Wikipedia:Reference desk/Mathematics

Welcome to the mathematics section
of the Wikipedia reference desk.

skip to bottom

Select a section:

• Computing
• Entertainment
• Humanities
• Language
• Mathematics
• Science
• Miscellaneous
• Archives

Want a faster answer?

Main page: Help searching Wikipedia

   

How can I get my question answered?

• Select the section of the desk that best fits the general topic of your question (see the navigation column to the right).
• Post your question to only one section, providing a short header that gives the topic of your question.
• Type '~~~~' (that is, four tilde characters) at the end – this signs and dates your contribution so we know who wrote what and when.
• Don't post personal contact information – it will be removed. Any answers will be provided here.
• Please be as specific as possible, and include all relevant context – the usefulness of answers may depend on the context.
• Note:
  • We don't answer (and may remove) questions that require medical diagnosis or legal advice.
  • We don't answer requests for opinions, predictions or debate.
  • We don't do your homework for you, though we'll help you past the stuck point.
  • We don't conduct original research or provide a free source of ideas, but we'll help you find information you need.

Ready? Ask a new question!

How do I answer a question?

Main page: Wikipedia:Reference desk/Guidelines

• The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.

See also:

• Teahouse
• Help desk
• Village pump
• Help manual

January 16

Could someone check my work please?

"Find the equation of a tangent line to the graph of f(x) = cos x that can be used to approximate the value of cos(π/6 + 0.1). Then, find an approximation of cos(π/6 + 0.1)."

Tangent-line equation: f'(x) = -sin x

f(x₀) ≈ f'(c)(x₀ - c) + f(c)

where x₀ = the number whose f( ) value we're trying to approximate = π/6 + 0.1, and
c = a convenient number close to x₀ on the tangent line = π/6

cos (π/6 + 0.1) ≈ -(sin(π/6))(π/6)(π/6 + 0.1 - π/6) + cos(π/6)
cos (π/6 + 0.1) ≈ -(π/6)(1/2)(0.1) + (√3)/2
cos (π/6 + 0.1) ≈ -π/120 + (√3)/2
cos (π/6 + 0.1) ≈ -(π + 60)/120

But somehow it seems a little weird to have π, though it's a cornerstone of radian notation, in a y-value ... Am I on the right track? Thanks, anon.

cos (π/6 + 0.1) ≈ -(sin(π/6))(π/6)(π/6 + 0.1 - π/6) + cos(π/6)

Where did this extra factor of π/6 come from? Also, how did you manage to get rid of the √3 in the last step? —Keenan Pepper 02:14, 16 January 2008 (UTC)

Think of it this way: the slope of y = cos(x) at x = π/6 is -sin(π/6), which is -1/2. So the tangent line to y = cos(x) at x = π/6 is the line with slope -1/2 that passes through (π/6, cos(π/6)). The point-slope formula tells us that the equation of this line is

$y-\cos \left({\frac {\pi }{6}}\right)=-{\frac {1}{2}}\left(x-{\frac {\pi }{6}}\right)$

$\Rightarrow y=\cos \left({\frac {\pi }{6}}\right)-{\frac {1}{2}}\left(x-{\frac {\pi }{6}}\right)$

Now find the value of y when x = π/6 + 0.1 and you have an approximation to cos(π/6 + 0.1). Gandalf61 (talk) 07:42, 16 January 2008 (UTC)

Thanks Keenan for catching my mistakes and Gandalf for confirming my correction! —anon —Preceding unsigned comment added by 165.155.200.146 (talk) 16:11, 16 January 2008 (UTC)

About Pi

Where does the expression pi origionate from? —Preceding unsigned comment added by 207.224.29.240 (talk) 03:12, 16 January 2008 (UTC)

What do you mean, "the expression pi"? If you mean $\pi$, in its context as the ratio of a circle's circumference to its diameter, check out Pi - where the reason for its use is explained in the lead section. If you mean $\Pi$, in the context of an operator to multiply several things together, see Multiplication#Capital pi notation, and while it doesn't explain why Pi was chosen, it would be to stand for "product" just like the sigma of Sigma notation stands for "sum". For other uses of pi, see Pi (disambiguation). Confusing Manifestation(Say hi!) 03:30, 16 January 2008 (UTC)

Contraction Mapping exam question

In my Metric Spaces examination this morning, the following question was worth 4 marks out of 60:

Suppose that f and (S,p) [are a function and metric space fulfilling the requirements of] Banach's Contraction Mapping Principle and suppose that g:S->S is a function with the property that f(g(x)) = g(f(x)) for all x in S. Show that g has a unique fixed point.

Now, I was easily able to show that g has at least one fixed point (it being the same fixed point as the one f has, as guaranteed by B's CMP). However, I got stuck on the uniqueness part until I realised this:

• If g is the identity on S, f(g(x)) = f(x) = g(f(x)) as required.
• The identity has as many fixed points as there are points in S, because g(x) = i(x) = x by definition.

Unless I'm missing something... was the question wrong? If so, what would ensure g had a unique fixed point?

Rawling4851 12:06, 16 January 2008 (UTC)

To further complicate things, if we choose f to be $f(x)=x_{0}$ and g to be any function such that $g(x_{0})=x_{0}$ (possibly with many other fixed points), then the premises hold but not the conclusion. So it looks like the question is pretty much wrong. However, I think you may be able to prove something if you assume that f is injective and that g is not the identity. -- Meni Rosenfeld (talk) 13:38, 16 January 2008 (UTC)

Cheers. I ran it past our genius postgrad too and he agreed, so we'll see what the lecturer has to say... Rawling4851 00:25, 17 January 2008 (UTC)

Crossing Non-Vertical Asymptotes

In the graph of a rational function, I understand that the vertical asymptotes can never be crossed, and the one possible non-vertical asymptote can be crossed "near" the y-axis.

Is there a limit to number of times the non-vertical asymptote can be crossed?

I'm hoping that there is such a limit and that it is related to the degree of the denominator...or at least the degree of the denominator once all common factors have been eliminated.

