Why Use a Dummy Variable in the Fundamental Theorem of Calculus?

[mathwonk]

From reading Matt's posts, it seems it is I who do not understand the concept of a bound variable.

I.e. an example more germane than the one i gave would be this:

(for all x)( x equals x implies that [for some x, x =3 = 5])."

now it is rather difficult, for the reader to decide which quantifier to refer to for the meaning of the symbol x in "x+3 = 5", since the quantifiers are nested.

so it seems indeed those are correct who suggest using a different variable under the integral.

of course i could always argue that the positions of the symbols in an integral allows them to contain more information than a simple logical nested statement, but that would be insincere fudging.

still i think the notation for integrals can be rather confusing, and one must simply refer to whatever definition was given for it.

in particular the notation dx is confusing, unless a variable has been specified for the range of integration. i.e. dx is really an operator on intervals, or more properly, on tangent vectors, and hence really should involve a third variable, one that represents not points of the interval but tangent vectors at such points.

for instance if the variable for the interval is t, and x is a function of t, then

the integral from a to b of dx, would mean the inetgral of dx/dt dt from t=a to t=b.

hence if one wrote it as simply: " integral from a to b of dx", without displaying the variable t for the interval it would be mistakenly understood as meaning (b-a).

this is relevant to matt's post in 48, where the integral thus has a different possible interpretation, although admittedly a wacky one, since x is undefined there. so the confusion stems from the dual role of the letter x as a function on the real line, and as a variable. in matts post 48 it refers to the function x.

i.e. one gets a diffrent meaning by substituting sin and writing "integral from a to b of dsin"

the x at the top of the integral in post 1 however is a variable.

of course in some sense a variable is a function but one must know its range of validity.

but no doubt this is way outside the range of the intended discussion, whatever that was. help I'm going nuts! (again)

 

[matt grime]

ah, that's one i'[ve not mentioned, just to confuse you even more cyrus..

let f be a function, then df(x) is f'(x)dx so that d3 is indeed zero setting f(x)=3 as the constant function.

 

[mathwonk]

now we 've got that ball rolling! let's see now how much mroe damage i can do...


recalling that the limit we take of riemann sums requires fixing the value of f but allowing the value of the x in dx to have two values we could write f(t) d(x,t) where d(x,t) means something l;ike x-t, then we are takign sums of expressions like f(t)(x-t) where x has various values, so we could write the integral as;

integral from t= a to t= b of f(t)d(x,t). that would show the difference between the two roles of x!

heaven be praised , it is thundering here and i must stop this foolishness.

 

[Cyrus]

ah, that's one i'[ve not mentioned, just to confuse you even more cyrus..

let f be a function, then df(x) is f'(x)dx so that d3 is indeed zero setting f(x)=3 as the constant function.

So your saying then that the integral would be:

F(3)=\int^3_a f(3)d3 = \int^3_a f(3)*0 = 0

My god, such a simple question has turned into a freakin mess...

 

[matt grime]

*cough* for the umpteenth time, *that* is not an integral in any conventional sense. possessing an integral sign does not make it an integral. the notion of substituting 3 for x was supposed to highlight this. please try and get this: what you've written has no conventional meaning in mathematics, if you want to give it one then feel fre as long as it is consistent, which it isn't. I am not saying what it is, i am telling you that it isn't anything!

 

[mathwonk]

lets go back to post 1.

look suppose you had a function f(n) defined on all natural numbers,a nd you defiejnd another function F(n) to equal f(1) + f(2) + f(3)+...f(n), the sum of the first n values of f.

would you want to write that as summation of f(n) as n goes from 1 to n?

wouldn't that be confusing?

thats what your original question asked.

i.e. F(x) there was the integral of the values f(t) of the function f, as t goes from a to x.

wouldn't it be confusing to say "integral of the values f(x) as x goes from a to x"?

uh oh, now we are cycling over again.

 

[Cyrus]

*cough* for the umpteenth time, *that* is not an integral in any conventional sense. possessing an integral sign does not make it an integral. the notion of substituting 3 for x was supposed to highlight this. please try and get this: what you've written has no conventional meaning in mathematics, if you want to give it one then feel fre as long as it is consistent, which it isn't. I am not saying what it is, i am telling you that it isn't anything!

OK! So far so good. I TOTALLY agree with you on this! THANK GOD! .

Now, previously you said that "d3" is meaningless, and then you said it is equal to zero. Could I say that "d3" itself is not meaningless, in fact, it is equal to zero, but the conseqence of it being equal to zero makes the INTEGRAL meaningless. Does that jive well with you?

 

[fourier jr]

lets go back to post 1.

look suppose you had a function f(n) defined on all natural numbers,a nd you defiejnd another function F(n) to equal f(1) + f(2) + f(3)+...f(n), the sum of the first n values of f.

would you want to write that as summation of f(n) as n goes from 1 to n?

wouldn't that be confusing?

thats what your original question asked.

i.e. F(x) there was the integral of the values f(t) of the function f, as t goes from a to x.

wouldn't it be confusing to say "integral of the values f(x) as x goes from a to x"?

uh oh, now we are cycling over again.

yes! to mathwonk you listen!
http://images.google.com/images?q=tbn:Vf6vLFcDbaEJ:http://www.formfunctionemotion.net/mt-static/images/yoda.jpg

 

[Cyrus]

What the bleep??...

 

[mathwonk]

the bleep? we really love you abdollahi, it just seems not sometimes. please have patience with us too.

 

[fourier jr]

What the bleep??...

i think it was in empire strikes back when obi-wan tells luke not to go try save leia/han/chewie/etc & that it's a trap, he should wait until his training is done before he faces darth vader, & yoda says "yes! to obi-wan you listen!"

edit: start at the beginning. can you tell us what is wrong (if anything) with writing
\sum_{n=0}^n f(n)

 

[Cyrus]

Sure, your index starts at n=0, and goes to n. That would be like going from 0 to 0. n only ever takes on one single value.

 

[fourier jr]

how do you fix it & explain why it works the correct way

 

[Cyrus]

I love all of you! No, honestly, I am sorry for being a pain in your rear. I am thankful for all your help! (to mathwonk)

 

[Cyrus]

Well, to make it work you would have to change the n on top of the sigma to some other value. Then n can increment from zero to, let's say r. ( if we change that n on top of sigma to r.)

\sum_{n=0}^r f(n) But I am not sure about the n inside the f(n), would it always remain at zero, or would it change? I think it would change. It would increment until it reaches the value of r and stops.

 

[matt grime]

the d3 thing. the thing is that what the person who first poted it was trying to get across was something silly. ok? no, if g is any function then the symbol dg(x) is the same as g'(x)dx. right? but the original use of this was not to refer to 3 as a constant function but to simply a number. it is moot what the intention was. I *can* give it meaning, but that isnt' necessarily what was intended. maths isn't abuot some set of things that exist and come with notation already.

 

[fourier jr]

^^ stop confusing cyrus

Well, to make it work you would have to change the n on top of the sigma to some other value. Then n can increment from zero to, let's say r. ( if we change that n on top of sigma to r.)

\sum_{n=0}^r f(n) But I am not sure about the n inside the f(n), would it always remain at zero, or would it change? I think it would change. It would increment until it reaches the value of r and stops.

yeah, & it's similar with \int_{t=a}^{x} g(t)dt

just like you can give \sum_{n=0}^r f(n) a name like F(r), we can give \int_{t=0}^{x} g(t)dt a name like G(x), where G'(x) = g(x). the 'dummy variable' t 'increments' from 0 until it reaches the 'value' x.