}{\sqrt{x}-3}
=\lim_{x \to 9} \sqrt{x}+3
=\sqrt{9}+3
= 6
Visit me at Ftbhrygvn (Talk|Contribs|Log|Userboxes) 15:15, 23 February 2008 (UTC)

Ftbhrygvn, your desire to help is appreciated, but we prefer not to give full solutions to what may be homework problems. -- Meni Rosenfeld (talk) 15:24, 23 February 2008 (UTC)
An alternative and lazier way to solve this problem is to apply l'Hôpital's rule. Differentiate the numerator and the denominator, so that you get \lim_{x \to 9} 2\sqrt{x}, substitute x for 9 and you get 6. --Taraborn (talk) 18:17, 23 February 2008 (UTC)

[edit] question about functions

can we say that any known function,f(x)=y,satisfies the condition,limit[f(x)\x]→1,when x,either approches to infinity or to afixed value,x0?Husseinshimaljasimdini (talk) 11:44, 23 February 2008 (UTC)

I'm not sure I understand the question. Are you asking if there exists such a function? If so, then yes, f(x)=x has that property. If you're asking if all functions have that property, then no, take f(x)=0, for example. --Tango (talk) 13:58, 23 February 2008 (UTC).

no sir.I mean are the all known functions,f(x)=y.I mean all of the functions,satisfy the above properties?in another word.is there any function doesnot satisfies the above properties?Husseinshimaljasimdini (talk) 10:08, 24 February 2008 (UTC)

Yes, as Tango pointed out, f(x) = 0 does not satisfy your condition. There are infinitely many other examples just using f(x) = c where c is a constant. Using y as a variable has no real meaning in this, the properties are fully determined by f(x). -mattbuck (Talk) 10:22, 24 February 2008 (UTC).
OK OK IGOT IT.ok sir,i got it.now can we say that all the functions except f(x)=0,satisfy the above properties?and one more thing about f(x)=0,dont you think that lim[f(x)\g(x)],when both f(x)&g(x)=0,then the limit has aknown value when x→some point.namely,LOHOPITAL RULE.Husseinshimaljasimdini (talk) 10:28, 24 February 2008 (UTC).thank you indeed.thank all of you guys very much.Husseinshimaljasimdini (talk) 11:00, 24 February 2008 (UTC)
\lim_{x\rightarrow 0}\frac{0}{x}=0, you can check that by l'Hopital or just by observing that \frac{0}{x}=0,\,\forall x\ne 0. If you want another example, try f(x)=-x, that also doesn't satisfy your property for any values of x (including infinity). You might like to check out fixed point theorem, it's closely related to what you ask, although not exactly the same. If a function has a fixed point other than 0, your property will be satisfied. --Tango (talk) 12:17, 24 February 2008 (UTC)

[edit] alittle bit of help

i read here about that figure which has an infinite surface but afinite volume though.can i get some help to locate the article?Husseinshimaljasimdini (talk) 11:49, 23 February 2008 (UTC)

An example of such a surface is Gabriel's Horn. -- Xedi (talk) 13:50, 23 February 2008 (UTC)
A standard example of the lower dimensional case of a finite area and infinite perimeter is any landmass - assuming the coastline is effectively random at every scale, if you measure it with a smaller and smaller ruler you'll get a larger and larger answer (because you take into account smaller and smaller deviations from a straight line) and in the limit, the answer is infinite. --Tango (talk) 13:56, 23 February 2008 (UTC)
That's a good point, there's more than one way to get it. An abstract example of what Tango said is the Koch snowflake. Similar fractals can be defined in three dimensions, giving something with infinite surface and not only finite volume but bounded extent. Black Carrot (talk) 19:18, 24 February 2008 (UTC)

[edit] Help needed in understanding 1925 Biometrika paper

I'm reading the paper Tippet, LHC. On the Extreme Individuals and the Range of Samples Taken From a Normal Population. Biometrika vol 17, 364-387, 1925., temporarily available here.

