Wikipedia:Reference desk/Archives/Mathematics/2006 December 20

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December 20

Probability of finding DNA sequences

Hello folks. This one will be easy for you, but for a poor mathematically challenged biologist its proving frustrating. Consider a genome where the sequence of bases can be A, T, C or G at any given position (and assume there is no bias in the sequence). Here is what I wish to know: If you select 20 different 10-base sequences at random, what will the mean distance between them be in the genome? Also, what will the distance be given 60 different sequences? Thanks in advance! Rockpocket 02:33, 20 December 2006 (UTC)

I'm not sure I understand the question. The 10-base sequences might occur multiple times (which instances do you pick?) or even not at all (does that make the distance infinite?). —Keenan Pepper 03:25, 20 December 2006 (UTC)

Oh, I thought about it a little harder and now I see what you're getting at. Say you have a sequence of n+9 bases, where n is some large number. Then there are n 10-base sequences (starting at positions 1 thru n), and each of them has an equal probability of being any of the 4¹⁰ possible ones, so the average number of matching sequences is ${\frac {20n}{4^{10}}}$ and the average distance between their starting points is ${\frac {4^{10}}{20}}$ = 52428.8. If there are three times as many sequences, then they occur three times as often and the average distance is three times smaller: 17476. Is that what you're asking? —Keenan Pepper 03:36, 20 December 2006 (UTC)

Thank you, I think that calculation is what I am looking for. The answer is certainly in the ballpark I would expect based upon the experimental procedure this is based on. I should add, before someone else does, that I am also assuming the DNA is single stranded, and your calculation appears to assume that also. Thanks again! Rockpocket 04:37, 20 December 2006 (UTC)

Special matrix

Is there anything interesting about this matrix? --HappyCamper 05:04, 20 December 2006 (UTC)

1  i
1 -i

Its inverse is half its conjugate transpose. It follows that the matrix is normal. I don't know if that is interesting.  --Lambiam^Talk 10:28, 20 December 2006 (UTC)

It has nothing special beyond that of any other 2 by 2 matrix - I really have to say no. (did you see anything special in it?)87.102.4.227 11:53, 20 December 2006 (UTC)

[The following answer is not taken from a reputable source, and so may violate WP:V and WP:NOR.] Take a look at its cube. Gandalf61 16:25, 20 December 2006 (UTC)

Its eigenvalues have the same magnitude and their arguments are both rational multiples of pi. Therefore, some power of it is a scaled identity matrix. —Keenan Pepper 17:48, 20 December 2006 (UTC)

Oops...sorry, I missed something. There's supposed to be a factor of ${\frac {1}{\sqrt {2}}}$ in front of the matrix. I came across it in the context that this was a coordinate transformation from "linear polarization" to "circular polarization" but it's not immediately obvious to me how this is accomplished. --HappyCamper 18:33, 20 December 2006 (UTC)

You mean Jones calculus? —Keenan Pepper 04:36, 21 December 2006 (UTC)

In that case its inverse is equal to its conjugate transpose; in other words, it is a unitary matrix.  --Lambiam^Talk 00:43, 21 December 2006 (UTC)

We do have an article on circular polarization that may help a little. Reading about Lissajous curves might help even more.

Maxwell's equations for electromagnetism describe the propagation of an electromagnetic wave as an oscillation between an electrical component and a magnetic component. This pattern fits the same mathematics as simple harmonic motion, a periodic sinusoidal variation. In geometric terms, we think of the waves as oscillating in a plane perpendicular to the direction in which the radiation is travelling. The subtlety is that in a plane we have two independent directions, call them x and y, both perpendicular to z, and we can also have two independent oscillations.

Here's where the Lissajous curves come in. The two oscillations must occur at the same frequency, but they may have different strengths, and they need not reach their peaks and valleys at the same time. But suppose they are synchronized, and are equally strong. Then whatever the swing in x may be at any instant, the swing in y is exactly the same; this traces out a path in the plane that stays on the line x = y.

