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February 18

Making n the subject

If A= P(1+r)n^{Superscript text} Then make n the subject of formula —Preceding unsigned comment added by 66.36.215.172 (talk) 10:21, 18 February 2009 (UTC)[reply]

Assuming you mean A = P(1+r)ⁿ look into power rules etc. 89.240.5.5 (talk) 10:38, 18 February 2009 (UTC)[reply]

(added header) Consider taking logarithms of both sides. Confusing Manifestation(Say hi!) 22:34, 18 February 2009 (UTC)[reply]

A rant about inequalities

Don't get them at all, no sir.

Find all x that satisfy:

${\frac {-2x+5}{x+3}}<1$

Now I was told I have to consider two cases, x > 3 and x < 3. x = 3 is obviously undefined.

x > 3:

$-2x+5<x+3\,$

$-3x<-2\,$

$x>{\frac {2}{3}}$

So x > 3 and $x>{\frac {2}{3}}$ . I guess that means $x>{\frac {2}{3}}$ ?

x < 3:

$-2x+5>x+3\,$

(Since x+3 must be negative)

$-3x>-2\,$

$x<{\frac {2}{3}}$

We get the same kind of logic as before. This is how it was explained to me. I won't even start on the fact that there's a numerator that has an x in it as well, not to mention that you can probably rearrange simple inequalities as rational ones and no doubt create a world of pain for yourself anyway. Someone is playing a joke on me.

Anyhow I was told that the method I'm using sucks and that I should solve the inequality by getting the right hand side to zero first.

${\frac {-2x+5}{x+3}}-1<0$

${\frac {-3x+2}{x+3}}<0$

Okay, so above is negative for x > 0, but also for x < -3. That's what I've ascertained from examining the equation with my head, which is not a nice algorithmic process for solving problems, and there's probably something I've overlooked. I'm skeptical now because all the other answers in this useless book are of the form a < x < b. But I wouldn't be surprised if it's part of the elaborate joke to make all the even-numbered questions of a completely different type to the odds. Ha. Ha. HA. Help me before I kill myself. 82.32.49.186 (talk) 10:56, 18 February 2009 (UTC)[reply]

You are approaching the problem correctly. Both of the methods you described work and can be used to solve this problem, but unfortunately you made numerical mistakes in both approaches.

In your first approach: under what conditions is x+3 positive? You use that x+3 is positive if and only if x > 3, but that is incorrect. Similarly for negative.

Also in your first approach, you have inadvertantly switched "and" and "or". When is it true that both x > 3 and x > 2/3? The answer is x > 3, not x > 2/3. (However, it was supposed to be x > -3 anyhow.)

Although your first approach does indeed work, I agree that the second approach is easier. You correctly reach the inequality

{\frac {-3x+2}{x+3}}<0

, but then incorrectly state that that holds for x > 0 and x < -3. As you haven't explained your reasoning in detail, I can't help you find your mistake, but maybe you should try x = 1/2 as an example.

Finally, the answer will not be of the form a < x < b. Please let us know if you would further help with this problem; it looks to me like you are understanding this material well, and just making easy-to-make numerical errors. Eric. 131.215.158.184 (talk) 11:41, 18 February 2009 (UTC)[reply]

Through 5 or 6 mintues of tedious trial and error, I examined the points where the sign of the function (the rational expression, treat it as a function) switched. The function, according to these calculations is negative for all x < -3 and all x > (2/3), if I got it right. I suspect that the trick is to examine the cases less than or equal to where the denominator is undefined, and where the numerator is = 0. Does this trick hold for every rational, polynomial expression? Brute force is not my idea of a good time 82.32.49.186 (talk) 12:10, 18 February 2009 (UTC)[reply]

You are (roughly) correct: for any rational expression a / b (whether polynomial or otherwise), the sign of a / b can be determined by knowing the signs of a and b. Thus the "critical points" to examine are when the numerator a or denominator b are zero, for those are the only points where the expression a / b can change sign. I am sure others can pick it up from here if you have further questions, I am going to bed. (And by the way, your answer is correct.) Eric. 131.215.158.184 (talk) 12:34, 18 February 2009 (UTC)[reply]

