## Integrals with sec and tan when the power of tan is odd

We went through an example in class today which was

$\int tan^6\theta \sec^4\theta d\theta$

In this case we took out two powers of sec and then converted all the other $\sec$ into $latex\ tan$, which left a function of tan times $sec^2\theta d\theta$. We wanted to do this because the derivative of $\tan$ is $\sec^2$ and so we can do a simple substitution. If we have an odd power of $\tan$, we can employ a different trick. Let’s look at:

$I=\int \tan^5\theta\sec^7\theta d\theta$.

Here, sec is an odd power and so we can’t employ the same trick as before. Now we want to convert everything to a function of $\sec$ and have only a factor which is the derivative of $\sec$ left over. The derivative of $\sec$ is $\sec\tan$, so let’s try and take this out:

$I=\int \tan^5\theta\sec^7\theta d\theta=\int \tan^4\theta\sec^6\theta (\sec\theta\tan\theta)d\theta$.

Now convert the $\tan$ into $\sec$ by $\tan^2\theta=\sec^2\theta-1$:

$I=\int (\sec^2\theta-1)^2\sec^6\theta (\sec\theta\tan\theta)d\theta=\int (\sec^{10}\theta-2\sec^8\theta+\sec^6\theta) (\sec\theta\tan\theta)d\theta$

where here we have just expanded out the bracket and multiplied everything out.…

## Fundamental theorem of calculus example

We did an example today in class which I wanted to go through again here. The question was to calculate

$\frac{d}{dx}\int_a^{x^4}\sec t dt$

We spot the pattern immediately that it’s an FTC part 1 type question, but it’s not quite there yet. In the FTC part 1, the upper limit of the integral is just $x$, and not $x^4$. A question that we would be able to answer is:

$\frac{d}{dx}\int_a^{x}\sec t dt$

This would just be $\sec x$. Or, of course, we can show that in exactly the same way:

$\frac{d}{du}\int_a^{u}\sec t dt=\sec u$

That’s just changing the names of the variables, which is fine, right? But that’s not quite the question. So, how can we convert from $x^4$ to $u$? Well, how about a substitution? How about letting $x^4=u$ and seeing what happens. This is actually just a chain rule. It’s like if I asked you to calculate:

$\frac{d}{dx} g(x^4)$.

You would just say: Let $x^4=u$ and then we have:

$\frac{d}{dx} g(x^4)=\frac{du}{dx}\frac{d}{du}g(u)=4x^3 g'(u)$.…

## PDE: Physics, Math and Common Sense. Part I: Conservation Law

Source: CFDIinside blog

INTRODUCTION

The course of Partial differential equations (PDEs) usually is a tough one. There is a number of factors contributing to this toughness:

• PDE course combines the knowledge from calculus, algebra, ordinary differential equations (ODEs), complex analysis and functional analysis. Simply put, there is a lot that you need to know about!
• PDE methods often (or should I say, mostly?) come from physics, but this aspect is not always emphasized and, as a result, the intuition is lost.
• There is lots of abstraction in the PDE course material: characteristics, generalized functions (distributions), eigenfunctions, convolutions and etc. Many of these concepts actually have simple interpretations, but again, this is not emphasized.
• PDEs themselves are tough. In contrast to ODEs, there are no general methods for all kinds of PDEs. The field is young and a bit messy.

This series of posts aims to demystify PDEs and show some general way of handling PDE problems by combining physical intuition and mathematical methods.…

## Using integration to calculate the volume of a solid with a known cross-sectional area.

Hi there again, I have not written a post in while, here goes my second post.

I would like us to discuss one of the important applications of integration. We have seen how integration can be used to solve the area problem, in this post we are going to see how we can use a similar idea to solve the volume problem. I suggest that we start by looking at the solids whose volume we know very well. You should be able to calculate the volumes of the cylinders below (yes,  they are all cylinders.)

