quasi wrote: > On 15 Jan 2006 16:48:19 -0800, "Kenneth Bull" <kenneth.b...@gmail.com> > wrote:
> >john_rams...@sagitta-ps.com wrote: > >> quasi wrote: > >> > On 15 Jan 2006 15:31:01 -0800, "Kenneth Bull" <kenneth.b...@gmail.com> > >> > wrote:
> >> > >john_rams...@sagitta-ps.com wrote: > >> > >> Kenneth Bull wrote: > >> > >> > On what interval(s) is f(x) = x^3 increasing?
> >> > >> > a) (-oo, 0) U (0, oo)
> >> > >> > b) (-oo, oo)
> >> > >> > [Can someone prove their answer?]
> >> > >> > I pick a because:
> >> > >> > f'(x) = x^2 and f'(x) > 0 on (-oo,0)U(0,oo) thus f is increasing on > >> > >> > that interval
> >> > >> > f'(0) = 0 so f is not increasing at 0.
> >> > >> > Agreed?
> >> > >> Before seeing the other replies I'd have unhesitatingly agreed > >> > >> (apart from f'(x) being 3.x^2 ! ), believing that a function is not > >> > >> increasing at a stationary point; but now I'm confused.
> >> > >> Obviously there's a distinction between "increasing at a point" > >> > >> and "increasing over an interval", despite the same word being > >> > >> used in each phrase.
> >> > >> I think we need to wait for twenty or more people to reply, and > >> > >> take a vote on the number of (a)s versus (b)s ;-P
> >> > >> (Also, Rob's definition of "increasing" is what I'd call > >> > >> "non-decreasing", > >> > >> reserving "increasing" for what he calls "strictly increasing"; but > >> > >> that's > >> > >> a side-issue.)
> >> > >Try pondering on which greatest set of real numbers f(x) = x^4 is > >> > >convex on (this is closely related to the original question).
> >> > Same answer (-oo, oo).
> >> > The definition of convexity is also not based on derivatives.
> >> That I do agree with, and as it's easy to see that a function > >> need not be convex in an interval in which it is strictly > >> increasing I'm not entirely sure that Kenneth is hinting at.
> >f'(x) = 4x^3
> >My book defines concavity of a function using f' (convex if f' > 0 , > >concave if f' < 0)
> You mean f'', not f', but even so ...
> Are you sure you're referencing the actual definition? It might be > that what you are calling the definition is not the definition, but > the following proposition:
> If f''>0 for all x, then f is convex.
> If f''<0 for all x, then f is concave.
> In other words, the implication is in one direction only, not the > direction you're claiming.
> Just to be sure, can you tell us which text you're using?
Correction, it defines concave up and down (concave and convex) using a vague sentence "if the graph lies above all its tangents on an interval, it is concave up on the interval" and similar for concave down.
Right after this it presents the Concavity Test which uses f'' to test for concavity.
But I realize what you are saying about the one direction implication.
>quasi wrote: >> On 15 Jan 2006 16:48:19 -0800, "Kenneth Bull" <kenneth.b...@gmail.com> >> wrote:
>> >john_rams...@sagitta-ps.com wrote: >> >> quasi wrote: >> >> > On 15 Jan 2006 15:31:01 -0800, "Kenneth Bull" <kenneth.b...@gmail.com> >> >> > wrote:
>> >> > >john_rams...@sagitta-ps.com wrote: >> >> > >> Kenneth Bull wrote: >> >> > >> > On what interval(s) is f(x) = x^3 increasing?
>> >> > >> > a) (-oo, 0) U (0, oo)
>> >> > >> > b) (-oo, oo)
>> >> > >> > [Can someone prove their answer?]
>> >> > >> > I pick a because:
>> >> > >> > f'(x) = x^2 and f'(x) > 0 on (-oo,0)U(0,oo) thus f is increasing on >> >> > >> > that interval
>> >> > >> > f'(0) = 0 so f is not increasing at 0.
>> >> > >> > Agreed?
>> >> > >> Before seeing the other replies I'd have unhesitatingly agreed >> >> > >> (apart from f'(x) being 3.x^2 ! ), believing that a function is not >> >> > >> increasing at a stationary point; but now I'm confused.
>> >> > >> Obviously there's a distinction between "increasing at a point" >> >> > >> and "increasing over an interval", despite the same word being >> >> > >> used in each phrase.
>> >> > >> I think we need to wait for twenty or more people to reply, and >> >> > >> take a vote on the number of (a)s versus (b)s ;-P
>> >> > >> (Also, Rob's definition of "increasing" is what I'd call >> >> > >> "non-decreasing", >> >> > >> reserving "increasing" for what he calls "strictly increasing"; but >> >> > >> that's >> >> > >> a side-issue.)
>> >> > >Try pondering on which greatest set of real numbers f(x) = x^4 is >> >> > >convex on (this is closely related to the original question).