If we pass the wave through a polarizing filter that allows only x oscillation, then all the y energy is lost, and now that curve just swings back and forth along the x axis. If we diminish the y component, but neither extinguish it nor disrupt its synchronization, we don't get pure x, but we still get a oscillation along a line. In fact, by varying the relative strength of the two components we can rotate that line of excursion to any angle we like, and we can make the size of the excursion however small or large we like.

But there is another possibility. Suppose we pass a wave of light through a quarter-wave plate of calcite. This common mineral has a useful optical property because of its crystal structure: light oscillating in one direction can travel faster than light oscillating in another (perpendicular) direction. (Other minerals such as iolite have different colors in different directions!) The thickness of a quarter-wave plate is carefully controlled so that one direction of oscillation is delayed a quarter of a wavelength; the light emerges with the two directions out of synchronization.

Explicitly, the oscillation are now

${\begin{aligned}x&{}=A\sin({\frac {\pi }{2}}+\omega t)\\&{}=A\cos(\omega t)\\y&{}=A\sin(\omega t)\end{aligned}}$

In other words, the oscillation traces out a circle in the xy plane. Depending on how we handle the delay, the circle may be traced either clockwise or the opposite. This is circular polarization.

But what has this to do with the complex-valued matrix you found? The connection is the famous equation

$e^{i\theta }=\cos \theta +i\sin \theta .\,\!$

In dealing with sinusoidal oscillations, it is often convenient to work with complex numbers and a complex exponential rather than with separate sine and cosine components, or with amplitude and phase. And it happens that left and right circular polarization form a vector space basis for all possible oscillation patterns just as good as the more obvious xy basis. The matrix given is nothing fancier than such a change of basis. Now would be a good time to read about Jones calculus. --KSmrq^T 07:25, 21 December 2006 (UTC)

solution of triangles

Dear Sir/Madame,

Could you please help me solve this problem:

Prove that in any plane triangle,the value of tanBtanC+tanCtanA+tanAtanB cannot lie between zero and nine.

Thank you.

-Sruthi

For the upper figure of 9, I suggest you consider an equilateral triangle where this value is obtained, then by considering movement away from equality convince yourself that there will be an increase. A negative value can be got only if one angle exceeds 90 degrees, of course.81.154.107.121 18:27, 20 December 2006 (UTC)

This looks like a typical students' competition problem, and unless you convince us that you do not intend to use the answers for cheating, I would advise against answering at all.--80.136.157.88 17:26, 21 December 2006 (UTC)

Permutations and odds of matching elements in a multiset

With multiset {1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5}, what are the odds that two random permutations will have n matching elements (i.e., the permutations match at n indexes)? I'm trying to determine the odds that someone tested for ESP using Zener cards will have a certain number of correct guesses. The problem would be much easier (see Checking if a coin is fair) if the test allowed every drawn card to be put back in the deck. — Elembis 21:43, 20 December 2006 (UTC)

There was a related question the other day at Wikipedia talk:WikiProject Mathematics#Representing probability. Experimentally, using 75000000 trials with a random generator, I obtained (n = # of correct matchings, Fn = fraction of occurrences of n):

 n   Fn
-- -------
 0 0.00429
 1 0.02548
 2 0.07337
 3 0.13655
 4 0.18428
 5 0.19193
 6 0.16033
 7 0.11008
 8 0.06328
 9 0.03088
10 0.01290
11 0.00465
12 0.00145
13 0.00039
14 0.00010
15 0.00002
16 0.00000
.. .......
25 0.00000

 --Lambiam^Talk 10:20, 21 December 2006 (UTC)

Thanks! — Elembis 00:27, 22 December 2006 (UTC)

Have I done my derivative correctly?

