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Selected Solutions to My Pure Complex Class (1)

八月 19, 2009

Lesson 1:
Page 26
\begin{aligned} \sum_{k = 1}^{n} \cos^2(k \theta) &= \sum_{k = 1}^{n} \frac{ 1 + \cos(2 k \theta) }{2} \\ &= \frac{n}{2} + \frac{1}{2}\sum_{k = 1}^{n} \cos(2 k \theta) \\ &= \frac{n}{2}  + \frac{\sin(n \theta) \cos((n + 1) \theta)}{2\sin(\theta)}, \end{aligned} where the last line follows from part a(ii).

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Solutions to selected problems for my Olympiad Class

八月 19, 2009

Book 1

Page 1
Answer \binom{3}{2} = 3 which can be looked up from the Pascal Triangle.

Page 2
Answer: \binom{4}{2} = 6.

Page 2
Answer: \binom{6}{3} = 20.

Page 4
Answer: \binom{4}{2} = 6 because we have to travel 4 steps in total, and we need only to decide the 2 horizontal steps.

Alternatively, one can count and add along the way to see that the Pascal triangle appears again as the number of paths.

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Concept Check for my Linear Algebra Students

八月 6, 2009

Let (E) be the following system:

a_{11}x + a_{12}y + a_{13}z = b_1\\a_{21}x + a_{22}y + a_{23}z = b_2 \\a_{31}x + a_{32}y + a_{33}z = b_3

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Review Problems for 324 Final

五月 21, 2009

Chapter 14 Review, pg 946. Q37, 39, 47, 48.

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Mid-Term 2 Solution

五月 19, 2009
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Math 324 Mid-Term 1 Solution

四月 18, 2009

1) (a) V(S) = \int_{-1}^{1}\int_{-\sqrt{1 - x^2}}^{\sqrt{1 - x^2}} \int_{x^2 + y^2}^{\sqrt{x^2 + y^2}}1 \, dz \, dy \, dx.

(b and c) \begin{aligned} V(S) &= \int_{0}^{2\pi}\int_{0}^{1}\int_{r^2}^{r}r \, dz \, dr \, d\theta\\ &= 2\pi \int_{0}^{1} r^2 - r^3 dr \\ &= 2\pi (\frac{1}{3} -\frac{1}{4}) = \frac{\pi}{6}.\end{aligned} 閱覽全文 »

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Tricks of Integration (1) – The Koop Subs

四月 8, 2009

In this article, I will talk about some of my favorite substitutions for integration, which I will named them as the “Koop subs”. They are particularly useful in mathematics competitions such as the Putnam competition.

Example 1) Compute the integral: I = \int_{0}^{1} \frac{x^3}{1 - 3x + 3x^2} dx. 閱覽全文 »

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Solution to “Possible HKAL Pure Problems”

四月 1, 2009

1(a) Expanding along the last row, we have:

f(x) = 1  \det{\begin{pmatrix}x &  x^3\\y &  y^3 \end{pmatrix}} - z \det{\begin{pmatrix} 1 & x^3 \\1 & y^3 \end{pmatrix}} + z^3 \det{\begin{pmatrix} 1 &  x \\1 &  y \end{pmatrix}}.

Therefore, f(z) is a cubic in z with leading coefficient \det \begin{pmatrix} 1 & x \\1 & y \end{pmatrix} = y - x. 閱覽全文 »

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Possible HKAL Pure Problem (2)

四月 1, 2009

Prove by contradiction or otherwise, show that \log(x) is not a polynomial with real coefficients.

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A possible HKAL Pure problem?

三月 5, 2009

Let A = \begin{pmatrix} 1 & x & x^3\\ 1 & y & y^3 \\1 & z & z^3 \end{pmatrix}

(a) By expanding the determinant along the last row (or otherwise), show that det(A) is a cubic polynomial in z, and we call det(A) = f(z). (1 point)

(b) Show that f(x) = f(y) = 0, and hence conclude that f(z) = (y - x)(z - x)(z - y)(z - r) for some r in terms of x and y. (5 points)

(c) Using the relationship between the roots and coefficients of a polynomial (or otherwise), show that r = -x - y. (3 points)