Thanks, Stableyr —Preceding unsigned comment added by 66.100.0.42 (talk) 13:12, 16 January 2008 (UTC)

Indeed, assuming the function is not identically equal to its asymptote, the number of crossing points cannot be greater than the degree of the denominator minus 1. -- Meni Rosenfeld (talk) 13:22, 16 January 2008 (UTC)

I believe the cutoff would be either the degree of the numberator minus 1 or the degree of the denominator minus 1, whichever is greater. Take (x^2-3x+2)/x = (x-1)(x-2)/x, for instance. Its asymptote is x. The difference between the curve and the asymptote is (-3x+2)/x, which is zero precisely when x=2/3. Meni's rule would predict no intersection. Black Carrot (talk) 19:31, 16 January 2008 (UTC)

Take the function to be (ax^n + P(x))/(bx^m + Q(x)), where P(x) is a polynomial of degree less than n and Q(x) is a polynomial of degree less than m. Its asymptote will be (a/b)x^(n-m). The difference between the two will be one minus the other, (P(x) - ((a/b)x^(n-m))Q(x))/(bx^m + Q(x)). This will be zero no more often than the numerator is zero, and examples can be produced where the numerator is zero only when the denominator isn't, so we can take that as the upper bound. P(x) - ((a/b)x^(n-m))Q(x) = 0, or bP(x)x^m - aQ(x)x^n = 0, with some fiddly details if x = 0. Breaking it into cases, where m<n, m=n, and m>n, there are no more zeros than respectively n-1, n-1=m-1, and m-1. Black Carrot (talk) 19:40, 16 January 2008 (UTC)

No. The asymptote for your first example is $x-3$, not x. More generally, an asymptote is a curve whose distance from the function tends to 0. In our case we are interested in a polynomial asymptote, so when we calculate the difference all terms from $x^{m}$ and up cancel, and we are left with a polynomial of degree at most $m-1$. -- Meni Rosenfeld (talk) 20:16, 16 January 2008 (UTC)

You're right, I don't know what I was thinking. Black Carrot (talk) 20:22, 16 January 2008 (UTC)

I do. You were thinking about the notion of asymptoticity that only requires the ratio of the functions to tend to 1. Unfortunately, this doesn't fit the context of the question. -- Meni Rosenfeld (talk) 20:33, 16 January 2008 (UTC)

January 17

Two calculus questions

1. I'm trying to find the limit as theta approaches 0 of (1 - cos theta)/(2 sin^2 theta). Direct substitution yields an indeterminate form, so I did the following:

= 1/2 lim as theta approaches 0 of (1 - cos theta)/(1 - cos^2 theta)

Then I did something pretty sketchy involving taking the reciprocal of what I was finding the limit of so that I could get it into a form of the derivative, and then I tried doing something different a second time, but it's hardly worth transcribing all my erroneous work—can anyone point me in the right direction? The right answer is given as 1/4, but I'm having a hell of a time getting there.

2. If f(x) = sin x/2, then there exists a number c in the interval π/2 < x < 3π/2 that satisfies the conclusion for the Mean Value Theorem. I have to find a possible value for c. I tried plugging into the formula f'(c) = (f(b) - f(a))/(b - a), but that just gave me 0, since the numerator = sin 3π/4 - sin π/4 = 0. But apparently the answer is π. I can see that π = 2π/2, the mean of π/2 and 3π/2, but I thought the Mean Value Theorem had more to do with the formula I gave ... again, any help is much appreciated ... thanks, anon. —Preceding unsigned comment added by 70.19.22.49 (talk) 00:19, 17 January 2008 (UTC)

• 1 - note that $\sin ^{2}\theta =1-\cos ^{2}\theta =(1-\cos \theta )(1+\cos \theta )$
• 2 - I'm not entirely sure what exactly it is you want to show. - mattbuck 00:31, 17 January 2008 (UTC)

I think you have to show that there exists a value c between a and b such that f'(c) = (f(b) − f(a))/(b − a).  --Lambiam 01:11, 17 January 2008 (UTC)

That's what we have from the MVT. It appears that here we need to find such a c. As the OP has correctly observed, this must be a c such that f'(c)=0. All that remains is to differentiate sin(x/2) and solve. Algebraist 01:37, 17 January 2008 (UTC)

My belated gratitude. Owing to your hints, I was able to solve both problems and put them up on the board in my calc class. Thanks again! —anon —Preceding unsigned comment added by 70.18.17.227 (talk) 05:01, 20 January 2008 (UTC)

Solving Equations: Rational and Exponential Intersection points

Dear Wikipedia, I am faced with a math problem and you help would be much appreciated, thanks.

y=2^(-x+3) y=(1/8)(1/(x+2))+9

Intersection, therefore y=y, solve for x.

2^(-x+3)=(1/8)(1/(x+2))+9, Note: there ARE two real solutions.

LS: 2^(-x+3) RS:(1/8)(1/(x+2))+9

   8*2^(-x)             1/(8x+16)+9
                        1/(8x+16)+(9(8x+16)/(8x+16)
                        1/(8x+16)+(72x+144)/(8x+16)
                        (1+72x+144)/(8x+16)
                        (72x+145)/(8x+16)

Combine 8*2^(-x)=(72x+145)/(8x+16) 8*2^(-x)=(72x+145)/(8(x+2) multiply both sides by 8 2^6*2^(-x)=(72x+145)/(x+2) simplify 2^(6-x)=(72x+145)/(x+2) multiply both sides by (x+2) (x+2)(2^(6-x))=72x+145 2x^(-x+6) + 2^(-x+7)-72x=145 and I'm stuck... please help, what is next step? did i mess up along the way? thanks —Preceding unsigned comment added by 99.241.96.136 (talk) 02:55, 17 January 2008 (UTC)

The equation cannot be solved by using algebra. You can instead solve it numerically by using a root-finding algorithm. The two functions of x are approximately equal when x = −0.1809; the value of y is then about 9.0687.  --Lambiam 08:44, 17 January 2008 (UTC)

Also when $x\approx -1.99454,\ y\approx 31.879$. -- Meni Rosenfeld (talk) 09:15, 17 January 2008 (UTC)

Put the equation on the form f(x)=0, rather than on the form LS(x)=RS(x). Your result 8·2^−x=(72·x+145)/(8·x+16) leads to the equation f(x)=0 where f is defined by f(x)=(72·x+145)·(2^x)−64·x−128. In addition to the two solutions given above by Lambiam and Meni Rosenfeld, there are also an infinite number of non-real complex solutions. Substitute 2^x=e^x·ln(2) and use the power series for the exponential function to obtain polynomial approximations to f(x). Solve the corresponding equations using the Durand-Kerner method. Bo Jacoby (talk) 22:08, 18 January 2008 (UTC).

conversion to metric

What is 4.8 miles/second in metric? —Preceding unsigned comment added by 67.58.207.35 (talk) 04:37, 17 January 2008 (UTC)

Metric what? 4.8 (miles per second) = 7.7248512 kilometers per second. You can do this with google. 70.162.25.53 (talk) 04:45, 17 January 2008 (UTC)

The metric system? There really is only one way to represent that metrically (ignoring prefixes and forgetting about metric time), the question is very unambiguous. Anyway, see this for an example of how to do it with google. risk (talk) 10:33, 17 January 2008 (UTC)

I guess 70.162.25.53 just wanted to note that in the metric system you can have kilometers/second or meters/second among many others. --Taraborn (talk) 13:00, 17 January 2008 (UTC)

Bad calculator!

