Hello, this may seem like a trivial post, but I would be grateful for any assistance.
Is it correct that a smooth function is a function that has a derivative of all orders, that it can be diferentiated an infinite number of times, and for a function of several real variables, the definition is that it has all its partial derivatives of all orders?
Assuming the above is the case, and I have a function and differentiate it to get the zero function, e.g. f(x) = 1, f'(x)=0 am I right in thinking that the derivative of this is zero and so on, i.e. f''(x)=0, f'''(x) = 0...?
A smooth function has all (partial) derivatives defined, but can these be defined as zero, like in the above case? Is f(x) = x^2 a smooth function, i.e. f'(x) = 2x, f''(x) = 2, f'''(x) = 0, and all other derivatives are zero? Or are smooth functions restricted to functions like trig and exponentials etc?
I am just unsure as to whether the derivatives being zero equates to the derivatives not being defined.
Thanks in advance,
Anthony
I still think I struggle with maths more than Einstein, I suspect he was just being modest...
> Is it correct that a smooth function is a function that has a > derivative of all orders, that it can be diferentiated an infinite > number of times, and for a function of several real variables, the > definition is that it has all its partial derivatives of all orders?
I guess most of the time it goes like you outlined. But if I'm not mistaken, *sometimes* a function is also called "smooth" if the first order derivative exists and is continuous.
> Assuming the above is the case, and I have a function and > differentiate it to get the zero function, e.g. f(x) = 1, f'(x)=0 am > I right in thinking that the derivative of this is zero and so on, > i.e. f''(x)=0, f'''(x) = 0...?
yes.
> A smooth function has all (partial) derivatives defined, but can these > be defined as zero, like in the above case? Is f(x) = x^2 a smooth > function, i.e. f'(x) = 2x, f''(x) = 2, f'''(x) = 0, and all other > derivatives are zero? Or are smooth functions restricted to functions > like trig and exponentials etc?
No, a constant function like f(x) = k is so smooth that it almost hurts. They don't come any smoother.
> I am just unsure as to whether the derivatives being zero equates to > the derivatives not being defined.
Being zero is perfectly defined, zero being not just another number, but still a perfectly acceptable one ;-)
> Thanks in advance,
> Anthony
> I still think I struggle with maths more than Einstein, I suspect he > was just being modest...
In article <8a9174dc.0209171326.24eaa...@posting.google.com>, anthonymbu...@hotmail.com (anthony burke) wrote:
> Hello, this may seem like a trivial post, but I would be grateful for > any assistance.
> Is it correct that a smooth function is a function that has a > derivative of all orders, that it can be diferentiated an infinite > number of times, and for a function of several real variables, the > definition is that it has all its partial derivatives of all orders?
It will usually mean that, but it may, ocasionally, only mean it has continuous first derivatives. Which meaning is usually clear from context
> Assuming the above is the case, and I have a function and > differentiate it to get the zero function, e.g. f(x) = 1, f'(x)=0 am > I right in thinking that the derivative of this is zero and so on, > i.e. f''(x)=0, f'''(x) = 0...?
Once you get a zero function, all further derivatives are automatically zero functions too.
> A smooth function has all (partial) derivatives defined, but can these > be defined as zero, like in the above case? Is f(x) = x^2 a smooth > function, i.e. f'(x) = 2x, f''(x) = 2, f'''(x) = 0, and all other > derivatives are zero? Or are smooth functions restricted to functions > like trig and exponentials etc?
Smooth functions include all polynomial functions (in any number of variables) and rational functions with no poles (quotients of polynomials whose divisor polynomials are never zero).
> I am just unsure as to whether the derivatives being zero equates to > the derivatives not being defined.
Zero is perfectly well defined real number, and a function whose value is always zero is just a special case of a constant function. Such functions are exremely smooth, since their derivatives of all orders exist and are always zero functions.
On 17 Sep 2002 14:26:20 -0700, anthonymbu...@hotmail.com (anthony
burke) wrote: >Hello, this may seem like a trivial post, but I would be grateful for >any assistance.
>Is it correct that a smooth function is a function that has a >derivative of all orders, that it can be diferentiated an infinite >number of times, and for a function of several real variables, the >definition is that it has all its partial derivatives of all orders?
The word "smooth" is used with various meanings (check out what "smooth" means in Zygmund "Trigonometric Series" for example!) but this is one of the most common meanings, yes.