I have been trying to teach myself differential calculus using Wikipedia and the Calculus Wikibook. After a bit of reading and working through some examples, I tried to find the derivative of $f(x)=x^{4}$, which gave me the answer of $4x^{3}$, which is apparently correct. The steps I used to obtain this are listed below:

$f'(x)\,$	$=\lim _{{\Delta x}\to 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x}}$
	$=\lim _{{\Delta x}\to 0}{\frac {(x+\Delta x)^{4}-x^{4}}{\Delta x}}$
	$=\lim _{{\Delta x}\to 0}{\frac {x^{4}+4x\Delta x+\Delta x^{4}-x^{4}}{h}}$
	$=\lim _{{\Delta x}\to 0}{\frac {4x\Delta x+\Delta x^{4}}{\Delta x}}$
	$=\lim _{{\Delta x}\to 0}(4x+\Delta x^{3})=4x^{3}.$

While this gives the correct answer, I don't like the fact that the second to last line gives the last line. Have I done this correctly? --80.229.152.246 11:28, 21 December 2006 (UTC)

You have the right approach, but your algebra is not quite correct. You need to check your expansion of $(x+\Delta x)^{4}$ in line 3. Use the binomial theorem. You should get $x^{4}+4x^{3}\Delta x$ + stuff in higher powers of $\Delta x$. Gandalf61 11:57, 21 December 2006 (UTC)

You've multiplied out the bracket incorrectly, and your last limit, had it been correct, would have given you an answer of $4x$, not $4x^{3}$. If you expand the bracket properly, you will get the following.

$\lim _{{\Delta x}\to 0}{\frac {(x+\Delta x)^{4}-x^{4}}{\Delta x}}=\lim _{{\Delta x}\to 0}{\frac {x^{4}+4x^{3}(\Delta x)+6x^{2}(\Delta x)^{2}+4x(\Delta x)^{3}+(\Delta x)^{4}-x^{4}}{\Delta x}}$

$=\lim _{{\Delta x}\to 0}{4x^{3}+6x^{2}(\Delta x)+4x(\Delta x)^{2}+(\Delta x)^{3}=4x^{3}}$

Readro 12:12, 21 December 2006 (UTC)

Thanks very much you two. I understand it now. Thanks again. --80.229.152.246 12:22, 21 December 2006 (UTC)

So for $f(x)=x^{5}$, the derivative would be:

$\lim _{{\Delta x}\to 0}{\frac {(x+\Delta x)^{5}-x^{5}}{\Delta x}}=\lim _{{\Delta x}\to 0}{\frac {x^{5}+5x^{4}\Delta x+10x^{3}(\Delta x)^{2}+10x^{2}(\Delta x)^{3}+5x(\Delta x)^{4}+(\Delta x)^{5}-x^{5}}{\Delta x}}=$

$\lim _{{\Delta x}\to 0}5x^{4}+10x^{3}\Delta x+10x^{2}(\Delta x)^{2}+5x(\Delta x)^{3}+(\Delta x)^{4}=5x^{4}$

This is correct, is it not (I know why they use the shortcut now)? --80.229.152.246 12:37, 21 December 2006 (UTC)

I've fixed some trivial copying errors (4 remaining instead of 5), and bracketed most occurrences of Δx. Although, as it is used here, Δx is just a variable, my instinct is to read Δx² as Δ(x²), a change in x². Although the use of Δx may help initial understanding, for practical work I prefer a single letter, as in:

$f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}.$

The power of the differential calculus is in the use of rules like: if f(x) = xⁿ, then f'(x) = nxⁿ⁻¹, which allows you to bypass the definition of limit.  --Lambiam^Talk 14:19, 21 December 2006 (UTC)

You have learned well, and seem to have a correct grasp on the essentials. However, calculus builds on algebra, and you do need to be more careful with the algebra, as others have pointed out.

It may be of interest to see a more exotic approach to derivatives, one that actually is closer to the thinking of Newton. Results in pure mathematical logic allow us to extend the real numbers with some extra quantities. Specifically, we can include a Λ that is larger than any integer, and also include all arithetic combinations using it, such as 2Λ and Λ−1 and so on. Of particular interest here are reciprocals, such as ¹⁄_Λ. This is a quantity which is positive (so not zero), but smaller than any standard real number. We have a "fuzzball" around every standard real number consisting of sums with such infinitesimals.