My [Sharp EL-506V], when calculating 3.5 x 7.6x10^-14 7.6x10^-10 displays the answer 0.000000002 BUT the actual answer is 2.66x10^-9 so it should say exactly that and at the very least, it should round properly and give 0.000000003 (still not acceptable in my opinion). Please check the output of your favourite calculator and enlighten me. :) ----Seans Potato Business 17:27, 17 January 2008 (UTC)

The result of this calculation is actually $2.66\cdot 10^{-13}$. For the value $2.66\cdot 10^{-9}$, this depends on the display mode of the calculator. The ones I know have two "normal" modes, where one of them only uses scientific notation when there is no other choice. Try changing it to the other option (I don't know how with this particular calculator). -- Meni Rosenfeld (talk) 17:38, 17 January 2008 (UTC)

Edit conflict. I was just about to come back and change 14 to 0 (typo). My calculator is in 'degree' mode but I don't suppose you're talking about that. Alobng the top it gives four modes: normal, CPLX, 3-VLE and STAT ----Seans Potato Business 17:43, 17 January 2008 (UTC)

I've looked at the manual, and it seems you need to switch to scientific notation mode by pressing "2ndF" "FSE" a few times, and then perhaps set the number of digits by pressing "2ndF" "Tab" 9. The relevant information is at the bottom-left corner of the PDF file. -- Meni Rosenfeld (talk) 17:50, 17 January 2008 (UTC)

Okay, thank you very much. I've changed it and will bear that in mind (and keep a copy of the manual). I do think that the calculator is wrong to do this, even when not in scientific mode - don't you? Does your calculator do the same thing when not in scientific mode? The answer it gives is basically incorrect in my opinion. ----Seans Potato Business 19:52, 17 January 2008 (UTC)

Yes, my nearest calculator displays the same when in "Normal 2" mode. Which part are you disagreeing with? If it's using the fixed-point notation (thus hiding some digits) then, well, that's exactly what not being in scientific mode means, and you need scientific mode for calculations where this is significant. If it's about rounding down rather than up in this case, I don't have a good answer why that is so. Note that $1+2.66\cdot 10^{-9}$ is rounded up on my calculator, perhaps in yours as well. The reason is probably technical, but again, when using fixed-point mode it is assumed that values of $10^{-9}$ or less are insignificant for you. -- Meni Rosenfeld (talk) 20:03, 17 January 2008 (UTC)

Well, I think that a well designed scientific calculator would switch to scientific display mode as soon as it meant that it could display more useful digits of the answer. Even if you disagree with this, I can see that you don't disagree regarding the incorrect rounding. Mine also rounds correctly when I add one. Having scientific mode activated all the time looks weird and makes me uncomfortable... --Seans Potato Business 20:36, 17 January 2008 (UTC)

You are basically saying the calculator should decide what is important to you and what is not. This is not entirely unplausible, but at the very least it would introduce additional levels of complexity while calculators are meant to be simple. I'd suggest you use fixed-point normally, and in those rare circumstances when it is unsatisfactory switch to scientific mode. -- Meni Rosenfeld (talk) 20:58, 17 January 2008 (UTC)

I definitely prefer my calculator telling me that 1/7 is 0.142857143 rather than 1.428571429 × 10⁻¹. —Bkell (talk) 21:53, 17 January 2008 (UTC)

Would you also prefer 0.000000002 instead of 2.66x10^-9? That would have lost me points in an exam. --Seans Potato Business 10:20, 18 January 2008 (UTC)

I don't use a calculator, but, like Meni, I would prefer mine to do what it's told and not think for itself excessively. Algebraist 11:21, 18 January 2008 (UTC)

But if you see 0.000000002, then you know there are hidden significant digits and it's "battlescientific mode" time. -- Meni Rosenfeld (talk) 12:29, 18 January 2008 (UTC)

I would of course prefer 2.66 × 10⁻⁹ to 0.000000002; I was simply making a counterargument to your claim that "a well designed scientific calculator would switch to scientific display mode as soon as it meant that it could display more useful digits of the answer." Somewhere there's a point at which scientific notation will be more useful than standard decimal notation, and that point is somewhere between the two extremes of 0.142857143 and 0.000000002. Exactly where that point should be set, though, is a matter of opinion. Like many others here, I would prefer that point to be in accordance with my own opinion, not the calculator manufacturer's opinion, so I have no problem explicitly switching the calculator to scientific mode if I want scientific notation. —Bkell (talk) 12:51, 18 January 2008 (UTC)

January 18

Getting the function back from a table

This probably sounds weird, but are there any Wikipedia articles on the topic of "reconstructing" the function used in a table of values? For example, if I had a table:

x	8	10	12
y	71	97	123

Just in case you didn't know, it means that f(8)=71, f(10)=97, and f(12)=123. I'm trying to find out what the function (f) is. I can then use a certain method to find that the function is $y=x^{2}-3x+10$. Is there any more information out there on this topic (the "reconstruction")?

--wj32 ^t/c 02:12, 18 January 2008 (UTC)

It is, in general, impossible, because there are an infinite number of possible functions that satisfy those criteria. However, if you have further information, generally on the class of function that it comes from, then it would be possible to perform a regression to determine the function. Confusing Manifestation(Say hi!) 02:43, 18 January 2008 (UTC)

Ah, what I mean is that the function must be polynomial (forgot to say that) and that I know what degree the polynomial is. I just don't know the coefficients. Regression is kind of what I'm looking for, except that I'm looking for the exact function. I'll just give you some background info: I know that function is quadratic, so to solve that previous problem I can use algebra:

• 64a + 8b + c = 71
• 100a + 10b + c = 97
• 144a + 12b + c = 123

And solve for a, b, and c. Then (obviously) I put those values back into $ax^{2}+bx+c$ to get the original function. If I generalize the process a bit I can get formulas for a, b, and c:

$a={\frac {f(n)-2f(n+z)+f(n+2z)}{2z^{2}}}$

$b={\frac {4f(n+z)(n+z)-f(n)(2n+3z)-f(n+2z)(2n+z)}{2z^{2}}}$

$c={\frac {f(n)(n^{2}+3nz+2z^{2})-2nf(n+z)(n+2z)+nf(n+2z)(n+z)}{2z^{2}}}$

In these formulas n means the "start" of the data. If I was solving the previous example n would be 8. z means the step between each input of the function. So in that case z would be 2. Obviously, we know f(n) (aka f(8)), f(n+z) (aka f(10)), and f(n+2z) (aka f(12)). So, we can work out the values of a, b, and c.