>Assuming the above is the case, and I have a function and >differentiate it to get the zero function, e.g. f(x) = 1, f'(x)=0 am >I right in thinking that the derivative of this is zero and so on, >i.e. f''(x)=0, f'''(x) = 0...?
Of course.
>A smooth function has all (partial) derivatives defined, but can these >be defined as zero, like in the above case? Is f(x) = x^2 a smooth >function, i.e. f'(x) = 2x, f''(x) = 2, f'''(x) = 0, and all other >derivatives are zero?
Yes.
>Or are smooth functions restricted to functions >like trig and exponentials etc?
No.
>I am just unsure as to whether the derivatives being zero equates to >the derivatives not being defined.
anthonymbu...@hotmail.com (anthony burke) writes: > Is it correct that a smooth function is a function that has a > derivative of all orders, that it can be diferentiated an infinite > number of times, and for a function of several real variables, the > definition is that it has all its partial derivatives of all orders?
As others have pointed out, "smooth" can mean several things, depending on context; however, the above definition of smooth is not quite correct (in all the times I have seen the word used). Instead, there is a subtly different definition: that all of the partial derivatives of all orders exist *and* are continuous. (For functions of one variable, the existence of derivatives implies continuity [by the existence of the next derivative and the fact that differentiability implies continuity]; however, it makes a big difference for multivariate functions.)
> I am just unsure as to whether the derivatives being zero equates to > the derivatives not being defined.
Zero exists just like any other number; the function that is identically zero is a perfectly acceptable function. If derivative=0 then derivative exists.
>> Is it correct that a smooth function is a function that has a >> derivative of all orders, that it can be diferentiated an infinite >> number of times, and for a function of several real variables, the >> definition is that it has all its partial derivatives of all orders?
>As others have pointed out, "smooth" can mean several things, >depending on context; however, the above definition of smooth is not >quite correct (in all the times I have seen the word used). Instead, >there is a subtly different definition: that all of the partial >derivatives of all orders exist *and* are continuous. (For functions >of one variable, the existence of derivatives implies continuity [by >the existence of the next derivative and the fact that >differentiability implies continuity]; however, it makes a big >difference for multivariate functions.)
Holy cow, what's-his-name is right, I should be fired. I'm willing to admit the following only because I've recently read that nobody of any consequence reads sci.math:
Duh. An hour ago if you'd asked me I would have said that while of course the existence of, say, first-order partials does not imply continuity, but if you assume _enough_ partials exist that _does_ imply continuity, so if all the partials exist then they are all continuous. But that's nonsense. For the benefit of anyone who knows a little about what sort of things a smooth function can do, but who nonethless has the same gaping hole in his understanding that I did (if there is such a person other than me):
It's easy to give an example. For example, say
V = {(x,y) : |x| <= |y|/2} union {(x,y) : |y| <= |x|/2},
and let
D = {(x,y) : (x,y) <> (0,0) and |x| = |y|}.
It's clear that there exists an f which is smooth except at the origin and which equals 0 at every point of V and 1 at every point of D. And now the fact that f vanishes on V shows that every single friggin partial of f at the origin exists (and equals 0); this is clear by induction on the order of the partial, because all the partials of f vanish at every point (x,0) with x <> 0 and every point (0,y) with y <> 0.
What an idiot I've been all my life. Just read a book "Everything You Know is Wrong" - I think I should send this to the editor for inclusion in the next edition.
Thanks for clearing _that_ up. Maybe there's some hope - at least it didn't take me long to decide that your obviously absurd statement was in fact correct. Oh well. Remind me not to believe _anything_ that's obviously true unless I actually know how to prove it. (And please don't show this to the department head here at OSU.) I wonder what other idiocies are lurking in this sorry substitute for a brain. Let's see, the fundamental theorem of calculus, no I can prove that...
>> I am just unsure as to whether the derivatives being zero equates to >> the derivatives not being defined.
>Zero exists just like any other number; the function that is >identically zero is a perfectly acceptable function. If derivative=0 >then derivative exists.
> >Hello, this may seem like a trivial post, but I would be grateful for > >any assistance.
> >Is it correct that a smooth function is a function that has a > >derivative of all orders, that it can be diferentiated an infinite > >number of times, and for a function of several real variables, the > >definition is that it has all its partial derivatives of all orders?
