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Suppose that the first derivative of f(x) is g(x).
What the hell is the derivative of f(f(x))?
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On Mon, 16 Nov 2009 13:39:50 +0100, Invisible <voi### [at] dev null> wrote:
> Suppose that the first derivative of f(x) is g(x).
>
> What the hell is the derivative of f(f(x))?
http://www.google.com/search?q=What+is+the+derivative+of+a+composite+function%3F
http://en.wikipedia.org/wiki/Chain_rule
Basic high-school level math. I am slightly surprised that you did not
know it, though I am not in the least surprised that you did not bother to
google for it.
--
FE
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Fredrik Eriksson wrote:
> Basic high-school level math. I am slightly surprised that you did not
> know it,
What, you mean given that I've never been tought mathematics?
> though I am not in the least surprised that you did not bother
> to google for it.
Oh, I did Google it. I just didn't comprehend the answer. (In
particular, I've already read the Wikipedia entry.)
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On Mon, 16 Nov 2009 13:51:11 +0100, Invisible <voi### [at] devnull> wrote:
>
> Oh, I did Google it. I just didn't comprehend the answer. (In
> particular, I've already read the Wikipedia entry.)
What is so difficult to understand about it? I thought function
composition would be one of the areas of math that you were actually good
at, given your fanatical obsession with a certain functional programming
language.
Anyway:
The derivative of f( g(x) ) is f'(g(x))*g'(x).
Substitute the inner function into the derivative of the outer function,
and then multiply by the derivative of the inner function.
In your example, we want the derivative of f(f(x)) (i.e. the outer and
inner functions are the same), and we know that f'(x) = g(x) (try not to
mix up all the f's and g's).
df(f(x))/dx = f'(f(x))*f'(x) = g(f(x))*g(x)
--
FE
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>> Oh, I did Google it. I just didn't comprehend the answer. (In
>> particular, I've already read the Wikipedia entry.)
>
> What is so difficult to understand about it?
One formula appears to say the derivative is f(x) * g(x), and another
appears to say it's f'(g(x)) * g'(x), which is different, but these are
supposed to be different notations for the same formula...?
> I thought function
> composition would be one of the areas of math that you were actually
> good at, given your fanatical obsession with a certain functional
> programming language.
Function composition I get. It's what that does to the derivating I
can't quite wrap my brain around.
> Anyway:
>
> The derivative of f( g(x) ) is f'(g(x))*g'(x).
>
> Substitute the inner function into the derivative of the outer function,
> and then multiply by the derivative of the inner function.
> df(f(x))/dx = f'(f(x))*f'(x) = g(f(x))*g(x)
Right. So basically, the function I'm looking for is f'(f(x))*f'(x)?
So presumably the derivative of f(f(f(x))) is going to be something like
f'(f(f(x))) * f'(f(x)) * f'(x)? And in general, the Nth composition will
be ever more complex as N increases?
Oh dear.
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Fredrik Eriksson wrote:
> Basic high-school level math. I am slightly surprised that you did not
> know it, though I am not in the least surprised that you did not bother
> to google for it.
...what I *should* have done of course is
http://www.wolframalpha.com/input/?i=derivative+f(f(x))
*facepalm*
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Invisible wrote:
> So presumably the derivative of f(f(f(x))) is going to be something like
> f'(f(f(x))) * f'(f(x)) * f'(x)? And in general, the Nth composition will
> be ever more complex as N increases?
>
> Oh dear.
Ooo, but wait... The Nth derivative of a polynomial will be zero if N
exceeds the order of the polynomial. Hmm.
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>> So presumably the derivative of f(f(f(x))) is going to be something
>> like f'(f(f(x))) * f'(f(x)) * f'(x)? And in general, the Nth
>> composition will be ever more complex as N increases?
>>
>> Oh dear.
>
> Ooo, but wait... The Nth derivative of a polynomial will be zero if N
> exceeds the order of the polynomial. Hmm.
...which doesn't help, because the derivative of the Nth iterate will be
f'(x) * f'(f(x)) * f'(f(f(x)) * f'(f(f(f(x))) * ...
which involves only the first derivating of f and the iterates of f.
