Wikipedia:Reference desk/Archives/Mathematics/2007 June 25

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June 25

Additional Mathematics Project Work 2007 (Form 5)

Question: Ang, Bakar and Chandran are friends and they have just graduated from a local university. Ang works in a company with a starting pay of RM2000 per month. Bakar is a sales executive whose income depends solely on the commission he receives. He earns a commission of RM1000 for the 1st month increases by RM100 for each subsequent month. On the other hand, Chandran decides to go into business. He opens a cafe and makes a profit of RM100 in the first month. For the first year, his profit in each subsequent month is 50% more than that of the previous month.

In the second year, Ang receives a 10% increment in his monthly pay. On the other hand, the commission received by Bakar is reduced by RM50 for each subsequent month. In addition, the profit made by Chandran is reduced by 10% for each subsequent month.

1. a) How much does each of them receive at the end of the 1st year? (2 or more method are required for this ques.) [max: besides progressions method, what can i use?? show me..]

b) What is the percentage change in their total income for the 2nd year compared to the 1st year? Comment on the answers.

c) Ang, Bakar and Chandran, each decided to open a fixed deposit account of RM10000 for 3years without any withdrawal. - Ang keeps the amount at an interest rate of 2.5% per annum for a duration of 1month renewable at the end of each month. - Bakar keeps the amount at an interest rate of 3% per annum for a duration of 3months renewable at the end of every 3months. - Chandran keeps the amount at an interest rate of 3.5% per annum for a duration of 6months renewable at the end of every 6months.

(i) Find the total amount each of them will receive after 3years. (ii) Compare & comment on the difference in the interests received. If you were to invest RM10000 for the same period of time, which fixed deposit account would you prefer? Give your reasons.

Further Exploration

2. a) When Chandran’s 1st child, Johan is born, Chandran invested RM300 for him at 8% compound interest per annum. He continues to invest RM300 on each of Johan’s birthday, up to and including his 18th birthday. What will be the total value of the investment on Johan’s 18th birthday?

b) If Chandran starts his investment with RM500 instead of RM300 at the same interest rate, calculate on which birthday will the total investment be more than RM25000 for the 1st time.

PLEASE GIVE ME THE FULL ANSWERS A.S.A.P..THANK YOU..

Go away and do your homework. Come back when you at least have started some of the problems.—Preceding unsigned comment added by 203.49.208.227 (talk • contribs)

Non zero y so that x.y = 0

Hello.

My question is as follows:

If $k\geq 2$ and x is in R^k, prove that there exists y in R^k such that y is non zero but x.y = 0. Is this true if k = 1?

The way I have thought is as follows: R is an integral domain so the result is not true for non zero x if k is 1. For $k\geq 2$ we want that for non zero x,

$\sum _{i=1}^{k}x_{i}y_{i}=0$ there exists a non zero $(y_{1},\ y_{2}\cdots y_{k})$. But this is obvious as the set $\{x_{1},\ \cdots x_{k}\}$ is linearly dependent in R. Hence the result. What I want to know is there some other simpler way of constructing y. Thanks.--Shahab 09:41, 25 June 2007 (UTC)

Without giving it away, consider what it means for a sum to be zero. Some of the summands must be negative. This tells you the sign of some of the components of the vector y. Make another manipulation to easily force the sum to be zero. That will handle the case of k even. The case where k is odd is handled similarly, but with a minor twist.

Seems to me the simplest and clearest approach for k>1 is to construct a specific example. Set $y_{1}=y_{2}=...=y_{k-1}=1$. What value does $y_{k}$ have to take to make x.y equal 0 ? Is the vector y in R^k? If so, you are done. Gandalf61 11:30, 25 June 2007 (UTC)

This approach is pretty much equivalent also.

Gandalf's solution, as presented, fails in the case $x_{k}=0$. Of course, for a nonzero x, at least one coordinate is nonzero, and you can adapt the argument accordingly. -- Meni Rosenfeld (talk) 13:36, 25 June 2007 (UTC)

Ah, yes. Let's just say "details are left to the reader" ! Gandalf61 14:22, 25 June 2007 (UTC)

I'd just like to point out that it's not necessarily true that if a sum is zero, then some term must be negative. Tesseran 06:44, 26 June 2007 (UTC)

Yes, excluding the case where every summand are zero, however some summand(s) must be negative, if we are talking about a field of characteristic zero (which, in this case, I believe we are).

The vectors (1,0) and (0,1) are orthogonal over any field, of any characteristic. The sum to compute their dot product is 1x0 + 0x1 = 0, in which no summand is negative. Tesseran 17:12, 26 June 2007 (UTC)

Every summand is zero! That is the exclusion case I provided.

help me in my add maths project work

how to solve the questions.pls help me.i am lack of time.

Wikipedia's reference desk is not a service to do people's homework, so you should not waste what little time you have asking us to do so. However, if there is anything you do not understand and wish us to clarify, we will be more than happy to help. -- Meni Rosenfeld (talk) 15:08, 25 June 2007 (UTC)

Ellipse

I'm trying to create a drawing program much like paint and the circle drawing tool can obviously draw ellipses from knowing where the mouse's last click position was and where it is now. So how to find out an ellipe's two focus points from it's height and width? I saw nothing about it in the ellipse article, or it was too obscure to decipher. 62.20.156.137 16:51, 25 June 2007 (UTC)

Why do you want to find the foci? The easiest way to draw an axis-parallel ellipse is with the parametric equations $x=a\cos {t}$ and $y=b\sin {t}$, where a is the horizontal radius (half the width), b is the vertical radius (half the height), and t varies from 0 to 2π. However, if you insist on finding the foci, the relevant information is in the first section ("Eccentricity") of the ellipse article - the distance of each from the center, assuming $a>b$, is $a\epsilon$, where $\epsilon ={\sqrt {1-{\frac {b^{2}}{a^{2}}}}}$ is the eccentricity. If $a<b$ you need to reverse their roles. -- Meni Rosenfeld (talk) 17:21, 25 June 2007 (UTC)

counter intuitive probability

A friend of mine blew my mind with a bit of probability the other day, but she did a lousey job explaining it. The scenario is of a game show where one is offered three curtains, one of which is a prize and the others are something else. After one is picked, one of the false curtains is revealed and the contestant is offered the chance to change thier curtain of choice. She said that one should always change curtains, but couldn't articulate why. Is there a name or an article for this scenario?