> The word "smooth" is used with various meanings (check out what > "smooth" means in Zygmund "Trigonometric Series" for example!) > but this is one of the most common meanings, yes.
> >Assuming the above is the case, and I have a function and > >differentiate it to get the zero function, e.g. f(x) = 1, f'(x)=0 am > >I right in thinking that the derivative of this is zero and so on, > >i.e. f''(x)=0, f'''(x) = 0...?
> Of course.
> >A smooth function has all (partial) derivatives defined, but can these > >be defined as zero, like in the above case? Is f(x) = x^2 a smooth > >function, i.e. f'(x) = 2x, f''(x) = 2, f'''(x) = 0, and all other > >derivatives are zero?
> Yes.
> >Or are smooth functions restricted to functions > >like trig and exponentials etc?
> No.
> >I am just unsure as to whether the derivatives being zero equates to > >the derivatives not being defined.
> No. If f' = 0 then f has a derivative.
> >Thanks in advance,
> >Anthony
> >I still think I struggle with maths more than Einstein, I suspect he > >was just being modest...
The process of natural radioactive is described as follows: Number of atoms decaying at any instant of time (or dN) is equal to the number of atoms (N) (balance) in the mass or dN=F(N). Although the process natural radioactive decay is described in terms of number of atoms, as if 'mass' is atomistic, 'number' that describes the process, is never used in any calculation. The initial number of atoms is replaced by 'mass'(=1) so that we do not have to restrict number of half lives (decay continues indefinitely) as if mass is continuous, not atomistic. The process of natural radioactive decay is represented by a smooth function - any number of successive differentiation of this function will not give a constant. 1/2+1/2^2+1/2^3------------1/2^n------up to INFINITY is not a smooth function. The sum of this series doesnot constitute a continuous mass. The moment we say that 'duration' (space or time) between consecutive terms of the above series is constant (constant half-life), the sum of the series or the 'mass' becomes continuous and does not remain atomistic. The process represents an exponential function which is 'time' both, base and exponent. Every smooth function must be a function of time.
> > Is it correct that a smooth function is a function that has a > > derivative of all orders, that it can be diferentiated an infinite > > number of times, and for a function of several real variables, the > > definition is that it has all its partial derivatives of all orders?
> As others have pointed out, "smooth" can mean several things, > depending on context; however, the above definition of smooth is not > quite correct (in all the times I have seen the word used). Instead, > there is a subtly different definition: that all of the partial > derivatives of all orders exist *and* are continuous. (For functions > of one variable, the existence of derivatives implies continuity [by > the existence of the next derivative and the fact that > differentiability implies continuity]; however, it makes a big > difference for multivariate functions.)
> > I am just unsure as to whether the derivatives being zero equates to > > the derivatives not being defined.
> Zero exists just like any other number; the function that is > identically zero is a perfectly acceptable function. If derivative=0 > then derivative exists.
> Kevin.
I do not think 'smooth' can mean several things. It is associated with our sense of touch of, say, an edge or a surface. Therefore we have to associate 'smooth' with CONTINUOUS lines and surfaces of geometry. A continuous straight line or a plain flat surface (a plane in geometry) is smooth. A continuous curve (open or closed) is smooth. A parabola or sphere is smooth. But any discontinuous line or surface is not 'smooth' although it may be possible to CONSTRUCT a 'smooth' line or surface by putting them together. But the process of 'putting things together' is in expressible in symbolic mathematics. This is the reason why we cannot consider a hyperbola expressed as XY=1 is smooth. Here we assign values to X and calculate the value of Y. By this process we can never fill all the points (they will be infinite in number) and make the curve 'smooth'. The values of TanA (A= 0 to 90 degrees) cannot become continuous by joining (by putting together) different values of TanA by a line. If we want to make values of TanA continuous then we have to express TanA as the function of time. TanA as the function of time would allow us to draw all the open curves but not any closed curve, like a circle or an ellipse. The formula for all possible closed curve is(dA/dL)*(dL/dT)=constant (here dA is differential of 'angle' and dA/dL is instantaneous curvature). The expression represents Keppler's law. The 'smooth'is an integral constituted of a continuous variable in space-time (we can never know when a pleasent warmth has become a painful heat) or something always constant in time (sleep), both are inexpressible.
