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[kamil9876]

Positive Definiteness: and  if

remember that this should be iff. Hence you are left to prove that if then

(as usual, the most difficult axiom to verify for an inner product)

[/0]

Positive Definiteness: and   if

remember that this should be iff. Hence you are left to prove that if then

(as usual, the most difficult axiom to verify for an inner product)

Ah I see, well thankfully not too hard in this case. (at least i think)

I referenced wikipedia for the axioms, and it didn't have iff! Apparently it is incomplete!

Hmmm.. yeah other sites have 'iff', thanks for telling me

[kamil9876]

        with equality only for .

I guess they must be pretty quick at trolling VN and correcting their shit, good on em

[/0]

Doesn't "only" just mean "if", not "iff"?


Gah ok I get it, I misread

[Ahmad]

The 'if' part comes for free from linearity, which implies .

I do think that Spivak book is too advanced. However Spivak's book "Calculus" is fantastic and of an appropriate level for you, that is what I used to learn basic real analysis (aka rigorous calculus).

[Ahmad]

Oh, and I feel obliged to congratulate you on finishing Stewart. You're doing fantastically well.

[/0]

The 'if' part comes for free from linearity, which implies .

I do think that Spivak book is too advanced. However Spivak's book "Calculus" is fantastic and of an appropriate level for you, that is what I used to learn basic real analysis (aka rigorous calculus).

I thought calculus on manifolds was the only calculus book spivak wrote Cool I'll try to get a hold of "calculus"

Oh, and I feel obliged to congratulate you on finishing Stewart. You're doing fantastically well.

thanks! I was fortunate to learn a lot of it from umep maths lectures. I pretty much had to cover the rest of vector calc, multiple integration and sequences/series
(and i researched differential equations as part of the spesh application task)

[TrueTears]

Oh, and I feel obliged to congratulate you on finishing Stewart. You're doing fantastically well.

isnt his name adrian?

over9000 is adrian

[assef]

Oh, and I feel obliged to congratulate you on finishing Stewart. You're doing fantastically well.

isnt his name adrian?

over9000 is adrian

No he is not.

[/0]

Oh, and I feel obliged to congratulate you on finishing Stewart. You're doing fantastically well.

isnt his name adrian?

over9000 is adrian

No he is not.

I'm not adrian, so it must be over9000

[Mao]

You have finished Stewart's? =O

I've completed second year maths and I haven't bothered trying to finish Stewart's yet... That's dedication right there man. You're gonna go far.

[/0]

You have finished Stewart's? =O

I've completed second year maths and I haven't bothered trying to finish Stewart's yet... That's dedication right there man. You're gonna go far.

Well it's not like I did every exercise
I read through the theory and selected a few exercises from each chapter to do
But yeah, it was exciting

[zzdfa]

I personally did not like spivak's calculus. It's may be a good introduction for someone who's never seen calculus before, but I don't think he does a good job of telling us why we care about rigor.

From the preface of Stephen Abbotts' "Understanding Real Analysis":


Spivak was the first 'real' maths book I picked up, and it did 2) quite well. I learnt how to do read and write rigorous proofs. But I also find it quite boring: it did not satisfy 1) and 3). I stopped after chapter 11 (before all the fun starts?) and decided that analysis was just an 'elaborate reworking of standard introductory calculus'. Fortunately, I decided to try another book: Pugh's 'Real Mathematical Analysis', it has a lot of cool stuff - for example, the fact that there exists continuous paths that hit every point on the unit square.

So if you get bored of Spivak, try Abbotts' or Pugh's book. Both do a good job of motivating the rigor, but apparently Abbott is better suited for someone who has not done proofs before .

[addikaye03]

You can try them both on this, compare and contrast,



there's also a really clever and well known trick for this one, that I wasn't clever enough to discover on my own but you might want to try,

here's a hint: let and , consider both bJ + aI and aJ - bI.

Where int. F(x)/Q(x) dx

F(x)=AQ(x)+BQ'(x)

Where Q'(x) is the derivative. Is that the method of integration you're referring to? So then you Solve for A & B and then divide by Q(x) to give Int. F(x)/Q(x)= int. {A+Q'(x)/Q(x)} dx = Ax+ln(Q(x))

I've seen that indirect method somewhere. It's good for trig integration where double angle rules would be incedibly tedious

EDIT: No wait, i see what you did there.

Let a function be I and let a function be P.

I+P=something

I-P=something else

I=[(I+P)+(I-P)]/2=I

This is good for such Q as int. x^2cos^2x dx, letting P= int. x^2sin^2x

[/0]

So if you get bored of Spivak, try Abbotts' or Pugh's book. Both do a good job of motivating the rigor, but apparently Abbott is better suited for someone who has not done proofs before .

Thank zzdfa, I'll might investigate those books
And also addikaye03, Yeah seems like it could work for those functions too, I'll give it a go



Another question, for the dirac delta function,

and

The derivative is meant to be defined as:



Just wondering, does this make sense mathematically? The above expression can be derived with integration by parts, but the function isn't even continuous :/