Damnit, I really suck at math. :'{
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Invisible wrote:
> Damnit, I really suck at math. :'{
D f[n](x) = D f[n-1](x) * f'(f[n-1](x))
Maybe this isn't so bad...
Or maybe I just need to GROW A FRIGGIN' BRAIN! >_<
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On Mon, 16 Nov 2009 14:14:27 +0100, Invisible <voi### [at] devnull> wrote:
>
> One formula appears to say the derivative is f(x) * g(x), and another
> appears to say it's f'(g(x)) * g'(x), which is different, but these are
> supposed to be different notations for the same formula...?
The first one is obviously wrong, unless you mixed up all the f's, g's and
little circles. It is difficult to tell without context.
>> df(f(x))/dx = f'(f(x))*f'(x) = g(f(x))*g(x)
>
> Right. So basically, the function I'm looking for is f'(f(x))*f'(x)?
Yes.
> And in general, the Nth composition will be ever more complex as N
> increases?
Yes.
> ...what I *should* have done of course is
> http://www.wolframalpha.com/input/?i=derivative+f(f(x))
Except that it is the wrong answer, or rather the wrong question. Wolfram
thought you meant (f <dot> f)(x), when what you actually meant was (f
<circle> f)(x).
Just one more reason why it is better to learn the formulas yourself
instead of relying on a computer to help you all the time.
--
FE
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On Mon, 16 Nov 2009 14:32:00 +0100, Invisible <voi### [at] devnull> wrote:
>
> Damnit, I really suck at math. :'{
From what I can tell, you basically suffer from a lack of education. That
can be remedied, if you are willing to make the effort.
--
FE
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>> One formula appears to say the derivative is f(x) * g(x), and another
>> appears to say it's f'(g(x)) * g'(x), which is different, but these
>> are supposed to be different notations for the same formula...?
>
> The first one is obviously wrong, unless you mixed up all the f's, g's
> and little circles. It is difficult to tell without context.
Well, one formula says something like dy/dx = dy/du * du/dx, while the
other says (f.g)'(x) = f(g(x)) * g(x).
>>> df(f(x))/dx = f'(f(x))*f'(x) = g(f(x))*g(x)
>>
>> Right. So basically, the function I'm looking for is f'(f(x))*f'(x)?
>
> Yes.
Nice to know I got something right today...
>> And in general, the Nth composition will be ever more complex as N
>> increases?
>
> Yes.
Hmm.
>> ...what I *should* have done of course is
>> http://www.wolframalpha.com/input/?i=derivative+f(f(x))
>
> Except that it is the wrong answer, or rather the wrong question.
Yes. The query should be f[f[x]] instead of f(f(x)). If you ask for a
higher iterate, it quickly becomes obvious that something is wrong.
> Just one more reason why it is better to learn the formulas yourself
> instead of relying on a computer to help you all the time.
Oh, sure. But I find it useful to check against the computer. If the
answers are the same, you probably did it right. If the answers are
different, either you asked the computer the wrong thing or you stuffed
up the math.
(For example, (Sqrt(3) a)^3 = 3 a^2. Or NOT, in fact - as Wolfram
rightly pointed out...)
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Fredrik Eriksson wrote:
> From what I can tell, you basically suffer from a lack of education.
Well, I _did_ go to a school for stupid people.
> That can be remedied, if you are willing to make the effort.
I'm more concerned about the time and cost aspect of it, but sure.
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On 16-11-2009 14:42, Invisible wrote:
> Fredrik Eriksson wrote:
>
>> From what I can tell, you basically suffer from a lack of education.
>
> Well, I _did_ go to a school for stupid people.
>
>> That can be remedied, if you are willing to make the effort.
>
> I'm more concerned about the time and cost aspect of it, but sure.
It will cost time but not so much money. It is for your personal
education, so you don't need to enroll in university. And even time is
relative here, as long as you are trying to compute these things _not_
studying it will cost you more time.
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On Mon, 16 Nov 2009 14:41:12 +0100, Invisible <voi### [at] devnull> wrote:
>
> Well, one formula says something like dy/dx = dy/du * du/dx
Yes, but note the use of 'u', which in this case basically represents the
inner function.