The paper defines the function

\overline{w}(n)=\int_{-\infty}^{+\infty} (1-(1-\alpha)^n-\alpha^n) \, dx ,

where \, \alpha_p initially was defined as \int_{-\infty}^{x_p}\phi(x)dx, where \int_{-\infty}^{+\infty}\phi(x)dx = 1, i.e. it is a cumulative distribution function, and n is an integer greater than one. The function returns the expected value of the range of a sample of size n, taken from the probability distribution defined by \, \phi(x) . If \, \phi(x)=\Phi(x) is the standard normal distribution, w is known as the control chart constant d2 in the field of statistical process control.

I posted a related question recently about this function. I have found that the function

d_2(n)=\int_{-\infty}^{+\infty} (1-(1-\Phi(x))^n-\Phi(x)^n) \, dx

is easily evaluated numerically, using a naive brute force approach such as the following C++ implementation:

double d2(int n)
{
    double x, dx;
    dx = 0.01;
    double sum = 0.0;
    for (x = -12; x <= 12; x = x + dx)
    {
        double alpha = cumulative_normal_distribution(x);
        sum = sum + (1 - pow(1-alpha,n) - pow(alpha,n))*dx;
    }
    return sum;
}

I know that there are far more efficient and elegant ways of evaluating integrals numerically. Nevertheless, this implementation works, and mirrors my simple-minded mental image of what is going on, so I'll stick with it for now. Tippett then defines as equation (10) the following:

\, \mu_2 = \sigma^2 = 2 \int_{-\infty}^{+\infty} \int_{-\infty}^{x_1}(1 - \alpha_1^n - (1 - \alpha_n)^n - (\alpha_1 - \alpha_n)^n)dx_1 dx_n - \overline{w}^2

and states (page 372) that

"The second moment, and hence the standard deviation of the distribution, was determined in several cases by equation (10). The work is very laborious, as it involves cubature, and even so, the result can only be given to a few figures. It is believed that the values given in Table IV are correct to the last figure."

The return values square root of \, \mu_2 should correspond to the control chart constant d3, and Table IV confirms that this is indeed the case. My problem is that I haven't been able to evaluate \, \mu_2 , and I suspect I've misunderstood something, possibly something pretty elementary about double integrals (I have no mathematical training beyond this, and it's a long time ago). Here's my implementation, again using a similar brute force approach. I've highlighted in red the part that I was most unsure of:

//NOTE: Comments in green added by OP after the question was posted, based on Meni Rosenfeld's answers.

double d3(int n)
{ 
    double x_1, x_n, dx_1, dx_n;
    dx_1 = dx_n = 0.01;
    double sum = 0.0;
    for (x_n = -12; x_n <= 12; x_n = x_n + dx_n)
    {
        double alpha_n = cumulative_normal_distribution(x_n); // WRONG - should be doing alpha_1 in the outer loop!
        // Making x_1 (-inf ... +inf) the outer loop, and x_n (-inf ... x_1) the inner loop fixes the problem.
        for (x_1 = -12; x_1 <= x_n; x_1 = x_1 + dx_1) // WRONG - see the code in red: x_1 should be the limit of integration. 
        {
            double alpha_1 = cumulative_normal_distribution(x_1); // WRONG - should do alpha_n in the inner loop!
            sum = sum + (1 - pow(alpha_1,n) - pow(1 - alpha_n,n) - pow(alpha_1 - alpha_n, n))*dx_1*dx_n;
                    // As Meni pointed out there's an error here ^ . The minus sign should be plus.
        }
    }
    double variance = 2*sum - pow(d2(n),2);
    return sqrt(variance);
}


The return values of this function are way off the correct values. My question is: can someone spot the misunderstanding(s) in my interpretation of the paper, the maths, and/or the error(s) in my implementation? Your help would be much appreciated. Thanks in advance, --NorwegianBlue talk 16:13, 23 February 2008 (UTC)

