infinite turning points without trig…possible?
$begingroup$
Is it possible to make a function with INFINITE turning points using ONLY
REAL exponentials (and logs)
REAL powers of x
powers,roots, sums, products, quotients, compositions etc of the above
(NO trigs, piecewise, mod)
My feeling is not on the grounds that such a function could be broken down into smaller units each of which will only have a finite number of TP and as such a combination of finite TP cannot produce infinite TP. Can anyone suggest a formal approach which I can use as a starting point
Thanks for your views.
analysis
asked Jan 5 at 23:09
Quadratica MPhilQuadratica MPhil
274
$endgroup$
|
show 19 more comments
$begingroup$
Is it possible to make a function with INFINITE turning points using ONLY
REAL exponentials (and logs)
REAL powers of x
powers,roots, sums, products, quotients, compositions etc of the above
(NO trigs, piecewise, mod)
My feeling is not on the grounds that such a function could be broken down into smaller units each of which will only have a finite number of TP and as such a combination of finite TP cannot produce infinite TP. Can anyone suggest a formal approach which I can use as a starting point
Thanks for your views.
analysis
asked Jan 5 at 23:09
Quadratica MPhilQuadratica MPhil
274
$endgroup$
4
$begingroup$
$f(x) = 0.hphantom{}$
$endgroup$
– Ennar
Jan 5 at 23:11
$begingroup$
@Ennar That's clearly cheating. Any other examples?
$endgroup$
– Matt Samuel
Jan 5 at 23:18
2
$begingroup$
@Matt Samuel, sure if you'll let piecewise defined functions. But if you insist on $f = f_1circldots circ f_n$, where $f_i$ is elementary, non-constant, non-trigonometric, then no, by induction.
$endgroup$
– Ennar
Jan 5 at 23:20
$begingroup$
@Ennar I think that last one is what the op is looking for.
$endgroup$
– Matt Samuel
Jan 5 at 23:21
2
$begingroup$
As I said in an earlier comment $f(x) = x$ and $g(x) = -x$ both have exactly one root but $h(x) = f(x) + g(x) = 0$ has infinitely many roots. So first of all, you need to come up with a true conjecture.
$endgroup$
– Rob Arthan
Jan 6 at 0:14
|
show 19 more comments
$begingroup$
Is it possible to make a function with INFINITE turning points using ONLY
REAL exponentials (and logs)
REAL powers of x
powers,roots, sums, products, quotients, compositions etc of the above
(NO trigs, piecewise, mod)
My feeling is not on the grounds that such a function could be broken down into smaller units each of which will only have a finite number of TP and as such a combination of finite TP cannot produce infinite TP. Can anyone suggest a formal approach which I can use as a starting point
Thanks for your views.
analysis
asked Jan 5 at 23:09
Quadratica MPhilQuadratica MPhil
274
$endgroup$
Is it possible to make a function with INFINITE turning points using ONLY
REAL exponentials (and logs)
REAL powers of x
powers,roots, sums, products, quotients, compositions etc of the above
(NO trigs, piecewise, mod)
My feeling is not on the grounds that such a function could be broken down into smaller units each of which will only have a finite number of TP and as such a combination of finite TP cannot produce infinite TP. Can anyone suggest a formal approach which I can use as a starting point
Thanks for your views.
analysis
analysis
asked Jan 5 at 23:09
Quadratica MPhilQuadratica MPhil
274
asked Jan 5 at 23:09
Quadratica MPhilQuadratica MPhil
274
edited Jan 6 at 0:43
Quadratica MPhil
asked Jan 5 at 23:09
Quadratica MPhilQuadratica MPhil
274
asked Jan 5 at 23:09
Quadratica MPhilQuadratica MPhil
274
asked Jan 5 at 23:09
Quadratica MPhilQuadratica MPhil
274
274
4
$begingroup$
$f(x) = 0.hphantom{}$
$endgroup$
– Ennar
Jan 5 at 23:11
$begingroup$
@Ennar That's clearly cheating. Any other examples?
$endgroup$
– Matt Samuel
Jan 5 at 23:18
2
$begingroup$
@Matt Samuel, sure if you'll let piecewise defined functions. But if you insist on $f = f_1circldots circ f_n$, where $f_i$ is elementary, non-constant, non-trigonometric, then no, by induction.
$endgroup$
– Ennar
Jan 5 at 23:20
$begingroup$
@Ennar I think that last one is what the op is looking for.
