Wikipedia:Reference desk/Archives/Mathematics/2006 November 7

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November 7

algebra 2 word problem

the sum of two digits is nine. 12 times the ten digit equals the number. —The preceding unsigned comment was added by 68.37.97.86 (talk • contribs) .

Assuming the number has two digits (this is important), try representing the tens digit as x and the ones digit as y, and the number as n. Thus 10x+y=n. First, why is the equation I wrote correct? (Try it with some numbers if it is not clear). Next, can you represent the two statements you made with algebraic equations similar to the one I wrote? Try solving those equations for n. Best wishes, --TeaDrinker 01:02, 7 November 2006 (UTC)

Why not use better symbols, like T for Tens and S for Singles ? (I wouldn't use O for Ones, since it looks like a zero). StuRat 05:56, 7 November 2006 (UTC)

Surely we are meant to assume a two digit number. The first statement implies the number is a multiple of 9. The second statement implies it is also a multiple of 4. This leaves only two possible candidates for the number, only one of which works. The suggestion of TeaDrinker is a more algebraic way of thinking, and so is likely what the problem intends. --KSmrq^T 03:27, 7 November 2006 (UTC)

To eliminate the other candidate, note that the two conditions imply that the ten digit must be divisible by 3.--gwaihir 09:49, 7 November 2006 (UTC)

Drag coefficient

This is more of a physics question, but since it pertains to formulas, I will list it here. How can the drag coefficient be found for objects? The drag coefficient page only lists values for $C_{d}$ for certain objects, and the formula at the top doesn't seem to help. Thanks. (Note: If this is in the wrong place, please move it.) PullToOpen ^talk 04:28, 7 November 2006 (UTC)

You need to use a wind tunnel, where the amount of drag is measured precisely relative to the wind speed, size, and shape of the object. You might also try a Google search, as the particular material you're interested in may already have a known value. StuRat 05:49, 7 November 2006 (UTC)

top X random, with weights

How can I get a list of X random entries out of a larger total, where each entry in the list has a different weight?

Eg: ad banners. I have space for 5 ads, but have sold 20. Each ad is given a weight, and I'd like the total number of views for each ad to match their weight.

My first idea was: randomly pick 1 entry from the list, using the weights. Remove that entry from the list, then pick the next one, using the new weights in the now shorter list.

But I'm concerned that this is not actually correct - because now the weights in the shorter list don't take into account the weight from the removed entry. Ariel. 05:19, 7 November 2006 (UTC)

Most random number generators return a number between 0 and 1. So, add all the weights up, and divide each weight by that number to get the "unitized weights". Here's an example:

Weight A =  2
Weight B =  1
Weight C =  5
Weight D =  2
           --
Total    = 10

Unitized Weight A =  2/10 = 0.2
Unitized Weight B =  1/10 = 0.1
Unitized Weight C =  5/10 = 0.5
Unitized Weight D =  2/10 = 0.2

Range A = 0.0 - 0.2
Range B = 0.2 - 0.3
Range C = 0.3 - 0.8
Range D = 0.8 - 1.0

StuRat 05:46, 7 November 2006 (UTC)

Thanks for answering, but that's for getting a single random entry - what if I want to pick 2 of them out of your list of 4? But, which 2 I pick is random, but weighted. Ariel. 06:08, 7 November 2006 (UTC)

I don't see the problem. Use the random number generator once, if it gives you 0.753, then run ad C. Then use the random number generator again, if it gives you 0.332, then run ad C again. It will eventually work out very close to the desired weight for each ad. Are you saying you need it to work out EXACTLY even over some period ? If so, then readjust the weighting each time. StuRat 22:58, 7 November 2006 (UTC)

Don't forget to add the restriction that the same ad cannot be show simultaneous on different ad spots. 211.28.178.86 10:05, 7 November 2006 (UTC)

I'm not sure I understand the problem. Are you repeatedly presenting N = 5 out of 20 ads, so that in the long run ad_i is presented proportionally often to given weight w_i? Assuming the weights have been normalized to sum to 1, that means ad_i is presented on the average N·w_i times per round, or, in a total of R rounds, R·N·w_i times. You could call this the due number. A possible approach is to keep track of the difference between the due number and the actual number of presentations, and to select those with the largest deficit first. Here is an algorithm in pseudocode:

• Initialize the deficit d_i := 0.5 for all ads (a somewhat arbitrary choice);
• Repeat, for each round:

• Increment d_i := d_i + N·w_i for i ranging over all ad numbers;
• Set the ad selection for this round to empty;
• For s = 1, ..., N:

• Find the value of i maximizing d_i (in case of a tie, pick any);
• Add ad_i to this round's selection;
• Decrement d_i := d_i − 1.

This can lead to an ad being selected more than once in a round, but only if N·w_i > 1. In that case, not showing it more than once every now and then leads to an ever-increasing deficit.  --Lambiam^Talk 10:27, 7 November 2006 (UTC)

Expanding fractional indices

Is it possible to expand bracketed expressions with fractional indices? How is it done? If not, are there any specific cases where it can be done? --Awesome 09:15, 7 November 2006 (UTC)

Yes you can, see Binomial theorem, although for non-integer exponents you get an infinite expansion. This is also related to certain aspects of Taylor series. Confusing Manifestation 09:28, 7 November 2006 (UTC)

Of course, there are also "expansions" like $(a+b)^{3/2}=(a+b)(a+b)^{1/2}=a(a+b)^{1/2}+b(a+b)^{1/2}$.--gwaihir 09:46, 7 November 2006 (UTC)

By "index", do you mean exponent?  --Lambiam^Talk 10:30, 7 November 2006 (UTC)

That's what I assumed. It's a term sometimes used, especially (IIRC) when first learning about "index rules" (like a^(b+c) = a^b a^c). Confusing Manifestation 10:52, 7 November 2006 (UTC)

College Algebra

I am needing help with my homework on quadratic equations and formulas. I cant seem to understand how to do them.I am kind of needing step by step help. If someone can help i would appreciate it very much.