--
FE
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On 16-11-2009 14:15, Invisible wrote:
> Fredrik Eriksson wrote:
>
>> Basic high-school level math. I am slightly surprised that you did not
>> know it, though I am not in the least surprised that you did not
>> bother to google for it.
>
> ...what I *should* have done of course is
>
> http://www.wolframalpha.com/input/?i=derivative+f(f(x))
>
> *facepalm*
no, because that is a derivative with respect to f not x.
your answer might be
http://www.wolframalpha.com/input/?i=derivative+f%20f%20x%20with%20respect%20to%20x
but then again it might not, depending on what you really want.
Many of my students are (hopefully: were) unable to perform a long
division or even a long multiplication. They weren't though at school
because 'everybody has a calculator at hand anyway'.
Someday in the near future all the rest of mathematics and calculus will
not be though anymore because 'everybody has an internet connection at
hand anyway'. That will be a sad day for our culture.
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>> ...what I *should* have done of course is
>>
>> http://www.wolframalpha.com/input/?i=derivative+f(f(x))
>>
>> *facepalm*
>
> no, because that is a derivative with respect to f not x.
Yes, I figured that out already. (It's actually quite nice that
Mathematica has an unambiguous notation for this - but you have to
remember to use it!)
> Many of my students are (hopefully: were) unable to perform a long
> division or even a long multiplication. They weren't though at school
> because 'everybody has a calculator at hand anyway'.
At my school, we did nothing *but* long division! Pages and pages and
pages of it, for years on end. From when I was 14 until I left school,
_all_ we did in our maths lessons was long division. (Not to mention the
homework, which was... long division.)
> Someday in the near future all the rest of mathematics and calculus will
> not be though anymore because 'everybody has an internet connection at
> hand anyway'. That will be a sad day for our culture.
I'm not sure they teach calculus *currently* in schools. Then again, I'm
not sure because I didn't attend a normal school... I think in this
country, if you want to learn about mathematics (not just arithmetic),
you have to take a special course. I think it's outside the normal
school curriculum.
Then again, how many people will ever need to know how to compute the
integral of a polynomial? If you can't add, you're going to have a bit
of a problem doing your weekly shopping, but there's not much need for
higher math unless you're working in some specialist industry somewhere.
(Or you're an academic, which nobody is.)
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>> Well, one formula says something like dy/dx = dy/du * du/dx
>
> Yes, but note the use of 'u', which in this case basically represents
> the inner function.
This is one of the more confusing aspects of calculus. The duality
between "functions" and "variables that share a relationship". That and
the multiple notations for the same concept...
I still don't entirely understand what you're saying, but presumably
you're right.
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On Mon, 16 Nov 2009 15:11:13 +0100, Invisible <voi### [at] devnull> wrote:
>
> I still don't entirely understand what you're saying
(Excuse the mixing of Lagrange notation and ASCII-fied Leibniz notation)
First look at y = f(x). Then f'(x) = dy/dx.
Now try y = f( g(x) )
Let u = g(x). Then y = f( u ).
In f'(x) = dy/dx, we can trivially introduce 'u' thus:
f'(x) = dy/dx * (du/du) = (dy/du) * (du/dx)
dy/du is obviously the derivative of 'y' with respect to 'u', i.e. f'(u).
Similarly, du/dx is the derivative of 'u' with respect to 'x', i.e. g'(x).
f'(x) = (dy/du) * (du/dx) = f'(u) * g'(x)
Substitute u = g(x) to get:
f'(x) = f'( g(x) ) * g'(x)
--
FE
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> Then again, how many people will ever need to know how to compute the
> integral of a polynomial?
This is why it's not taught to everyone, and is reserved until when people
*choose* they want to do maths (when you're 16 in the UK).
> If you can't add, you're going to have a bit of a problem doing your
> weekly shopping, but there's not much need for higher math unless you're
> working in some specialist industry somewhere.