>> >> > Same answer (-oo, oo).
>> >> > The definition of convexity is also not based on derivatives.
>> >> That I do agree with, and as it's easy to see that a function >> >> need not be convex in an interval in which it is strictly >> >> increasing I'm not entirely sure that Kenneth is hinting at.
>> >f'(x) = 4x^3
>> >My book defines concavity of a function using f' (convex if f' > 0 , >> >concave if f' < 0)
>> You mean f'', not f', but even so ...
>> Are you sure you're referencing the actual definition? It might be >> that what you are calling the definition is not the definition, but >> the following proposition:
>> If f''>0 for all x, then f is convex.
>> If f''<0 for all x, then f is concave.
>> In other words, the implication is in one direction only, not the >> direction you're claiming.
>> Just to be sure, can you tell us which text you're using?
>Correction, it defines concave up and down (concave and convex) using a >vague sentence "if the graph lies above all its tangents on an >interval, it is concave up on the interval" and similar for concave >down.
>Right after this it presents the Concavity Test which uses f'' to test >for concavity.
>But I realize what you are saying about the one direction implication.
> >> >> > >john_rams...@sagitta-ps.com wrote: > >> >> > >> Kenneth Bull wrote: > >> >> > >> > On what interval(s) is f(x) = x^3 increasing?
> >> >> > >> > a) (-oo, 0) U (0, oo)
> >> >> > >> > b) (-oo, oo)
> >> >> > >> > [Can someone prove their answer?]
> >> >> > >> > I pick a because:
> >> >> > >> > f'(x) = x^2 and f'(x) > 0 on (-oo,0)U(0,oo) thus f is > >> >> > >> > increasing on > >> >> > >> > that interval
> >> >> > >> > f'(0) = 0 so f is not increasing at 0.
> >> >> > >> > Agreed?
> >> >> > >> Before seeing the other replies I'd have unhesitatingly agreed > >> >> > >> (apart from f'(x) being 3.x^2 ! ), believing that a function is > >> >> > >> not > >> >> > >> increasing at a stationary point; but now I'm confused.
> >> >> > >> Obviously there's a distinction between "increasing at a point" > >> >> > >> and "increasing over an interval", despite the same word being > >> >> > >> used in each phrase.
> >> >> > >> I think we need to wait for twenty or more people to reply, and > >> >> > >> take a vote on the number of (a)s versus (b)s ;-P
> >> >> > >> (Also, Rob's definition of "increasing" is what I'd call > >> >> > >> "non-decreasing", > >> >> > >> reserving "increasing" for what he calls "strictly increasing"; > >> >> > >> but > >> >> > >> that's > >> >> > >> a side-issue.)
> >> >> > >Try pondering on which greatest set of real numbers f(x) = x^4 is > >> >> > >convex on (this is closely related to the original question).
> >> >> > Same answer (-oo, oo).
> >> >> > The definition of convexity is also not based on derivatives.
> >> >> That I do agree with, and as it's easy to see that a function > >> >> need not be convex in an interval in which it is strictly > >> >> increasing I'm not entirely sure that Kenneth is hinting at.
> >> >f'(x) = 4x^3
> >> >My book defines concavity of a function using f' (convex if f' > 0 , > >> >concave if f' < 0)
> >> You mean f'', not f', but even so ...
> >> Are you sure you're referencing the actual definition? It might be > >> that what you are calling the definition is not the definition, but > >> the following proposition:
> >> If f''>0 for all x, then f is convex.
> >> If f''<0 for all x, then f is concave.
> >> In other words, the implication is in one direction only, not the > >> direction you're claiming.
> >> Just to be sure, can you tell us which text you're using?
> >Correction, it defines concave up and down (concave and convex) using a > >vague sentence "if the graph lies above all its tangents on an > >interval, it is concave up on the interval" and similar for concave > >down.
> >Right after this it presents the Concavity Test which uses f'' to test > >for concavity.
> >But I realize what you are saying about the one direction implication.
> Ok, so x^4 lies above all its tangents, right?
> quasi
And below all its secant segments (joining two points of the curve). This test has the advantage of not needing anything about tangents, which need not exist at all points of convexity.
E.g., f(x) = (x^4)^(1/3) is convex everywhere but has no tangent at x=0.
> > >> >> > >> > f'(x) = x^2 and f'(x) > 0 on (-oo,0)U(0,oo) thus f is > > >> >> > >> > increasing on > > >> >> > >> > that interval
> > >> >> > >> > f'(0) = 0 so f is not increasing at 0.
> > >> >> > >> > Agreed?
> > >> >> > >> Before seeing the other replies I'd have unhesitatingly agreed > > >> >> > >> (apart from f'(x) being 3.x^2 ! ), believing that a function is > > >> >> > >> not > > >> >> > >> increasing at a stationary point; but now I'm confused.