Let ρ be one such infinitesimal quantity, say the reciprocal of Λ. Now consider the ratio of input change to output change of x⁴ when the amount of change is ρ.

${\frac {(x+\rho )^{4}-x^{4}}{(x+\rho )-x}}={\frac {(x+\rho )^{4}-x^{4}}{\rho }}$

We can expand a power of a sum of two terms using the binomial theorem, which is due to (surprise!) Newton. The expansion of (a+b)ⁿ always has the form aⁿ+naⁿ⁻¹b+(higher order terms in b). In the case under consideration, we have

$(x+\rho )^{4}=x^{4}+4x^{3}\rho +6x^{2}\rho ^{2}+4x\rho ^{3}+\rho ^{4}.\,\!$

We're subtracting off the first term, therefore our ratio simplifies to

${\frac {(x+\rho )^{4}-x^{4}}{\rho }}=4x^{3}+6x^{2}\rho +4x\rho ^{2}+\rho ^{3}.\,\!$

All but the first term is infinitesimal "fuzz". Discard the fuzz to get a standard real answer, 4x³. This is the derivative. Given our observation about the nature of binomial expansions, we can immediately generalize to say that the derivative of xⁿ will be nxⁿ⁻¹.

Newton used his Method of Fluxions in theoretical physics, to obtain equations of motion. Even today it is not uncommon to see physics reasoning "discard infinitesimals" instead of using limits. --KSmrq^T 21:26, 21 December 2006 (UTC)

Isn't your "infinitesimal fuzz" just an alternate definition of the derivative?

$f'(x)=\lim _{{x}\to a}{\frac {f(x)-f(a)}{x-a}}$

--Ķĩřβȳ♥ŤįɱéØ 00:12, 24 December 2006 (UTC)

I'm not sure how to respond. The topological machinery needed for limits is not at all the same as the logical machinery needed for a non-standard model of the reals. Yet we will be very disappointed if we do not reach the same conclusion for this example!

Here's a summary of what I described: Take a question about standard real numbers. Do some calculations using a non-standard model that includes infinitesimals. (Sophisticated arguments in mathematical logic guarantee the existence of such a model.) Note that the result is usually not a standard real result. Discard the non-standard part. Take the standard real part as the answer.

The "fuzz" is part of the non-standard model which we used as a tool; it is not a part of the derivative.

The definition you gave depends on a concept called a "limit". It, too, is a technical tool we use in our calculations, then discard. The derivative of x⁴ is 4x³, an answer that hides all evidence of how we computed.

Consider a function for a semicircle, the top half of x²+y² = 1. Algebraically, the function is f(x) = (1−x²)^1/2. If we define a derivative as a tangent to the curve, we can use geometry to learn about the answer. A tangent to a circle is perpendicular to a radius there. Therefore a tangent at

$(x,{\sqrt {1-x^{2}}})\,\!$

will be in the direction

$({\sqrt {1-x^{2}}},-x),\,\!$

so the ratio of vertical displacement to horizontal displacement is

$-{\frac {x}{\sqrt {1-x^{2}}}}.$

This is, in fact, the derivative. We found it without limits and without infinitesimals. We expect that we will get the same result if we use either of those methods.

The usual method for this expression uses no geometry, no limits, no infinitesimals. Instead it assembles known facts without regard to how they were found. The first fact is the general rule about the derivative of a power, which holds even when the power is not an integer, because of the binomial theorem. Thus the derivative of x² is 2x, and the derivative of x^1/2 is ¹⁄₂x^−1/2. The second fact concerns composition of functions. If we know that the ratio of output change to input change of g(x) is α, and that the ratio for h(x) is β, then for f(x) defined as g(h(x)) the ratio is the product, αβ. The third fact concerns weighted sums of functions, with constant weights; the rule is that we can work on each term independently. Assembly of the facts is left as an exercise for the reader.