I'm a bit of a newb beyond the basic math we learn at school. So, now I've got three questions:

• Has this stuff been researched before?
• Is there a general rule for these formulas that can extend to polynomials of the third or higher degrees?
• Do you even understand what I'm saying (mmm)?

--wj32 ^t/c 03:05, 18 January 2008 (UTC)

Yay I found it (although it's a bit complicated...)! Polynomial interpolation --wj32 ^t/c 03:11, 18 January 2008 (UTC)

You might find System of linear equations more helpful. With the three equations you get from three data points, you can solve to find the three coefficients of your polynomial, as you have done. Interpolation is sort of related, but I think you more want to know how to find the polynomial itself. Yes, this sort of thing is very well researched and applies to polynomials of any degree, and all sorts of other problems. As the number of equations get large, we have more "automated" ways of solving a system of equations, such as Gaussian elimination. - Rainwarrior (talk) 05:33, 18 January 2008 (UTC)

Thanks, I was just frustrated I couldn't find any info on what I was looking for... --wj32 ^t/c 06:28, 18 January 2008 (UTC)

Guys, you've got it all wrong. What the OP is looking for is Lagrange polynomial. It is signficantly easier than solving a system of equations. -- Meni Rosenfeld (talk) 12:32, 18 January 2008 (UTC)

Ages ago, I, too, was interested in these kinds of formulae. Of course, I had no knowledge of the Lagrange polynomial at the time (and probably little knowledge about Gaussian elimination). I tried to find the formulae using a brute-force "ad-hoc" method, and succeeded for the case of a quadratic polynomial. It looks like you have only derived the formulae for the case that the points are placed equidistantly, I happened to do it for the general case. Then I tried my luck with the cubic case. The involved calculations were enormous, I had to staple several pieces of paper in a long row just to be able to write a single line of the calculation. Eventually I gave up. Only years later have I found that all this work was unnecessary. -- Meni Rosenfeld (talk) 12:46, 18 January 2008 (UTC)

Wow, thanks for the link! BTW how did you derive a formula for general cases? That's amazing (or at least for me)! --wj32 ^t/c 01:04, 19 January 2008 (UTC)

I don't remember exactly. There's a slight chance I have kept my notes from back then, if I ever find them I'll enlighten you. I suppose it was more or less equivalent to solving the system of linear equations, but probably in a roundabout way. -- Meni Rosenfeld (talk) 12:08, 19 January 2008 (UTC)

The reason I mentioned regression is because I am *fairly* sure that it will return exactly the right answer, since it aims to minimise the error between the fitted curve and the given points, which as long as you get the degree right should be when you retrieve the original equation, so it should be equivalent to the other methods mentioned. Confusing Manifestation(Say hi!) 02:16, 19 January 2008 (UTC)

No doubt, but it's not "the right tool for the job". Not to mention that Regression is a dab, and Nonlinear regression doesn't seem to contain the relevant practical information. -- Meni Rosenfeld (talk) 12:08, 19 January 2008 (UTC)

Fair enough. I've obviously been in the statistical world for too long. Although technically polynomial fitting is still a type of linear regression (where the "independent" variables are 1, x, x^2, ..., x^n). Confusing Manifestation(Say hi!) 22:44, 20 January 2008 (UTC)

You're right, I didn't think about it this way. -- Meni Rosenfeld (talk) 09:35, 21 January 2008 (UTC)

probability.. helppp..!!

first of al this isnt my homework question..!! so plz help me solve this question of fuji film.. it is for my own personal understanding....!! plz try and help me answer all questions..i have been trying to solve it since 3 days but aint gettin newhere m typing down the whole case here..

In the early 1990s, Fuji Photo Film, USA, joined forces with four of its rivals to create the Advanced Photo System (APS), which is hailed as the first major development in the film industry since 35 - millimeter technology was introduced. In February 1996, the new 24 millimeter system, promising clearer and sharper pictures, was launched. By the end of the year, the lack of communications and a limited supply of products made retailers angry and consumers baffled. Advertising was almost nonexistent. Because the product was developed by five industry rivals, the companies had enacted a secrecy agreement in which no one outside of company management, including the company's sales force, would know details about the product until each company introduced its APS products on the same day. When the product was actually introduced, it came with little communication to retailers about the product, virtually no training of sales representatives on the product ( so that they could demonstrate and explain the features ), and a great underestimation of demand for the product. Fortunately, Fuji pressed on by taking an "honesty is the best policy" stance and explaining to retailers and other costumers what had happened and asking for patience. In addition, Fuji increased its research to better ascertain market positioning and size. By 1997, Fuji had geared up production to meet the demand and was increasing customer promotion. APS products were on the road for success. By 1998, APS cameras owned 20% of the point and shoot camera market.

1. As stated, by 1998 APS cameras owned 20% of the point and shoot camera market. Now it is the year 2003 and the market share might be nearer to 40%. Suppose 30 cusomers from the point and shoot camera market are randomly selected. If the market share is really .40 , what is the expected number of point and shoot camera customers who purchase an APS camera? What is the probability that six or fewer purchases and APS camera? Suppose you actually got six or fewer APS customers in the sample of 30. Based on the probability just calculated, is this enogh evidence to convince you that the market share is 40% Why or why not?

2. Suppose customer complaints on the 24 millimeter film are poison distributed at an avegare rate of 2.4 complaints/100,000 rols sold. Suppose further that Fuji is having trouble with shipments being late and one batch of 100,000 rolls yields seven complaints from customers. Assuming that it is unacceptable to management for the average rate of complaints to increase, is this enough evidence to convince management that the average rate of complaints has increased, or can it be written off as a random occurence that happens quite frequently? Produce the Poisson distribution for this question and discuss its inplication for this problem.