> > >> >> > >> Obviously there's a distinction between "increasing at a point" > > >> >> > >> and "increasing over an interval", despite the same word being > > >> >> > >> used in each phrase.
> > >> >> > >> I think we need to wait for twenty or more people to reply, and > > >> >> > >> take a vote on the number of (a)s versus (b)s ;-P
> > >> >> > >> (Also, Rob's definition of "increasing" is what I'd call > > >> >> > >> "non-decreasing", > > >> >> > >> reserving "increasing" for what he calls "strictly increasing"; > > >> >> > >> but > > >> >> > >> that's > > >> >> > >> a side-issue.)
> > >> >> > >Try pondering on which greatest set of real numbers f(x) = x^4 is > > >> >> > >convex on (this is closely related to the original question).
> > >> >> > Same answer (-oo, oo).
> > >> >> > The definition of convexity is also not based on derivatives.
> > >> >> That I do agree with, and as it's easy to see that a function > > >> >> need not be convex in an interval in which it is strictly > > >> >> increasing I'm not entirely sure that Kenneth is hinting at.
> > >> >f'(x) = 4x^3
> > >> >My book defines concavity of a function using f' (convex if f' > 0 , > > >> >concave if f' < 0)
> > >> You mean f'', not f', but even so ...
> > >> Are you sure you're referencing the actual definition? It might be > > >> that what you are calling the definition is not the definition, but > > >> the following proposition:
> > >> If f''>0 for all x, then f is convex.
> > >> If f''<0 for all x, then f is concave.
> > >> In other words, the implication is in one direction only, not the > > >> direction you're claiming.
> > >> Just to be sure, can you tell us which text you're using?
> > >Correction, it defines concave up and down (concave and convex) using a > > >vague sentence "if the graph lies above all its tangents on an > > >interval, it is concave up on the interval" and similar for concave > > >down.
> > >Right after this it presents the Concavity Test which uses f'' to test > > >for concavity.
> > >But I realize what you are saying about the one direction implication.
> > Ok, so x^4 lies above all its tangents, right?
> > quasi
> And below all its secant segments (joining two points of the curve). > This test has the advantage of not needing anything about tangents, > which need not exist at all points of convexity.
> E.g., f(x) = (x^4)^(1/3) is convex everywhere but has no tangent at x=0.
>> >> >> > >john_rams...@sagitta-ps.com wrote: >> >> >> > >> Kenneth Bull wrote: >> >> >> > >> > On what interval(s) is f(x) = x^3 increasing?
>> >> >> > >> > a) (-oo, 0) U (0, oo)
>> >> >> > >> > b) (-oo, oo)
>> >> >> > >> > [Can someone prove their answer?]
>> >> >> > >> > I pick a because:
>> >> >> > >> > f'(x) = x^2 and f'(x) > 0 on (-oo,0)U(0,oo) thus f is >> >> >> > >> > increasing on >> >> >> > >> > that interval
>> >> >> > >> > f'(0) = 0 so f is not increasing at 0.
>> >> >> > >> > Agreed?
>> >> >> > >> Before seeing the other replies I'd have unhesitatingly agreed >> >> >> > >> (apart from f'(x) being 3.x^2 ! ), believing that a function is >> >> >> > >> not >> >> >> > >> increasing at a stationary point; but now I'm confused.
>> >> >> > >> Obviously there's a distinction between "increasing at a point" >> >> >> > >> and "increasing over an interval", despite the same word being >> >> >> > >> used in each phrase.
>> >> >> > >> I think we need to wait for twenty or more people to reply, and >> >> >> > >> take a vote on the number of (a)s versus (b)s ;-P
>> >> >> > >> (Also, Rob's definition of "increasing" is what I'd call >> >> >> > >> "non-decreasing", >> >> >> > >> reserving "increasing" for what he calls "strictly increasing"; >> >> >> > >> but >> >> >> > >> that's >> >> >> > >> a side-issue.)
>> >> >> > >Try pondering on which greatest set of real numbers f(x) = x^4 is >> >> >> > >convex on (this is closely related to the original question).
>> >> >> > Same answer (-oo, oo).
>> >> >> > The definition of convexity is also not based on derivatives.
>> >> >> That I do agree with, and as it's easy to see that a function >> >> >> need not be convex in an interval in which it is strictly >> >> >> increasing I'm not entirely sure that Kenneth is hinting at.
>> >> >f'(x) = 4x^3
>> >> >My book defines concavity of a function using f' (convex if f' > 0 , >> >> >concave if f' < 0)
>> >> You mean f'', not f', but even so ...
>> >> Are you sure you're referencing the actual definition? It might be >> >> that what you are calling the definition is not the definition, but >> >> the following proposition:
>> >> If f''>0 for all x, then f is convex.