I am trying to show that the idea of a derivative transcends the particular tool(s) we use to define or calculate it. --KSmrq^T 05:09, 24 December 2006 (UTC)

Yes, I agree with you. I'm just saying that, if you replace the infinitesimal p with a variable, and then place $\lim _{{p}\to 0}$ in front, we would reach the same result. --Ķĩřβȳ♥ŤįɱéØ 07:01, 25 December 2006 (UTC)

Actually, my infinitesimal is not "p" (lower case Latin "pee"), but "ρ" (lower case Greek "rho"), for "reciprocal" (of Λ for "large"). But yes, our infinitesimal can be as small as we like. The idea of "taking the limit" as ρ goes to zero mimics the act of "discarding the fuzz". Early calculus did not use limits; that approach was introduced partly as a response to a criticism by Bishop Berkeley, who ridiculed infinitesimals:

• “And what are these Fluxions? The Velocities of evanescent Increments? And what are these same evanescent Increments? They are neither finite Quantities nor Quantities infinitely small, nor yet nothing. May we not call them the Ghosts of departed Quantities?”

The basic technical challenge troubles students to this day. If ρ is zero, then we cannot divide by it; but if ρ is not zero, then we cannot get the clean answer we want. --KSmrq^T 11:40, 25 December 2006 (UTC)

Fermat's Last Theorem

I know that a^6+b^6=c^6 admits no non-trivial solution in the integers.But this can be written as (a^3)^2+(b^3)^2=(c^3)^2 which leads us to the form of Pythaogars' identity A^2+B^2=B^2 where namely A=a^3,B=b^3,and C=c^3....This has intinitely many integer solutions.How can this be? Am i missing something?

"This has infinitely many integer solutions". No it hasn't. You could pick A and B at random, but C would have to be a cube of an integer. yandman 16:37, 21 December 2006 (UTC)

Yes, there are infinitely many integer solutions to Pythagoras' equation, but there are none in which A, B and C are themselves cubes. Gandalf61 17:32, 21 December 2006 (UTC)

The Koch Snowflake

Hello, I am a student at school. I was practicing questions from some reference books (Probably completing an assessed task for their IB Diploma internal assessment as this task is used regularly - the answer will be claimed as their own). I came across the following difficulty, The question was based on the Koch snowflake. It says that as the number of stages go on the Area also changes. I found a relation between the area and the number of stages. It goes like this-

 An=An-1 + [(1/3)*(4/9)n-1 * A0]

Where A_n is the area at the n stage and A₀ is the area at the 0 stage or first stage. The area at the first stage is the squareroot of 3 divided by 4. Now how do I prove this relation using PMI i.e. the Principle of Mathemathical Inducton. If you think that I have explained vaguely, please visit the site below. Please could somebody reply soon.

http://mathworld.wolfram.com/KochSnowflake.html

You could start with the true statement that after the first stage or two, the added areas decrease by a factor of 4/9 with every stage. Then you could use induction to relate the added area at a given stage to some initial area (I don't recommend A0) through a power of (4/9).

A second use of induction would take the formula for the area added at a given stage and then prove the validity of a formula expressing the total area at a given stage.

Since geometric progressions are well-understood, in some sense the induction has already been done for you by long-dead mathematicians, and you can move between steps without thinking about induction at all. It depends on what exactly you want to study. If you're not comfortable with induction yet, it might be good practice to flesh out my comments. If you're just studying geometry, it might be better to think about possible shortcuts. Melchoir 20:04, 21 December 2006 (UTC)

Sir, thank you for your kind attention and time. But I could not understand the meaning of your solutions. I mean could you please show me mathematically. I am comfortable with induction and the question specifically says that we have to prove it by induction. The following is the question:

In 1904 Helge Von Koch identified a fractal that appeared to model the snowflake. The fractal is built by starting with the equilateral triangle and removing the inner third of each side, building another equilateral triangle at the location where the side was removed, and then repeating the process indefinitely. The process is illustrated above, showing the original triangle at stage 0 and the resulting figures after one, two and three iterations.

Method: Let N_n = the number of sides, l_n = the length of a single side, P_n = the length of the perimeter, and A_n = the area of the snowflake, at the nth stage.