The first step would be to read Quadratic equation and tell us if there is anything there you don't understand, or if you have an exercise which it doesn't help you solve. -- Meni Rosenfeld (talk) 17:27, 7 November 2006 (UTC)

If you get stuck on a specific problem and need the theory to solve it, you could ask that (as compared to asking for the answer). Again, I'm differentiating that from asking a homework question. Ask a theory question. --AstoVidatu 20:46, 7 November 2006 (UTC)

Show us what the maths problem is. Show us what you have done so far. Then show us the location where you have got stuck. We shall explain where you went off the track and suggest a way forward for you. 202.168.50.40 20:58, 7 November 2006 (UTC)

A question related to the polynomial remainder theorem

If you divide a polynomial by (x – a) and get a remainder b, and you divide the same polynomial by (x – c) and get a remainder d, what happens when you divide the same polynomial by (x – a)(x – c)? Is the remainder somehow related to b and d?

Your conditions essentially say that $p(a)=b$ and $p(c)=d$, and the remainder must be a polynomial of degree at most 1 satisfying these same conditions.--gwaihir 00:39, 8 November 2006 (UTC)

To spell this out, and assuming a ≠ c, the remainder upon division by (x – a)(x – c) is then:

${\frac {b-d}{a-c}}x+{\frac {ad-bc}{a-c}}.$

If you have all integer coefficients, these fractions are also integers.  --Lambiam^Talk 08:14, 8 November 2006 (UTC)

A math question

This is outside of school... Anyway, here it is:

Find all integers k such that k is between -8 and 8 inclusive for which x²+kx+2k+4 is factorable over the integers. For this one, can anyone think of a faster way to do it than just plugging them in and solving? --AstoVidatu 23:39, 7 November 2006 (UTC)

Hint: first solving, then plugging in?--gwaihir 01:06, 8 November 2006 (UTC)

The first sentence of the following may be the same hint gwaihir has given. Factorability of this quadratic form requires that its discriminant k²−8k−16 be a square, which is somewhat easier to test. Considerations of modular arithmetic show that k is then of the form 3n+1, and also of one of the two forms 5m and 5m+3. So, taken together, k mod 15 is either 10 or 13. In the given range that leaves only two possibilities, both of which actually give a solution.  --Lambiam^Talk 09:46, 8 November 2006 (UTC)

Yeah, I had forgotten about the quadratic equation when I wrote this. That's the piece I was missing. Thanks for the help Lambian. --AstoVidatu 11:48, 8 November 2006 (UTC)

Just had a thought - dont know why.

What is the infinith root of infinity?--Light current 23:38, 7 November 2006 (UTC)

It would be infinity unless I'm missing something. The root of infinity is infinity. And the root of that is infinity. And the root of that is... (I think you get the picture) ;) --AstoVidatu 23:41, 7 November 2006 (UTC)

The infinitith root of infinity would have to mean

$\infty ^{\frac {1}{\infty }}$

One way to make sense of this question would be to say what if we had some functions g(x) and h(x) which both go to infinity as x increases, and we ask about the behavior of the function defined by

$f(x)=g(x)^{\frac {1}{h(x)}}$

In that case, we could take logarithms and note that

$\ln f(x)={\frac {\ln g(x)}{h(x)}}$

This function, in its limit as x increases, could equal anything from zero to infinity, depending on the relative rates of increase of $\ln g(x)$ and $h(x)$. This would make f(x) equal anything from 1 to infinity. So, in a sense, the answer to your question is "anything you want it to be, as long as it's at least 1". -GTBacchus^(talk) 00:03, 8 November 2006 (UTC)

For example,

$\lim _{n\rightarrow \infty }(n)^{\frac {1}{n}}=1$

If I may interject, it's interesting (if technically unrelated) that $\max _{x\in (0,\infty )}x^{1/x}$ (or ${\sqrt[{x}]{x}}$) occurs for $x=e$. --Tardis 18:49, 8 November 2006 (UTC)

but

$\lim _{n\rightarrow \infty }(2^{n})^{\frac {1}{n}}=2$

and

$\lim _{n\rightarrow \infty }(n)^{\frac {1}{\ln(n)}}=e$.

Gandalf61 00:32, 8 November 2006 (UTC)

Nice ones, Gandalf. One more perhaps...

$\lim _{n\rightarrow \infty }(e^{n})^{\frac {1}{\sqrt {n}}}=\infty$

-GTBacchus^(talk) 00:42, 8 November 2006 (UTC)

Thanks to all. My math knowledge is now increased by $\infty ^{\frac {1}{\infty }}$ percent. 8-)--Light current 02:04, 8 November 2006 (UTC)

This is an example of what is called an indeterminate form.  --Lambiam^Talk 07:45, 8 November 2006 (UTC)