You've mixing your y(x)'s and f(x)'s. The f(x) I gave had nothing to do with what you're talking about.
The expansion you've given is for f(x+a) via a Taylor series.
You don't seem to even think coherently.
The Structural analysis
- Thread starter mishin05
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The expansion you've given is for f(x+a) via a Taylor series.
I'm not sure I agree with this. Conceptually, if you simply think of the indefinite integral as not yielding a function, but rather as relating a function to an equivalence class of functions (the equivalence being that they differentiate to the same function), with +, -, etc. overloaded accordingly to such equivalence classes, then you don't have to worry about constants in your formulation of the rules at all, until you want to remove the $$\int\,\cdot\, dx$$ anti-differentiation operator. And then, a simple, single $$+C$$ (to indicate the general case) will do. Voilá, no mucking about with different "constants".
A large part of the problem is that the nomenclature "constant" makes it sound like there's a uniquely defined constant for a given indefinite integral, which really, really isn't the case. I strongly disagree with AN's stance that the "constant" of integration is defined by the limits on the integral. Rather, what happens is that any actual constant vanishes when we're dealing with a definite integral, which is why we are free to use any representative of the equivalence class above, to give the exact value of a definite integral by the fundamental theorem of calculus.
It already starts to be beyond human patience!
If expression is integrated:
$$a+b+c+k=p$$, That it becomes as follows:
$$\int adx+\int bdx+\int cdx+\int kdx=\int pdx.$$
For example:
$$\frac{2}{3}x^2+6x^2+2x^2-\frac{1}{3}x^2=\frac{25}{3}x^2;$$
$$\int\frac{2}{3}x^2dx+\int 6x^2dx+\int 2x^2dx-\int\frac{1}{3}x^2dx=\int\frac {25}{3}x^2;$$
$$\frac{2}{9}x^3+2x^3+\frac{2}{3}x^3-\frac{1}{9}x^3=\frac{25}{9}x^3.$$
Make, please so for your example!
Me your proof of a rule doesn't interest: $$ \int \frac {dk} {dx} {dx}$$ because it isn't defined yet for any "k":
$$\frac{2}{9}x^3+2x^3+\frac{2}{3}x^3-\frac{1}{9}x^3=\frac{25}{9}x^3.$$
$$\int\frac{d(\frac{2}{9}x^3)}{dx}dx+\int\frac{d(2x^3)}{dx}dx+\int\frac{d(\frac{2}{3}x^3)}{dx}dx-\frac{1}{9}x^3=\frac{25}{9}x^3.$$
Just to make my point above a little clearer:
$$
\int \frac{df}{dy} dy = \int \frac{d(f+C)}{dy}dy
$$ (because obviously $$ \frac{df}{dy} =\frac{d(f+C)}{dy}$$)
But once we look at definite integrals, we find that there's no need for the C:
$$
\int_a^x \frac{df}{dy} dy = \int_a^x \frac{d(f+C)}{dy}dy = (f(x)+C) - (f(a)+C) = f(x) - f(a)
$$
A central point about the fundamental theorem of calculus is that you can use any representative function f and not have to worry about the fact that the antiderivative isn't uniquely defined, because any constant disappears. The C has no place being anywhere when we're talking about definite integrals.
I emphatically disagree. I'd say that mathematics very much says that $$\int 0 dx = C$$. We agree that $$\int f(x) dx = F(x) + C$$ where $$F(x)\frac{d}{dx} = f(x)$$, right? Well, any constant function works as $$F$$, here, and it doesn't get rid of the $$+C$$.
Any limits on the integral will not be sufficient to uniquely determine C. That's why the notation is there in the first place! If it were the case that the antiderivative was a uniquely defined function, we'd simply write f on the RHS!