1. Using an initial side length of 1, create a table that shows the values of Nn, ln, Pn and An for n=0, 1, 2 and 3. Use exact values in your results. Explain the relationship between successive terms in the table for each quantity N_n, l_n, P_n and A_n.

2. Using your general expression for A_n and the iterative relationship between A_n+1 and A_n already established in step 1 above, prove the general expression for A_n by induction.

could you please tell me the answer mathemathically?

I'm not sure what you need more help with, since the page you link to has a fairly complete set of computations. Perhaps my esteemed colleague has already told you all you need to know, though it seems not. So, as usual, I will speculate and ramble; I hope it helps.

First, look at the initial triangle, its area, and one side. The step to the next stage replaces a side which is a single line segment of length s, say, with four line segments each of length ¹⁄₃s. The area, A, that was inside the triangle is undiminished; rather, new area is added on each side in the form of an equilateral triangle with ¹⁄₃ the side and thus (¹⁄₃)² the area of the original.

Now our task is to handle all the bookkeeping.

• We count stages with n, where the initial triangle is n = 0. At each stage n increases by 1.
• We count sides with N_n, where initially we have N₀ = 3. At each stage N increases by a factor of four, as noted above. Thus N_n = N₀·4ⁿ = 3·4ⁿ.
• We count side length as s_n. At each stage s decreases by a factor of ¹⁄₃, also noted above. Thus s_n = s₀·(¹⁄₃)ⁿ.
• We count area as A_n. In passing from stage A_n−1 to stage A_n, we add N_n−1 new triangles, each with area A₀·(^s_n⁄_s0)². Thus A_n = A_n−1+A₀·(3·4ⁿ⁻¹)·((¹⁄₃)ⁿ)².

Obviously the area is not in the form we would prefer (though it provides the answer you requested); instead, we have a summation. Simplifying ((¹⁄₃)ⁿ)² to (¹⁄₉)ⁿ, we have

${\begin{aligned}A_{1}&{}=A_{0}+3A_{0}4^{0}\left({\frac {1}{9}}\right)^{1}\\A_{2}&{}=A_{1}+3A_{0}4^{1}\left({\frac {1}{9}}\right)^{2}\\&{}=A_{0}+3A_{0}4^{0}\left({\frac {1}{9}}\right)^{1}+3A_{0}4^{1}\left({\frac {1}{9}}\right)^{2}\\A_{3}&{}=A_{2}+3A_{0}4^{2}\left({\frac {1}{9}}\right)^{3}\\&{}=A_{0}+3A_{0}4^{0}\left({\frac {1}{9}}\right)^{1}+3A_{0}4^{1}\left({\frac {1}{9}}\right)^{2}+3A_{0}4^{2}\left({\frac {1}{9}}\right)^{3}\\&\vdots \\A_{n}&{}=A_{0}+3A_{0}\sum _{i=1}^{n}4^{i-1}\left({\frac {1}{9}}\right)^{i}\\&{}=A_{0}+{\frac {3}{4}}A_{0}\sum _{i=1}^{n}\left({\frac {4}{9}}\right)^{i}\end{aligned}}$

Although it is not obvious if you have never seen it done before, we can eliminate the summation. In fact, we can show that as n goes to infinity the area approaches exactly ⁸⁄₅ that of the original triangle. Let ρ = ⁴⁄₉. Then the sum is

$S=\rho +\rho ^{2}+\cdots +\rho ^{n-1}+\rho ^{n}.\,\!$

Multiplying this by ρ shifts everything right one place.

$\rho S=0+\rho ^{2}+\rho ^{3}+\cdots +\rho ^{n}+\rho ^{n+1}\,\!$

Subtracting ρS from S cancels almost all the terms, leaving

$S-\rho S=\rho -\rho ^{n+1};\,\!$

then factoring the left-hand side and dividing gives

$S={\frac {\rho -\rho ^{n+1}}{1-\rho }}.$

Pretty, eh? Now since ρ is less than 1 (but positive), ρⁿ⁺¹ becomes smaller and smaller as n grows larger and larger. Thus the sum approaches simply ρ/(1−ρ).

