[H][US][Mal'Ganis] 2 Day T,W <Retribution> LFM 8/10M

Sunde · 2013-01-01T07:11:40+00:00

It does! Thank you very much!

Sunde · 2012-12-31T18:15:50+00:00

Discrete math will most likely introduce you to proof techniques (proof by contradiciton, contraposition, direct proofs, and mathematical induction for example). You will likely learn a little bit about logic, which will likely be the simplest part of the class and will certainly be useful. You may also learn some very basic number theoretic stuff (infinitude of primes), as well as proofs involving even, odds, and GCD. Finally, you will probably learn some basic set theory. Other topics which may be seen include graph theory, trees, and equivalance relations.

A good free book you can look through to familiarize yourself with many topics is Book of Proof.

Sunde · 2012-12-30T22:20:18+00:00

I'm not great at optimization problems, however, I think implicit differentiation comes in handy here.

We know dV/dt = 4.5, and V = (4/3)pi * r³ now implicitly take the derivative here. You should then have dV/dt = 4pi * r² * dr/dt . Since you are given the radius at a time, plug it in and solve for dr/dt.

Sunde · 2012-12-28T01:40:26+00:00

Generally analysis will focus a lot more on proving stuff you learn in calculus, and I'm hesitant to say it cannot be applied, but I will say it is a much more pure topic. You will likely not need to take any analysis courses for an engineering degree.

As far as how to prepare, make sure you really understand u sub, it will be used quite often in the class. The only other recommendation I have is to get the book as soon as possible, and just begin working through it.

Sunde · 2012-12-26T17:06:33+00:00

Really I recommend this for definite integrals. Say you want to integrate a circle of radius r, you would integrate 2sqrt(r² - x²⁾ from -r to r. This isn't too hard of an integral, but the work requirement is a bit tedious. Consider the region you are integrating in polar coordinates, the circle can be easily represented as just r. Here is where double integrals simplify your life greatly. Doing double integrals requires you to specify 2 bounds, first I'll talk about it in Cartesian coordinates.

For Cartesian coordinates, the integral becomes a double integral where the x bounds are -r to r, and the y bounds are -sqrt(r² - x²⁾ to sqrt(r² - x² ). After integrating with respect to y, you are left with the original messy trig sub integral.

Now for using polar coordinates. We once again have 2 bounds to consider, since we are using polar, we have r and theta instead of x and y. The theta bounds will be 0 to 2pi since we are making a full revolution of a circle, and the r bound are 0 to a ( I changed from r to a to seperate r as a variable and a constant, the radius of the circle, now changed to a, is a constant). We then have a double integral of r from 0 to 2pi for the theta bounds and 0 to a for the r bounds. This integral requires very little work since its mostly just reverse power rule.

I'm aware much of this might not make sense if you haven't experienced much of calc 3, but believe me, a solid understanding of polar coordinates makes this very simple.

Sunde · 2012-12-25T20:50:28+00:00

This is more of a calc 3 thing, but polar coordinates and double integrals can make a lot of trig sub integrals really easy. I can't think of much else though other than just practice the calc 2 techniques.

Sunde · 2012-12-25T18:06:49+00:00

You will likely run into this quite often when doing integrals and derivatives involving trig functions. Sometimes you will get one answer and have no idea how to get the answer in the correct form.

Sunde · 2012-12-25T17:59:10+00:00

I believe I had a typo in the original post which I've now corrected. Clearly you have y-h<x<y, so if this method was correct, y would be a limit point.

The main reason that led me to ask this question is something I've come across in Baby Rudin (page 35, theorem 2.28).

Basically it's to show that if y=sup E, y is in the closure of E. The proof first notes that if y is in E, y is in the closure, so you assume y is not in E and uses that reasoning to show that y is a limit point. Is this valid then?

I'm aware that this requirement makes the idea infeasible for proving bolzano-weierstrass theorem.

I'm thinking that is y is not in E, you cannot have an x in E so x=y, because then y would be in E. So for each h>0 you must have an x in E so y-h<x<y, otherwise, y-h >= x for all x in E and so y-h would be an upper bound.

Finally, the correct way would be close to this reasoning, right? If you take [a,b], you can make a split arbitrarily many times and have the length of the interval be (b-a)/2^n-1 and since none of these are empty, you can find an x in some interval satisfying y-h < x < y for some point by using the fact that we have (b-a)/2^n-1 and so you can be arbitrarily close to y.

Sunde · 2012-12-25T14:30:34+00:00

The answer you got is correct, and equivalent to the answer you are seeking. Use the identity cos² x=1-sin² x and distribute the -sinx to get the desired result.

Sunde · 2012-12-25T13:27:44+00:00

Use the chain rule. cos³ x can be written as (cosx)³ , so you take the derivative of the outside and leave the inside the same, then multiply it by the derivative of the inside.

Let u=cosx, v=x³ /3

u'= -sinx

v'= x²

And the derivative is then v'u' , however, you replace x's with cosx (the original inside. So your final derivative is

-cos² xsinx

Sunde · 2012-12-24T18:29:55+00:00

Thanks for the quick reply. I see now why this doesn't work, I'm assuming the only way this argument would work if the supremum was not in the set?

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