You'd be surprised how many industries require specialist knowledge :-) For
example, I suspect the structural engineer that designed your building did
plenty of calculations using calculus to estimate stresses based on the
loading (simple cases can be looked up in tables, but anything unusual needs
to be worked out manually). Ditto for an electrical engineer who designed
the power supply for your computer, the DCDC converter in your mobile phone
and numerous other circuits - without an understanding of calculus you're
going to be totally lost.
Of course computers can simulate and calculate stuff for you, but you're
going to look a right idiot if you need to run ten 12 hour simulations to
decide the correct structure or capacitor size when your colleague can work
it out exactly in a few minutes with a piece of paper.
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Invisible <voi### [at] devnull> wrote:
> I'm not sure they teach calculus *currently* in schools. Then again, I'm
> not sure because I didn't attend a normal school... I think in this
> country, if you want to learn about mathematics (not just arithmetic),
> you have to take a special course. I think it's outside the normal
> school curriculum.
Depends what you mean by 'school'. Derivatives (including the chain rule!),
integrals, and even differential equations were all on the maths A-level
syllabus that I did (I would hope that they still are!). I did my A-levels at
school, but I think many people go to college to study A-levels if their school
does not offer them.
Note that an A-level college is usually *not* the same thing as a university
(cue lengthy debate on schooling ages/terminology in different countries!).
> (Or you're an academic, which nobody is.)
I believe there are many regular posters here who might disagree with that! :-)
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> Depends what you mean by 'school'. Derivatives (including the chain
> rule!),
> integrals, and even differential equations were all on the maths A-level
> syllabus that I did (I would hope that they still are!).
They were when I did my maths A levels (1999).
> Note that an A-level college is usually *not* the same thing as a
> university
> (cue lengthy debate on schooling ages/terminology in different
> countries!).
Hehe yes, I went to a sixth form college to do my A levels because schools
in my area tended to finish at 16 and not offer A levels at all. Then I
left this college at 18 to go to another college, which was part of a
university - grande confusione!
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Invisible wrote:
> Suppose that the first derivative of f(x) is g(x).
>
> What the hell is the derivative of f(f(x))?
Unrelated to your question, but when I read the topic I first thought
you were after the n-th derivative of some function...
For example, deriving n times f(x)*g(x) has a fun formula :-)
--
Vincent
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>> Depends what you mean by 'school'. Derivatives (including the chain
>> rule!),
>> integrals, and even differential equations were all on the maths A-level
>> syllabus that I did (I would hope that they still are!).
>
> They were when I did my maths A levels (1999).
I don't *have* any A-levels, so I wouldn't know what is or isn't
included there.
(You get to choose which ones you want to do though, right?)
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Invisible <voi### [at] devnull> wrote:
> >> Depends what you mean by 'school'. Derivatives (including the chain
> >> rule!),
> >> integrals, and even differential equations were all on the maths A-level
> >> syllabus that I did (I would hope that they still are!).
> >
> > They were when I did my maths A levels (1999).
>
> I don't *have* any A-levels, so I wouldn't know what is or isn't
> included there.
>
> (You get to choose which ones you want to do though, right?)
Yep. As scott said, any education beyond 16 is optional in the UK. If you were
so inclined, there'd be nothing to stop you doing a maths A-level now - many
people do. I suspect you'd find much of it relatively easy, and you'd definitely
find the calculus interesting and useful! No idea what it might cost tho.
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"scott" <sco### [at] scottcom> wrote:
> > Depends what you mean by 'school'. Derivatives (including the chain
> > rule!),
> > integrals, and even differential equations were all on the maths A-level
> > syllabus that I did (I would hope that they still are!).
>
> They were when I did my maths A levels (1999).
I think the structure of A-levels changed in 2000 (more continuous assessment
instead of oodles of exams after 2 years), but I expect it's broadly the same.
> Hehe yes, I went to a sixth form college to do my A levels because schools
> in my area tended to finish at 16 and not offer A levels at all. Then I
> left this college at 18 to go to another college, which was part of a
> university - grande confusione!
Indeed. And of course many universities have the word 'college' in their names,
especially the younger ex-polytechnics, confusing the issue still further!