3. One study of 52 product launches found that those undertaken with revenue growth as tha main objective are more likely to fail than those undertaken to increase customer satisfaction or to create a new market such as the APS system. Suppose of the 52 products launched, 34 were launched with revenue growth as the main objective and the rest were launched to increase cusotmer satisfaction or to create a new market. Now suppose only 10 of these products were successful ( the rest failed) and seven were products that were launched to increase customer satisfaction or to create a new market. What is the probability of this result occuring by chance? What does this probability tell you about the basic premise regarding the importance of the main objective? —Preceding unsigned comment added by 220.225.79.210 (talk) 14:00, 18 January 2008 (UTC)

• For not being a homework problem, this reads an awful lot like one (especially because I was easily able to search and find it asked in the past elsewhere). That being said, you say you have attempted this for three days... what have you gotten so far? --Kinu ^t/_c 22:16, 18 January 2008 (UTC)

  Curiously enough, that posting of 10 months ago contains exactly the same spelling errors: "cusomers", "enogh", "poison", "avegare", "rols", "occurence", "inplication". "tha", "cusotmer", "occuring". The probability of this result occuring by chance is less than that of a million monkeys producing the correct answer by chance. You would expect an instructor distributing this not to make so many errors and to have corrected them by now.  --Lambiam 00:25, 19 January 2008 (UTC)

conversion from kilobytes and megabytes to gigabytes

I need to know how many gigabytes is 153.44 megabytes and any many gigabytes is 4812 kilobytes? Thanks!

71.145.168.69 (talk) 17:24, 18 January 2008 (UTC)

That all depends on where the numbers come from. Strictly, a gigabyte is 1000 megabytes so 153.44 megabytes = 0.15344 gigabytes, and 4812 kilobytes = 4.812 megabytes = 0.00418 gigabytes. This will probably be correct for capacity of hard discs. However, computer memory has traditionally been measured, not in the SI units mentioned above, but in mebibytes where one mebibyte = 1024 kilobytes or kibibytes and in gibibytes where one gibibyte = 1024 mebibytes. In those circumstances, you might have to divide by 1024 each time, giving 153.44 megabytes = 0.1498437 gibibytes and 4812 kilobytes = 0.004589 gibibytes.

Complications continue, because it is just possible that your 153.44 megabytes are really 153.44 mebibytes, which is 153.44 x 1024 x 1024 bytes, and converting this gives 0.1608935 gigabytes. Because usage varies, even between manufacturers, the only sure way is to try it and see! Sorry the answer is not simple. dbfirs 17:47, 18 January 2008 (UTC)

In 99% of cases of common usage, the word "kilobyte" will refer to 1024 bytes (and so on), so it is safe to assume that for whatever purpose the OP is after, this is the relevant interpretation. Regarding the actual questions, Google is your friend. -- Meni Rosenfeld (talk) 18:13, 18 January 2008 (UTC)

For RAM, yes, but for capacities of hard disk drives as quoted on data sheets by hard disk manufacturers usually not. There 1 GB is 1 billion bytes, not all of which may be user-accessible to boot.  --Lambiam 00:07, 19 January 2008 (UTC)

But it's been a long time since hard drive capacities were measured in kilobytes. I think it's true that a kilobyte, specifically, is virtually guaranteed to be 1024 bytes.

Here are some rules of thumb:

• For RAM (DIMMs that plug into your motherboard), mega and giga always mean 2²⁰ and 2³⁰. RAM is always sold in power-of-two sizes (or occasionally small multiples thereof), so the binary units are a lot more convenient.
• For hard drive capacities reported by the manufacturer, mega and giga always mean 10⁶ and 10⁹. For hard drive capacities reported by software, they almost always mean 2²⁰ and 2³⁰. I think the manufacturers have the right idea here and the software should be considered broken; binary units are inconvenient and useless for hard drives since they aren't sold in power-of-two sizes. People sometimes describe the smaller number reported by the software as the "formatted capacity", but that's wrong; it's the result of using different units. There is some filesystem overhead, but the "total size" reported by the software is typically the full drive size as reported by the manufacturer.
• Flash drives (CF, SD, USB keys, etc.) follow the same rules are hard drives. Bizarrely, flash memory devices usually use power-of-two multipliers (256 megabytes, 8 gigabytes), even though that makes no sense in front of a decimal unit. A gigabyte of RAM is exactly twice as much as 512 megabytes of RAM, but a 1-gigabyte flash card is only 1000/512 ≈ 1.95 times as large as a 512-megabyte flash card. This does strike me as deliberately deceptive marketing.

-- BenRG (talk) 10:14, 19 January 2008 (UTC)

I actually never knew that (probably because I find specifications of disk capacities less interesting than sizes reported by software). -- Meni Rosenfeld (talk) 12:16, 19 January 2008 (UTC)

I see someone knows enough about computers to point out that to the computer, a Megabyte is 1048576=1024^2 bytes and a Gigabyte is 1073741824=1024^3 bytes, when a Hard Drive manufacturer or other electronic data storage device maker advertises say 250GB, they almost always mean 250000000000 bytes=2.50x10^11 bytes (often slightly more to make the number a power of 2, my hard drive is a 250GB drive and it has 250048479232 bytes of storage space (possibly not including the File Allocation Table which tells the computer where the data for a file is.) I would also agree that it is deliberately deceptive marketing to call a drive which has 232GB (as the computer sees it) a 250GB drive. A math-wiki (talk) 07:45, 24 January 2008 (UTC)

I don't see how one can argue that a computer "sees" in gibibytes. The only disk size units you'd normally deal with in software are bits, bytes, sectors, cylinders, or clusters. There's nothing which always comes in integral multiples of gibibytes, so if you did measure something in gibibytes you'd need to include fractional bits, at which point you're really storing a byte count (or whatever) and calling it something else. Metric prefixes are for human beings who aren't comfortable dealing with lots of zeros. If there is a use for prefixes like mebi- and gibi- it's in human-human or human-computer interactions, and I can't see a reason in the world to use them there except maybe for sizes that tend to be exact powers of two (like RAM and hash tables and circular buffers). I can remember many cases in which I had to waste time multiplying or dividing a size by 1024 or 1048576 in order to figure out how it compared to another size. I can't remember any case in which the use of the binary prefixes saved me time or was helpful. I can't even imagine any such situation. -- BenRG (talk) 15:36, 24 January 2008 (UTC)

