Fudamental Theorem of Calculus

[mathwonk]

look back at post 20 and ask yourself what it means if say f is the "greatest integer" function. i.e. hypotheses really do matter. do some thinking!

 

[Cyrus]

Im sorry, I don't understand what you mean mathwonk.

$$\frac{d}{dx} \int f(x)dx= f(x)$$


That is all that is in post number 20.

See, Crosson said to evalute:

$$F(x)=\int_a ^x \frac{Sin (x)}{x} dx$$

You would replace all the x's by 3's if you choose to do the function at the point 3, giving you:

$$F(3)=\int_a ^3 \frac{Sin (3)}{3} d3$$

But what I was asking matt about was the fact that crosson wrote d3, which does not make sense, it should have been left as dx.

I don't see how post 20 comes into play here.

 

[cepheid]

Correct me if I'm wrong (I didn't read everything), but I think that it has been mentioned already that Crosson was just giving an example to show the pitfalls of *failing* to distinguish between the variable that is the limit of integration and the dummy variable.

See, Crosson said to evalute:

$$F(x)=\int_a ^x \frac{Sin (x)}{x} dx$$

You would replace all the x's by 3's if you choose to do the function at the point 3

NO! You wouldn't! Or, if you did, you would be making a mistake. You wouldn't replace all the x's. What should be replaced with 3? Whatever variable the function F depends on. What variable is that? Well, F has been defined as an integral, so F changes when the integral changes. The integral changes as its upper limit varies. So the upper limit, x, is the variable on which F depends, an x that has NOTHING to do whatsoever with the other x's that are inside the integral. That suggests that the x's inside the integral must not be called x's at all, for if they are, you might naively be tempted to do this when you set x=3 to evaluate F(3):

$$F(3)=\int_a ^3 \frac{Sin (3)}{3} d3$$

resulting in utter nonsense. Yes, the above is utter nonsense because you replaced variables with 3 that were not supposed to be replaced. Yes, d3 is nonsense, and the integrand should still be a function. That is the whole point that Crosson was making in the first place, by showing what sort of error could potentially be made if these two very distsinct variables are not represented using two different symbols. So to prevent any chance of this error occurring, you would write it like this:

$$F(x) = \int_a^x \frac{\sin t}{t}dt$$

$$F(3) = \int_a^3 \frac{\sin t}{t}dt$$

So, Crosson nicely answered the question of why we do the above, which, if I'm not mistaken, was the whole point of the thread.

 

[Cyrus]

YES! Thank you Cephid, your the man. Thanks for helping me. Same goes to mathwonk, matt, steve, and the others! For some reason I picture you guys shaking your heads as you stand in a circle beating me with your shoe in your hand becuase I've asked the same question so many times.

 

[mathwonk]

take the integral of an integrable function f from a to x. ask yourself if that new function has a derivative. in fact it does not always have one. but it has one wherever the integrand is continuous.

so what does it mean to say that the derivative of the integral is the original function when the original function may not be continuous?

of course i realize now that in post 20 your notation may not refer to the integral, but to the antiderivative. still, what does it eman to write "antiderivative of f" when maybe f does not have an antiderivative?

i.e. there are many f's. like the greatest integer function, that have integrals everywhere, but not antiderivatives everywhere.


i.e. integration and differentiation are not really inverse operations unless you understand what the exceptions are and how much they matter.


the answer is, for riemann integration,l that a function which is rieman integrable, is also continuous almost everywhere. its indefinite integral, i.e. from a to x, is then lipschitz continuous, and differentiable almost everywhere. and the derivative doeas equal the original function everywhere the originjal function is continuous.

but how do you recognize an "antiderivative: in this sense? i.e. when does F have the properety that F(b) - F(a) equals the integral?

it happens if F is any lipschitz continuous function with derivative equal to the original function f wherever f is continuous. But it is not enough to say merely that F is contin with dertivative equal to f where f is continuous. so you cannot use this method of "antiderivatives" to evaluate all integrals, unless you know about lipschitz continuity.

i./e. it is not always true that the integral equals F(b)-F(a), just because f is integrable and F is an antiderivative wherever the integral HAS a derivative.

oh, and concdrning d3 and so on, there is a notion in logic of constants and variables. constants are symbols that are reserved as having an assigned meaning and variables are other symbols. so the symbol after d is supposed to be a variable. now it can be a variable that has already been used, so strictly speaking you could write the integral from a to x of f(x)dx, because in the context of f(x)dx, x is what is called a "bound" variable, i.e. having meaning only within the confines of the expression f(x)dx. but it is better to use not only a variable but a new variable, like t. and write integral from a to x of f(t)dt, as it is less confusing. but it is not allowed to use d3, since 3 already is reserved for the natural number, so d3 would mean zero.

i.e. a variable x is supposed to be able to stand for any number, but 3 can only stand for 3, unless you say otherwise, but that would really be coinfusing, such as my saying that every letter i just typed actually stands for the letter that follows it in the alphabet, (and z stands for a).

