Learn how to think the way mathematicians do – a powerful cognitive process developed over thousands of years. Mathematical thinking is not the same as doing mathematics – at least not as mathematics is typically presented in our school system. School math typically focuses on learning procedures to solve highly stereotyped problems. Professional mathematicians think a certain way to solve real problems, problems that can arise from the everyday world, or from science, or from within mathematics itself. The key to success in school math is to learn to think inside-the-box. In contrast, a key feature of mathematical thinking is thinking outside-the-box – a valuable ability in today’s world. This course helps to develop that crucial way of thinking.
Tutorial for Assignment 10
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From the course by Stanford University
Introduction to Mathematical Thinking
1029 оценки
From the lesson
Week 8
In this final week of instruction, we look at the beginnings of the important subject known as Real Analysis, where we closely examine the real number system and develop a rigorous foundation for calculus. This is where we really benefit from our earlier analysis of language. University math majors generally regard Real Analysis as extremely difficult, but most of the problems they encounter in the early days stem from not having made a prior study of language use, as we have here.
Meet the Instructors
Dr. Keith DevlinCo-founder and Executive Director
H-STAR institute
Okay. Well, we are at the end of the course now.
So in keeping with my basic philosophy behind the course I'm only going to answer
one or two of the questions from, from assignment ten since there's no more
lectures to come, no more material to come in this course.
There's nothing that's going to depend upon on anything now.
You'll only need this stuff for future courses or for using it in the real world.
So I'm going to leave most of the, the questions on assignment 10, for you to do
on your own time and to, to resolve them with colleagues and so forth.
But I just wanted to, just to sort of set the ball rolling and, and show good faith,
if you like. Okay.
So, this one, I'll do assignment 10.1 number 1, the intersection of 2 intervals
is a given interval. It's sort of easy if you think in terms of
diagrams. I mean, what are the possibilities you
could have two intervals that could sort of overlap.
Or one could be inside one of another. They could overlap the other way around.
They could be disjoined and completely separate.
Uh,[unknown] can overlap another way, one could be completely inside another one or
they could be completely disjoined. But let's just do it in, in a symbolic
fashion, let's just let's A, B. Well, I'll take the case for open
intervals, an analogy[UNKNOWN] will work for closed intervals.
So, let's A be AB and let C With the interval CD, then by definition A,
intersection C is the set of all X, so[UNKNOWN] is less than X is less than B
intersected with the set of all X, this is X[UNKNOWN], such that, C is less than X is
less than D. Alright, and because of the way
conjunction works that's a set of X such that the maximum of A and C is less than
X, is less than the minimum. Of, of b and d.
Okay, I think I've used the right letters, there you are that's correct.
Okay. Which is an interval.
It's in fact the interval, open interval, maximum of a and c to the minimum of b and
d. And that's an interval.
It may be empty. You know, if it's disjoint, it's empty.
The empty set is an interval. It's the, it's still the set of numbers
between two points. Okay.
So, The other thing is, what[UNKNOWN] with that one.
And let me just say. Similarly, for closed intervals.
And for half, half open intervals. Half open half closed.
Okay so That's really it and in the case of unions its false for unions.
For example if I take the open interval 0, 1 and I, form union with the open interval
3, 4, then that's not an interval. Okay.
And that's that one. Okay, Which one shall I do next?
Let me do, oh, what shall I do next? Hm, oh Let's do question five in 10.1.
As in question five asks us to, to, to verify this alternative definition of
least upper-bound. Okay?
So let's see, well, first of all, A just stays at B is an upper bound.
Okay? So let me write that.
A says b is an upper bound. So the issue is, does condition b.
Say that it's at least 1. Okay.
Well let's see what's, what that amounts to.
B is at least upper bound if and only if, no c less than b is an upper bound.
Okay, that's what the word, the least, means, right, there isn't a smaller one.
So, that's the original concept, there's nothing small[UNKNOWN] lower bound and
upper bound, okay. Well, that's true if and only if for any C
less than B C is not an upper bound. That's just another way of saying the same
thing. If and only if, for any C less than B.
If it's not an upper bound, that means there is an A In A such that it's not the
case that A is less than or equal to C. Okay?
You can find an A for which C. It's not bigger than it, it's not an upper
bound. If and only if, and I'm really laying this
one, one with incredible detail here, for any C less than B, there is an A in A.
