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Topic: Is (t^2-9)/(t-3) defined at t=3?
Replies: 166   Last Post: Oct 30, 2013 9:41 AM

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Hetware

Posts: 148
Registered: 4/13/13
Re: Is (t^2-9)/(t-3) defined at t=3?
Posted: Oct 19, 2013 7:17 PM
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On 10/19/2013 2:15 PM, Peter Percival wrote:

> You really see no difference between
>
> Let f be a continuous, twice differentiable function...
>
> and
>
> Let f be defined by <some formula written here> and
> f is continuous because I say so.
>
> ?
>
>


The notion of "formula" must be taken as extremely broad. Any means of
determining a set of ordered pairs (x, f(x)) will be a formula.

The following is Mathematica code for a function F[x, C, r] where x is
some real value, C is a Boolean value representing the assertion of
continuity, and r is some expression mapping elements of the domain to
elements of the range.


ContinuousQ[r_, x_] := True /; Quiet[NumberQ[r[x]] &&
Element[r[x], Reals] &&
(Limit[r[t], t -> x] == r[x])]
ContinuousQ[r_, x_] := False

RemovableQ[r_, x_] := True /; Quiet[! NumberQ[r[x]] &&
Element[Limit[r[t], t -> x], Reals]]
RemovableQ[r_, x_] := False

(* if continuity of F not asserted return r[x]*)
F[x_, C_, r_] := r[x] /; ! C

(* else if r[x] is continuous return r[x].*)
F[x_, C_, r_] := r[x] /; ContinuousQ[r, x]

(* else if r[x] is a removable discontinuity return Limit[r[t],t->x].*)
F[x_, C_, r_] := Limit[r[t], t -> x] /; RemovableQ[r, x]

(* else asserting F[x,True,r] leads to a contradiction.*)
F[x_, C_, r_] := False

(*The following are examples of using F[x,C,r] to define function of the
form (x, f(x)) where x and f(x) are real numbers.*)
r[x_] := x/x
f[x_] := F[x, True, r]
Print["r[x_]:=x/x"]
Print["f[x_]:=F[x,True, r]"]

Print["f[1]=", f[1]]
Print["f[0]=", f[0]]

g[x_] := Quiet[F[x, False, r]]
Print["g[x_]:=F[x,False,r]"]
Print["g[1]=", g[1]]
Print["g[0]=", g[0]]

s[x_] := 1/x
Print["s[x_]:=1/x"]

h[x_] := F[x, True, s]
Print["h[x_]:=F[x,True, s]"]
Print["h[1]=", h[1]]
Print["h[0]=", h[0]]

i[x_] := Quiet[F[x, False, s]]
Print["i[x_]:=F[x,False, s]"]
Print["i[1]=", i[1]]
Print["i[0]=", i[0]]

(* And the output *)

r[x_]:=x/x

f[x_]:=F[x,True, r]

f[1]=1

f[0]=1

g[x_]:=F[x,False,r]

g[1]=1

g[0]=Indeterminate

s[x_]:=1/x

h[x_]:=F[x,True, s]

h[1]=1

h[0]=False

i[x_]:=F[x,False, s]

i[1]=1

i[0]=ComplexInfinity

(*End Output*)


So given the mapping r[x_] := x/x we define a function f[x_] := F[x,
True, r] which uses the definition of continuity and r[x] to find a
suitable value for f[x]. If r[x] results in an indeterminate form, f[x]
has a rule which returns the value of Limit[r[t], t -> x]. If r[x] is
not a suitable candidate, that is to say it has one or more irremovable
discontinuities, then f[x] returns False, meaning that the definition
resulted in a contradiction.

Notice that not asserting continuity gives a different result. In fact,
g[x_]:=F[x,False,r] is just g[x]=r[x].

Is the proposition 'the mapping m is continuous' meaningful?


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9/28/13
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