 

[Cyrus]

so d3 would mean zero.

Are you sure, I thought d3 is meaningless. For example, when you evaluate a definite integral, you would have the limits from a to b. and a function, f(x)dx. And x would vary between a and b. You would NOT plug in, f(a)d(a) +...f(b)d(b) as x takes on those values, becasue d(a) or d(b) makes no sense. That is one of the things that I thought about, because you guys said that it would be f(3)d3, but when you do the definite integral I just wrote above, you clearly would NOT write f(a)d(a). You would write
f(a)dx+...f(b)dx , like a rieman sum approximation. You would leave it alone as dx. Where dx~= b-a/n. Dont the rules of integration say you leave the dx alone?

As for post 20, that was not my post mathwonk, someone else wrote that one.

 

[mathwonk]

my remarks about post 20 apply even more accurately to your post 1.

crosson does not seem to understand the concept of "bound variable". i.e. there are certain places where a symbol appears, i.e. where it occurs as a bound variable, that it is not to be replaced by the value.

for instance in the statement: " (for every x, x = x) implies that (for some x, 2x = 3)" the two occurrences of x are bound separately, hence this statement is exactly the same as the statement "(for every y, y = y) implies that (for some x, 2x = 3)".

i.e. there is no need at all, in the original statement, for the first symbol x to represent the same number as the last one. as i said earlier, this is not normally done, because many people find it confusing, even if correct.

however one often sees the notation used by crosson and in fact no one ever substitutes in for all the x's occurring there, because it would be obviously wrong. thus indeed many people have the right intutition but are mistaken in thinking it cannot also be justified rigorously.

 

[Cyrus]

But would you agree, that it should have been written as: $$F(3) = \int^3_a f(3) dx$$ ?

Where the dx stayed an x, and not a 3. I can see where cepheid is comming from in saying that you MIGHT be TEMPTED to put a 3 in there, but then I could argue that you might be tempted to put in f(a)d(a) +...f(b)d(b) when doing a riemann sum. It is just something you do not do, period. So I don't really see why you would be tempted to do such a thing in the first place.

 

[mathwonk]

no, the last two occurrences are both x's. the only place the three goes is at the top. to see this notice that another way to write the same thign is to put simply f under the integral sign. no x no dx.

 

[mathwonk]

one variation that is possible is to put f(x)dg(x) under the integral sign.

 

[mathwonk]

note too the most precise notation, which does not allow x everywhere, would have an x out at the bottom and top of the integral sign, i.e. the limits of integration, using t instead of x, would then run from t=a to t=x, and there would be f(t)dt under the integral sign.

this is the one i use in my classes as it leaves the least possibilioty for confusion.'

it also makes it clear that the avriable of F is the x at the top of the integral sign, and that the t is merely an index, i.e. "dummy variable", for describing the range of integration.

 

[Cyrus]

Right, I agree that the 3's have no business inside the integral. Thats fine. I am no longer in dispute about that fact; however, what I am saying is that let's assume you mistakenly left the variable inside the integral as an x and dx. When you came back an tried to evaluate.

$$F(x) = \int^x_a f(x) dx$$

at the value, x=3, wouldent you get a number equal to,

$$F(3) = \int^3_a f(3) dx = f(3)*(3-a)$$

Because once x takes on a value, the function f(x) now becomes a constant function.
It looks like there is a built in contradition, because x is a single value, so dx should be zero. At the same time, you could argue that dx should equal to 3-a/n, because those are the limits of the integral. Does that make any sense?

 

[matt grime]

For the love of God cyrus, crosson said the integral from a to 3 of f(3) d3 and that is what he meant!

$$\int_a^3 f(3)dx$$

actually makes sense and is exactly f(3)(3-a)

but almost the entire thread has been wasting bandwidth discussing nonsense! you realize that you are asking which of these incorrect notations is more correct? who cares? it is just wrong. why debate something that is false and therefore may be made whatever we feel like?

 

[Cyrus]

You are right, I appologize matt. From talking with you, I got the impression that:

$$F(x)=\int^x_af(x)dx = f(x)(x-a)$$

And it is a mathematically correct integral,

the only problem is that it does not do what we are after, and so we have to use a dummy variable, since f is constant function at a particular value for x. But apparently it is not true that $$F(x)=\int^x_af(x)dx = f(x)(x-a)$$.

 

[matt grime]

You are right, I appologize matt. From talking with you, I got the impression that:

$$F(x)=\int^x_af(x)dx = f(x)(x-a)$$

And it is a mathematically correct integral,

no it isn't. this is the exact thing you first posted and it was explained to you then.

the only problem is that it does not do what we are after, and so we have to use a dummy variable, since f is constant function at a particular value for x. But apparently it is not true that $$F(x)=\int^x_af(x)dx = f(x)(x-a)$$.

i am now even more confused as to what you want to talk about, and we're at 50 posts now.

 

[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. :rofl: 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.