<folti...@math.utexas.edu> wrote: >David C. Ullrich <ullr...@math.okstate.edu> writes:
>> Just read a book "Everything You Know is Wrong"
>You sure about that?
Yes. If you assume enough (mixed) partials exist then it's easy to concoct an argument showing that f is continuous, and it follows that I just read that book.
>> > Is it correct that a smooth function is a function that has a >> > derivative of all orders, that it can be diferentiated an infinite >> > number of times, and for a function of several real variables, the >> > definition is that it has all its partial derivatives of all orders?
>> As others have pointed out, "smooth" can mean several things, >> depending on context; however, the above definition of smooth is not >> quite correct (in all the times I have seen the word used). Instead, >> there is a subtly different definition: that all of the partial >> derivatives of all orders exist *and* are continuous. (For functions >> of one variable, the existence of derivatives implies continuity [by >> the existence of the next derivative and the fact that >> differentiability implies continuity]; however, it makes a big >> difference for multivariate functions.)
>> > I am just unsure as to whether the derivatives being zero equates to >> > the derivatives not being defined.
>> Zero exists just like any other number; the function that is >> identically zero is a perfectly acceptable function. If derivative=0 >> then derivative exists.
>> Kevin.
>I do not think 'smooth' can mean several things.
Doesn't matter whether you think that or not. In mathematics words mean _exactly_ what their definitions say they mean, no more and no less. And the word "smooth" is given various different definitions in different mathematical contexts.
> > > Is it correct that a smooth function is a function that has a > > > derivative of all orders, that it can be diferentiated an infinite > > > number of times, and for a function of several real variables, the > > > definition is that it has all its partial derivatives of all orders?
> > As others have pointed out, "smooth" can mean several things, > > depending on context; however, the above definition of smooth is not > > quite correct (in all the times I have seen the word used). Instead, > > there is a subtly different definition: that all of the partial > > derivatives of all orders exist *and* are continuous. (For functions > > of one variable, the existence of derivatives implies continuity [by > > the existence of the next derivative and the fact that > > differentiability implies continuity]; however, it makes a big > > difference for multivariate functions.)
> > > I am just unsure as to whether the derivatives being zero equates to > > > the derivatives not being defined.
> > Zero exists just like any other number; the function that is > > identically zero is a perfectly acceptable function. If derivative=0 > > then derivative exists.
> > Kevin.
> I do not think 'smooth' can mean several things. It is associated with > our sense of touch of, say, an edge or a surface. Therefore we have to > associate > 'smooth' with CONTINUOUS lines and surfaces of geometry. A continuous > straight line or a plain flat surface (a plane in geometry) is smooth. > A continuous curve (open or closed) is smooth. A parabola or sphere is > smooth. But any discontinuous line or surface is not 'smooth' although > it may be possible to CONSTRUCT a 'smooth' line or surface by putting > them together. But the process > of 'putting things together' is in expressible in symbolic > mathematics. This is the reason why we cannot consider a hyperbola > expressed as XY=1 is smooth. Here we assign values to X and calculate > the value of Y. By this process we can never fill all the points (they > will be infinite in number) and make the curve 'smooth'. The values of > TanA (A= 0 to 90 degrees) cannot become continuous by joining (by > putting together) different values of TanA by a line. If we want to > make values of TanA continuous then we have to express TanA as the > function of time. TanA as the function of time would allow us to draw > all the open curves but not any closed curve, like a circle or an > ellipse. The formula for all possible closed curve > is(dA/dL)*(dL/dT)=constant (here dA is differential of 'angle' and > dA/dL is instantaneous curvature). The expression represents Keppler's > law. The 'smooth'is an integral constituted of a continuous variable > in space-time (we can never know when a pleasent warmth has become a > painful heat) or something always constant in time (sleep), both are > inexpressible.
According to Wolfram Smooth Function A smooth function is a function that has continuous second-order derivatives over some domain. A function can therefore be said to be smooth over a restricted interval such as (a, b) or [a, b]. http://mathworld.wolfram.com/SmoothFunction.html,
Unfortunately a math definition cannot agree with everyone's unique vision of the context of the word.
David C. Ullrich <ullr...@math.okstate.edu> writes:
> On 19 Sep 2002 11:21:38 -0500, Kevin Foltinek > <folti...@math.utexas.edu> wrote:
> >David C. Ullrich <ullr...@math.okstate.edu> writes:
> >> Just read a book "Everything You Know is Wrong"
> >You sure about that?
