The maths thread

[Geoff]

It took me a few seconds to get what you were saying there but it just clicked. It still seems rather counter intuitive to me

Takes practice I think. You basically have to find a way to involve a "perfect square" in the expression, then mess about with the left-over bits.

[Xamonas Chegwé]

OK - hint time!

Try this one for a laugh. For what negative values of x does the function f(x) = x^x (that's x to the power x) have values that are: -
(a) Positive and real?
(b) Negative and real?
(c) Imaginary?
(d) Undefined?

There are infinite solutions but they can be expressed quite succinctly.

Clue: Think rationally!

You can also try differentiating the function as well - not especially difficult but needs a little head-jugglery.

The key to this is to take logs of both sides.

y=xx is equivalent to loge(y) = loge(xx) for positive values of x. (The function is not differentiable for negative x.)

Which gives loge(y) = xloge(x)

so, reversing the taking of logs, we get y = e(xloge(x))

I'll leave you to apply the product and chain rules to finish it off.

[CJ]

[JimC]

OK - hint time!

Try this one for a laugh. For what negative values of x does the function f(x) = x^x (that's x to the power x) have values that are: -
(a) Positive and real?
(b) Negative and real?
(c) Imaginary?
(d) Undefined?

There are infinite solutions but they can be expressed quite succinctly.

Clue: Think rationally!

You can also try differentiating the function as well - not especially difficult but needs a little head-jugglery.

The key to this is to take logs of both sides.

y=xx is equivalent to loge(y) = loge(xx) for positive values of x. (The function is not differentiable for negative x.)

Which gives loge(y) = xloge(x)

so, reversing the taking of logs, we get y = e(xloge(x))

I'll leave you to apply the product and chain rules to finish it off.

XC, you are officially the forum's maths guru! That goes one step beyond my level. I get it up to and including:

loge(y) = xloge(x) (taking logs is useful in solving indicial equations),

then I got lost...

[Xamonas Chegwé]

OK - hint time!

Try this one for a laugh. For what negative values of x does the function f(x) = x^x (that's x to the power x) have values that are: -
(a) Positive and real?
(b) Negative and real?
(c) Imaginary?
(d) Undefined?

There are infinite solutions but they can be expressed quite succinctly.

Clue: Think rationally!

You can also try differentiating the function as well - not especially difficult but needs a little head-jugglery.

The key to this is to take logs of both sides.

y=xx is equivalent to loge(y) = loge(xx) for positive values of x. (The function is not differentiable for negative x.)

Which gives loge(y) = xloge(x)

so, reversing the taking of logs, we get y = e(xloge(x))

I'll leave you to apply the product and chain rules to finish it off.

XC, you are officially the forum's maths guru! That goes one step beyond my level. I get it up to and including:

loge(y) = xloge(x) (taking logs is useful in solving indicial equations),

then I got lost...

The next step is just antilogs - or raising e to the power of the expression on both sides - again, this is permissible because we are only considering positive values of x by now. I missed a half-step which I thought would be obvious. Does this help?

loge(y) = xloge(x)

==> eloge(y) = e(xloge(x))

and since eloge(y) is just y

==> y = e(xloge(x))

Is that clearer? (Fuck it is hard to write maths without dedicated fonts / software!)

And don't claim you can't apply the product and chain rules - finish it off for me!

[Chinaski]

Would any of the mathematicians on here be willing to help me with matrices? I'm missing something...

[Xamonas Chegwé]

Would any of the mathematicians on here be willing to help me with matrices? I'm missing something...

I used to be pretty hot on matrices - I may need to revise a bit but if you have any specific areas of missing, I can probably assist.

[Geoff]

[Chinaski]

Ok so the problem is the following (for now): X is a 2x2 matrix where a=1, b=1, c=1 and d=1. Y is a 2x2 matrix where a=1, b=-1, c=-1 and d=1 (a to d are upper left to lower right). I need to find ways to express the general statement for X^n, Y^n, and (X+Y)^n. I have no fucking idea where to start. I'm guessing it's got something to do with geometric series, but my memory of that is reaaally hazy.

[Xamonas Chegwé]

Ok so the problem is the following (for now): X is a 2x2 matrix where a=1, b=1, c=1 and d=1. Y is a 2x2 matrix where a=1, b=-1, c=-1 and d=1 (a to d are upper left to lower right). I need to find ways to express the general statement for X^n, Y^n, and (X+Y)^n. I have no fucking idea where to start. I'm guessing it's got something to do with geometric series, but my memory of that is reaaally hazy.