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>>>> Depends what you mean by 'school'. Derivatives (including the chain
>>>> rule!),
>>>> integrals, and even differential equations were all on the maths A-level
>>>> syllabus that I did (I would hope that they still are!).
>>> They were when I did my maths A levels (1999).
>> I don't *have* any A-levels, so I wouldn't know what is or isn't
>> included there.
>>
>> (You get to choose which ones you want to do though, right?)
>
> Yep. As scott said, any education beyond 16 is optional in the UK. If you were
> so inclined, there'd be nothing to stop you doing a maths A-level now - many
> people do. I suspect you'd find much of it relatively easy, and you'd definitely
> find the calculus interesting and useful! No idea what it might cost tho.
...all of which seems to support my assessment of "calculus is no longer
part of standard school education".
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Invisible wrote:
> What the hell is the derivative of f(f(x))?
LMATFY
http://www.wolframalpha.com/input/?i=derivative+of+f(f(x))
--
Darren New, San Diego CA, USA (PST)
I ordered stamps from Zazzle that read "Place Stamp Here".
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Fredrik Eriksson wrote:
> Except that it is the wrong answer, or rather the wrong question.
> Wolfram thought you meant (f <dot> f)(x), when what you actually meant
> was (f <circle> f)(x).
Which just goes to show the problem I have with 90% of all matehmatical
notation. It's so utterly inconsistent that even something like (f(f(x))) is
ambiguous.
--
Darren New, San Diego CA, USA (PST)
I ordered stamps from Zazzle that read "Place Stamp Here".
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Darren New wrote:
> Invisible wrote:
>> What the hell is the derivative of f(f(x))?
>
> LMATFY
>
> http://www.wolframalpha.com/input/?i=derivative+of+f(f(x))
FAIL.
The correct query is of course
http://www.wolframalpha.com/input/?i=derivative+of+f[f[x]]
Nice try tho. :-P
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Darren New wrote:
> Which just goes to show the problem I have with 90% of all matehmatical
> notation. It's so utterly inconsistent that even something like
> (f(f(x))) is ambiguous.
Several millennia of mathematical discoveries, all made by different
people in different places, and apparently several of them discovered
the same or similar things, but gave them different names - or gave them
names which clash with existing but inrelated things they didn't know about.
Just for giggles: how many meanings can you find for "normal"?
There's the normal distribution, normal vectors, a normed space...
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Invisible wrote:
> Darren New wrote:
>> Invisible wrote:
>>> What the hell is the derivative of f(f(x))?
>>
>> LMATFY
>>
>> http://www.wolframalpha.com/input/?i=derivative+of+f(f(x))
>
> FAIL.
Yes. I didn't see there were 50 responses before I posted that. :-)
> http://www.wolframalpha.com/input/?i=derivative+of+f[f[x]]
Apparently so. One of the reasons I hate mathematical notation.
--
Darren New, San Diego CA, USA (PST)
I ordered stamps from Zazzle that read "Place Stamp Here".
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>> FAIL.
>
> Yes. I didn't see there were 50 responses before I posted that. :-)
>
>> http://www.wolframalpha.com/input/?i=derivative+of+f[f[x]]
>
> Apparently so. One of the reasons I hate mathematical notation.
Wolfram at least have come up with a consistent, non-ambiguous notation.
In their notation, a(b)c unambiguously means the product a*b*c, and
a[b]c unambiguously means that b is the argument of function a,
multiplied by c. This is partly how Mathematica is able to determine
precisely what you meant.
Of course, nobody else uses this notation. And while *Mathematica*
rigorously follows it, *Alpha* attempts to "guess" what you mean. So in
the case above, it takes "f(f(x))" and guessees that you maybe mean "f *
f[x]", which isn't quite right. (This is what happens when computers try
to guess things...)
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Invisible wrote:
> Just for giggles: how many meanings can you find for "normal"?
Sure. But even differences between things like
sin x^-1 vs sin^-1 x
where the two "^-1" mean entirely different operations. Similarly with
lg^2 X vs lg X^2
Drove me nuts, as I learned computers before I learned all this higher-level
math.
--
Darren New, San Diego CA, USA (PST)
I ordered stamps from Zazzle that read "Place Stamp Here".