Cylinders are nice, we only need to multiply the cross-sectional area by the height/length to find the volume. This is because they have two identical flat ends and the same cross-section from one end to the other. Unfortunately, not all the solid figures that we come across everyday are cylinders. The figures below are not cylinders.…

## An integral expression for n!

I gave a challenge question at the end of class a week or so ago. Here I will give the solution and show that it gives us something rather strange and surprisingly useful.

I wrote down the following, and asked you to prove it:

$\int_0^\infty e^{-t} t^N dt=N!$

For $N\ge 0, N\in \mathbb{Z}$. Now, N! can be thought of as the number of different orderings of pulling N objects out of a bag (without replacement) when they are all different. If you have N things in a bag, then there are N possible things that you can pull out first. There are then N-1 ways of pulling out the next object, N-2 ways of pulling out the next, etc. and finally, when you’ve pulled out N-1 objects there’s only a single possibility of pulling out the last. So:

$N!=N(N-1)(N-2)(N-3)...3.2.1$

And the number of ways of pulling no objects out of a bag is 1, because you just don’t pull anything out.…

Whilst reading Mícheál Ó Searcóid’s book Metric Spaces, I found out about a nuance in the definition of continuity that I was not previously aware of, and something which may be taught incorrectly at high schools. M. Searcóid states that a function such as tan(x) is continuous (read the page here). The definition of continuity at a point is based on the fact that the function has a value at that point (if a function is continuous at x = a, then f(a) has a value in the expression |f(x)-f(a)|<ϵ). However, following M. Searcóid’s line of thought about continuous functions, it does not make sense to consider points at which a function is not defined. If we were asked to prove that the function is ‘discontinuous’ at a point, we would need to show that the condition for continuity at that point is false. And the negation* of the condition for continuity would not make sense at a point where the function value is not defined.…

## On Convergent Sequences and Prime Numbers

Ever since Euclid first proved that there are infinitely many prime numbers, mathematicians have found ever more creative ways to prove the same result, and also various stronger theorems that imply it. Dirichlet’s Theorem, for example, states that if$m$ and $n$ are relatively prime integers, then there are infinitely many prime numbers of the form $mk + n$ for some integer $k$. It is also known that the sum of the reciprocals of the prime numbers diverges, that the sum

$\displaystyle \sum_{\substack{p \leq n \\ p \text{ prime}}} \frac{1}{p} \sim \log(\log(n))$

and that the number of prime numbers less than $n$ is asymptotically equal to $\displaystyle \frac{n}{\log(n)}$. In this blog post, we will continue this proud tradition by proving that there are infinitely many prime numbers which have your phone number somewhere in their digits, and which simultaneously have a prime number of digits.

To do so, we will look at the convergence of two different sums: that of the reciprocals of the primes with a prime number of digits, and that of the reciprocals of the natural numbers which do not contain your phone number amongst their digits.…

## Welcome to Reproducing Kernel Hilbert Space

In a series of posts I hope to introduce Mathemafrica readers to some useful data analysis methods which rely on operations in a little back-water of Hilbert space, namely Reproducing Kernel Hilbert Space (or RKHS).

We’ll start with the “classic” example. Consider the data plotted in figure 1. Each data point has 3 “properties”: an $x_1$ coordinate, an $x_2$ coordinate and a colour (red or blue). Suppose we want to be able to separate all data points into two groups: red points and blue points. Furthermore, we want to be able to do this linearly, i.e. we want to be able to draw a line (or plane or hyperplane) such that all points on one side are blue, all points on the other are red. This is called linear classification.

Figure 1: A scatter of data with three properties: an x_1 coordinate, an x_2 coordinate and a colour.