I would say that the first place to start would be to look at a few powers of these matrices and see if you can generalise from that. I am assuming that you know how to perform matrix multiplication and addition - so try raising the expressions to ^2, ^3, ^4, etc. and look for patterns. They should become obvious really quickly.

PM me if you want another hint.

[Xamonas Chegwé]

Try this one for a laugh. For what negative values of x does the function f(x) = x^x (that's x to the power x) have values that are: -
(a) Positive and real?
(b) Negative and real?
(c) Imaginary?
(d) Undefined?

There are infinite solutions but they can be expressed quite succinctly.

You can also try differentiating the function as well - not especially difficult but needs a little head-jugglery.

Answers after the Leeds meet up.

I forgot all about this! I will talk you through the solution, which is actually rather interesting (in a mathsy sort of way! )


Let's look at negative integers first, as that is the simplest case.

-1-1 = 1 / -1 = -1
-2-2 = 1 / (-2)2 = 1/4
-3-3 = 1 / (-3)3 = -1/27
-4-4 = 1 / (-4)4 = 1/256

The pattern continues in this way, with odd, negative integers yielding negative, real values and even, negative integers yielding positive real values. Although this shows that the function is defined for some negative values at least, it tells us very little else (although the alternation of positive and negative values hints that the function may not be continuous.)

If we then look at rational numbers, there are several cases:

(i) Odd denominator and odd numerator in the lowest form.
(ii) Odd denominator and even numerator in the lowest form.
(iii) Even denominator and odd numerator in the lowest form.

(NB. If the numerator and denominator are both even, the rational number is not in the lowest form, as it may be divided by 2 top and bottom!)


Now let us look at what the numerator and denominator mean in terms of fractional powers.

The numerator says "raise the number to this power."
The denominator says "take this root of the number."

eg. X3/5 means "the fifth root of X cubed" (NB. it doesn't make a difference if you perform the powering or the cubing first, the result is identical.)

But we are dealing with negative powers. What the minus sign does, is to take the inverse of the power - eg. X-3/5 = 1 / X3/5 = "one over the fifth root of X cubed".

This inversion however, has no effect on whether the result is positive, negative, real or imaginary - as the inverse of any of those cases will always have the same characteristic.


So where does that leave us? We have a negative number which we are going to raise to a certain power, take another certain root of, and then invert.

Let us look at the three cases above one by one:


(i) Odd denominator and odd numerator in the lowest form.

We first raise this number by an odd power. A fact about odd powers is that they preserve the sign of the number being raised to that power. Hence the result is still negative.
We next take an odd root of this negative result. A fact about odd roots of negative numbers is that there is always a real, negative solution.
As I mentioned above, inverting has no effect on either reality or sign.

So an odd denominator and odd numerator in the lowest form yields a real, negative solution. This corresponds to (b) in the original problem.


(ii) Odd denominator and even numerator in the lowest form.

We first raise this number by an even power. A fact about even powers is that the sign of the result is always positive.
We next take an odd root of this positive result. A fact about odd roots of positive numbers is that there is always a real, positive solution.
As I mentioned above, inverting has no effect on either reality or sign.

So an odd denominator and even numerator in the lowest form yields a real, positive solution. This corresponds to (a) in the original problem.


(iii) Even denominator and odd numerator in the lowest form.

As in (i), we first raise this number by an odd power. A fact about odd powers is that they preserve the sign of the number being raised to that power. Hence the result is still negative.
We next take an even root of this negative result. A fact about even roots of negative numbers is that there are only ever imaginary results!
As I mentioned above, inverting has no effect on either reality or sign.

So an even denominator and odd numerator in the lowest form yields only imaginary solutions. This corresponds to (c) in the original problem.


Irrational powers have no simple definition. They are instead defined as having a value between arbitrarily close rational powers, on either side, for continuous functions only.

As it is always possible to find an example of (iii) between any two given examples of either (i) or (ii) when x is negative, we can be sure that y=xx is not continuous over any range when x < 0. Thus y is undefined for irrational, negative values of x. This is (d) in the original question.


I will complete the differentiation at another time. I'm mathsed out!

[Durro]

What is half of two plus two ?

[Feck]

or half of two , plus two ?

[Durro]

As it's written.

Most people answer "2", but of course, the answer is "3".

Multiplying and dividing take precedence over addition and subtraction, so unless there are brackets, the sum should read like :-

what is half of 2......plus 2 ?

Almost nobody out of the dozens of people I've given it to get the right answer.

[Feck]

BODMAS I remember