$endgroup$
– Matt Samuel
Jan 5 at 23:21
2
$begingroup$
As I said in an earlier comment $f(x) = x$ and $g(x) = -x$ both have exactly one root but $h(x) = f(x) + g(x) = 0$ has infinitely many roots. So first of all, you need to come up with a true conjecture.
$endgroup$
– Rob Arthan
Jan 6 at 0:14
|
show 19 more comments
4
$begingroup$
$f(x) = 0.hphantom{}$
$endgroup$
– Ennar
Jan 5 at 23:11
$begingroup$
@Ennar That's clearly cheating. Any other examples?
$endgroup$
– Matt Samuel
Jan 5 at 23:18
2
$begingroup$
@Matt Samuel, sure if you'll let piecewise defined functions. But if you insist on $f = f_1circldots circ f_n$, where $f_i$ is elementary, non-constant, non-trigonometric, then no, by induction.
$endgroup$
– Ennar
Jan 5 at 23:20
$begingroup$
@Ennar I think that last one is what the op is looking for.
$endgroup$
– Matt Samuel
Jan 5 at 23:21
2
$begingroup$
As I said in an earlier comment $f(x) = x$ and $g(x) = -x$ both have exactly one root but $h(x) = f(x) + g(x) = 0$ has infinitely many roots. So first of all, you need to come up with a true conjecture.
$endgroup$
– Rob Arthan
Jan 6 at 0:14
4
4
$begingroup$
$f(x) = 0.hphantom{}$
$endgroup$
– Ennar
Jan 5 at 23:11
$begingroup$
$f(x) = 0.hphantom{}$
$endgroup$
– Ennar
Jan 5 at 23:11
$begingroup$
@Ennar That's clearly cheating. Any other examples?
$endgroup$
– Matt Samuel
Jan 5 at 23:18
$begingroup$
@Ennar That's clearly cheating. Any other examples?
$endgroup$
– Matt Samuel
Jan 5 at 23:18
2
2
$begingroup$
@Matt Samuel, sure if you'll let piecewise defined functions. But if you insist on $f = f_1circldots circ f_n$, where $f_i$ is elementary, non-constant, non-trigonometric, then no, by induction.
$endgroup$
– Ennar
Jan 5 at 23:20
$begingroup$
@Matt Samuel, sure if you'll let piecewise defined functions. But if you insist on $f = f_1circldots circ f_n$, where $f_i$ is elementary, non-constant, non-trigonometric, then no, by induction.
$endgroup$
– Ennar
Jan 5 at 23:20
$begingroup$
@Ennar I think that last one is what the op is looking for.
$endgroup$
– Matt Samuel
Jan 5 at 23:21
$begingroup$
@Ennar I think that last one is what the op is looking for.
$endgroup$
– Matt Samuel
Jan 5 at 23:21
2
2
$begingroup$
As I said in an earlier comment $f(x) = x$ and $g(x) = -x$ both have exactly one root but $h(x) = f(x) + g(x) = 0$ has infinitely many roots. So first of all, you need to come up with a true conjecture.
$endgroup$
– Rob Arthan
Jan 6 at 0:14
$begingroup$
As I said in an earlier comment $f(x) = x$ and $g(x) = -x$ both have exactly one root but $h(x) = f(x) + g(x) = 0$ has infinitely many roots. So first of all, you need to come up with a true conjecture.
$endgroup$
– Rob Arthan
Jan 6 at 0:14
|
show 19 more comments
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4
$begingroup$
$f(x) = 0.hphantom{}$
$endgroup$
– Ennar
Jan 5 at 23:11
$begingroup$
@Ennar That's clearly cheating. Any other examples?
$endgroup$
– Matt Samuel
Jan 5 at 23:18
2
$begingroup$
@Matt Samuel, sure if you'll let piecewise defined functions. But if you insist on $f = f_1circldots circ f_n$, where $f_i$ is elementary, non-constant, non-trigonometric, then no, by induction.
$endgroup$
– Ennar
Jan 5 at 23:20
$begingroup$
@Ennar I think that last one is what the op is looking for.
$endgroup$
– Matt Samuel
Jan 5 at 23:21
2
$begingroup$
As I said in an earlier comment $f(x) = x$ and $g(x) = -x$ both have exactly one root but $h(x) = f(x) + g(x) = 0$ has infinitely many roots. So first of all, you need to come up with a true conjecture.
$endgroup$
– Rob Arthan
Jan 6 at 0:14