Good math books on real analysis and calc

[PartyGirlUK]

Looks like I might be needing to learn these over the next 4 months. Looking for books for someone with minimal knowledge of these fields, but lots of ability. Something with a bunch of problems/solutions would be awesome.

D

[blah_blah]

Walter Rudin's 'Princples of Mathematical Analysis' is the standard analysis text at most elite universities. Do you not know any calculus at all? If no, you'll need to learn some before you're ready to tackle analysis - there are many more or less equivalent texts (Stewart et al), and some texts which are halfway between real analysis and calculus (Spivak, Apostol).

[Neuge]

You could also consult MIT Courseware (http://ocw.mit.edu/index.html) for lecture notes, problems, and other course materials.

[PartyGirlUK]

Ye I dont really have any calc, did some at high school, integration by parts, chain rule......

[wooly_chicken]

Spivak's calculus (the word calculus may be misleading here as this book is not at all like high school calc) is probably a good starting point.

[PartyGirlUK]

OK so I have this book

http://www.amazon.com/Advanced-Calculus-Patrick-M-Fitzpatrick/dp/0534376037

international edition. It has a bunch of exercises, which look useful, but no solutions, so I dont know how I am doing. wtf?

[PartyGirlUK]

Like is says prove that 1^3 + 2^3 +.... n^3 == (1 + 2 +.... n) ^2

Im pretty sure its a simple proof that Id really like to know, but the solution is eluding me right now.......grrrrrr

[molotom]

[ QUOTE ]
Like is says prove that 1^3 + 2^3 +.... n^3 == (1 + 2 +.... n) ^2

Im pretty sure its a simple proof that Id really like to know, but the solution is eluding me right now.......grrrrrr

[/ QUOTE ]

induction is golden for anything like that

[PartyGirlUK]

OK how about:

the above obviously holds true for n=1

now assume it holds true for n. Let LHS = q. Then the bracketed part on the right side = q^.5

ok now lets see if it holds for n+1

the RHS increases to q + (n+1)^3. or q + n^3 + 2(n^2) + 4n +1 (eq. 1)

let s = n+1

the rhs is now (sqrt q + s)^2

or

q + s^2 + 2*s*sqrtq (eq. 2)

now note that that sqrtq = 1+2+....n which we know equals n(n+1)/2

so substituting in s= n+1 and sqrtq = n(n+1)/2 into (eq. 2) yields, after multiplying out (eq. 1)

So by induction we have proved the above.

Good?

[molotom]

[ QUOTE ]
OK how about:

the above obviously holds true for n=1

now assume it holds true for n. Let LHS = q. Then the bracketed part on the right side = q^.5

ok now lets see if it holds for n+1

the RHS increases to q + (n+1)^3. or q + n^3 + 2(n^2) + 4n +1 (eq. 1)

let s = n+1

the rhs is now (sqrt q + s)^2

or

q + s^2 + 2*s*sqrtq (eq. 2)

now note that that sqrtq = 1+2+....n which we know equals n(n+1)/2

so substituting in s= n+1 and sqrtq = n(n+1)/2 into (eq. 2) yields, after multiplying out (eq. 1)

So by induction we have proved the above.

Good?

[/ QUOTE ]

That's pretty much it, you can tidy up the induction step a little but it amounts to the same argument:

Assume 1 + 2 + ... + n³ = (1 + 2 + ... + n)²
Then

1 + 2 + ... + n³ + (n+1)³
= (1 + 2 + ... + n)² + (n+1)³ (by our induction hypothesis)
= [½n(n+1)]² + (n+1)³
= (n+1)²[¼n² + (n+1)]
= ¼(n+1)²[n² + 4n + 4]
= ¼(n+1)²(n+2)²
= [½(n+1)(n+2)]²
= (1 + 2 + ... + (n+1) )²

as required

Sometimes it's possible to spot a neat/clever proof for things like this, but induction is the most reliable approach, because it mechanically reduces the problem to something much simpler, rather than involving a 'leap' of reasoning.

Re your original post, for real analysis, you don't really need much calculus. There are many problems in real analysis that have nothing to do with calculus, and those than do involve calculus typically involve giving rigorous proofs of relatively simple results in calculus.

For example, you probably learnt the sign change rule in high-school: if f(a) < 0 < f(b), then there exists x between a,b with f(x) = 0 , with the caveat that f must be 'smooth' or 'nice'. In real analysis we give our intuitive concept of a 'smooth' function (one that has no 'jumps') a concrete mathematical definition, and show why the sign change rule then holds, etc.

It often seems totally 'obvious' that many of the results in real analysis/calculus are true, like Rolle's Theorem, the Mean Value Theorem, the Intermediate Value Theorem, etc, even to someone with a rudimentary grasp of calculus... but the difficulty is actually proving them!

That said, there are also plenty of results in analysis that are bizarre and counter-intuitive, but that part of what makes it interesting.