The problem of differential calculus
A discussion of geometry and its lack of basis in phenomenal reality reminded me of this problem, mentioned in passing by an authority on the matter, to which I elucidated a little.
The problem of differentiation from first principles: It appears at first that division by 0 is a problem (cue dunning-kruger scoffing), though mathematicians (or those who use maths) tend to brush aside this and tell you to go and learn about “limits” and there is no problem. This says (presumably hard mathematicians write this stuff, Penguin dictionary of mathematics):
“Differential calculus is concerned with the rates of change of functions with respect to changes in the independent variable. It came out of problems of finding tangents to curves”
A change in the independent variable necessitates a non-zero difference between 2 points on a curve, hence we have a chord rather than a tangent. A tangent to the curve seems satisfactory to provide an exact value for the slope of the curve at one point, whereby there is a change of ‘direction’ of points entering and leaving the point in question. However, this is exposed as fallacious: we require an infinitesimal approach of neighbouring points, there can be no variation in the direction of such points.
They go on to say:
“In the 1820s, Cauchy put the differential and integral calculus on a more secure footing by using the concept of a limit. Differentiation he defined by the limit of a ratio”
Limit is defined as:
“A value that can be approached arbitrarily closely by the dependent variable when some restriction is placed on the independent variable of a function”
The example is given of ‘the limit of 1/x as x tends to infinity is 0’.
As is clear in this example, the limit will not be reached by the dependent variable. “Arbitrary” is used to describe the lack of a determined value of separation between the DV and the limit, it’s just really really close. So we do seem to merge, alas fudge, the necessity of a change in the IV while also reducing the delta to 0 in the algebra. Not sure about a ‘sure footing’. Perhaps this is why the idea of ‘linear approximation’ is used. The arbitrary limit is not actually reducing the delta to 0 as the algebra would suggest.
1&onlybillyshears,
I think maybe the problem you are having is that you are focused on infinitismal quantities. While the notion of infinitismals was indeed used by Leibniz in his original development of calculus, the modern definition of limit has completely eliminated the idea. If f is any function, then the limit as x->c of f(x) is equal to L if and only if for all possible values epsilon (which can be arbitrarily small, but not infinitismal) there some value delta such that |f(c+delta)-L|
Let's examine in PlainSpeak this property -
"Magnitudes are said to have a ratio to one another which can, when multiplied, exceed one another."
an axiom we are not obliged to adhere to, nonetheless let's accept it for the sake of argument. Let us also, if agreeable have an underpinning of classical logic. Probably there will be no agreement, see below re skepticism in education, "Others still can prove that 1/3 0.333..., but, upon being confronted by the fractional proof, insist that "logic" supersedes the mathematical calculations."... yes, indeed it does if we are to avoid straight-jacketing. The offence is ofc continuity and its opposite discontinuity; rejection of infinitesimals and its opposite in the reduction of delta to 0; indefinite continuation and its opposite termination etc etc.
I see I am not alone. All power to the student body, the student ultimately surpasses the master. Emperor's clothes etc.
https://en.wikipedia.org/wiki/0.999...#I...
Students of mathematics often reject the equality of 0.999... and 1, for reasons ranging from their disparate appearance to deep misgivings over the limit concept and disagreements over the nature of infinitesimals. There are many common contributing factors to the confusion:
Students are often "mentally committed to the notion that a number can be represented in one and only one way by a decimal". Seeing two manifestly different decimals representing the same number appears to be a paradox, which is amplified by the appearance of the seemingly well-understood number 1.[g]
Some students interpret "0.999..." (or similar notation) as a large but finite string of 9s, possibly with a variable, unspecified length. If they accept an infinite string of nines, they may still expect a last 9 "at infinity".[51]
Intuition and ambiguous teaching lead students to think of the limit of a sequence as a kind of infinite process rather than a fixed value since a sequence need not reach its limit. Where students accept the difference between a sequence of numbers and its limit, they might read "0.999..." as meaning the sequence rather than its limit.[52]
These ideas are mistaken in the context of the standard real numbers, although some may be valid in other number systems, either invented for their general mathematical utility or as instructive counterexamples to better understand 0.999...; see § In alternative number systems below.