I don't see a problem with the way you are calculating the double integral. However, there is something terribly wrong about the function you are trying to integrate. To explain, I'll introduce some notation:
a(n,x_1,x_n)=1-\Phi(x_1)^n-(1-\Phi(x_n))^n-(\Phi(x_1)-\Phi(x_n))^n\;\!
b(n,x_1)=\int_{-\infty}^{x_1}a(n,x_1,x_n)\ dx_n
c(n)=\int_{-\infty}^{\infty}b(n,x_1)\ dx_1
d_3(n)=\sqrt{2c(n)-d_2(n)^2}
In order for any of this to be meaningful, b must exist for any x1. For this to happen, at the very least we must have \lim_{x_n \to - \infty}a(n,x_1,x_n)=0 for every x1. This is clearly not the case, because \Phi(-\infty)=0 and so a(n,x_1,-\infty)=1-2\Phi(x_1)^n. Thus the function is not integrable. Check that you have copied the equations exactly as they appear in the paper, and that the notation means exactly what you think it does; If so, there is possibly a mistake in the paper. -- Meni Rosenfeld (talk) 12:10, 24 February 2008 (UTC)
Hmm. Well, I've checked the equations, and believe my reproduction is correct. I also very much doubt that there is a mistake in the paper, as it was widely cited, and equation (10) was used by subsequent authors to tabulate d3. Therefore, the most likely explanation is that I've misunderstood the notation. I prepared a small .PNG file from the paper with the relevant equations and a figure which explains the notation. x1 refers to the highest value in a sample of n values, and xn refers to the lowest, as is shown in the figure. The .PNG is here, the paper is here. Thanks again! --NorwegianBlue talk 13:20, 24 February 2008 (UTC)
Note the step from equation (9) to (10), where +(\alpha_1-\alpha_n)^n\;\! turns into -(\alpha_1-\alpha_n)^n\;\!. The plus is correct, and everything falls into place nicely if you make this correction. So it turns out even widely cited papers can have errors. Also, the table you linked to has some errors, for example d_2(1)=0\neq 1. -- Meni Rosenfeld (talk) 14:36, 24 February 2008 (UTC)
Thanks a lot! The error in the step between equations (9) and (10) was glaringly obvious, once I became aware of it. I've looked through the steps from equation (4) to (9) for more errors, but unfortunately, most of it exceeds my mathematical capabilities. I'm also confused about the notation. What does the S-notation mean? Is appears in places which might suggest summation, is it the same as sigma notation?
Unfortunately, everything did not fall nicely into place. I corrected the minus sign to plus, but the function still does not work correctly. The entire program I've used is here. The section of the paper immediately preceding equation (10) might suggest that equation (10) only works for even values of x. Here's what I found, with the correction applied:
            n   program   correct
            ---------------------
            2    23.94    0.8525
            3     8.65    0.8884
            4    23.90    0.8794
            5    10.01    0.8641
            6    23.86    0.8480
            7    10.70    0.8332
            8    23.82    0.8198
            9    11.15    0.8078
Even values slowly decreasing from 23.84, odd values increasing from 8.65. Strange. A standard deviation of >23 for the range of samples of n=2 drawn from a standard normal distribution is obviously not right! Any suggestions of what else might be wrong? --NorwegianBlue talk 16:58, 24 February 2008 (UTC)
Well, there aren't any problems with the mathematics. I have calculated the sum in Mathematica and it gives the correct result. A step of 0.01 is enough to get a few correct digits, and an integration bound of 12 is overkill - 5 will be more than enough. So you need to debug your code - it looks generally okay but I don't speak C++ fluently. Try to define intermediate functions as I have done above, calculate some values of them, and either check them for plausibility or put them here so I can have a look.
The S notation is unfamiliar to me - it is probably, indeed, an archaic version of the Sigma notation. -- Meni Rosenfeld (talk) 17:40, 24 February 2008 (UTC)
Or start throwing cout to test intermediate values of variables in between every single line. x42bn6 Talk Mess 19:49, 24 February 2008 (UTC)
Thanks a million, Meni!!! I implemented your a, b, c and d3 functions exactly as you wrote them, and the result was reasonably close to the correct one (overestimated it slightly). It turned out that the step size was important. The smaller I made it, the closer I got to the values listed as correct above. I was able to achieve 3 digits using a step size of 1/2048.