January 20

Quick calculus question

In a problem I'm trying to solve, I am shown the graph of the derivative of a function g. The derivative is continuous on the shown interval, which includes the x-value a. Since a y-value on the graph of the derivative is given for the x-value a, I can infer that g is differentiable at x = a. But I don't know how to infer from the graph of g‘ whether g is continuous at x = a. Does differentiability imply continuity? Thanks, anon. —Preceding unsigned comment added by 70.18.17.227 (talk) 05:00, 20 January 2008 (UTC)

• You bet it does... the section on "Continuity and differentiability" at Derivative provides an excellent explanation on why this is so, so I'll omit it here, for the sake of not being redundant. :) --Kinu ^t/_c 05:11, 20 January 2008 (UTC)

January 21

Differential Equation

This is not a homework problem - I thought of it while I was playing with my cat. If you dangle a string and spin it the right way, it forms a sort of standing wave. There are points along it, usually at most 3, that are almost stationary, and the rest looks like a squished sine wave spinning around its x-axis. In trying to figure out its shape, I got stuck with an equation I couldn't solve. To start, I assume no air resistance, that the string is the same mass all the way along and completely flexible, and that it has attained its ideal shape - a stationary planar wave spinning around a vertical axis. Call the axis of rotation the x-axis, and a perpendicular line the y-axis, with the y-axis spinning with the string. Since I wasn't sure how to solve it directly, I made the string massless with point masses along it at equal intervals, and drew a free body diagram for each mass. The first mass, at the bottom end of the string, has only two forces on it: mg downwards, and F₁ along the string. Every other mass, say the nth along from the bottom, has three forces: mg downwards, and F_n-1 and F_n along the string. Equating X-components, F_n,x = nmg, because no mass is moving up or down. The Y-components, then, must add up to the actual acceleration (a_n) of the point. Since the point moves in a circle at constant angular velocity, acceleration is towards the x-axis, and since all points move with the same angular velocity, their accelerations are proportional. That is, there's some constant a such that a_n = ay_n for all n, where y_n is the y-coordinate of the point mass. Equating Y-components, then, $F_{n,y}=ma\sum _{k=1}^{n}y_{k}$. In other words, each bit of the string must support the weight of everything below it, and keep everything below it moving. Since the force on each mass must be in the direction of the next mass, that is along the string, F_n,y/F_n,x = (y_n+1-y_n)/(x_n+1-x_n). Since the masses are at even intervals along the string, there's some constant d such that d² = (y_n+1-y_n)²+(x_n+1-x_n)² for all n. Substituting the first two equations into the third, I get ${\frac {a\sum _{k=1}^{n}y_{k}}{ng}}={\frac {y_{n+1}-y_{n}}{x_{n+1}-x_{n}}}$ for all n. Once I had the shape of the problem, I shifted it to differentials. n becomes a continuous independant variable measuring distance along the string, x and y the dependant variables. If f(n)=y, ${\frac {a\int _{0}^{n}f(s)\,ds}{g}}={\frac {y'}{x'}}$ and 1 = (y')²+(x')². I differentiate the first, so (a/g)y = (x'y"-y'x")/(x')². Plugging in the other equation and its derivative to get rid of x, (a/g)y²(1-(y')²)³ = (y")². Choosing b=a/g, I'm left with by²(1-(y')²)³ = (y")² Unfortunately, I'm not sure how to solve this for y. Any ideas? Black Carrot (talk) 00:09, 21 January 2008 (UTC)

Decide whether y' shall mean dy/dx or dy/ds. The differential equation must be supplemented by boundary conditions: the length of the chain (S=integral ds), the fixed end of the chain (x(0)=y(0)=0), and the loose end of the chain (x"(S)=y"(S)=0). Note the trivial (unstable) solution that the chain is straight and vertical, (x(s)=s, y(s)=0). The other solutions are characterized by the number of crossings of the x-axis. It's a nice problem. You may substitute some power series of s (or is it x?) for y in the differential equation and thus transform it into a difference equation in the coefficients. Have fun! Bo Jacoby (talk) 02:19, 21 January 2008 (UTC).

y' is dy/dn, and x' is dx/dn. s is a dummy variable in the integration, like k was in the summation. I've started pushing a power series through it, but it's a bit messy with a degree-six polynomial. Black Carrot (talk) 02:49, 21 January 2008 (UTC)

I've hit a bit of a difficulty with the power series. When I expand it around some points, it doesn't form a wave - all the odd-degree terms are zero. Specifically, if the x term is zero, corresponding to one of the peaks of the wave, all the odd-degree terms are as well. This means that if y is analytic anywhere, it isn't for very far. Is there a way to pull some bounds on that out of the equation? Black Carrot (talk)05:22, 21 January 2008 (UTC)

Odd-degree terms being zero does not prevent a power series from being a wave - cosine is a wave with odd-degree terms zero. But cos(x) written parametric as (x(s),cos(x(s))) where (dx/ds)^2+(dy/ds)^2=1 is not handy. You might prefer considering the differential equation for the function y(x) rather than for (x(s),y(s)). And you might want to study the catenary article for a similar, but somewhat simpler, problem. Observe that your problem is related to the quantum mechanics of hydrogen-like atoms where boundary conditions also give raise to discrete quantum numbers labelling the solutions of the differential equation, but quantum mechanics is easier because the differental operator of the Schrödinger equation is linear. Bo Jacoby (talk) 08:14, 21 January 2008 (UTC).