No, that's not right: You're mixing two fundamentally different concepts. The C is a constant for f as an antiderivative of $$\frac{df}{dy}$$, in the sense that you have
$$
\int \frac{df}{dy} dy = \int \frac{d(f+C)}{dy}dy
$$ (because obviously $$ \frac{df}{dy} =\frac{d(f+C)}{dy}$$)
But once we look at definite integrals, we find that there's no need for the C:
$$
\int_a^x \frac{df}{dy} dy = \int_a^x \frac{d(f+C)}{dy}dy = (f(x)+C) - (f(a)+C) = f(x) - f(a)
$$
A central point about the fundamental theorem of calculus is that you can use any representative function f and not have to worry about the fact that the antiderivative isn't uniquely defined, because any constant disappears. The C has no place being anywhere when we're talking about definite integrals.
You didn't learn in the textbook, what record $$ y=x^3 $$ is concrete record $$ y=f (x) $$???!!!
I won't answer more idiocies!:m::m::m:
My gut feeling is that adding a +c to every indefinite integral is an unnecessary complication. When we write:
$$\int f(x)~dx = F(x)$$
it is understood that F(x) is any antiderivative of f(x). Since antiderivatives can differ only by a constant, the constant is implied. If we then write F(x) explicitly, we need to add the constant.
As for the linearity condition
$$\int af(x)~dx = a\int f(x)~dx$$
it is perhaps simpler to say that this does apply when a=0. Or, alternatively, we simply change that particular condition to
$$\int af(x)~dx = a\int f(x)~dx + c$$
I don't think the additivity rule needs any modification, since again it should be understood that $$\int f(x)~dx$$ always means an antiderivative of f(x). That is, the following rule needs no modification:
$$\int \left[f(x)+g(x)\right]~dx = \int f(x)~dx+ \int g(x)~dx$$
As that page looks right now (and I assume you're not the sole author), it looks overly complicated with all the explicit constants everywhere. As I say, my hunch is that careful definition of what we mean by the indefinite integral notation solves most of the problems.
I agree.
In fact, the constant can't be built in by defining something like:
$$\int f(x)~dx \equiv \int_c^x f(t)~dt$$
because an antiderivative of f(x) does not necessarily have a range that encompasses all real numbers (or all complex numbers if you're talking about complex integrals).
Those guys who developed and created calculus haven't guessed very simple thing:
$$ f ' (x) = \lim_ {\Delta x\to 0} \frac {f (x_0 +\Delta x)-f (x_0)} {\Delta x} $$
At substitution: $$ f ' (x) =f (y); x_0=C=const; \Delta x=y $$
will look so:
$$f(y)=\frac{f(y+C)-f(y)}{C}|_{C=0}!$$
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You made a few mistakes there.
For example, you replaced a $$\Delta x$$ with a $$C$$ when you should have replaced it with a $$y$$. Also, you forgot to take the limit. Finally, the biggest problem is that you've created a recursive formula out of a simple definition, so your substitutions make no sense at all.
I don't think the adults here need you in this thread any longer. We've moved on to discussing some real mathematics, and don't need your nonsense any more.
Thanks for getting the thread started though. Now move aside and let the big kids play.
Unless you address post #101 directly, I'm unlikely to waste my time talking to you. You might amuse me a little from time to time, but I won't take you seriously.
Thank You!
Right, and this is exactly what they taught me back in introductory calculus. But there's some subtlety to the question of adding in that one constant - the "naive" rule that people seem to use there is that you only add a constant when explicitly evaluating an indefinite integral. That's precisely the method through which mishin05's earlier logic produced the expression $$\int 0 ~dx = 0$$.
Which is of course completely valid if both sides of that equation are understood to refer to equivalence classes (i.e., there's still an implied final step of adding a constant to get the final answer, once all manipulations are finished). But it does not get understood that way by the unfamiliar, since they expect to see a $$+C$$ wherever an antiderivative is written out (according to the convention). The inclusion of a $$+C$$ is the notation for an equivalence class, after all, so it seems backwards to suppress that notation when what you want to emphasize is the fact the you're dealing in equivalence classes.
So the correct rule here is that you have to inspect the final expression and add in whatever constant may be required for it to make sense as an equivalence class relation. Of course, those of us who've already mastered the subject aren't bothered by that, but it seems too much to ask of those who haven't (which is effectively the entire audience for indefinite integrals, given how rarely they're used outside of introductory calculus). Rather, a useful convention should guarantee that following concrete rules will produce expressions with all of the appropriate constants in the appropriate places. Otherwise, the convention can't be used to identify and locate errors in inferences that employ it, which is a huge handicap for anyone trying to learn the subject.
