discrete mathematics - How to teach mathematical induction?

Some students are not convinced that a proof by mathematical induction is a proof. I have given the analogy of dominoes toppling but still some remain unconvinced. Is there very convincing way of introducing mathematical induction? I need something which will have an impact. Are there any real life applications of induction?

Answer

Use the analogy of dominoes, and then give the real-life example of the sinking of the Titanic: The management of the Titanic realized that the ship was doomed when they realized that the bulkhead that was being flooded would be completely flooded, and that when a given bulkhead was completely flooded, the next bulkhead would undergo the same fate.

Talk about impact...

real analysis - A convergent sequence that is defined recursively

I would like to get a hint on how to establish the convergence of the following sequence:

$$a_{n+1}= a_n + \frac{\sqrt{\vert a_n \vert }}{n^2}$$

where $a_1$ is arbitrary. This is an increasing sequence, so if I could show that it was bounded above I would be done. I cannot figure out how to do that. Any help would be appreciated.

Answer

We have $a_n=\sum_{j=1}^{n-1}\frac{\sqrt{|a_j|}}{j^2}+a_1$. Let $M$ such that $M>|a_1|+\frac{\pi^2}6\sqrt M$. Assume that $|a_j|\leqslant M$ for all $1\leqslant j\leqslant n-1$. Then
$$|a_n|\leqslant \sqrt M\sum_{j=1}^{n-1}\frac 1{j^2}+|a_1|As $|a_1|

algebra precalculus - What is $sqrt{-4}sqrt{-9}$?

I assumed that since $a^c \cdot b^c = (ab)^{c}$, then something like $\sqrt{-4} \cdot \sqrt{-9}$ would be $\sqrt{-4 \cdot -9} = \sqrt{36} = \pm 6$ but according to Wolfram Alpha, it's $-6$?

Answer

The property $a^c \cdot b^c = (ab)^{c}$ that you mention only holds for integer exponents and nonzero bases. Since $\sqrt{-4} = (-4)^{1/2}$, you cannot use this property here.

Instead, use imaginary numbers to evaluate your expression:

$$
\begin{align*}

\sqrt{-4} \cdot \sqrt{-9} &= (2i)(3i) \\
&= 6i^2 \\
&= \boxed{-6}
\end{align*}
$$

real analysis - Limit of sequence is indeterminate.

We have to find the limit of the sequence $a_k = \left(\dfrac{k^4 11^k + k^9 9^k}{7^{2k} +1}\right)$.

Here is my attempt:

$$\left(\frac{k^4 11^k + k^9 9^k}{7^{2k} +1}\right) = \frac{k^4 11^k + k^9 9^k}{49^k +1} = \frac{\frac{\left(\frac{11}{49}\right)^k}{k^5} + \left(\frac{9}{49}\right)^k}{\frac{1}{k^9} + \frac{1}{k^9 49^k}}$$

But as $k \to \infty $, both the numerator and the denominator tend to zero. So I am left with an indeterminate form. How can I avoid this?

Answer

Guide:

$$\left|\frac{k^4 11^k + k^9 9^k}{7^{2k} +1}\right| \le \frac{k^4 11^k + k^9 9^k}{49^k } = k^4 \left( \frac{11}{49}\right)^k + k^9 \left( \frac{9}{49}\right)^k$$

You can focus on finding $\lim_{k \to \infty}k^4 \left( \frac{11}{49}\right)^k$ and $\lim_{k \to \infty}k^9 \left( \frac{9}{49}\right)^k$

or in general $\lim_{k \to \infty} k^a r^k$ where $|r|<1$ and $a \ge 0$.

How to prove that $sum_{n=1}^infty frac{1}{n^2}=frac{pi^2}{6}$ without using Fourier Series

Can we prove that $\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$ without using Fourier series?

Answer

Yes. The most common way to do this is attributed to Euler. It does still require Maclaurin series, however.

Consider the Maclaurin polynomial for $\frac{\sin x}{x}$:

$$\frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \cdots$$

However, note that this is a polynomial $p(x)$ with zeroes $\{\pm k\pi\;|\;k \in \Bbb N\}$, and for which $p(0) = 1$. These two properties mean that

$$\frac{\sin x}{x} = \left(1 + \frac{x}{\pi}\right)\left(1 - \frac{x}{\pi}\right)\left(1 + \frac{x}{2\pi}\right)\left(1 - \frac{x}{2\pi}\right)\cdots$$

And by multiplying adjacent terms,

$$\frac{\sin x}{x} = \left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{4\pi^2}\right)\left(1 - \frac{x^2}{9\pi^2}\right)\cdots$$

Equating the $x^2$ terms in the Maclaurin polynomial and its factored form yields

$$-\frac{x^2}{3!} = -x^2\left(\frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \cdots\right)$$

And multiplying both sides by $-\frac{pi^2}{x^2}$ gives us

$$\frac{\pi^2}{6} = 1 + \frac{1}{4} + \frac{1}{9} + \cdots$$

complex numbers - how to compute $|a-ib|^2$ if given $(a-ib)^3$

If i know that, for example, $(a-ib)^3=5+4i$ - how can i compute the value of $|a-ib|^2$ ?

I can take modulus of $5+4i$ which is $\sqrt{5^2+4^2}$ but i don't know what i'm getting here.

I don't know what's a third root of a complex number here.

Any help?

Thanks.

Answer

We know that $|z\cdot w|=|z|\cdot|w|$, and so in general $|z^n|=|z|^n$.

Therefore, since you know $z^3$, you can take its modulus to get $|z^3|=|z|^3$, and raising that to the $2/3$ power gives $|z|^2$.

Mathematical Induction

I have to prove my mathematical induction that $5^n - 1$ is divisible by 4. for all non negative $n ≥ 0$. My solution is the following.

Base case: When $n = 0,
5^n-1 = 5^0-1 = 0$.
Base case holds for $n = 0.$

Induction Hypothesis: Assume the property holds for $n = k$, i.e. $5^k-1$ is divisible by $4$.

Induction Step: When $n = k + 1$, we must prove that $5^{k+1}-1$ is divisble by $4$.

$5^{k+1}-1 = 5 * 5^k-1$

From the hypothesis we know that $5^k-1$ is divisible by $4$. Any number divisible by $4$ and multiplied by $5$ is divisible by $4$.

Thus $5^{k+1}-1$ is divisible by $4$.

The actual answer booklet offers a solution that seems unnecessarily complex.

$5^{k+1} -1 = 5*5^{k-1}$

$= 5 *(5^k-1+1)-1$

$= 5 * (5^k - 1 ) + 5 - 1$

$= 5 * (5^k - 1) + 4$

By the induction hypothesis, $5k - 1$ is divisible by $4$. Clearly $4$ is also divisible by $4$ and therefore $5 ∗ (5 k − 1) + 4$ is divisible by $4$ and the induction step is proven.

Is my way of doing it correct or is it not complete enough?

Thanks.