My book gives -3 < x < 2/3 as the answer. What the hell? 82.32.49.186 (talk) 12:48, 18 February 2009 (UTC)[reply]

Did you give us the problem correctly? Then the book is wrong. Try x = 0 if you have doubts. Eric. 131.215.158.184 (talk) 18:54, 18 February 2009 (UTC)[reply]

An obvious thing to do to get a quick feel whether your results are the right ones is to use a spreadsheet to calculate and graph f(x) against x. In your case maybe plot values from x = -5 to +5 in steps of 0.1? and just see which values of x have f(x) < 1. That will help you spot gross blunders (eg +3 instead of -3 and so on easily. -- SGBailey (talk) 12:46, 18 February 2009 (UTC)[reply]

If you say x > 2/3 AND x > 3, then that means x > 3. You cannot ignore the fact that the word AND means AND, nor the fact that 3 > 2/3.

Next, where you say x < 2/3 AND x < 3, then that means x < 2/3.

So the solution is x > 3 OR x < 2/3. Any number greater than 3 is a solution; any number less than 2/3 is a solution; no other numbers are solutions.

Your other method also works. You have

{\frac {-3x+2}{x+3}}<0

and then you can divide both sides by −3, which requires changing "<" to ">" since what you're dividing by is negative:

{\frac {x-(2/3)}{x+3}}>0.

Now either x > 3 or x < 2/3 or x is either between those two or equal to one of them. If x > 3 then both the numerator and denominator are positive, so you've got a solution. If x < 2/3 then both are negative, so you've got a solution. If x is between the two, then the denominator is negative and the numerator is positive, so you haven't got a solution. Michael Hardy (talk) 21:38, 18 February 2009 (UTC)[reply]

An alternative approach involves the rule that "multiplying both sides of an inequality by a strictly positive value does not affect the inequality". Instead of testing individual cases, multiply everything by $(x+3)^{2}$ . For $x\neq 3$ , this is guaranteed to be positive, and you can then rearrange the inequality to give you (quadratic) > 0. Then a factorisation of the quadratic and a quick check of where it's positive should spit out the answer (just remember to check for any funny business around the x=3 point as well). Confusing Manifestation(Say hi!) 22:32, 18 February 2009 (UTC)[reply]

Problem from Linear Algebra

The following question is from Kluwer's Foundations of Linear Algebra and has been driving me crazy for the last day. I imagine the solution is simple and that I am just missing it, whatever the case, any help would be deeply appreciated:

    Let A be a nonsingular 4 x 4 matrix so and B, C, D, and E be 2 x 2 matrices with A = (B C)
                                                                                         (D E)
    can BE - DC, BE - CD, EB - DC, and EB - CD all be singular?

The best result I can get is that if B and D commute or C and E, then no; and if the commutator of B and E or of D and C is singular, then no (not 100% sure that I didn't make a mistake on the last one though.) Again, thanks in advance for any help:) Phoenix1177 (talk) 11:18, 18 February 2009 (UTC)[reply]

With the help of the partial results you provided, I found a counterexample with a little trial and error (i.e., a nonsingular four by four matrix A such that BE - DC etc. are all singular). However, in my example the commutator of B and E and of D and C are singular, so I'd suggest checking your proof of that fact. I'll leave the joy of hunting for counterexamples to you. Eric. 131.215.158.184 (talk) 11:51, 18 February 2009 (UTC)[reply]

I assumed that I made a mistake there, it's late here. At any rate, I was hoping that it wouldn't come down to hunting out a counterexample...66.202.66.78 (talk) 12:16, 18 February 2009 (UTC)[reply]

I (or someone else) can give you a hint in the morning if you want one. As usual, when there exist counterexamples, there is often a very simple counterexample. Eric. 131.215.158.184 (talk) 12:38, 18 February 2009 (UTC)[reply]

Thank you, I actually found a counterexample as soon as you mentioned there was one. This is more a case of bad textbook logic, on my part, than anything else; since this was in a chapter on determinants, and finding a counterexample didn't seem to have much to do with determinants, I assumed that a proof was being looked for...obviously, the answer is in the negative, so no proof. Thank you :) P.S. The stuff about the commutators was assuming one of them was invertible. Phoenix1177 (talk) 13:15, 18 February 2009 (UTC)[reply]