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>> Just for giggles: how many meanings can you find for "normal"?
>
> Sure.
Because it's not just the notations that are ambiguous - it's the
language as well!
> But even differences between things like
>
> sin x^-1 vs sin^-1 x
>
> where the two "^-1" mean entirely different operations.
The first one, at least, is unambiguous. But the second one? Now do you
suppose that's the arcsine of x? Or the reciprocol of the sine of x?
Good luck with that one.
> Similarly with
>
> lg^2 X vs lg X^2
>
> Drove me nuts, as I learned computers before I learned all this
> higher-level math.
And then of course, people will write "log x". Wanna take a guess which
base that is? Now, sometimes it actually doesn't matter which base. And
if it does, it *probably* means the natural logarithm. Probably...
Speaking of which, the base of natural logarithms is "e". And sometimes
"e" means 1 + 1/1 + 1/2 + 1/3 + 1/4... And sometimes "e" is just another
variable.
Hell, I've seen formulas where pi does *not* refer to the well-known
transcendental number!
Doomed, DOOMED, WE'RE ALL DOOMED! >_<
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Invisible wrote:
> But the second one? Now do you
> suppose that's the arcsine of x? Or the reciprocol of the sine of x?
That's exactly my point.
> if it does, it *probably* means the natural logarithm.
Well, that makes sense, yes.
> Doomed, DOOMED, WE'RE ALL DOOMED! >_<
I asked an exchange professor from Greece once, and he said they use the
same capital-sigma notation for summation and delta-for-change and such that
we use in America, except all their variables are also greek letters. *That*
must suck. :-)
--
Darren New, San Diego CA, USA (PST)
I ordered stamps from Zazzle that read "Place Stamp Here".
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Darren New wrote:
> I asked an exchange professor from Greece once, and he said they use the
> same capital-sigma notation for summation and delta-for-change and such
> that we use in America, except all their variables are also greek
> letters. *That* must suck. :-)
Heh. Mathematics, not content with featuring a zoo of custom symbols
ranging from "+" and "/" right up to those weird curly arrows, *also*
uses letters from at least the Latin, Greek and Hebrew alphabets. All at
once!
Sqrt(-1) = i [So that's Latin.]
Exp(1) = e [Latin again - except when people use Epsilon instead...]
Asin(1) = pi [Definitely Greek.]
|Z| = Alpha0 [So Hebrew then - with Arabic subscripts just for good
measure.]
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Invisible wrote:
> |Z| = Alpha0 [So Hebrew then - with Arabic subscripts just for good
> measure.]
...assuming you can SPELL. >_<
The cardinality of the set of integers is of course Aleph-null.
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Invisible <voi### [at] devnull> wrote:
> And then of course, people will write "log x". Wanna take a guess which
> base that is? Now, sometimes it actually doesn't matter which base. And
> if it does, it *probably* means the natural logarithm. Probably...
IIRC, 'log x' with no base usually means base 10, and 'ln x' is the natural log.
But, as you say, depends what the local conventions are.
> Hell, I've seen formulas where pi does *not* refer to the well-known
> transcendental number!
Now that's just careless.
> Doomed, DOOMED, WE'RE ALL DOOMED! >_<
Eat more pi.
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On Mon, 16 Nov 2009 17:12:29 +0100, Darren New <dne### [at] sanrrcom> wrote:
>
> Which just goes to show the problem I have with 90% of all matehmatical
> notation. It's so utterly inconsistent that even something like
> (f(f(x))) is ambiguous.
Although I understand (and to some extent agree with) the point you are
trying to make, (f(f(x))) is not ambiguous in mathematical notation. The
problems come when you remove parentheses or interpret them in a
non-standard manner. Wolfram Alpha does those things because mathematical
notation does not translate easily into ASCII.
--
FE
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"Fredrik Eriksson" <fe79}--at--{yahoo}--dot--{com> wrote:
> Basic high-school level math. I am slightly surprised that you did not
> know it, though I am not in the least surprised that you did not bother to
> google for it.
They teach derivatives at high school?
