Calculus for the Ambitious

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Thread starter 4 years ago
#1
I'm in year 13 and I have been reading a book, Calculus for the Ambitious. Up till now I have understood most of it but I have reached a point where I understand the maths that's going on but I'm not sure about the idea behind it, like the part about it being unambiguous. And the notation "I" wasn't used at any point up till now, and I get that it means integral but why use it instead of the summa notation? If anyone could shed some light on these pages(all the help I can get is welcome, those questions I asked are just examples) I'd be very grateful;
in the attached images the last two are the ones I am talking about and the first two are the pages before those pages for context.
Thanks!
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4 years ago
#2
There's only two pages, and no mention of any I..
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Thread starter 4 years ago
#3
Okay I have edited my post, I was having trouble uploading the images sorry, first post.
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4 years ago
#4
Ah ok. The book wants to understand what happens when you integrate functions which are negative, when all it currently knows is how to integrate strictly positive functions. It says the idea is to make negative functions by considering the difference of two positive functions. But, the big issue is this leaves choices, and choices are bad unless you prove they are safe. As an example, we can write the function as for any positive integer n, say. There are infinitely many ways of presenting f as a difference of two positive functions - the issue is this choice might make a difference to the result. We need to prove an integral defined this way is well-defined, which simply means any arbitrary choices we make don't matter.

Calling the new operator I was just a way to recognise it's not the integral you've been using so far - it's not yet defined for negative functions. But then it immediately gives you in the exercise that looking to special cases, it coincides with the original integral, and uses it in this more general case.
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4 years ago
#5
(Original post by FireGarden)
Ah ok. The book wants to understand what happens when you integrate functions which are negative, when all it currently knows is how to integrate strictly positive functions. It says the idea is to make negative functions by considering the difference of two positive functions. But, the big issue is this leaves choices, and choices are bad unless you prove they are safe. As an example, we can write the function as for any positive integer n, say. There are infinitely many ways of presenting f as a difference of two positive functions - the issue is this choice might make a difference to the result. We need to prove an integral defined this way is well-defined, which simply means any arbitrary choices we make don't matter.

Calling the new operator I was just a way to recognise it's not the integral you've been using so far - it's not yet defined for negative functions. But then it immediately gives you in the exercise that looking to special cases, it coincides with the original integral, and uses it in this more general case.
WOW
1
Thread starter 4 years ago
#6
(Original post by FireGarden)
Ah ok. The book wants to understand what happens when you integrate functions which are negative, when all it currently knows is how to integrate strictly positive functions. It says the idea is to make negative functions by considering the difference of two positive functions. But, the big issue is this leaves choices, and choices are bad unless you prove they are safe. As an example, we can write the function as for any positive integer n, say. There are infinitely many ways of presenting f as a difference of two positive functions - the issue is this choice might make a difference to the result. We need to prove an integral defined this way is well-defined, which simply means any arbitrary choices we make don't matter.

Calling the new operator I was just a way to recognise it's not the integral you've been using so far - it's not yet defined for negative functions. But then it immediately gives you in the exercise that looking to special cases, it coincides with the original integral, and uses it in this more general case.
Thanks a lot! So basically you're showing it's well defined by taking two examples of gt - ht which is ht, and equating them to each other and showing that their integrals are also equal? I would have just assumed that we know that without needing to prove it since we already proved that
is this because we assume nothing is given until we prove it mathematically, even if it looks obvious? I'm just asking because I should try and prepare myself for that Maths university mentality
Thanks for the help though, really helped. I also don't understand the significance of the inequality rule for integration?
Especially the first part of exercise 2.2.6, how is it different to the second page I sent? I keep getting a bit confused on these two pages..
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4 years ago
#7
(Original post by H.A.)
Thanks a lot! So basically you're showing it's well defined by taking two examples of gt - ht which is ht, and equating them to each other and showing that their integrals are also equal? I would have just assumed that we know that without needing to prove it since we already proved that
is this because we assume nothing is given until we prove it mathematically, even if it looks obvious? I'm just asking because I should try and prepare myself for that Maths university mentality
Thanks for the help though, really helped. I also don't understand the significance of the inequality rule for integration?
Especially the first part of exercise 2.2.6, how is it different to the second page I sent? I keep getting a bit confused on these two pages..
Yeah I got confused on this part too. I think the author does this on purpose to try and make you think really hard about what is going on. If you think really hard you will eventually understand it. If there's any exercise you're stuck on you can look at the outline solutions on the authors homepage https://www.dpmms.cam.ac.uk/~twk/ Also look at the corrections because the book has a lot of mistakes so if one part is particularly confusing you it might be because it's wrong.

The one part that I couldn't understand for ages was the bottom of page 32 when he went from "f(t) + g(t) ≤ (Mr(f) + Mr(g))s" to "Mr(f+g) ≤ Mr(f) + Mr(g)". The trick is to remember that by definition (Mr(f) + Mr(g))s is the smallest possible multiple of s greater than or equal to f(t) + g(t). Since Mr(f+g)s is greater than or equal to f(t) + g(t) then it is either equal to (Mr(f) + Mr(g))s or greater than (Mr(f) + Mr(g))s. And from this the result follows.

To answer your question about the inequality rule for integration he says on page 35 "We shall make use of this result over and over again in the rest of this book". To name a few: page 36, 38, 41, 46. It's also used to show a really important result called taylors theorem in chapter 6.

If you have any more questions to ask about that chapter i'd be happy to answer
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Thread starter 4 years ago
#8
(Original post by Oblogog)
Yeah I got confused on this part too. I think the author does this on purpose to try and make you think really hard about what is going on. If you think really hard you will eventually understand it. If there's any exercise you're stuck on you can look at the outline solutions on the authors homepage https://www.dpmms.cam.ac.uk/~twk/ Also look at the corrections because the book has a lot of mistakes so if one part is particularly confusing you it might be because it's wrong.

The one part that I couldn't understand for ages was the bottom of page 32 when he went from "f(t) + g(t) ≤ (Mr(f) + Mr(g))s" to "Mr(f+g) ≤ Mr(f) + Mr(g)". The trick is to remember that by definition (Mr(f) + Mr(g))s is the smallest possible multiple of s greater than or equal to f(t) + g(t). Since Mr(f+g)s is greater than or equal to f(t) + g(t) then it is either equal to (Mr(f) + Mr(g))s or greater than (Mr(f) + Mr(g))s. And from this the result follows.

To answer your question about the inequality rule for integration he says on page 35 "We shall make use of this result over and over again in the rest of this book". To name a few: page 36, 38, 41, 46. It's also used to show a really important result called taylors theorem in chapter 6.

If you have any more questions to ask about that chapter i'd be happy to answer
Thanks a lot man! Yeah I noticed a few errors myself in the beginning. Okay I'll go through the solutions again and hopefully that will help. Thanks!
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Thread starter 4 years ago
#9

But then why do the solutions show this instead of, say,
f(t) + g(t) = f1(t) - f2(t) + g1(t) - g2(t)

And get the result from there? Why the extra steps, is there something I'm missing?
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