Calculus Review

Chapter 0.

The requirement of our college is that the process should be written in English...

The content of this essay is very very trivial, so should you find yourself with some leisure time, you might find it mildly amusing to peruse. And what I truly concern is whether I can get full score in the test...

Chapter 1. Function

1.1 Odd function and even function

Example 1. Let \(f(x)\) be a function defined on \((-l,l)\). Prove that \(f(x)\) can be written as the sum of an even function and an odd function.

Proof:

It is a classical construction problem... And it is easy to find that:

\[g(x)=\left(f(x)+f(-x)\right) / 2 \]

\[h(x) = \left(f(x)-f(-x)\right) / 2 \]

are the functions that meet the conditions. \(\quad \square\)

1.2 The convertion between polar equation and Cartesian equation

\[\begin{cases} x = r\cos \theta \\ y = r \sin \theta \\ x ^ 2 + y ^ 2 = r ^ 2 \end{cases} \]

There are not many noticable problems.

1.3 Some inequalities

Example 2. To prove that:

\[\prod _ {k = 1} ^ {n} \dfrac{2k-1}{2k} < \dfrac{1}{\sqrt{2n+1}} \]

Proof:

It is obvious that:

\[\prod _ {k = 1} ^ {n} \dfrac{\sqrt{(2k-1)(2k+1)}}{2k} < 1 \]

Thus we obtain the original inequality. \(\quad \square\)

Example 3. To prove that:

\[\sum _ {k = 1} ^ {n} \dfrac{1}{\sqrt{k}} < \sqrt{n} \quad (n \ge 2) \]

Proof:

\[\sum _ {k = 1} ^ {n} \dfrac{1}{\sqrt{k}} < \sum _ {k=1}^{n} \dfrac{1}{\sqrt{n}}<\sqrt{n}. \quad \square \]

Chapter 2. Limit and Continuity

2.1 Squeeze Theorem

Example 1. To prove that:

\[\lim _ {n \to \infty} n ^ {1/n} = 1 \]

Proof:

Let \(n ^ {1/n} = 1 + \alpha _ n\). We obtain:

\[n = ( 1 + \alpha _ n ) ^ n \ge \dfrac{n(n-1)}{2} \alpha _ n ^ 2 \quad (\forall n \ge 2), \]

which yields that \(0 \le \alpha _ n \le \sqrt{2/(n-1)}\).

Therefore, \(\lim _ {n \to \infty} \alpha _ n = 0\), which implies that:

\[\lim _ {n \to \infty} n ^ {1/n} = 1. \quad \square \]

2.2 Monotonic Bounded Principle

We see that if a generating sequence converges to \(A\), it will satisfy that \(A=G(A)\).

Thus, if we prove that the sequence is convergent by Mnotonic Bounded Principle, we will be able to use the property to obtain the limit.

Example 2. \(a _ {n+1} = \sqrt{ 6 + a _ n}\), and \(a _ 1 = 10\).

Solution:

To prove monotonicity, we consider the induction on \(n\).

For the base case, we know that \(a_2 = \sqrt{6+ a _ 1 } = 4\), which gives \(a _ 2 - a _ 1 < 0\).

Suppose that $\forall n \le k : a _ n - a _ {n-1} < 0 $, from which we obtain that

\[a _ {n+1} - a_n = \sqrt{6 + a_ n} - \sqrt{6 + a_{n-1}} = \dfrac{a_n - a_ {n-1}}{\sqrt{6 + a_ n} + \sqrt{6 + a_{n-1}} } < 0. \]

By the induction, we yields \(a _ {n+1} - a_ n < 0\) is true for all \(n > 0\).

Since \(a_ n > 0\), it follows that \(\left\{ a_ n\right\}\) is convergent frome Monotonic Bounded Principle.

Without loss of generality, suppose that \(\lim _ {n \to \infty} a _ {n+1}= A\).

Since

\[\lim _ {n\to \infty} a_{n+1} = \lim _ {n\to \infty} \sqrt{6 + a _ n} = A, \]

it implies that \(A = \sqrt{6+A}\), which yields \(A = 3\).

Thus, \(\lim _ {n\to\infty} a_n = 3\). \(\quad \square\)