Suppose for each data point we generate a representation of the data point $\phi(x)=[x_1, x_2, x_1x_2]$.…

## Fibonacci and the Golden Ratio

The Fibonacci numbers $1, \ 1, \ 2,\ 3, \ 5,\ 8, 13,\ 21,\ 34, \ 55,\ 89,\ 144, \ \cdots$ are defined by the recurrence relation

$F_1 = F_2 = 1, \ F_{n+2} = F_{n+1} + F_n \ \textrm{ for } n \ge 1.$

Consider now the ratios $\dfrac{F_{n+1}}{F_n}$:

$\dfrac 11, \ \dfrac 21, \ \dfrac 32, \ \dfrac 53, \ \dfrac 85, \ \dfrac {13}8, \dfrac {21}{13}, \ \dfrac {21}{13}, \ \dfrac {34}{21}, \dfrac {55}{34}, \dfrac {89}{55}, \ \dfrac {144}{89}, \ \cdots \ .$

The sequence of fractions appears to rise and fall alternately. This observation can be confirmed by reference to the equation

$F_{n+1}^2 - F_nF_{n+2} = (-1)^n$

which, when divided by $F_{n+1}F_n$ gives

$\dfrac{F_{n+1}}{F_n} - \dfrac {F_{n+2}}{F_{n+1}} = \dfrac {(-1)^n}{F_{n+1}F_n } \ ,$

showing that the difference between successive terms in the sequence of ratios is alternately positive and negative.

Calculating the first few terms suggest that successive ratios converge to 1.618 … . This may be established by using Binet’s formula (see the previous post):

$F_n = \dfrac 1{\sqrt 5}\Big(\Big(\dfrac{1+\sqrt 5}2 \Big)^n - \Big(\dfrac{1-\sqrt 5}2 \Big)^n\Big) \$

from which we have

$\dfrac {F_{n+1}}{F_n} = \dfrac {\Big(\dfrac{1+\sqrt 5}2 \Big)^{n+1} - \Big(\dfrac{1-\sqrt 5}2 \Big)^{n+1}} {\Big(\dfrac{1+\sqrt 5}2 \Big)^n - \Big(\dfrac{1-\sqrt 5}2 \Big)^n}$

Now look at the second terms in the numerator and denominator.

Since

$-1< \dfrac{1-\sqrt 5}2 < 1$,

these terms tend to zero as $n$ tends to infinity, so are insignificant in comparison with the first terms. It follows that

$\frac {F_{n+1}}{F_n} \rightarrow \frac{1+\sqrt 5}2 \ \ - \ \ \textrm{the Golden Ratio}.$

Check out the link between the golden spiral and the Fibonacci sequence here.

————

Originally by John Webb

 How clear is this post?

## Fibonacci and Binet

The Fibonacci numbers $1, \ 1, \ 2,\ 3, \ 5,\ 8, 13,\ 21,\ 34, \ 55,\ 89,\ 144, \ \cdots$

are defined by a recurrence relation

$F_1 = F_2 = 1, \ F_{n+2} = F_{n+1} + F_n \ \textrm{ for } n \ge 1.$

This definition allows one to find any Fibonacci number, but for large values of $n$ it would be time-consuming to compute $F_n$ this way. Worse, any error of calculating a term would be repeated and magnified in every subsequent term. So a natural question to ask is whether there is a simple formula, in terms of $n$, which enables one to calculate $F_n$ without having to find all the earlier Fibonacci numbers.

In 1843 the French mathematician Jacques Philippe Marie Binet announced that he had found a Fibonacci Formula. Although it was later revealed that other mathematicians, including Leonhard Euler and Abraham de Moivre, had worked out the same formula a hundred years earlier, the formula is today still labelled Binet’s Formula.

Deriving the Binet’s formula starts with a simple little quadratic equation $x^2 = x + 1.$

Multiplying the equation by successive powers of $x$, and simplifying at each step, gives

$x^3 = x(x^2) = x(x+1) = x^2 + x = (x+1) + x = 2x + 1$

$x^4 = x(x^3) = x(2x+1) = 2x^2 + x = 2(x+1) + x = 3x+2$

$x^5 = x(x^4) = x(3x+2) = 3x^2 + 2x = 3(x+1) + 2x = 5x + 3$

$x^6 = x(x^5) = x(5x+3) = 5x^2 + 3x = 5(x+1) + 3x = 8x + 5$

and so on.…