142.33.70.60 17:45, 25 June 2007 (UTC)

Yes, the article is called Monty Hall problem. --mglg_(talk) 17:52, 25 June 2007 (UTC)

And you are welcome back if you want more persuasion that it's actually true. :-) —Bromskloss 18:48, 25 June 2007 (UTC)

Thanks :) Actually, it turns out that the arguement we were having revolved around a glossed over detail, covered in the article--that it all depends on Monty's rules(If he -always- has to show a goat curtain or not). I was positive that it was 50/50, but now that I know Monty always has to show the curtain, I understand what's going on.142.33.70.60 20:12, 25 June 2007 (UTC)

solve the equation

find the value of x and y where x^y+y^x=17 and x^x+y^y=31

Well I notice that in both equations, we're looking at examples of symmetric functions, meaning that we can exchange $x$ and $y$ without changing the equations. So if there are any solutions (and there are), they'll come in pairs. I'm not sure if that helps at all, though. Donald Hosek 17:59, 25 June 2007 (UTC)

Guess some simple values of x and y and try them. Then keep in mind Donald's tip. --mglg_(talk) 19:01, 25 June 2007 (UTC)

Finding a solution to this problem by guessing is pretty straightforward (even easier if you have a grapher which can graph the equations, the domains of the functions are kind of interesting, actually). I'm starting to wonder if there's any way of leveraging the symmetry to be able to generalize this to solving the systems $x^{y}+y^{x}=k$, $x^{x}+y^{y}=m$. Is there anything interesting to the values of $(k,m)$ which have integral solutions in $(x,y)$? Donald Hosek 19:50, 25 June 2007 (UTC)

That they have integral solutions? Black Carrot 08:09, 26 June 2007 (UTC)

Finding patterns graphically I find interesting. Why don't you plot the $(k,m)$ that satisfy the criterion (and show us too)? —Bromskloss 14:23, 26 June 2007 (UTC)

Solution: x = 2, y = 3; y = 2, x = 3. Probably the only trivial solution. Graphing, numerical approximations or "guessing" will obtain the other solutions, if there are any. Keep in mind that x and y are both nonzero, and that x and y can't both be negative. Also, y ≠ x under any circumstances.

Here's my advice on graphing: sketching the first curve is difficult. It does not touch either of the coordinate axes (in fact they're both asymptotes). The curve exists only in the 1st, 2nd and 4th quadrants. Also plot some symmetric points (reflected in the line y = x, which is also a diagonal asymptote), including (2, 3) and (3, 2). As someone said above, it's an interesting curve.

Now attack the next curve. The y-intercept of the next curve is somewhere between 3 and 4 (as the function y^y is monotone increasing). The x-intercept is the exact same value, again, somewhere between 3 and 4. Note that the points are also reflected in the line y = x, which is an asymptote. Unlike the other curve, this curve touches the coordinate axes. This curve, also, exists only in the 1st, 2nd or 3rd quadrants.

The local extrema of both curves can not be determined by elementary means, so don't bother with them. You should use this information to roughly plot both curves, and from there you can find approximate solutions. You can then use Newton's method to produce more accurate and more approximate solutions. The graph-sketching will be very hard as the curve is discontinuous at so many points and oscilliates. The values at the extreme points of the axes of the curve explode between large positive and negative values.

Actually unsigned, your arithmetic is a bit off. Neither curve can touch the axes. For the first curve, set $x=0$. Then you have $0^{y}+y^{0}=1$ for all non-zero values of $y$. The behavior of exponentiation is not well-defined on the reals when the base is less than zero, so unless we make a definition of exponentiation which will depend on a branch of the complex log function and will by necessity exclude a ray from 0 in its domain (usually the negative reals), we can't talk intelligently about the graph of this function outside of the domain of positive reals, so our graphs will only be in the first quadrant if we assume $x,y\in \mathbb {R}$. There may be complex solutions to this system, but I'd hate to even consider them. Donald Hosek 18:28, 27 June 2007 (UTC)

Good point there, Donald, I had a lapse in concetration and gave $0^{0}$ a numerical definition, when it doesn't have one. I am still not convinced that the two curves are exclusively in the first quadrant though.

The problem is that you need to expand yourself to complex numbers to get out of positive reals, and once you do that, you're faced with the difficulty of defining just what you mean by, say, $-1^{1/2}$ (two possibilities) or worse still $-1^{\sqrt {2}}$ (infinite possibilities). Defining $c^{z}$ requires choosing a branch of the complex log function, since log is a multi-valued function over $\mathbb {C}$. Most of the time, the "standard" domain chosen for log over $\mathbb {C}$ specifically excludes the ray $(-\infty ,0]\subset \mathbb {R}$. We could potentially try and talk about values of the curves over $\mathbb {Z} \setminus \{0\}$, I suppose, since we do have a well-defined approach to integer exponents. One we expand to non-integral exponents, then we start dealing with the fact that we get multi-valued functions which don't have the obvious "right" choice that comes out of staying in the positive reals. It's possible to do, but not practical. Donald Hosek 17:19, 4 July 2007 (UTC)