You're right, it doesn't. That was silly. I've made more progress with differentials, though. I can reduce it to a first-order equation: 1=(-(b^1/2/2)y²+c)(1-(y')²)^1/2, where c is an arbitrary constant. This is seperable, giving the hideous (but hopefully accurate) integral $\int _{}^{}{\sqrt {\frac {hy^{2}+c}{hy^{2}+c-1}}}\,dy=\int _{}^{}\,dn$, where h=-b^1/2/2 and c is an arbitrary constant. Another sticking point. Black Carrot (talk) 19:26, 21 January 2008 (UTC)

Exponential naming

"Square" and "cube" are apparently used to describe the exponents 2 & 3 because of their relationship to geometrical forms. Are there any forms that could lend their names to higher powers? Retarius | Talk 02:47, 21 January 2008 (UTC)

I read a book a bit ago that called four-dimensional figures "planoplanes". I think it was the Arithmetica Infinitorum. Black Carrot (talk) 02:53, 21 January 2008 (UTC)

Actually, in that direction, many n-dimensional objects are named after 2 or 3-dimensional objects they resemble, with some modifier (hyperplane, hypersphere, hypercube, n-ball, etc.) Black Carrot (talk) 02:55, 21 January 2008 (UTC)

Apparently quartic functions can be called "biquadratic", so maybe the fourth power could be a bisquare. Black Carrot (talk) 03:03, 21 January 2008 (UTC)

I was attempting to explain powers of numbers to someone I was tutoring when the question arose: "If three is a cube, what does four make?" (Golden Rule: To discover how little you know about something try explaining it to someone who knows nothing about it.) I tried to imagine a four-dimensional form to use as a name-source but, of course, that fourth dimension is Time in conventional discourse. Retarius | Talk 03:12, 21 January 2008 (UTC)

That would be a better opportunity to teach the theory of language than math. It shows up in all sequences that people started naming before they realized they couldn't stop. First, second, third, then nth. Eleven, twelve, then n-teen. Linear, quadratic, cubic, quartic, quintic, n-degree. Ten, hundred, thousand, million, n-illion. Black Carrot (talk) 03:27, 21 January 2008 (UTC)

Now I've had a chance to look at some of the concepts you've referred to I think a valid terminology could be derived from the table shown in the article on the hypercube. Although "Two hexeracted" might lose in a cost/benefit analysis against "Two to the sixth". Many thanks for your guidance, Black Carrot. Retarius | Talk 04:00, 21 January 2008 (UTC)

That's a good word. I wish I had a copy of that book still, I think it had some other odd words for powers. That was back when algebra was some newfangled toy, and I think he made up a lot of stuff. For instance, according to his article he was the first person to use the infinity symbol in print. Black Carrot (talk) 05:26, 21 January 2008 (UTC)

I usually refer to fourth powers as Quarts, (for Quartic) and Quints for 5ths and so forth. This is mainly due to my having studied Polynomials a fair amount on my own. A math-wiki (talk) 08:04, 21 January 2008 (UTC)

See zenzizenzic and zenzizenzizenzic. Gandalf61 (talk) 14:52, 21 January 2008 (UTC)

Apparently a question of long standing...Zenzizenzic? Now that's a word for quiz nights! Retarius | Talk 01:44, 22 January 2008 (UTC)

free shell account for math?

I have only a very slow computer with very slow dialup access, any chance I could do mathematical computation (matlab, mathematica, etc) on a normal free shell account somewhere? I don't know where to begin.

I may start a math degree this fall. —Preceding unsigned comment added by 212.51.122.8 (talk) 03:19, 21 January 2008 (UTC)

What OS do you have? Why not download mathematica and run it? --wj32 ^t/c 03:53, 21 January 2008 (UTC)

Well, Mathematica isn't free. If you're going to download something, you might try SAGE. —Bkell (talk) 05:44, 21 January 2008 (UTC)

Oops, I was thinking of Maxima, not Mathematica. Stupid names. --wj32 ^t/c 06:18, 21 January 2008 (UTC)

You definitely don't need Matlab for a math degree. Mathematica can be useful but is far from being necessary (and can even be harmful if you over-rely on it). -- Meni Rosenfeld (talk) 09:28, 21 January 2008 (UTC)

Numerical methods

The measured heights are 2950cm and 35 cm, while the true values are 2945cm and 30 cm respectively. compare the absolute and relative errors —Preceding unsigned comment added by 210.212.161.105 (talk) 06:43, 21 January 2008 (UTC)

Think about this, those two numbers are vastly different in size, so for example being 5cm off on the larger measurement is really not a big deal for practical purposes, but on the smaller one it could be a real issue, this is where absolute error and relative error differ. One of these would directly compair the difference between (expected and observed) or (true and measured) in our case the other would account for the fact that the two true values are different in size, and obtain a percentage for the error. Can you guess which one is which?? A math-wiki (talk) 08:02, 21 January 2008 (UTC)

Okay, I've compared them. -- Meni Rosenfeld (talk) 09:25, 21 January 2008 (UTC)

Also see Approximation error. GromXXVII (talk) 12:55, 21 January 2008 (UTC)

Limit question

I'm trying to find the limit as x approaches 0 of (1 - cos 4x)/x. I rewrote it as (cos 4(0) - cos 4x)/(0 - x) and took the derivative of the subtrahend in the numerator to get -4 sin 4x. Substituting 0 in for x, that gives me 0, which was indeed given as the right answer. But my question is, was I justified in rewriting
lim as x approaches 0 of (1 - cos 4x)/x

as
lim as x approaches 0 of (cos 4(0) - cos 4x)/(0 - x)

given that that seems to change the sign of the denominator? Is it acceptable in this case because x is approaching 0 and the sign before 0 doesn't matter? Thanks, anon. —Preceding unsigned comment added by 70.19.20.251 (talk) 19:30, 21 January 2008 (UTC)

No, that rewriting is not justified. However, you could have written lim(1 - cos 4x)/x=lim(-(cos 4(0) - cos 4x)/(0 - x))=-lim(cos 4(0) - cos 4x)/(0 - x)=-0=0. Algebraist 20:16, 21 January 2008 (UTC)

Or use l'Hôpital's Rule, which would give you $\lim _{x\to 0}{\dfrac {4\sin {4x}}{1}}={\dfrac {0}{1}}=0$. --Kinu ^t/_c 20:28, 21 January 2008 (UTC)

Gotcha. Thanks! —anon —Preceding unsigned comment added by 70.19.20.251 (talk) 20:38, 21 January 2008 (UTC)

Evaluating a definite integral

I'm trying to evaluate ∫₀²(2x³ + 3)dx. Here's what I have so far:

∆x = 1/n

c = a + i∆x = 0 + i(1/n) = i/n

lim [as n --> infinity] of ∑ [from 1 to n] of (2(i³/n³) + 3)(1/n)

lim [as n --> infinity] of ∑ [from 1 to n] of (2(i³/n⁴) + 3/n)

lim [as n --> infinity] of ((2/n⁴)∑ [from 1 to n] of i³ + (1/n)∑ [from 1 to n] 3)

lim [as n --> infinity] of ((2/n⁴)((n²)(n + 1)²/4) + (1/n)3n)

lim [as n --> infinity] of (1/2 + 3)