>> >> If f''<0 for all x, then f is concave.
>> >> In other words, the implication is in one direction only, not the >> >> direction you're claiming.
>> >> Just to be sure, can you tell us which text you're using?
>> >Correction, it defines concave up and down (concave and convex) using a >> >vague sentence "if the graph lies above all its tangents on an >> >interval, it is concave up on the interval" and similar for concave >> >down.
>> >Right after this it presents the Concavity Test which uses f'' to test >> >for concavity.
>> >But I realize what you are saying about the one direction implication.
>> Ok, so x^4 lies above all its tangents, right?
>> quasi
>And below all its secant segments (joining two points of the curve). >This test has the advantage of not needing anything about tangents, >which need not exist at all points of convexity.
Well, I prefer the usual algebraic definition but the version you describe using secants is also fine since it's equivalent.
>E.g., f(x) = (x^4)^(1/3) is convex everywhere but has no tangent at x=0.
Right.
But given that his text defines a convex function as a function whose graph lies above its tangents, he can at least get the correct answer for x^4.
>> > >> >> > >> > f'(x) = x^2 and f'(x) > 0 on (-oo,0)U(0,oo) thus f is >> > >> >> > >> > increasing on >> > >> >> > >> > that interval
>> > >> >> > >> > f'(0) = 0 so f is not increasing at 0.
>> > >> >> > >> > Agreed?
>> > >> >> > >> Before seeing the other replies I'd have unhesitatingly agreed >> > >> >> > >> (apart from f'(x) being 3.x^2 ! ), believing that a function is >> > >> >> > >> not >> > >> >> > >> increasing at a stationary point; but now I'm confused.
>> > >> >> > >> Obviously there's a distinction between "increasing at a point" >> > >> >> > >> and "increasing over an interval", despite the same word being >> > >> >> > >> used in each phrase.
>> > >> >> > >> I think we need to wait for twenty or more people to reply, and >> > >> >> > >> take a vote on the number of (a)s versus (b)s ;-P
>> > >> >> > >> (Also, Rob's definition of "increasing" is what I'd call >> > >> >> > >> "non-decreasing", >> > >> >> > >> reserving "increasing" for what he calls "strictly increasing"; >> > >> >> > >> but >> > >> >> > >> that's >> > >> >> > >> a side-issue.)
>> > >> >> > >Try pondering on which greatest set of real numbers f(x) = x^4 is >> > >> >> > >convex on (this is closely related to the original question).
>> > >> >> > Same answer (-oo, oo).
>> > >> >> > The definition of convexity is also not based on derivatives.
>> > >> >> That I do agree with, and as it's easy to see that a function >> > >> >> need not be convex in an interval in which it is strictly >> > >> >> increasing I'm not entirely sure that Kenneth is hinting at.
>> > >> >f'(x) = 4x^3
>> > >> >My book defines concavity of a function using f' (convex if f' > 0 , >> > >> >concave if f' < 0)
>> > >> You mean f'', not f', but even so ...
>> > >> Are you sure you're referencing the actual definition? It might be >> > >> that what you are calling the definition is not the definition, but >> > >> the following proposition:
>> > >> If f''>0 for all x, then f is convex.
>> > >> If f''<0 for all x, then f is concave.
>> > >> In other words, the implication is in one direction only, not the >> > >> direction you're claiming.
>> > >> Just to be sure, can you tell us which text you're using?
>> > >Correction, it defines concave up and down (concave and convex) using a >> > >vague sentence "if the graph lies above all its tangents on an >> > >interval, it is concave up on the interval" and similar for concave >> > >down.
>> > >Right after this it presents the Concavity Test which uses f'' to test >> > >for concavity.
>> > >But I realize what you are saying about the one direction implication.
>> > Ok, so x^4 lies above all its tangents, right?
>> > quasi
>> And below all its secant segments (joining two points of the curve). >> This test has the advantage of not needing anything about tangents, >> which need not exist at all points of convexity.
>> E.g., f(x) = (x^4)^(1/3) is convex everywhere but has no tangent at x=0.
>f'(x) = [ 4 * x^(1/3) ] / 3
>f'(0) = 0
>Tangent at x=0 has slope 0
Right.
Ok, how about this example instead:
f(x)=exp(|x|)
f is convex on (-oo, oo) but not differentiable at x=0.
>>> > >> >> > >> > f'(x) = x^2 and f'(x) > 0 on (-oo,0)U(0,oo) thus f is >>> > >> >> > >> > increasing on >>> > >> >> > >> > that interval
>>> > >> >> > >> > f'(0) = 0 so f is not increasing at 0.
>>> > >> >> > >> > Agreed?
>>> > >> >> > >> Before seeing the other replies I'd have unhesitatingly agreed >>> > >> >> > >> (apart from f'(x) being 3.x^2 ! ), believing that a function is >>> > >> >> > >> not >>> > >> >> > >> increasing at a stationary point; but now I'm confused.