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On Mon, 16 Nov 2009 18:22:34 +0100, Nicolas Alvarez
<nic### [at] gmailcom> wrote:
>
> They teach derivatives at high school?
They did at my high school, though possibly not to all students. I do not
recall which course introduced derivatives, but there were 5 math courses
in total, and only those studying natural sciences or technology took all
five. Most only took the first two.
--
FE
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On 16-11-2009 17:15, Invisible wrote:
> Darren New wrote:
>
>> Which just goes to show the problem I have with 90% of all
>> matehmatical notation. It's so utterly inconsistent that even
>> something like (f(f(x))) is ambiguous.
>
> Several millennia of mathematical discoveries, all made by different
> people in different places, and apparently several of them discovered
> the same or similar things, but gave them different names - or gave them
> names which clash with existing but inrelated things they didn't know
> about.
>
> Just for giggles: how many meanings can you find for "normal"?
>
> There's the normal distribution, normal vectors, a normed space...
Wednesday I have again an opportunity to give my talk on deriving
programs from specifications. One of the things I mention is that the
humble '=' has at least five and possibly more meanings depending on
context, interpretation, and the type op the object that it is applied
to*. And I will not even mention that the general use is inconsistent in
the context of A=B+C where A,B,and C are matrices. Here '+' is
pointwise, while '=' has an implicit summation**. As far as I know there
is not even an generally accepted symbol to express pointwise equality.
I know of people that try the opposite and make the summation explicit,
but that breaks most of other mathematical uses.
* equivalence, equality, definition, EXNOR, assignment and perhaps one
or more that don't have names. The type plays a role in A=B=C which is
OK if they are all booleans (or A or C is) but not if they are all
integers. In A=5 it can be an expression of equality, if A was undefined
it can be a binding/assignment, but it can also be a boolean that is
false everywhere except at 5.
** See also the concept of '=' in OO languages. Are two objects the same
if all fields are the same?
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On 16-11-2009 18:00, Bill Pragnell wrote:
> Invisible <voi### [at] devnull> wrote:
>> And then of course, people will write "log x". Wanna take a guess which
>> base that is? Now, sometimes it actually doesn't matter which base. And
>> if it does, it *probably* means the natural logarithm. Probably...
>
> IIRC, 'log x' with no base usually means base 10, and 'ln x' is the natural log.
> But, as you say, depends what the local conventions are.
>
>> Hell, I've seen formulas where pi does *not* refer to the well-known
>> transcendental number!
>
> Now that's just careless.
Not really, if you have a scheme where s->sigma t->tau then p->pi.
The one I remember (though I forgot the exact formula) is for the energy
per particle in an electric field. Energy is 'E', per particle it is 'e'
the expression contained an exponential function that depended on the
charge of an electron.
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>> And then of course, people will write "log x". Wanna take a guess which
>> base that is? Now, sometimes it actually doesn't matter which base. And
>> if it does, it *probably* means the natural logarithm. Probably...
>
> IIRC, 'log x' with no base usually means base 10, and 'ln x' is the natural log.
> But, as you say, depends what the local conventions are.
Some people use log to mean base-10. Some use it to mean an unspecified
base. But (arguably) *most* geniune maths sources use it to mean the
natural logarithm.
>> Hell, I've seen formulas where pi does *not* refer to the well-known
>> transcendental number!
>
> Now that's just careless.
It's a standard convention.
http://en.wikipedia.org/wiki/Prime-counting_function
--
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*
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>> If you can't add, you're going to have a bit of a problem doing your
>> weekly shopping, but there's not much need for higher math unless
>> you're working in some specialist industry somewhere.
>
> You'd be surprised how many industries require specialist knowledge :-)
> For example, I suspect the structural engineer that designed your
> building did plenty of calculations using calculus. Ditto for an
> electrical engineer who designed the power supply for your computer.
I'm going to go out on a limb and say that the number of people who
actually do that kind of things are a tiny minority.
As far as I can tell, the majority of people in the world have jobs
like... Telemarketing. Driving trucks. Working in an office doing
filing. Fitting central heating systems. Hotel receptionists.
