can help you win a girlfriend?

In my experience you cannot legally 'win' a girlfriend.

Maybe it is advice to help you improve your chat up line.

This probably only works well if the girl is keen on math.

If you don't know enough in advance about the girl, a more generally applicable version:

Bob

]]>You need to supply a more detailed algorithm.

eg.

Find minimum values

If only one, state its index.

If more than one, what?

(a) state the one in the middle

(b) state the one nearest to index 4

(c) state ..........?

Bob

]]>g(2x)=f(x) <=> g(x)=f(x/2)

The first statement is true for all x. So you can replace x by x/2 and it will still be true.

Similarly, if the second statement is true for all x, then replace x by 2x and it will still be true.

Here's a simple example.

f(x) = 6x + 7 = 3 times (2x) +7

So we can define g by:

g(2x) = 3 times (2x) + 7

which means g(2x) = f(x)

The definition of g may also be written:

g(x) = 3x + 7 = 6 times (x/2) + 7 = f(x/2)

So g(x) = f(x/2)

Bob

]]>Hope that helped and welcome to the forum.

]]>I've not seen this way of defining point symmetry before so I've had to do a bit of experimenting. When I come across a tricky bit of math, I try to simplify the numbers first, and also pick an example for what I'm trying to investigate.

For symmetry, I wouldn't choose a trig. function because the graphs have too many lines and points of symmetry, so its impossible to see what's going on.

I chose y = x^3, because it only has point symmetry around (0,0)

You'll find it is worth using the function grapher at

http://www.mathsisfun.com/data/function … c1=(1+x)^3

I've set this up with the function y = (1+x)^3. Try some values:

x = -3 , y = -8

x = -2 , y = -1

x = -1 , y = 0

x = 0 , y = 1

x = 1 , y = 8

You can see from this that the graph had point symmetry around (-1,0)

Try changing that '1' into other numbers and you'll see that, with a = -1

And f(a-x) = -f(a+x)

Now let's try adding a constant.

y = x^3 + 2

You'll see that this shifts the graph up by 2, so the symmetry point is moved to (0,2)

Then

Here I've put square brackets around the function and introduced an additional '2' (either + 2 or -2) to balance the equation

so in general

You don't need the square brackets here, but I thought it would help you to compare with the line above if I left them in.

This function as point symmetry around (0,b)

Finally put both bits together:

So y = (1+x)^3 + 2

x = -3 , y = -6

x = -2 , y = 1

x = -1 , y = 2

x = 0 , y = 3

x = 1 , y = 10

Try it on the grapher as well.

You'll see the point symmetry is now around (-1,2)

Again you can vary the value of the constant.

So if I put in square brackets for the function again:

y = [(1+x)^3 + 2]

a = -1 and b = 2

and

so in general

with the point of symmetry (-a,b)

Bob

]]>If the price increase one week is p pence, find expressions in terms of p for:

-the number N of items sold weekly;

-the production cost £C;

-the revenue £R from the sales.

-Show that the profit £P is given by

P = 250p – 10200 – p^2

]]>This problem has already been solved:

a(n)=(1/6)*(3*n^3+3*n)

http://oeis.org/search?q=1%2C5%2C15%2C3 … &go=Search

Post back if you can not wade through their work.

]]>Time to say, welcome to the forum!

]]>