So I went back to the original function, and compared the code, to see why it behaved differently. The problem was related to what I had highlighted in red. When I made x_1 the outer loop, and x_n the inner loop, and changed the test in the inner loop to x_n <= x_1, the modified original function gave the same results as the a, b, c and d3 functions. Thanks again, this was beginning to drive me nuts! By the way, did Mathematica give results with high accuracy? I'm particularly interested in a good approximation of d3(2). --NorwegianBlue talk 20:23, 24 February 2008 (UTC)
Glad to be of assistance, and sorry for not picking up earlier that it should be x_n \le x_1. Mathematica can give arbitrarily good accuracy, but since this is a double integral it can take some time. d3(2) = 0.8525024664 should be correct up to the digits specified. I'm running it now for more digits in case you'll need it. A few additional values (n=3 and up) are 0.888368, 0.879808, 0.864082, 0.84804, 0.833205, 0.819831, 0.807834, 0.797051. -- Meni Rosenfeld (talk) 22:04, 24 February 2008 (UTC)
Thank you again for taking the time. You have helped me tremendously, both in increasing my understanding of the maths underlying SPC, and in solving a tough practical problem. --NorwegianBlue talk 15:52, 25 February 2008 (UTC)
Gah! I looked at this yesterday, but posted nothing because I didn't see that ordering problem. Shame on the original paper (not the OP!) for writing a double integral as \int_{a_1}^{a_2}\int_{b_1}^{b_2}f(a,b)\,da\,db! Treating a differential (infinitesimal) as a first-class entity is fine, but writing the normal definite integral operation without nesting it properly leads to madness, since the \int doesn't get a variable name attached to it. --Tardis (talk) 16:39, 25 February 2008 (UTC)
As stated above, my mathematical training is limited, so I need to be spoonfed :-). You are saying that my approach, i.e. making the inner integral the inner loop, would normally be the way to go, right?
Would
\, \int_{-\infty}^{+\infty} \int_{-\infty}^{x_1}(1-\Phi(x_1)^n-(1-\Phi(x_n))^n {\color{red}+} (\Phi(x_1)-\Phi(x_n))^n)dx_n dx_1
be a more standard way of writing the integral? I was a bit confused about this, and that's why I flagged the x_1<=x_n test (which should have been x_n<=x_1) in the code. --NorwegianBlue talk 18:54, 25 February 2008 (UTC)
Yes. We are essentially calculating here the integral of a function which is itself an integral. The inner function is \int_{-\infty}^{x_1}(\cdots)\ dx_n, and thus it should appear as inner in the formula and be implemented as inner. We then calculate the integral of this, \int_{-\infty}^{+\infty}(\cdots)\ dx_1, thus the formula should be \int_{-\infty}^{+\infty}\int_{-\infty}^{x_1}(\cdots)\ dx_n\ dx_1. This would be even more critical if we wanted to calculate, say, \int_{-\infty}^{+\infty}\int_{-\infty}^{0}(\cdots)\ dx_n\ dx_1 - we need to know which is the variable that only goes up to 0, and flipping the order changes this. In our integral, we can understand from the context that \int_{-\infty}^{x_1} relates to xn (it wouldn't make much sense for x1 to have itself for an upper integration bound). In the implementation, it doesn't really matter in which order you make the loops, since you can change the order of integration as long as you keep the bounds right. The \int_{-\infty}^{x_1} bit clearly implies that the bound is x_n \le x_1, and it's unfortunate it took us so long to notice that. -- Meni Rosenfeld (talk) 23:14, 25 February 2008 (UTC)
Thanks again, and thanks to Tardis for making me aware of exactly why I tripped. I've learned a lot from this thread! --NorwegianBlue talk 10:05, 26 February 2008 (UTC)

[edit] Mathematics(Algebra)

Railways authorities increased the fare by 6.25%,hence the sale of tickets was decreased by 5%. Even then the income by selling tickets was increased by Rs 15000.Find the original income by selling of tickets?18:07, 23 February 2008 (UTC)117.195.16.73 (talk)

Welcome to the Wikipedia Reference Desk. Your question appears to be a homework question. I apologize if this is a misevaluation, but it is our policy here to not do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn how to solve such problems. Please attempt to solve the problem yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. Thank you. -- Meni Rosenfeld (talk) 18:22, 23 February 2008 (UTC)
Your first step is to formulate the problem in terms of equations. Give each unknown value a symbol (eg the original fare could be Fo and the new fare Fn, etc.), then rewrite all the information in the question as equations using the symbols. Once you've done that, it's just a matter of solving the equations. --Tango (talk) 19:16, 23 February 2008 (UTC)