Wikipedia:Reference desk/Archives/Mathematics/2009 May 17

= May 17 =

does anyone know how to solve these?
1. [(2+1)(22+1)(24 + 1)(28 + 1)(216 + 1) + 1]/ 233

2. a rectangle has its sides of length sinx and cosx for some x. what is the largest possible area it can have?

3. ROW 1 - 1 ROW 2 - 3 5

ROW 3 - 7 9 11

ROW 4 - 13 15 17 19 etc...

what's the number at the end of ROW 80?? —Preceding unsigned comment added by 122.50.137.185 (talk) 05:58, 17 May 2009 (UTC)


 * When asking questions on the reference desk, please tell us what progress you have made so far towards the solution and where you are stuck. This allows us to give answers that are most helpful, best directed towards your level, and takes less of our time guessing what it is you need help with.


 * For the first question, you need to expand $$(x + 1)(x^2 + 1)(x^4 + 1) \cdots (x^{2^k} + 1)$$. It may help to write this out by hand for a few small k to get an idea of what is going on.  To expand this for general k, I see two appraoches;  the easier one is to multiply by x - 1 (and then remember to divide back by x - 1 later), and the other uses the fact that any integer from 0 to $$2^{k + 1} - 1$$ inclusive can be uniquely written as a sum of powers of 2 up to $$2^k$$, a property of the binary writing system.


 * For the second question, what is the area of the rectangle with sides $$\sin x$$ and $$\cos x$$? From there you can solve the problem several different ways:  the most direct is to use calculus to maximize the function;  or observe that the area function is symmetric about a value of x, which must be its maximum;  or use a trigonometric identity to simplify the area function into an easier-to-use form with a very obvious maximum.  Or you can just guess the answer, as there's really only one reasonable guess.


 * For the third question, I'll need to know where you are stuck before I can help. Eric.  131.215.159.91 (talk) 08:45, 17 May 2009 (UTC)


 * For the third question, observe that all the numbers are odd, so add one to every number and divide by 2. Does that make the pattern easier to see? Can you then extend that pattern to row 80? Then you just double it and subtract 1. -- SGBailey (talk) 22:03, 17 May 2009 (UTC)


 * For the first question it will help to know that 20 = 1 and 21 = 2. Then you have some multiplying to do which can all be done by additions since 2a x 2b = 2a+b. Similarly if you need to divide then 2c / 2d = 2c-d.
 * For the second question draw a rectangle with a diagonal whose length is 1. Apply what you know about the definitions of sine and cosine for a right triangle.
 * For the third question write down and look at the rows in order on a single line [ROW1][ROW2][ROW3][ROW4]. Can you see 1) what happens at each step along the chain, and 2) how many numbers of the chain go to make up each row? BTW Row 4 is complete as given, the "etc." means there are successive Rows 5, 6 and so on. Cuddlyable3 (talk) 11:34, 19 May 2009 (UTC)

The literal answer to the OP is Yes! Cuddlyable3 (talk) 11:36, 19 May 2009 (UTC)

Fairly simple $$\nabla$$/inequality problem
Hi there -

I've got a very straightforward question here, it's short and simple but I still can't work out how to solve it and it's becoming very annoying, I was wondering if anyone could give me a hand (although it seems so short I wonder whether there's even anything to help or if I'm just missing the blindingly obvious!) -

Consider the cone C in $$\mathbb{R}^3$$ deﬁned by $$x_3^2 = x_1^2 + x_2^2$$, $$ x_3 > 0$$ Find a unit normal $$n = (n_1, n_2 , n_3 )$$ to C at $$x = (x_1 , x_2 , x_3 )$$ such that $$n_3 \geq 0$$ - fine so far, I used the grad function and just took the upwards-pointing normal.

Show that if $$p = (p_1, p_2 , p_3 )$$ satisﬁes $$p_3^2 \geq p_1^2 + p_2^2$$ and $$p_3 \geq 0$$ then $$p \cdot n \geq 0$$.

This is where I get stuck - intuitively I can see why it's true, since p is 'trapped' inside the cone, so must be at an angle < $$\frac{\pi}{2}$$ with the cone, but I'm not sure how to prove it formally. I'm trying to avoid using the $$a \cdot b = |a||b|cos \alpha$$ definition, rather the $$a \cdot b = \sum a_i b_i$$ one if possible - although naturally if the former is significantly simpler then I'd take either! It doesn't seem like it should take much work, I just can't see how to transfer the 'x's and 'p's and keep the inequality correct. Thanks a lot! Delaypoems101 (talk) 06:21, 17 May 2009 (UTC)


 * For the norm n we know $$n_1^2 + n_2^2 \leq n_3^2$$ (actually, we have equality, but we don't need that). So we have
 * $$n \cdot p = n_1p_1 + n_2p_2 + n_3p_3 \geq n_3p_3 - |n_1p_1 + n_2p_2| = n_3p_3 - |(n_1, n_2) \cdot (p_1, p_2) |$$.
 * We can do two different things here. One is to use the fact that the cone is rotationally symmetric to simply;  we can take p_2 = 0, for example, so we get $$n \cdot p \geq n_3p_3 - |n_1p_1|$$ and we are nearly done.  A more elegant approach is to use the Cauchy-Schwarz inequality to say
 * $$|(n_1, n_2) \cdot (p_1, p_2)| \leq \sqrt{n_1^2 + n_2^2} \sqrt{p_1^2 + p_2^2} \leq n_3p_3$$.
 * Eric. 09:05, 17 May 2009 (UTC) —Preceding unsigned comment added by 131.215.159.91 (talk)

LaTeX fraction spacing
If, I have, say:



\begin{array}{rclr} \dfrac{1}{2}&=&\dfrac{1}{4}+\dfrac{1}{4}\\ &=&\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8} \end{array} $$

You can see the fractions start to bunch themselves together over multiple lines.

Is there a way to fix this? I don't want to use  because I have a huge document and want to avoid finding out which lines are problematic (some lines, obviously, don't bunch together - only when   is involved). Is there a length command I can adjust via ? x42bn6 Talk Mess 23:08, 17 May 2009 (UTC)

You should be using "align", not "array", for this sort of thing. Here it is:

\begin{align} \dfrac{1}{2} & = \dfrac{1}{4}+\dfrac{1}{4} \\ & = \dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8} \end{align} $$

Also, one can add more space between lines as follows:

\begin{align} \dfrac{1}{2} & = \dfrac{1}{4}+\dfrac{1}{4} \\[6pt] & = \dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8} \end{align} $$ Michael Hardy (talk) 23:59, 17 May 2009 (UTC)
 * That was the last thing I wanted to hear... I suppose now I have to find a way of going through all 168 occurrences of the   environment in my document and analyse things...  Thanks, anyway.  I did find an alternative which is   which you can set and unset after each problematic environment (or do globally).  x42bn6 Talk Mess  02:03, 18 May 2009 (UTC)

Using the "array" environment, one can still add space between the lines:

\begin{array}{rclr} \dfrac{1}{2}&=&\dfrac{1}{4}+\dfrac{1}{4}\\[12pt] &=&\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8} \end{array} $$ Michael Hardy (talk) 10:36, 18 May 2009 (UTC)