Here endeth my rambling. If anything is still unclear, please let us know. --KSmrq^T 04:42, 22 December 2006 (UTC)

Sir i couldnt understand all the 8⁄5 type symbols in your answer. What do they stand for? And thanks for all the trouble tou took. But i really need to know how to do the second question by P.M.I i.e. The Principle of Mathemathical Induction!! Please sir it is urgent.

8/5 means: 8 divided by 5. It is a fraction, equal to 1.6 (or 1,6 if you use decimal commas). The relation between A_n and A_n−1 is not what you should prove using mathematical induction. What you should prove is the general formula for A_n, which is given by equation (16) on the page you linked to on MathWorld. It contains all the formulas you need. You must prove this equation (16), given (6) and (11) – or equivalently the relation between A_n and A_n−1 you gave above in your original posting, except that the last occurrence of n−1 must be an exponent, and I think it is easier to add 1 to n everywhere in the equation, so that you have:

A_n+1 = A_n + [(1/3) × (4/9)ⁿ × A₀].

Given this relation, the proof of (16) is not a particularly difficult proof. Have you read our article Mathematical induction? Look also at the example. If you don't understand it, or if you get stuck with the proof for Koch's snowflake, please show us what you have and come back with more questions.  --Lambiam^Talk 19:39, 22 December 2006 (UTC)

As Lambiam says, the notation you question is a fraction. Do you see the " ⁄ " character as a diagonal stroke resembling "/"? If not, perhaps your font or browser does not support this common Unicode character, U+2044. The font Arial does along with the browser Firefox, so perhaps you could try that combination. Also the "8" is marked up as a superscript so it is raised and made smaller, and the "5" is marked up as a subscript so it is lowered and made smaller. The special "fraction slash" character is designed to work in this way, where the basic "/" character (called a "solidus" in Unicode, code point U+002F) has the wrong angle and spacing to do so.

I did not give a proof by induction. If, as you say, you are "practicing questions from some reference books", I do not see why your need for an answer is urgent. As clearly stated at the top of this page, we will not do homework for you. Therefore I have deliberately avoided giving the details of such a proof. Even so, MathWorld and I have given you the geometric and algebraic pieces you need to assemble such a proof, and to check its conclusions.

Your questions suggest that you are not working on the proof yourself, but hoping someone else will do it for you. Show us the general formula you will try to prove by induction. Show us the base step. Show us that you understand what an induction step should look like. Tell us where you get stuck. Otherwise, we can tell you nothing more. --KSmrq^T 22:34, 22 December 2006 (UTC)

Pixels per inch

If you have a resolution of 300 dpi and the width is 600 pixels by 400 pixels high, what are the pixels in inches? Thank you.

Compare Pixels per inch and Dots per inch. When you print a typical raster image from a computer on a decent-quality printer, the printer resolution (in dpi) will be greater (finer) than the intended image resolution (in ppi). Printing software is well aware of this, and it doesn't force the pixels per inch to be tied to the dots per inch. So your question doesn't really have an answer, without more information. Melchoir 19:51, 21 December 2006 (UTC)

As I understood the question... 300dpi = 300 dots per inch - so divide 600 'dots' by 300 dots per inch gives 2 inches wide and 400/300 = 1 and one third inch high? Or did you mean the size of the pixel - which will be one three hundreth of an inch.83.100.255.234 19:52, 21 December 2006 (UTC)

The trouble is that pixels and dots aren't the same size, so 600 pixels doesn't imply 600 dots. Melchoir 20:05, 21 December 2006 (UTC)

"300 dpi" only means that an inch can hold 300 dots, and nothing else. You can't convert pixels directly to inches with this information alone because it is, in fact, meaningless in any practical application. Printers, unlike monitors, need a whole lot of little dots to make up a single color (as you can see from the above example image), so a dot can't be compared to a single pixel! That's why you also need to know at what size the image will be printed, in inches or cm. But then your question is trivial. — Kieff 20:19, 21 December 2006 (UTC)

big maths - infinite series?