These problems are all products of attempts to suppress notation for reasons of aesthetics/shorthand, which seems like a mismatched approach for a subject that is almost entirely the domain of newcomers.
Said another way:
JamesR said:
... it is my contention that such aspects should not be merely "understood" or "implied," but actually notated as such to avoid the evident confusion. I.e., we're mixing and matching regular functions and variables with equivalences classes of functions here, but without any consistent, explicit difference in notations. So when mishin05 arrives at a result like $$\int 0 ~dx = 0$$, it is not apparent to him that this is an equivalence class relation, and not an assertion that the antiderivative of the zero function is exactly the zero function.
If the expectation is that the conventions should always produce an appropriate $$+C$$ when an antiderivative is written out (which seems common), then we need to ensure that all formulations conform to that. The linearity rule can be patched by adding an explicit $$+C$$, but it's my contention that always including the $$+C$$ everywhere is guaranteed to avoid all such issues (which surely there must be some other examples of, apart from the linearity rule?). Why not just denote equivalence classes explicitly all the time, especially given the audience?
JamesR said:
On second thought, the defect in the additivity rule is a result of the defect in the linearity rule (from passing the -1 through the indefinite integral), so if you patch the linearity rule you don't have a problem here.
But this does beg the question of how one is to write the results of expressions like $$\int f(x)~dx - \int f(x)~dx$$. The desired final expression would be $$C$$, but the "naive" conventions only require us to add a constant when "evaluating" an indefinite integral - it bears emphasizing that "removing" an antiderivative operator includes algebraic cancellation as well as explicit evaluation. Supposing we're to avoid a convention with a final step along the lines of "add in whatever constant may be required to get the right answer," that is.
JamesR said:
Well, part of my contention here is that indefinite integrals are a pedagogical mess and should be mostly ignored once one has completed introductory calculus, after all.
JamesR said:
The lazy formulation handles most of the cases just fine to begin with. The concern here is getting it to work all the time (which I consider a reasonable expectation for math notation conventions), in order to patch the leaks that cranks like to squeeze through. Moreover I don't see much value in suppressing the notation here - given the audience, explicit notation is preferable (however ugly it may be).
Or, how about a compromise approach: by default, suppress the constants everywhere and simply deal in equivalence classes. If in doubt (or needing to demonstrate to someone not comfortable with equivalence classes), include the constants everywhere (including the indefinite integrals) to be totally explicit.
I feel I should explain myself a bit here.
Funk, I completely agree with the whole equivalence class thing. I was tempted to go down that road in one of my posts but I felt that it'd be completely wasted on the thread starter. An equivalence class commonly used in theoretical physics is the one related to the de Rham cohomology, that p-forms $$\alpha$$ and $$\beta$$ are in the same class if $$\alpha = \beta + d\eta$$ for some p-1 form $$\eta$$. Using $$d^{2}=0$$ you could rephrase this (in a not entirely equivalent but close way) as $$d\alpha = d\beta$$, which is pretty much what we're talking about, that two functions are related if their derivatives are equal, aka they differ by a constant.
In regards to the "The C comes from the limits term" my train of thought was that you can't say $$\int 0 dx = C$$ for an indefinite integral unless C=0, since $$\int 0 dx = 0 \int dx = 0$$ for any well defined integral, since you're multiplying by 0. I agree that C and 0 are in the same equivalence class but as I'm sure you know they aren't automatically equal as a result. The whole indefinite integral thing is often written with very sloppy notation, ie $$f(x) + C = \int \frac{df}{dx} dx$$ is bad notation, since you have a function dependent on x on the left yet x is integrated over on the right. As such you should really make the x on the RHS explicit, say as $$\int^{x} \frac{df}{dy} dy$$. If you put in some arbitrary limit as the other limit then you have $$\int_{a}^{x} \frac{df}{dy}dy = f(x) - f(a)$$. I acknowledge that I was also sloppy in saying "There's where C comes from" as not all C will be in the image of f for a in some specified domain, but I was more trying to get across to the thread starter where a constant term can arise. Yes, given an explicit f the possible f(a) in general only gives you part of the equivalence class but I'm sure the subtlety is lost on the thread starter. My comment about "You need an f" was that you can't get C from $$\int 0 dx$$, you have to have a non-zero function within the integral. 0 as the integrand is the only case where the indefinite integral is zero, if you view it as a definite integral whose limits you don't know. If you view it as a general equivalence class element then you should put in the +C.
