MaplePrimes - answers and comments on Question, The symmetry axe of a quadratic function?

For example

hirnyk — Sun, 12 Sep 2010 13:48:24 Z

For example, as follows

> p := 4*x^2-7*x+80; q := x^7-p;

                               4 x^2 - 7* x + 80

                            x ^7 - 4* x ^2 + 7 *x - 80
> AXE := proc (p) if degree(p, x) = 2 then a := coeff(p, x^2): b := coeff(p, x):
- (1/2)*b/a else undefined end if end proc:

Warning, `a` is implicitly declared local to procedure `AXE`
Warning, `b` is implicitly declared local to procedure `AXE`
> AXE(p);

                                      7
                                      -
                                      8
> AXE(q);
                                  undefined

PS. It seems like a homework. In that case you should present what you did to produce the result on your own.

PPS. If p is given as a function

p := x -> 4*x^2-7*x+80; ,

then the procedure AXE should be replaced by

AXEF := proc (p) if degree(p(x), x) = 2 then a := coeff(p(x), x^2): b := coeff(p(x), x):
-(1/2)*b/a else undefined end if end proc .

?extrema

Oliver K — Sun, 12 Sep 2010 17:01:45 Z

The x value of the symmetry axe is also an extremum, so you also could use ?extrema for this purpose. Or use completesquare to find x.

Isn't this a bit more straightforward ?

longrob — Sun, 12 Sep 2010 17:29:57 Z

Isn't this a bit more straightforward ?

axsym:=proc(f::mathfunc)
solve(D(f)(x)=0);
end proc:

Obviously this assumes that the input is a function, not an expression, and it's a quadratic, and the independent variable is x.

If the input is an expression, not a function, then this should work

axsym:=proc(f)
solve(diff(f,x)=0,x);
end proc;

another hint for that homework

Axel Vogt — Sun, 12 Sep 2010 18:49:59 Z

Say you have f := x -> 4*x^2-7*x+80; # or any other.

Then you want to have some s, which [for all x] satisfies f(s+x)=f(s-x).

And that means you have to solve for that s.

How would you do with paper and pencil? And then how you would do with Maple?

In that example you should find s = 7/8.

Yes, but

hirnyk — Sun, 12 Sep 2010 17:07:35 Z

the extrema command works not only for quadratic polynomials. The questioner wants a procedure.

Good

Oliver K — Sun, 12 Sep 2010 20:04:25 Z

This is probably the best (and intended) way to do this homework.

Unfortunately, now the main work is done by Axel.

Anyway, +rep