Many of these explanations were found by David Tall, who has studied characteristics of teaching and cognition that lead to some of the misunderstandings he has encountered with his college students. Interviewing his students to determine why the vast majority initially rejected the equality, he found that "students continued to conceive of 0.999... as a sequence of numbers getting closer and closer to 1 and not a fixed value, because 'you haven't specified how many places there are' or 'it is the nearest possible decimal below 1'".[23]
The elementary argument of multiplying 0.333... =
1
3
{\textstyle {\frac {1}{3}}} by 3 can convince reluctant students that 0.999... = 1. Still, when confronted with the conflict between their belief in the first equation and their disbelief in the second, some students either begin to disbelieve the first equation or simply become frustrated.[53] Nor are more sophisticated methods foolproof: students who are fully capable of applying rigorous definitions may still fall back on intuitive images when they are surprised by a result in advanced mathematics, including 0.999.... For example, one real analysis student was able to prove that 0.333... =
1
3
{\textstyle {\frac {1}{3}}} using a supremum definition but then insisted that 0.999... < 1 based on her earlier understanding of long division.[54] Others still can prove that
1
3
{\textstyle {\frac {1}{3}}} = 0.333..., but, upon being confronted by the fractional proof, insist that "logic" supersedes the mathematical calculations.
Mazur (2005) tells the tale of an otherwise brilliant calculus student of his who "challenged almost everything I said in class but never questioned his calculator", and who had come to believe that nine digits are all one needs to do mathematics, including calculating the square root of 23. The student remained uncomfortable with a limiting argument that 9.99... = 10, calling it a "wildly imagined infinite growing process".[55]
As part of the APOS Theory of mathematical learning, Dubinsky et al. (2005) propose that students who conceive of 0.999... as a finite, indeterminate string with an infinitely small distance from 1 have "not yet constructed a complete process conception of the infinite decimal". Other students who have a complete process conception of 0.999... may not yet be able to "encapsulate" that process into an "object conception", like the object conception they have of 1, and so they view the process 0.999... and the object 1 as incompatible. They also link this mental ability of encapsulation to viewing
1
3
{\textstyle {\frac {1}{3}}} as a number in its own right and to dealing with the set of natural numbers as a whole.[56]
The greeks might've entertained 0.999... = 1 for the reason that infinity is an attribute of the One, though equating seems a step too far. It's also ugly.
A strawman in any case. Instantaneous and change are mutually exclusive. The tangent is the linear approximation to the curve at that point, where a point disallows a delta (let's say a non-zero delta to emphasise the craziness) notwithstanding replacement by the roman d.
Have you derived trig from circles for yourself, or worked through the proofs for the exponential function, all that jazz?
The idea that a limit of a converging sequence exists should be inarguable.
What I think some here are stumbling on is how that limit is expressed.
Take a simple example:
9 + 9/10 + 9/100 + ... = 10.
We know exactly what 10 is because 10 is an algebraic number that can be defined exactly by algebraic processes. Nobody ever describes 10 as a limit of a converging sequence because it is unnecessary.
Now, take this series:
4 - 4/3 + 4/5 - 4/7 + 4/9 + ... = ???
The limit is pi, but pi can not be expressed exactly algebraicly. Hence humans will never know exactly what pi is, along with all transcendental numbers.
Basically, when anyone discusses pi they are actually describing a limit of a converging sequence.
So one should be able to recognize that limits of converging sequences work for all real numbers, both those we can express exactly and those we cannot.
The greeks might've entertained 0.999... = 1 for the reason that infinity is an attribute of the One, though equating seems a step too far. It's also ugly.
A strawman in any case. Instantaneous and change are mutually exclusive. The tangent is the linear approximation to the curve at that point, where a point disallows a delta (let's say a non-zero delta to emphasise the craziness) notwithstanding replacement by the roman d.
Let z=0.999Â… What is the value of 1-z? If they are not equal, then there must be some c<>0 such that c=1-z. Let (0.9)n be equal to a decimal point followed by n 9Â’s. Clearly for any n, (0.9)n
But c<>0 by assumption. Remembering how c was defined we take c>0 (If we donÂ’t think 0.999Â… is equal to 1, surely it should be <1 since all of the finite partial sums are). So we have c<10^-n for all n. Multiply both sides by 10^n and you get 10^n * c < 1 for all n. For any x, no matter how large, 10^n>x for sufficiently large n. By the Archimedean property there exists an x such that for any b>x, bc>1. We just shows that there exists n such that 10^n>x so this implies that for some n, 10^n *c >1.