> Yes. If you assume enough (mixed) partials exist > then it's easy to concoct an argument showing that > f is continuous, and it follows that I just read > that book.
Aren't you using the Axiom of Choice in that argument?
vgopa...@rediffmail.com (V.Gopal) writes: > I do not think 'smooth' can mean several things.
Your (evidently uninformed) opinion is irrelevant to the consensus definitions used by the mathematical community.
> It is associated with > our sense of touch of, say, an edge or a surface. Therefore we have to > associate > 'smooth' with CONTINUOUS lines and surfaces of geometry.
Sandpaper is continuous. Go rub 60 grit paper for an hour and then tell us if continuous things are smooth.
> [snip decreasingly meaningful sequence of words]
> > I do not think 'smooth' can mean several things.
> Your (evidently uninformed) opinion is irrelevant to the consensus > definitions used by the mathematical community.
> > It is associated with > > our sense of touch of, say, an edge or a surface. Therefore we have to > > associate > > 'smooth' with CONTINUOUS lines and surfaces of geometry.
> Sandpaper is continuous. Go rub 60 grit paper for an hour and then > tell us if continuous things are smooth.
> > [snip decreasingly meaningful sequence of words]
> Kevin.
Absolute value abs(x) is not smooth but it looks a *lot* smoother than y(x) = (1000000*x)^2 which *is* smooth. I'll sit on the abs peak, you can take the other one.
Bob Pease <bobpe...@concentric.net> wrote: >According to Wolfram >Smooth Function A smooth function is a function that has continuous >second-order derivatives over some domain. A function can therefore be said >to be smooth over a restricted interval such as (a, b) or [a, b]. >http://mathworld.wolfram.com/SmoothFunction.html,
>Unfortunately a math definition cannot agree with everyone's unique vision >of the context of the word.
Nor, apparently, need it agree with mathematicians' common "visions" of the word. Which is the mathematical community which would naturally take "smooth" to mean C^2? I would have expected C^1 and C^\infty to be more natural interpretations. I guess the family of C^2 functions is technically convenient for some tasks, but there are others which work better in other settings.
I like the MathWorld website a lot, but it's sort of unfortunate that some people believe it represents a practicing mathematician's understanding of the discipline. Sometimes yes, sometimes no. Ours is at least as specialized discipline as, say, law; but while some people have gotten into the habit of writing "IANAL but...", they seem less willing to write "I am not a mathematician but...".
> In article <amdd66$...@dispatch.concentric.net>, > Bob Pease <bobpe...@concentric.net> wrote:
> >According to Wolfram > >Smooth Function A smooth function is a function that has continuous > >second-order derivatives over some domain. A function can therefore be said > >to be smooth over a restricted interval such as (a, b) or [a, b]. > >http://mathworld.wolfram.com/SmoothFunction.html,
> >Unfortunately a math definition cannot agree with everyone's unique vision > >of the context of the word.
> Nor, apparently, need it agree with mathematicians' common "visions" of > the word. Which is the mathematical community which would naturally take > "smooth" to mean C^2? I would have expected C^1 and C^\infty to be more > natural interpretations. I guess the family of C^2 functions is > technically convenient for some tasks, but there are others which > work better in other settings.
To reprise myself, Ya gotta form a consensus, or at least a truce. James and James gives a working definition that is more precise than mathworld. It also might be less useful to a person studying CalcIII who just needs some guideline.
> I like the MathWorld website a lot, but it's sort of unfortunate that > some people believe it represents a practicing mathematician's > understanding of the discipline. Sometimes yes, sometimes no. > Ours is at least as specialized discipline as, say, law; but while > some people have gotten into the habit of writing "IANAL but...", they > seem less willing to write "I am not a mathematician but...".
I don't know that we have a disagreement here. Sometimes, however, flexibility is a need when communicating among varying levels of specialization or , indeed, among different levels of expertise.
<folti...@math.utexas.edu> wrote: >David C. Ullrich <ullr...@math.okstate.edu> writes:
>> On 19 Sep 2002 11:21:38 -0500, Kevin Foltinek >> <folti...@math.utexas.edu> wrote:
>> >David C. Ullrich <ullr...@math.okstate.edu> writes:
>> >> Just read a book "Everything You Know is Wrong"
>> >You sure about that?
