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Table of Contents

Return on investment (ROI) is a financial metric of profitability that is widely used to measure the return or gain from an investment. ROI is a simple ratio of the gain from an investment relative to its cost. It is as useful in evaluating the potential return from a stand-alone investment as it is in comparing returns from several investments.

In business analysis, ROI is one of the key metrics—along with other cash flow measures such as internal rate of return (IRR) and net present value (NPV)—used to evaluate and rank the attractiveness of a number of different investment alternatives. ROI is generally expressed as a percentage rather than as a ratio.

## How to Calculate ROI

The ROI calculation is a straightforward one, and it can be calculated by either of the two following methods.

The first is this:

$ROI = \frac{\text{Net\ Return \ on \ Investment}}{\text{Cost \ of \ Investment}}\times 100\%$

The second is this:

$ROI = \frac{\text{Final Value of Investment}\ -\ \text{Initial Value of Investment}}{\text{Cost of Investment}}\times100\%$

## Interpreting ROI

There are some points to bear in mind with regard to ROI calculations:

• As noted earlier, ROI is intuitively easier to understand when expressed as a percentage instead of a ratio.
• The ROI calculation has "net return" rather than "net profit or gain" in the numerator. This is because returns from an investment can often be negative instead of positive.
•  A positive ROI figure means that net returns are in the black, as total returns exceed total costs. A negative ROI figure means that net returns are in the red (in other words, this investment produces a loss), as total costs exceed total returns.
• To compute ROI with greater accuracy, total returns and total costs should be considered. For an apples-to-apples comparison between competing investments, annualized ROI should be considered.

## A Simple ROI Example

Let’s assume you bought 1,000 shares of hypothetical Worldwide Wicket Co. for $10 each. Exactly a year later, you sold the shares for$12.50. You earned dividends of $500 over the one-year holding period. You also spent a total of$125 on trading commissions when you bought and sold the shares. What is your ROI?

It can be calculated as follows:

$ROI\ =\ \frac{[(\12.50\ -\ \10.00)\ \times\ 1,000]\ +\ \500\ -\ \125\ \times\ 100\%}{\10.00\ \times\ 1,000}=\ 28.75\%$

Let's deconstruct this calculation resulting in a 28.75% ROI step by step.

1. To calculate net returns, total returns and total costs must be considered. Total returns for a stock arise from capital gains and dividends. Total costs would include the initial purchase price as well as commissions paid.
2. In the above calculation, the first term [($12.50 -$10.00) x 1,000] shows the gross capital gain (i.e., before commissions) from this trade. The $500 amount refers to the dividends received by holding the stock, while$125 is the total commission paid.
3. Dissecting the ROI into its component parts would result in the following:

\begin{aligned} &ROI\ =\ \text{Capital Gains (23.75\%)}\ +\ DY (5.00\%)\\ &\textbf{where:}\\ &DY=\text{Dividend yield} \end{aligned}

Why is this important? Because capital gains and dividends are taxed at different rates in most jurisdictions.