>>> > >> >> > >> Obviously there's a distinction between "increasing at a point" >>> > >> >> > >> and "increasing over an interval", despite the same word being >>> > >> >> > >> used in each phrase.
>>> > >> >> > >> I think we need to wait for twenty or more people to reply, and >>> > >> >> > >> take a vote on the number of (a)s versus (b)s ;-P
>>> > >> >> > >> (Also, Rob's definition of "increasing" is what I'd call >>> > >> >> > >> "non-decreasing", >>> > >> >> > >> reserving "increasing" for what he calls "strictly increasing"; >>> > >> >> > >> but >>> > >> >> > >> that's >>> > >> >> > >> a side-issue.)
>>> > >> >> > >Try pondering on which greatest set of real numbers f(x) = x^4 is >>> > >> >> > >convex on (this is closely related to the original question).
>>> > >> >> > Same answer (-oo, oo).
>>> > >> >> > The definition of convexity is also not based on derivatives.
>>> > >> >> That I do agree with, and as it's easy to see that a function >>> > >> >> need not be convex in an interval in which it is strictly >>> > >> >> increasing I'm not entirely sure that Kenneth is hinting at.
>>> > >> >f'(x) = 4x^3
>>> > >> >My book defines concavity of a function using f' (convex if f' > 0 , >>> > >> >concave if f' < 0)
>>> > >> You mean f'', not f', but even so ...
>>> > >> Are you sure you're referencing the actual definition? It might be >>> > >> that what you are calling the definition is not the definition, but >>> > >> the following proposition:
>>> > >> If f''>0 for all x, then f is convex.
>>> > >> If f''<0 for all x, then f is concave.
>>> > >> In other words, the implication is in one direction only, not the >>> > >> direction you're claiming.
>>> > >> Just to be sure, can you tell us which text you're using?
>>> > >Correction, it defines concave up and down (concave and convex) using a >>> > >vague sentence "if the graph lies above all its tangents on an >>> > >interval, it is concave up on the interval" and similar for concave >>> > >down.
>>> > >Right after this it presents the Concavity Test which uses f'' to test >>> > >for concavity.
>>> > >But I realize what you are saying about the one direction implication.
>>> > Ok, so x^4 lies above all its tangents, right?
>>> > quasi
>>> And below all its secant segments (joining two points of the curve). >>> This test has the advantage of not needing anything about tangents, >>> which need not exist at all points of convexity.
>>> E.g., f(x) = (x^4)^(1/3) is convex everywhere but has no tangent at x=0.
>>f'(x) = [ 4 * x^(1/3) ] / 3
>>f'(0) = 0
>>Tangent at x=0 has slope 0
>Right.
>Ok, how about this example instead:
>f(x)=exp(|x|)
>f is convex on (-oo, oo) but not differentiable at x=0.
>quasi
For that matter, a simpler example, also convex but not differentiable at x=0, is |x|.
However exp(|x|) is strictly convex, whereas |x|, though convex, is not strictly convex.
> >>> > >> >> > >> > f'(x) = x^2 and f'(x) > 0 on (-oo,0)U(0,oo) thus f is > >>> > >> >> > >> > increasing on > >>> > >> >> > >> > that interval
> >>> > >> >> > >> > f'(0) = 0 so f is not increasing at 0.
> >>> > >> >> > >> > Agreed?
> >>> > >> >> > >> Before seeing the other replies I'd have unhesitatingly agreed > >>> > >> >> > >> (apart from f'(x) being 3.x^2 ! ), believing that a function is > >>> > >> >> > >> not > >>> > >> >> > >> increasing at a stationary point; but now I'm confused.
> >>> > >> >> > >> Obviously there's a distinction between "increasing at a point" > >>> > >> >> > >> and "increasing over an interval", despite the same word being > >>> > >> >> > >> used in each phrase.
> >>> > >> >> > >> I think we need to wait for twenty or more people to reply, and > >>> > >> >> > >> take a vote on the number of (a)s versus (b)s ;-P
> >>> > >> >> > >> (Also, Rob's definition of "increasing" is what I'd call > >>> > >> >> > >> "non-decreasing", > >>> > >> >> > >> reserving "increasing" for what he calls "strictly increasing"; > >>> > >> >> > >> but > >>> > >> >> > >> that's > >>> > >> >> > >> a side-issue.)
> >>> > >> >> > >Try pondering on which greatest set of real numbers f(x) = x^4 is > >>> > >> >> > >convex on (this is closely related to the original question).
> >>> > >> >> > Same answer (-oo, oo).
> >>> > >> >> > The definition of convexity is also not based on derivatives.
> >>> > >> >> That I do agree with, and as it's easy to see that a function > >>> > >> >> need not be convex in an interval in which it is strictly > >>> > >> >> increasing I'm not entirely sure that Kenneth is hinting at.