But this is a contradiction since we found above that 10^n * c < 1 for all n. Our assumption that c<>0 is therefore incorrect and c=0. But c = 1-0.9999Â… , hence 1=0.9999Â….
The Greeks’ opinion is irrelevant. Proper logical inference from the axioms of real numbers is what matters, not the writings of people who lived over 2000 years ago. And BTW, calculus as a branch of mathematics does not incorporate the ideas of point or instantaneous change - these are descriptions of physical uses for the derivative. Calculus is formulated mathematically using only the concept of limit, real number, and function. The derivative of a given function is a new function that is defined as a limit of a quantity calculated using the given function. We regard it as a rate of change because we make sense of functions and their behavior by drawing graphs, looking at curves, etc. The deeivative happens to be the slope of the tangent line to a curve that represents the given function, but that is NOT what the derivative actually is. Even if we had no notion of analytic geometry and couldn’t draw curves and tangents to represent functions and derivatives, we would still be able to define derivatives. Any issues arising from that analytical geometrical representation are not problematic for calculus.
Let z=0.999Â… What is the value of 1-z? If they are not equal, then there must be some c<>0 such that c=1-z. Let (0.9)n be equal to a decimal point followed by n 9Â’s. Clearly for any n, (0.9)n
But c<>0 by assumption. Remembering how c was defined we take c>0 (If we donÂ’t think 0.999Â… is equal to 1, surely it should be <1 since
gotcha alright, already!
The Greeks’ opinion is irrelevant. Proper logical inference from the axioms of real numbers is what matters, not the writings of people who lived over 2000 years ago. And BTW, calculus as a branch of mathematics does not incorporate the ideas of point or instantaneous change - these are descriptions of physical uses for the derivative. Calculus is formulated mathematically using only the concept of limit, real number, and function. The derivative of a given function is a new function that is defined as a limit of a quantity calculated using the given function. We regard it as a rate of change because we make sense of functions and their behavior by drawing graphs, looking at curves, etc. The deeivative happens to be the slope of the tangent line to a curve that represents the given function, but that is NOT what the derivative actually is. Even if we had no notion of analytic geometry and couldn’t draw curves and tangents to represent functions and derivatives, we would still be able to define derivatives. Any issues arising from that analytical geometrical representation are not problematic for calculus.
Rate of change is fundamental ofc.
derivative
from The Penguin Dictionary of Mathematics
The rate of change of a function with respect to the independent variable...
That’s a useless definition, or at least an incomplete one. What does “rate of change of a function with respect to the independent variable” mean? We have a good intuitive grasp of what we think that means, but for formal mathematics that isn’t good enough. A precise definition is needed. That precise definition is
lim h->0 [f(x+h)-f(x)]/h].
I have defined limit already and shown that there is nothing problematic about that definition. It uses only logical quantifiers and operations on real numbers. The function f is taken as given, and the only remaining component of this definition is standard operations of arithmetic. Based on the definition of limit, these are valid because when defining the limit as h goes to zero, h is never allowed to actually be zero. Therefore division by h is not problematic. The modern formalism has been done specifically to deal with the issues you raised. The ideas of infinitismals no longer plays any role in calculus.
When I first encountered the Epsilon-Delta-Condition it boggled my mind. I had to keep repeating it like a mantra until it started to make sense to me. I think it's the first time many students are faced with defining a value as the number which satisfies a particular logical condition. I understand some classes now avoid Epsilon-Delta proofs and try to get students started doing Calculus through an intuitive, non-rigorous approach. I think this is a mistake. From this point on, mathematics abounds with complicated definitions of mathematical objects. As such definitions go, the Epsilon-Delta-Condition is one of the simplest. And one you can fall back on for proofs in many situations.
PairTheBoard
That’s a useless definition, or at least an incomplete one. What does “rate of change of a function with respect to the independent variable” mean? We have a good intuitive grasp of what we think that means, but for formal mathematics that isn’t good enough. A precise definition is needed. That precise definition is
lim h->0 [f(x+h)-f(x)]/h].
I have defined limit already and shown that there is nothing problematic about that definition. It uses only logical quantifiers and operations on real num
It's the same thing. The derivative is the rate of change of y wrt x (This is not a vague concept ofc). This is then defined as the limit of the derived expression as delta(x) approaches 0... so much for The Pledge.