Such that A, if it's not less than or equal to C, is bigger than C.
Now arguably, I could have deleted one of these lines, I'm really being very
pedantic about writing everything down, But, in my experience, both when I was a
student and from teaching this kind of material for many years, even though these
are very simple ideas it's very confusing at first dealing with the different
variations of, definitions of greatest lower bounds, least upper bounds, there's
something about the way the human brain work, that even though we recognize on 1
level that this has got to be really, really trivial it causes us problems.
So it's just something with the way the brain works that, that, that makes this a
challenge, at least for most of us. Okay, well that's that 1 Let's see what
we're going to do next. Well, I think I'll do just one more and
I'll pick one of those, limit questions, in, in number 4.
Okay, this is from assignment 10.2. And I'll pick this one, show that the
limiters of N over N plus 1 R squared turns to 1.
As N goes to infinity. And these all follow a standard pattern.
You start by, with a given epsilon greater than 0 and what we need to do is we need
to find an integer, N, such that for any natural number, N, bigger than or equal to
N It is the case that the absolute value of N over N plus 1, all squared, minus 1
is less than epsilon. From some point on in the sequence, the
difference between the, that term and 1 is less than epsilon.
Okay. So now let's just pull this apart and see
what this really wants us to find. We need to find, the integer N, such that
N greater equal to N implies that you write everything Over N plus 1 R squared
and I've got N squared minus N squared minus 2N minus 1.
That is less than epsilon, i.e., I need to find an N, so, it's just N bigger than or
equal to N implies The n squareds disappear, I've got a 2 n plus 1 over n
plus 1 squared, less than epsilon. Okay so the n squareds disappear.
I'm left with minus 2 n minus 1 over n plus 1 squared in an absolute value.
And taking the absolute value And so when I do check absolute value, the minus 2n
minus 1 just becomes 2n plus 1. And the n plus 1 squared is positive, of
course. So, that just reduces to that.
Okay. Now, we can see what to do.
What we need to do. Remember, we're looking for a big n.
So let's pick n. So big, that n plus 1 squared over 2n plus
1, is bigger than 1 over epsilon. We can always do that, Epsilon's given
it's a, it's a small number, we'll assume. I mean the, these really, they are, we use
the symbol Epsilon to sort of emphasize the fact that, that in reality Epsilon is
very small. That, that's not a logical restriction,
it's just our, our intuitions behind what the proof's doing.
Okay. So this is going to be a very large
number. But in this term, in, in this, this, this
quotient, the numerator is squared and the denominator is linear with N so,
eventually, the numerator dominates the denominator so we can make this as big as
we want and we can set it to make it bigger than 1 over epsilon, and when we do
that, we find of course, that if n is bigger than or equal to N Then 2 n plus 1
over n plus 1 squared is certainly less than or equal to big, 2 big n plus 1, over
big n plus 1 squared. Which is less than epsilon, because as we
pick big n[INAUDIBLE] to make that happen. Okay?
And that's it. Okay, we sort of, we just worked
backwards. We looked at the, the goal And we worked
backwards and, and indeed we can we can always find an n with this property.
Because the numerator grows faster than the denominator without, without bound.
Squared over, over linear. And so we get this expression.
This is very typical of, of limit verifications I wouldn't say there were
all like this, but the vast majority of them follow this pattern.
You work backwards and then you just, ask yourself where you have to look, how big
you have to, how far you have to go out in order to make something happen.
And you, you very often end up saying well pick something so big that it's bigger
than 1 over epsilon, very, very common for limits that involve quotients.
Okay. Well that's I think that's all I'm going
to do for, as I say I'm not going to do many on question, on assignment ten.
I'll leave it with those there are plenty more to, to work with and I hope you'll
enjoy working through those. I remember when I learnt this stuff for
the first time as a student. I actually enjoyed it, I thought it was a
lot of fun. 1st of all it's dealing with infinity in a
rigorous way and that's, that's pretty cool anyway.
Infinity, dealing with infinity and holding infinity in the palm of your hand
that's, that's really pretty cool. And, and it's just an intellectual
challenge, it, it's a fun game to play. Okay so enjoy it, have fun with these
things. And I'll, I hope you do okay in the, in
the final exam. Okay.
Bye for now.
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