> >>> > >> >f'(x) = 4x^3
> >>> > >> >My book defines concavity of a function using f' (convex if f' > 0 , > >>> > >> >concave if f' < 0)
> >>> > >> You mean f'', not f', but even so ...
> >>> > >> Are you sure you're referencing the actual definition? It might be > >>> > >> that what you are calling the definition is not the definition, but > >>> > >> the following proposition:
> >>> > >> If f''>0 for all x, then f is convex.
> >>> > >> If f''<0 for all x, then f is concave.
> >>> > >> In other words, the implication is in one direction only, not the > >>> > >> direction you're claiming.
> >>> > >> Just to be sure, can you tell us which text you're using?
> >>> > >Correction, it defines concave up and down (concave and convex) using a > >>> > >vague sentence "if the graph lies above all its tangents on an > >>> > >interval, it is concave up on the interval" and similar for concave > >>> > >down.
> >>> > >Right after this it presents the Concavity Test which uses f'' to test > >>> > >for concavity.
> >>> > >But I realize what you are saying about the one direction implication.
> >>> > Ok, so x^4 lies above all its tangents, right?
> >>> > quasi
> >>> And below all its secant segments (joining two points of the curve). > >>> This test has the advantage of not needing anything about tangents, > >>> which need not exist at all points of convexity.
> >>> E.g., f(x) = (x^4)^(1/3) is convex everywhere but has no tangent at x=0.
> >>f'(x) = [ 4 * x^(1/3) ] / 3
> >>f'(0) = 0
> >>Tangent at x=0 has slope 0
> >Right.
> >Ok, how about this example instead:
> >f(x)=exp(|x|)
> >f is convex on (-oo, oo) but not differentiable at x=0.
Yes. But the original example given by the other fellow was differentiable at x=0, that's why I responded to note that. I know that differentiability is not tied to the definition of convex/concave or increasing/decreasing. You don't have to convince me.
> >quasi
> For that matter, a simpler example, also convex but not differentiable > at x=0, is |x|.
|x| is not convex. If your notion of non-strict convex allows for the graph to lie on secants (instead of below), then that notion is weird.
That's as weird as the non-strict increasing that was described by another poster, which allows functions like y = 0.32 to be considered increasing.
Kenneth Bull wrote: > On what interval(s) is f(x) = x^3 increasing?
> a) (-oo, 0) U (0, oo)
> b) (-oo, oo)
> [Can someone prove their answer?]
> I pick a because:
> f'(x) = x^2 and f'(x) > 0 on (-oo,0)U(0,oo) thus f is increasing on > that interval
> f'(0) = 0 so f is not increasing at 0.
> Agreed?
The confusion stems from the fact that the idea of increasing/decreasing is tied to a function, yet you are trying to use the derivative of the function to investigate whether it is increasing or decreasing.
There are several ways to use the derivative of a function to gain insights into the functions. Say function f has derivative f'
it can be proved that if f'(x) > 0 on an interval A, then f is increasing on A [using the Mean Value Theorem] This requires the derivative to exist, and it requires an interval over which f'(x) > 0 is true. There is no notion of a "point" here because the MVT has no business with points. The converse of this is not generally true!!! Try to prove it, and you will hit major road bumps.
it can also be proved that if f'(x) = 0 on an interval B, then f is constant on B [using the Mean Value Theorem] This requires the derivative to exist, and it requires an interval over which f'(x) = 0 is true. There is also no notion of a "point" here. The converse of this result is true (easily proven). So you cannot apply the logic to a "single point" where f'(x) = 0. Ex: (0,0) of x^3 doesn't mean the function is not increasing at or around 0
Both of these results use the Mean Value Theorem, which works on pairs of points (any general pair of points in the intervals). This is why you can't gain insights into f when f' or higher derivative have special qualities at single isolated points - it's invervals that matter.
Now, it can also be proved that if f'(x) > 0 for all x in it's domain(which is a single interval) except at the non-adjacent points a, b, c, d ... etc where f'(a) =f'(b) = f'(c) = f'(d) = ...= 0 then f is increasing on it's domain.
Your function f(x) = x^3 falls under this.
But the best way to prove that f(x) is increasing it to deal with the function directly, instead of painfully trying to gain info into f(x) through f'(x).
On 15 Jan 2006 19:53:12 -0800, "Luke Wu" <LookSkywal...@gmail.com> wrote:
>quasi wrote: >> On Sun, 15 Jan 2006 22:31:13 -0500, quasi <qu...@null.set> wrote:
>> For that matter, a simpler example, also convex but not differentiable >> at x=0, is |x|.
>|x| is not convex. If your notion of non-strict convex allows for the >graph to lie on secants (instead of below), then that notion is weird.