The Turn, *in the limit* as delta *approaches* zero...
finally The Prestige, the delta *becomes* zero. Unless no application is ever made in which case there is no trick. The modern formalism is trash innit.
So 1 is the limit of the sequence 1, 1, 1, ..., but the limit of the sequence 1/1, 0.5/0.5, 0.25/0.25, ... is suspect?
Explain.
Limits heavy lifting straw men. Limits are not the problem.
It's the same thing. The derivative is the rate of change of y wrt x (This is not a vague concept ofc). This is then defined as the limit of the derived expression as delta(x) approaches 0... so much for The Pledge.
The Turn, *in the limit* as delta *approaches* zero...
finally The Prestige, the delta *becomes* zero. Unless no application is ever made in which case there is no trick. The modern formalism is trash innit.
Can you give some examples for what you're talking about here? When does the delta "become" zero. For what applications is this "trick" done?
PairTheBoard
It's the same thing. The derivative is the rate of change of y wrt x (This is not a vague concept ofc). This is then defined as the limit of the derived expression as delta(x) approaches 0... so much for The Pledge.
The Turn, *in the limit* as delta *approaches* zero...
finally The Prestige, the delta *becomes* zero. Unless no application is ever made in which case there is no trick. The modern formalism is trash innit.
I agree. It is the same thing. That’s because the formula I gave is the DEFINITION of rate of change wrt the independent variable. WITHOUT that definition, rate of change is indeed a vague and useless concept. If you don’t believe me, then give a clear, unambiguous definition without vague terms like “very close” or “very small”. The whole point of developing differential calculus was to provide a rigorous definition of the notion of rate of change. That is what I posted - the mathematically rigorous definition.
I agree. It is the same thing. That’s because the formula I gave is the DEFINITION of rate of change wrt the independent variable. WITHOUT that definition, rate of change is indeed a vague and useless concept. If you don’t believe me, then give a clear, unambiguous definition without vague terms like “very close” or “very small”. The whole point of developing differential calculus was to provide a rigorous definition of the notion of rate of change. That is what I posted - the mathematically ri
BTW, delta NEVER becomes zero. It doesn’t have to. The derivative is defined as a value that meets a certain logical condition. Namely the value for which it is true that the difference between the derivative and the d.q. can be made less than ANY epsilon for some delta, namely a delta sufficiently small. There is no need for delta to be zero, and the value so defined is unique.
For example, we know that if f(x)=x^2 then f’(x)=2x, for all x. We can prove this using the given definition. The differential quotient is [f(x+h)-f(x)]/h. We can expand the numerator as x^2+2xh-h^2-x^2 = 2xh-h^2. Therefore the differential quotient has the value of 2x+h for all x. Obviously the difference between the derivative and the d.q. is h. For ANY epsilon, then we can make the difference less than epsilon by making h
We can also show that the value 2x is the unique value for which the condition in the definition holds. Assume that there is some x such that f’(x) is some value other than 2x. This can be expressed as f’(x) = 2x+c for some nonzero value of c. The difference in this case is no longer equal to h, but rather c-h. There now exist values of epsilon such that for all h, the difference is greater than epsilon, namely epsilon
Thius our definition leads to a unique value for the derivative and does not require any mathematically invalid operations, so it is a well-founded definition.
Can you give some examples for what you're talking about here? When does the delta "become" zero. For what applications is this "trick" done?
PairTheBoard
say it ain't so PTB this is differentiation from 1st principles! Simply if the function is y=x^2, the derivative is ofc 2x. This limit cannot be derived unless we let delta(x) become 0. This applies to any derivative you care to find. Say the oxymoronic instantaneous rate of change, borrowing from the exact slope of the tangent which is nothing but an approximation to the curve. A point has no extension. To ignore a priori geometric truths is a requirement for the fudging going on above.
BTW, delta NEVER becomes zero. It doesn’t have to. The derivative is defined as a value that meets a certain logical condition. Namely the value for which it is true that the difference between the derivative and the d.q. can be made less than ANY epsilon for some delta, namely a delta sufficiently small. There is no need for delta to be zero, and the value so defined is unique.
For example, we know that if f(x)=x^2 then f’(x)=2x, for all x. We can prove this using the given definiti
Yeah exactly it's a fudge as is obvious. 2x is not 2x + delta(x), the former only occurs when delta(x)=0. See above re students complaining on intuitive terms why they will not accept it. The reason is simply that the derivative is approximate.