Also I claim the excuse of having replied while sitting on an enormously crowded train trying not to have people read over my shoulder.
Funk, I completely agree with the whole equivalence class thing. I was tempted to go down that road in one of my posts but I felt that it'd be completely wasted on the thread starter. An equivalence class commonly used in theoretical physics is the one related to the de Rham cohomology, that p-forms $$\alpha$$ and $$\beta$$ are in the same class if $$\alpha = \beta + d\eta$$ for some p-1 form $$\eta$$. Using $$d^{2}=0$$ you could rephrase this (in a not entirely equivalent but close way) as $$d\alpha = d\beta$$, which is pretty much what we're talking about, that two functions are related if their derivatives are equal, aka they differ by a constant.
In regards to the "The C comes from the limits term" my train of thought was that you can't say $$\int 0 dx = C$$ for an indefinite integral unless C=0, since $$\int 0 dx = 0 \int dx = 0$$ for any well defined integral, since you're multiplying by 0. I agree that C and 0 are in the same equivalence class but as I'm sure you know they aren't automatically equal as a result. The whole indefinite integral thing is often written with very sloppy notation, ie $$f(x) + C = \int \frac{df}{dx} dx$$ is bad notation, since you have a function dependent on x on the left yet x is integrated over on the right. As such you should really make the x on the RHS explicit, say as $$\int^{x} \frac{df}{dy} dy$$. If you put in some arbitrary limit as the other limit then you have $$\int_{a}^{x} \frac{df}{dy}dy = f(x) - f(a)$$. I acknowledge that I was also sloppy in saying "There's where C comes from" as not all C will be in the image of f for a in some specified domain, but I was more trying to get across to the thread starter where a constant term can arise. Yes, given an explicit f the possible f(a) in general only gives you part of the equivalence class but I'm sure the subtlety is lost on the thread starter. My comment about "You need an f" was that you can't get C from $$\int 0 dx$$, you have to have a non-zero function within the integral. 0 as the integrand is the only case where the indefinite integral is zero, if you view it as a definite integral whose limits you don't know. If you view it as a general equivalence class element then you should put in the +C.
Also I claim the excuse of having replied while sitting on an enormously crowded train trying not to have people read over my shoulder.
That's a very reasonable sentiment, and completely understandable.
Cool. Didn't know that.
Ah, so that's it. In that case I would point out my previous comment about +, - (and in this case $$\cdot$$) in the rules being overloaded to the equivalence classes, so the RHS 0 is not a real (complex, whatever) zero, but rather the equivalence class for the zero function.
Alright. I've just always thought of indefinite integration as the inverse of differentiation, rather than an underspecified definite integral, so my natural way of thinking of it is in term of functions, rather than values, in which case the 0 case works out just fine.
Hah! I hope you made your way alright - lots of snow, currently.Also I claim the excuse of having replied while sitting on an enormously crowded train trying not to have people read over my shoulder.
I tend to think of indefinite integrals more as a definite integral which has not had its limits stated, though I admit this might not be the best way to think about it as it is quite a 'high school point of view'. The equivalence class method is better, as it includes the C which are not expressible as f(a) and the f(a) is not actually the C anyway (as you previously commented on).
It would seem that is a demonstration of how the integral of a derivative does not automatically yield the original function in the sense I stated, ie there are pairings (f,C) for which $$f(x) + C = \int_{a}^{x} \frac{df}{dx'} dx'$$ is not true for any a. Its interesting to note that the equation, if always true, would be a stronger statement than the FT of calculus because it implies the FToC regardless of whether such an a exist, applying derivatives to both sides gives you the FToC for f'(x).