I participate in an after school math league, and while I understand most of it, some of the stuff on the test is over my head, since I'm only taking Algebra II/Trig this year. We were given 2 problems that I don't get, and I'd like to know how I would go about solving them:

First, a infinite series - it was presented on the test as a series with an ellipse, but Sigma notation should be easier:

For X from 2 through infinity: 1/(X(X+3))

Second one was a really long series:

((1/2-1/3)/(1/3-1/4))*((1/4-1/5)/(1/5-1/6))...((1/2002-1/2003)/(1/2003-1/2004))

Any ideas? ST47Talk 00:08, 22 December 2006 (UTC)

Read telescoping sum.--80.136.157.88 00:34, 22 December 2006 (UTC)

Thanks, that's helpful. That tells me that for n, 1 to infinity, 1/(n(n+3))=1(5/6)/3 = .611

When I subtract n=1 from that(1/4), I get the given answer. Any ideas on the second one? ST47Talk 01:54, 22 December 2006 (UTC)

In the above "5/6" should be "11/6".  --Lambiam^Talk 11:13, 22 December 2006 (UTC)

I guess 1(5/6) was supposed to mean 1+5/6 = 11/6.--80.136.177.142 11:15, 22 December 2006 (UTC)

Do read about telescoping series. Meanwhile, a little algebraic simplification goes a long way here.

1. Write the sum as

${\frac {1}{3}}\sum _{x=2}^{\infty }{\frac {3}{x(x+3)}},$

and use a partial fraction expansion to change this to

${\frac {1}{3}}\sum _{x=2}^{\infty }\left({\frac {1}{x}}-{\frac {1}{x+3}}\right).$

We have massive cancellation in the sum, leaving a much simpler problem.

2. Write the product as

$\prod _{i=2}^{2002}{\frac {1/i-1/(i+1)}{1/(i+1)-1/(i+2)}},$

and simplify the summand to reveal

$\prod _{i=2}^{2002}{\frac {i+2}{i}}.$

Again we have massive cancellation. --KSmrq^T 02:01, 22 December 2006 (UTC)

How are you simplifing #2? Shouldn't

${\frac {1/i-1/(i+1)}{i/(i+1)-1/(i+2)}}$

simplify to

${\frac {i+1-(i+2)}{i-(i+1)}}$

and to

${\frac {-1}{-1}}$

and thusly to

$\prod _{i=2}^{2002}1$

and give a result of 2001? (which is, in fact, the given answer, I just needed an explanation) I just don't get your simplification. ST47Talk 02:27, 22 December 2006 (UTC)

Hmm. I think one (or both) of us has the wrong product. What you initially wrote, without the fancy formatting, was, I think,

${\frac {1/2-1/3}{1/3-1/4}}\times {\frac {1/4-1/5}{1/5-1/6}}\times \cdots \times {\frac {1/2002-1/2003}{1/2003-1/2004}}.$

I notice that this is not quite the same as what I wrote, which would look like

${\frac {1/2-1/3}{1/3-1/4}}\times {\frac {1/3-1/4}{1/4-1/5}}\times \cdots \times {\frac {1/2002-1/2003}{1/2003-1/2004}}.$

That is, you stepped from one factor to the next with 1/2 becoming 1/4, while I went from 1/2 to 1/3. To adopt your version we must step by 2; the product reads

$\prod _{i=1}^{1001}{\frac {1/2i-1/(2i+1)}{1/(2i+1)-1/(2i+2)}},$

replacing i by 2i and adjusting the limits accordingly. This also simplifies dramatically, to

$\prod _{i=1}^{1001}{\frac {i+1}{i}}.$

Unfortunately, this product still yields something different than your stated result, namely 1002. So, rather than discuss the simplification, we should first try to establish exactly what we wish to simplify! --KSmrq^T 05:47, 22 December 2006 (UTC)