>That's as weird as the non-strict increasing that was described by >another poster, which allows functions like y = 0.32 to be considered >increasing.
The standard definition of convex function is as follows:
f is convex if for any two points x1, x2 and any t in [0,1],
f(t*x1 + (1-t)*x2) <= t*f(x1) + (1-t)*f*(x2)
In other words, the definition of convex function defaults to a non-strict version, whereas the definition of increasing defaults to a strict version.
Thus, by the standard definition, |x| is a convex function.
quasi wrote: > On 15 Jan 2006 19:53:12 -0800, "Luke Wu" <LookSkywal...@gmail.com> > wrote:
> >quasi wrote: > >> On Sun, 15 Jan 2006 22:31:13 -0500, quasi <qu...@null.set> wrote:
> >> For that matter, a simpler example, also convex but not differentiable > >> at x=0, is |x|.
> >|x| is not convex. If your notion of non-strict convex allows for the > >graph to lie on secants (instead of below), then that notion is weird.
> >That's as weird as the non-strict increasing that was described by > >another poster, which allows functions like y = 0.32 to be considered > >increasing.
> The standard definition of convex function is as follows:
> f is convex if for any two points x1, x2 and any t in [0,1],
> f(t*x1 + (1-t)*x2) <= t*f(x1) + (1-t)*f*(x2)
> In other words, the definition of convex function defaults to a > non-strict version, whereas the definition of increasing defaults to a > strict version.
So in the default sense: So y = 0.32 is convex, but not increasing?
>>>>>>>>On what interval(s) is f(x) = x^3 increasing?
>>>>>>>>a) (-oo, 0) U (0, oo)
>>>>>>>>b) (-oo, oo)
>>>>>>>>[Can someone prove their answer?]
>>>>>>>>I pick a because:
>>>>>>>>f'(x) = x^2 and f'(x) > 0 on (-oo,0)U(0,oo) thus f is increasing on >>>>>>>>that interval
>>>>>>>>f'(0) = 0 so f is not increasing at 0.
>>>>>>>>Agreed?
>>>>>>>Before seeing the other replies I'd have unhesitatingly agreed >>>>>>>(apart from f'(x) being 3.x^2 ! ), believing that a function is not >>>>>>>increasing at a stationary point; but now I'm confused.
>>>>>>>Obviously there's a distinction between "increasing at a point" >>>>>>>and "increasing over an interval", despite the same word being >>>>>>>used in each phrase.
>>>>>>>I think we need to wait for twenty or more people to reply, and >>>>>>>take a vote on the number of (a)s versus (b)s ;-P
>>>>>>>(Also, Rob's definition of "increasing" is what I'd call >>>>>>>"non-decreasing", >>>>>>>reserving "increasing" for what he calls "strictly increasing"; but >>>>>>>that's >>>>>>>a side-issue.)
>>>>>>Try pondering on which greatest set of real numbers f(x) = x^4 is >>>>>>convex on (this is closely related to the original question).
>>>>>Same answer (-oo, oo).
>>>>>The definition of convexity is also not based on derivatives.
>>>>That I do agree with, and as it's easy to see that a function >>>>need not be convex in an interval in which it is strictly >>>>increasing I'm not entirely sure that Kenneth is hinting at.
>>>f'(x) = 4x^3
>>>My book defines concavity of a function using f' (convex if f' > 0 , >>>concave if f' < 0)
>>You mean f'', not f', but even so ...
>>Are you sure you're referencing the actual definition? It might be >>that what you are calling the definition is not the definition, but >>the following proposition:
>>If f''>0 for all x, then f is convex.
>>If f''<0 for all x, then f is concave.
>>In other words, the implication is in one direction only, not the >>direction you're claiming.
>>Just to be sure, can you tell us which text you're using?
>quasi wrote: >> On 15 Jan 2006 19:53:12 -0800, "Luke Wu" <LookSkywal...@gmail.com> >> wrote:
>> >quasi wrote: >> >> On Sun, 15 Jan 2006 22:31:13 -0500, quasi <qu...@null.set> wrote:
>> >> For that matter, a simpler example, also convex but not differentiable >> >> at x=0, is |x|.
>> >|x| is not convex. If your notion of non-strict convex allows for the >> >graph to lie on secants (instead of below), then that notion is weird.
>> >That's as weird as the non-strict increasing that was described by >> >another poster, which allows functions like y = 0.32 to be considered >> >increasing.
>> The standard definition of convex function is as follows:
>> f is convex if for any two points x1, x2 and any t in [0,1],
>> f(t*x1 + (1-t)*x2) <= t*f(x1) + (1-t)*f*(x2)
>> In other words, the definition of convex function defaults to a >> non-strict version, whereas the definition of increasing defaults to a >> strict version.
>So in the default sense: So y = 0.32 is convex, but not increasing?