>> Yes. If you assume enough (mixed) partials exist >> then it's easy to concoct an argument showing that >> f is continuous, and it follows that I just read >> that book.
>Aren't you using the Axiom of Choice in that argument?
It's clear that AC _follows_ from the result that if f has partials of all orders then f is continuous.
>>According to Wolfram >>Smooth Function A smooth function is a function that has continuous >>second-order derivatives over some domain. A function can therefore be said >>to be smooth over a restricted interval such as (a, b) or [a, b]. >>http://mathworld.wolfram.com/SmoothFunction.html,
>>Unfortunately a math definition cannot agree with everyone's unique vision >>of the context of the word.
>Nor, apparently, need it agree with mathematicians' common "visions" of >the word. Which is the mathematical community which would naturally take >"smooth" to mean C^2?
My favorite is Zygmund "Trigonometric Functions", where smooth functions are functions in what everyone else calls the "little-oh Zugmund class" (f is continuous and
(f(x+h) - 2 f(x) + f(x-h))/h -> 0 as h -> 0.
No typo, it's h, not h^2. This is more than continuous but much less than C^1; just happens to be something with convenient characterizations in terms of Fourier series.)
>I would have expected C^1 and C^\infty to be more >natural interpretations. I guess the family of C^2 functions is >technically convenient for some tasks, but there are others which >work better in other settings.
>I like the MathWorld website a lot, but it's sort of unfortunate that >some people believe it represents a practicing mathematician's >understanding of the discipline. Sometimes yes, sometimes no. >Ours is at least as specialized discipline as, say, law; but while >some people have gotten into the habit of writing "IANAL but...", they >seem less willing to write "I am not a mathematician but...".
David C. Ullrich wrote: >On 17 Sep 2002 14:26:20 -0700, anthonymbu...@hotmail.com (anthony >burke) wrote:
>>Assuming the above is the case, and I have a function and >>differentiate it to get the zero function, e.g. f(x) = 1, f'(x)=0 am >>I right in thinking that the derivative of this is zero and so on, >>i.e. f''(x)=0, f'''(x) = 0...?
>Of course.
Assuming the OP meant what he wrote, and that the derivatives are the 0 function, and not that the value of the derivative at x just happens to be 0.
David C. Ullrich <ullr...@math.okstate.edu> writes:
> On 19 Sep 2002 15:47:07 -0500, Kevin Foltinek > <folti...@math.utexas.edu> wrote:
> >David C. Ullrich <ullr...@math.okstate.edu> writes:
> >> Yes. If you assume enough (mixed) partials exist > >> then it's easy to concoct an argument showing that > >> f is continuous, and it follows that I just read > >> that book.
> >Aren't you using the Axiom of Choice in that argument?
> It's clear that AC _follows_ from the result that if > f has partials of all orders then f is continuous.
I'm pretty sure that the above is true only if you assume that you did in fact read the book.
It seems, then, that AC is equivalent to you having read the book "Everything You Know is Wrong".
In retrospect, this should not have been at all surprising; the title of the book suggests explicitly assigning a truth value to the set of things you know, but implicit is the set of all consequences of things you know, and according to my copy of the book "Choice for Complete Idiots", discussing the consequences involves making some choices. Also, choices have consequences. QED.
<folti...@math.utexas.edu> wrote: >David C. Ullrich <ullr...@math.okstate.edu> writes:
>> On 19 Sep 2002 15:47:07 -0500, Kevin Foltinek >> <folti...@math.utexas.edu> wrote:
>> >David C. Ullrich <ullr...@math.okstate.edu> writes:
>> >> Yes. If you assume enough (mixed) partials exist >> >> then it's easy to concoct an argument showing that >> >> f is continuous, and it follows that I just read >> >> that book.
>> >Aren't you using the Axiom of Choice in that argument?
>> It's clear that AC _follows_ from the result that if >> f has partials of all orders then f is continuous.
>I'm pretty sure that the above is true only if you assume that you did >in fact read the book.
>It seems, then, that AC is equivalent to you having read the book >"Everything You Know is Wrong".
>In retrospect, this should not have been at all surprising; the title >of the book suggests explicitly assigning a truth value to the set of >things you know, but implicit is the set of all consequences of things >you know, and according to my copy of the book "Choice for Complete >Idiots", discussing the consequences involves making some choices. >Also, choices have consequences. QED.
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