1/2 + 3

7/2

But that's not one of the choices given; it says here that the answer is 14. Why is this? Thanks, anon. —Preceding unsigned comment added by 70.19.20.251 (talk) 20:47, 21 January 2008 (UTC)

Because you have computed (correctly) $\int _{0}^{1}2x^{3}+3dx$. You'll have to change things a bit to take account of the correct limits of integration. Algebraist 21:01, 21 January 2008 (UTC)

How do I do that? Should my ∆x be (b - a)/n? —Preceding unsigned comment added by 70.19.20.251 (talk) 21:13, 21 January 2008 (UTC)

Yes. Note that this correct ∆x is used both in the formula for c, and you multiply every term by it. I'll get you started:

$\Delta x={\frac {b-a}{n}}={\frac {2}{n}}$

$c_{i}=a+i\Delta x={\frac {2i}{n}}$

$\int _{0}^{2}(2x^{3}+3)dx=\lim _{n\to \infty }\sum _{i=1}^{n}(2c_{i}^{3}+3)\Delta x=\cdots$

-- Meni Rosenfeld (talk) 21:36, 21 January 2008 (UTC)

Thanks so much! I got it right now. —Preceding unsigned comment added by 70.19.20.251 (talk) 22:00, 21 January 2008 (UTC)

Intersection of two cones

There are two cones with intersecting axes. They don't necessarily have the same slope. That is, one may be pointier than the other (I don't know the way you're supposed to say it). I assume the intersection between them is a conic section. How do I find the plane the intersection is on? Specifically, I need a vector at right angles to it. — Daniel 21:49, 21 January 2008 (UTC)

I forgot to add: the cones are the same width where their axes intersect.

After a little experimentation, it looks like if the cones have the same slope, the normal of the plane they intersect on is the cross product of their axes. The plane doesn't go through the intersecting axes as I thought it would, so I need to find a point on the plane. If the cones don't have the same slope, they appear to intersect on a hyperbola stretched infinitely in the third dimension. For what I'm doing, it would probably be best just to approximate it as a plane by averaging the slopes and taking the first case. Can anyone tell me how to find the point I need for the first case? — Daniel 00:39, 22 January 2008 (UTC)

I don't think I understand your question. Could you be more specific, or sketch a picture or include a formula or something? Do you, for instance, include "oblique" cones? Black Carrot (talk) 01:21, 22 January 2008 (UTC)

Have I found the absolute minimum correctly?

I need to find the absolute minimum value of f(x) = 2x³ - 3x² - 12x, which is defined on the interval [-3, 2]. Relative extrema are marked by sign changes in the derivative, which is f'(x) = 6x² - 6x - 12 = 6(x² + x + 2) = 6(x - 2)(x + 1). So the possibilities are -3, -1, and 2. f(-3) = -45, f(-1) = 7, and f(2) = -20, so the answer is -45, isn't it? —Preceding unsigned comment added by 70.19.20.251 (talk) 22:07, 21 January 2008 (UTC)

You're fine, apart from a couple of typos (x² + x + 2 should be x² - x - 2). Algebraist 22:14, 21 January 2008 (UTC).

Whoops. Thanks!

Least-powers question

I'm looking for the critical numbers of f(x)= x√(16 - x²). I took the derivative: x•(1/2)(16 - x²)^1/2 + (16 - x²)^1/2. I factored out 16 - x², giving me (16 - x²)[(x/2) + 16 - x²]. So my critical numbers are ±4 and ... what? I tried running the contents of the other parenthesis through the quadratic formula, but I got the gobbledygook 1/4 ± √(257)/8 ... —Preceding unsigned comment added by 70.19.20.251 (talk) 22:33, 21 January 2008 (UTC)

You have an error in your calculation of the derivative. -- Meni Rosenfeld (talk) 22:38, 21 January 2008 (UTC)

You neglected the chain rule and you left 1/2 as the exponent where you needed (1/2) − 1. Michael Hardy (talk) 23:23, 21 January 2008 (UTC)

I see, I had typed it up wrong. Corrected: I'm looking for the critical numbers of f(x)= x√(16 - x²). I took the derivative: x•(1/2)(16 - x²)^-1/2 + (16 - x²)^1/2. I factored out (16 - x²)^-1/2, giving me (16 - x²)^-1/2[(x/2) + 16 - x²]. But my question still remains ... —Preceding unsigned comment added by 70.19.20.251 (talk) 23:46, 21 January 2008 (UTC)

It's still not correct. You still haven't applied the chain rule, which means that when differentiating ${\sqrt {16-x^{2}}}$ you need to multiply by the derivative of $16-x^{2}$. If this is unclear, take a look at the article. -- Meni Rosenfeld (talk) 23:51, 21 January 2008 (UTC)

Got it. Thanks so much! —Preceding unsigned comment added by 70.19.20.251 (talk) 00:09, 22 January 2008 (UTC)

Just another math problem

What is 16,777,216 to the 2,079,600th power? Tamashiihiroka (talk) 23:33, 21 January 2008 (UTC)

I don't know about calling that a math problem, but this is roughly $3.130311574455724557\cdot 10^{15024527}$, give or take a gazillion or two. -- Meni Rosenfeld (talk) 23:39, 21 January 2008 (UTC)

Hmm, interesting talk is it not? hydnjo talk 02:16, 22 January 2008 (UTC)

Warning. Visiting that talk page, which is 1,660,001 bytes, crashed my browser.  --Lambiam 04:15, 22 January 2008 (UTC)

I just get a completely blank page. Apparently the server refuses to serve any of it to me. —Keenan Pepper 04:19, 22 January 2008 (UTC)

January 22

Bra-ket notation in LaTeX

\left and \right are great, and work just as I expect for $\left\langle {\frac {big}{bras}}\right|$ and $\left|{\frac {big}{kets}}\right\rangle$. But when I put them together in a bracket, I'm at a loss what to do. Using an unadorned | gives me $\left\langle {\frac {this}{is}}|{\frac {no}{good}}\right\rangle$. Logically there should be something like \middle or \center, but the first doesn't exist and the second does something totally different. How do I tell LaTeX to automatically adjust the middle line to the correct size? —Keenan Pepper 04:17, 22 January 2008 (UTC)

can this diff eq be solved?

Does anyone have any suggestions on how to solve this non-linear second-order differential equation: $y^{\prime \prime }-axy^{\prime }+byy^{\prime }=0$ where a and b are positive constants, and y is a function of x. Thanks. --131.215.166.106 (talk) 04:20, 22 January 2008 (UTC)