> quasi wrote: > > >That's as weird as the non-strict increasing that was described by > > >another poster, which allows functions like y = 0.32 to be > > >considered increasing.
> > The standard definition of convex function is as follows:
> > f is convex if for any two points x1, x2 and any t in [0,1],
> > f(t*x1 + (1-t)*x2) <= t*f(x1) + (1-t)*f*(x2)
> > In other words, the definition of convex function defaults to a > > non-strict version, whereas the definition of increasing defaults > > to a strict version.
> So in the default sense: So y = 0.32 is convex, but not increasing?
Right!. But it is non-decreasing and not /strictly/ convex, which would require that
>Kenneth Bull wrote: >> On what interval(s) is f(x) = x^3 increasing?
>> a) (-oo, 0) U (0, oo)
>> b) (-oo, oo)
>> [Can someone prove their answer?]
>> I pick a because:
>> f'(x) = x^2 and f'(x) > 0 on (-oo,0)U(0,oo) thus f is increasing on >> that interval
>> f'(0) = 0 so f is not increasing at 0.
>> Agreed?
>Before seeing the other replies I'd have unhesitatingly agreed >(apart from f'(x) being 3.x^2 ! ), believing that a function is not >increasing at a stationary point; but now I'm confused.
>Obviously there's a distinction between "increasing at a point" >and "increasing over an interval", despite the same word being >used in each phrase.
I don't recall ever seeing a definition of "increasing at x". A function is increasing on a set if a < b implies f(a) < f(b). Honest. Even if there is such a thing as "increasing at x", the definition of "increasing on S" is positively _not_ "increasing at x, for all x in S."
Now, if I were going to _invent_ a definition of "f is increasing at x" I think the most reasonable choice would be "f(y) < f(x) for all y < x in some neighborhood of x and f(x) < f(y) for all y > x in some neighborhood of x". With that definition the funtion x^3 is increasing at 0.
What definition of "f is increasing at x" did you have in mind?
>I think we need to wait for twenty or more people to reply, and >take a vote on the number of (a)s versus (b)s ;-P
>(Also, Rob's definition of "increasing" is what I'd call >"non-decreasing", >reserving "increasing" for what he calls "strictly increasing"; but >that's >a side-issue.)
The DEFINITION of "increasing function" is that if x> y then f(x)> f(y). If x> y isn't x^3> y^3, for ALL x,y? It is true that "if f'(x)> 0 on an interval then f is increasing on that interval but that is a SUFFICIENT condition, not a NECESSARY condition.
>At 0, it could be said to be increasing; it depends on >your definition.
I don't think there is any definition of "increasing" that makes sense of increasing AT A SINGLE POINT. A function is increasing on an interval is whenever x> y then f(x)> f(y). It's certainly true that if x> y then x ^3> y^3 for x, y any real numbers.
<CB4e...@raptorsfans.com> wrote: > Rob wrote: > > f'(x) = 3x^2, not just x^2, but ya, f'(x) > 0 on (-oo,0)U(0,oo), and so > > f is increasing on this interval.
> > At 0, it could be said to be increasing; it depends on your definition. > > Here are what I think are usual definitions:
> > A function is said to be increasing on some interval A if for every x,y > > in A, x < y implies f(x) <= f(y). A function is said to be strictly > > increasing on some interval A if for every x,y in A, x < y implies f(x) > > < f(y).
> g(x) = { 4 on [-4,4] > { x - 4 on (4, inf) > { x + 4 on (-inf,-4)
> By the first definition you state, g(x) is increasing everywhere. > It would make little sense to call g(x) an increasing function (on it's > domain).
How do you figure his definition says g is increasing?
Ronald Bruck wrote: > In article <A5Byf.6498$xk1.81...@news20.bellglobal.com>, True Raptor > <CB4e...@raptorsfans.com> wrote:
>>Rob wrote:
>>>f'(x) = 3x^2, not just x^2, but ya, f'(x) > 0 on (-oo,0)U(0,oo), and so >>>f is increasing on this interval.
>>>At 0, it could be said to be increasing; it depends on your definition. >>> Here are what I think are usual definitions:
>>>A function is said to be increasing on some interval A if for every x,y >>>in A, x < y implies f(x) <= f(y). A function is said to be strictly >>>increasing on some interval A if for every x,y in A, x < y implies f(x) >>>< f(y).
>>g(x) = { 4 on [-4,4] >> { x - 4 on (4, inf) >> { x + 4 on (-inf,-4)
>>By the first definition you state, g(x) is increasing everywhere. >>It would make little sense to call g(x) an increasing function (on it's >>domain).
> How do you figure his definition says g is increasing?
> g(3) = 4 > g(4.01) = 0.01.
Sorry, should be 0 on [-4, 4]
g(x) = { 0 on [-4,4] { x - 4 on (4, inf) { x + 4 on (-inf,-4)