None of these seem to require knowledge of calculus. In fact, jobs that
*do* require such knowledge are seemingly so absurdly rare that I almost
find it difficult to believe they exist.
I mean, doing a job that requires mathematics is like being an olympic
athlete. Hypothetically, anybody can become an olympian. But seriously
guys, how many olympians have you personally met?
Exactly.
--
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*
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Orchid XP v8 <voi### [at] devnull> wrote:
> >> And then of course, people will write "log x". Wanna take a guess which
> >> base that is? Now, sometimes it actually doesn't matter which base. And
> >> if it does, it *probably* means the natural logarithm. Probably...
> >
> > IIRC, 'log x' with no base usually means base 10, and 'ln x' is the natural log.
> > But, as you say, depends what the local conventions are.
>
> Some people use log to mean base-10. Some use it to mean an unspecified
> base. But (arguably) *most* geniune maths sources use it to mean the
> natural logarithm.
Just looked at wikipedia - it seems you can tell a person's primary field by the
base they expect log x to be in!
http://en.wikipedia.org/wiki/Logarithm#Notations_of_bases_and_implicit_bases
> It's a standard convention.
>
> http://en.wikipedia.org/wiki/Prime-counting_function
I did not know that :-)
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>> Some people use log to mean base-10. Some use it to mean an unspecified
>> base. But (arguably) *most* geniune maths sources use it to mean the
>> natural logarithm.
>
> Just looked at wikipedia - it seems you can tell a person's primary field by the
> base they expect log x to be in!
>
> http://en.wikipedia.org/wiki/Logarithm#Notations_of_bases_and_implicit_bases
Crazy stuff, eh?
>> It's a standard convention.
>>
>> http://en.wikipedia.org/wiki/Prime-counting_function
>
> I did not know that :-)
I'm going to go out on a limb and guess that the people working on
number theory and investigating the properties of prime numbers didn't
need to know about the circumference of a circle. ;-)
I've also seen pi used more than once as a general variable, rather than
a function name or a mathematical constant. This is why you see phrases
like "e^(i x) where e is the base of natural logarithms and i is the
imaginary unit". Because otherwise it's horrifyingly ambiguous.
--
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*
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On 16-11-2009 22:22, Orchid XP v8 wrote:
> I've also seen pi used more than once as a general variable, rather than
> a function name or a mathematical constant. This is why you see phrases
> like "e^(i x) where e is the base of natural logarithms and i is the
> imaginary unit". Because otherwise it's horrifyingly ambiguous.
I am working in an environment where the imaginary unit is j (because i
is for current). I have trouble adapting.
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On 16-11-2009 21:40, Orchid XP v8 wrote:
>>> If you can't add, you're going to have a bit of a problem doing your
>>> weekly shopping, but there's not much need for higher math unless
>>> you're working in some specialist industry somewhere.
>>
>> You'd be surprised how many industries require specialist knowledge
>> :-) For example, I suspect the structural engineer that designed your
>> building did plenty of calculations using calculus. Ditto for an
>> electrical engineer who designed the power supply for your computer.
>
> I'm going to go out on a limb and say that the number of people who
> actually do that kind of things are a tiny minority.
>
> As far as I can tell, the majority of people in the world have jobs
> like... Telemarketing. Driving trucks. Working in an office doing
> filing. Fitting central heating systems. Hotel receptionists.
But for those who do there is a window in their development during which
they are most capable of learning maths. So you try to teach them in
that period. Besides it is part of our culture. OTOH I know there are
those cultural barbarians that think that it is cool not to know maths,
like politicians and most other public figures.
> None of these seem to require knowledge of calculus. In fact, jobs that
> *do* require such knowledge are seemingly so absurdly rare that I almost
> find it difficult to believe they exist.
So I don't exist? Nor do many of my collegues?
I think I disagree.
> I mean, doing a job that requires mathematics is like being an olympic
> athlete. Hypothetically, anybody can become an olympian. But seriously
> guys, how many olympians have you personally met?
none, but I can name a few who know a lot of them.
You could also have asked for e.g. opera singers. Most people also know
none of those. OTOH I think I have met more than 10.
So it is all just a matter of who you know and who they know.
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