The delta absolutely becomes 0, consult any text.
say it ain't so PTB this is differentiation from 1st principles! Simply if the function is y=x^2, the derivative is ofc 2x. This limit cannot be derived unless we let delta(x) become 0. This applies to any derivative you care to find. Say the oxymoronic instantaneous rate of change, borrowing from the exact slope of the tangent which is nothing but an approximation to the curve. A point has no extension. To ignore a priori geometric truths is a requirement for the fudging going on above.
What do you mean by "This limit" in the bolded sentence. Also, what do you mean by "delta(x)"? As Stremba shows above, the limit of the difference quotient for f(x)=x^2, [f(x+h)-f(x)]/h as h>0 goes to zero is 2x. So your bolded statement is just wrong unless you have another "limit" in mind. If by "delta(x)" you mean h in the difference quotient, you can see h is strictly greater than zero by definition of the limit. If you mean by delta(x) the delta in the delta epsilon proof you can see in Stremba's proof the delta depends on x and epsilon and is never 0 because epsilon is never 0.
I suspect by "This limit" you have some other idea of what a proper limit for a derivative ought to be but you can't clearly state it. Only you believe it requires a zero change in x, i.e. a zero delta, so is thus impossible. I think you should task yourself with clearly stating this other limit in rigorous mathematical terms. Until you do there's no use guessing what it is you're trying to say.
PairTheBoard
Yeah exactly it's a fudge as is obvious. 2x is not 2x + delta(x), the former only occurs when delta(x)=0. See above re students complaining on intuitive terms why they will not accept it. The reason is simply that the derivative is approximate.
The delta absolutely becomes 0, consult any text.
If 2x is the derivative and the derivative is approximate, what is it approximating?
PairTheBoard
What do you mean by "This limit" in the bolded sentence. Also, what do you mean by "delta(x)"? As Stremba shows above, the limit of the difference quotient for f(x)=x^2, [f(x+h)-f(x)]/h as h>0 goes to zero is 2x. So your bolded statement is just wrong unless you have another "limit" in mind. If by "delta(x)" you mean h in the difference quotient, you can see h is strictly greater than zero by definition of the limit. If you mean by delta(x) the delta in the delta epsilon proof you can see in St
To accuse moi of obfuscation here is rather rich isn't it? The derivative is 2x, i.e., this limit, is 2x.
Delta(x), meaning the change in x, not sure why this needs stating... there is no difference in the meaning of delta(x) in the delta-epsilon condition or any other context; the delta-epsilon condition is a get out of jail free card written on that paper you can buy that instantly combusts, not sure how this is doing any lifting at all. It requires variation, precisely the objection all along.
The limit as *delta(x)* approaches 0 in the expression (2x + delta(x)) is 2x. Hence, delta(x) becomes 0, or else you do not have an expression equal to 2x since there is another term in addition to it. This is the same limit as anybody is defining - I in plain terms, stremba in obfuscatory terms, nevertheless we are all on the same page here. Let us return to the OP for clarity which y'all had a good whinge about before agreeing that, yes, this is the definition. As the mathematicians have it:
Limit is defined as:
“A value that can be approached arbitrarily closely by the dependent variable when some restriction is placed on the independent variable of a function”.
That is really all there is to it. Epsilon-delta fudges the arbitrary approach to the limit, doing nothing in addition to the PlainSpeak definition cited above.
If 2x is the derivative and the derivative is approximate, what is it approximating?
PairTheBoard
in this case the derivative is the gradient of the tangent to the curve which approximates the gradient of the curve "at that point". There is no "gradient at that point" of a curve. Similar idea extended to whichever differentiable function takes fancy.
Yeah exactly it's a fudge as is obvious. 2x is not 2x + delta(x), the former only occurs when delta(x)=0. See above re students complaining on intuitive terms why they will not accept it. The reason is simply that the derivative is approximate.
The delta absolutely becomes 0, consult any text.
I think you have it backwards. For f(x)=x^2, the difference quotients, [f(x+h)-f(x)]/h with h's small but nonzero, are approximations. The number they are approximating is 2x, what we call the derivative of x^2 at x. This derivative, 2x, is exactly the slope of the tangent line to x^2 at x. We could write the equation of the tangent line to x^2 at p as
y = p^2 + 2p(x-p)
This is also the best linear approximation for x^2 at p.
PairTheBoard