Here's another way of calculating the ROI on your Worldwide Wicket Co. investment. Let's assume the following split of the $125 paid in total commissions—$50 when buying the shares and 75 when selling the shares. \begin{aligned} &IVI\ =\10,000+\50\ =\10,050\\ &FVI\ =\12,500+\500-\75\ =\12,925\\ &ROI\ =\frac{\12,925-\10,050}{\10,050}\times100\%\ =28.60\%\\ &\textbf{where:}\\ &IVI = \text{Initial value of investment (i.e. cost of investment)}\\ &FVI = \text{Final value of investment} \end{aligned} The slight difference in the ROI values (28.75% vs. 28.60%) occurs because, in the second instance, the commission of50 paid upon purchase of the shares was included in the initial cost of the investment. So while the numerator in both the equations was the same ($2,875), the slightly higher denominator in the second instance ($10,050 vs. 10,000) has the effect of marginally depressing the stated ROI figure. ## Annualized ROI The Annualized ROI calculation counters one of the limitations of the basic ROI calculation, which is that it does not consider the length of time that an investment is held (the "holding period"). Annualized ROI is calculated as follows: \begin{aligned} &\text{Annualized } ROI = [(1 + ROI) ^{1/n} - 1]\times100\%\\ &\textbf{where:}\\ &\begin{aligned} n=\ &\text{Number of years for which the investment}\\ &\text{is held} \end{aligned} \end{aligned} Assume you had an investment that generated an ROI of 50% over five years. What was the annualized ROI? The simple annual average ROI of 10% (obtained by dividing ROI by the holding period of five years) is only a rough approximation of annualized ROI because it ignores the effects of compounding, which can make a significant difference over time. The longer the time period, the bigger the difference between the approximate annual average ROI (ROI / holding period) and annualized ROI. \begin{aligned} &\text{From the formula above,}\\ &\text{Annualized ROI}=[(1+0.50)^{1/5}-1]\times100\%=8.45\% \end{aligned} This calculation can also be used for holding periods of less than a year by converting the holding period to a fraction of a year. Assume you had an investment that generated an ROI of 10% over six months. What was the annualized ROI? $\text{Annualized ROI}=[(1+0.10)^{1/0.5}-1]\times100\%=21.00\%$ (In the mathematical expression above, six months = 0.5 years). ## Comparing Investments and Annualized ROI Annualized ROI is especially useful when comparing returns between various investments or evaluating different investments. Assume your investment in stock X generated an ROI of 50% over five years, while your stock Y investment returned 30% over three years. What was the better investment in terms of ROI \begin{aligned} &AROIX=[(1+0.50)^{1/5}-1]\times100\%=8.45\%\\ &AROIY=[(1+0.30)^{1/3}-1]\times100\%=9.14\%\\ &\textbf{where:}\\ &AROIX = \text{Annualized ROI for stock }X\\ &AROIY = \text{Annualized ROI for stock }Y \end{aligned} Stock Y had a superior ROI compared to stock X. ## ROI with Leverage Leverage can magnify ROI if the investment generates gains, but by the same token, it can amplify losses if the investment proves to be a dud. In an earlier example, we had assumed that you bought 1,000 shares of Worldwide Wickets Co. for10 each. Let's further assume that you bought these shares on a 50% margin, which means that you put up $5,000 of your own capital and borrowed$5,000 from your brokerage as a margin loan. Exactly a year later, you sold the shares for $12.50. You earned dividends of$500 over the one-year holding period. You also spent a total of $125 on trading commissions when you bought and sold the shares. In addition, your margin loan carried an interest rate of 9%. What is your ROI? There are two key differences from the earlier example: • The interest on the margin loan ($450) should be considered in total costs.
• Your initial investment is now $5,000 because of the leverage employed by taking the margin loan of$5,000.

* This is the margin loan of $5,000 Thus, even though the net dollar return was reduced by$450 on account of margin interest, ROI is substantially higher at 48.50%, compared with 28.75% if no leverage was employed.

But instead of rising to $12.50, what if the share price fell to$8.00, and you had no choice but to cut your losses and sell the full position? ROI, in this case, would be:

\begin{aligned} \text{ROI}=&\frac{[(\8.00-\10.00)\times1,000]+\500-\125-\450}{(\10.00\times1,000)-(\10.00\times500)}\\ &\times100\%=-\frac{\2,075}{\5,000} =-41.50\% \end{aligned}

In this case, ROI of -41.50% is much worse than ROI of -16.25% that would have resulted if no leverage was employed.

## Unequal Cash Flows

When evaluating a business proposal, one often has to contend with unequal cash flows. This means that the returns from an investment will fluctuate from one year to the next.

The calculation of ROI in such cases is more complicated and involves using the internal rate of return (IRR) function in a spreadsheet or calculator.

Assume you have a business proposal to evaluate that involves an initial investment of $100,000 (shown under Year 0 in the "Cash Outflow" row in the following Table). The investment generates cash flows over the next five years, as shown in the "Cash Inflow" row. The "Net Cash Flow" row sums up the cash outflow and cash inflow for each year. What is the ROI? Using the IRR function, the calculated ROI is 8.64%. The final column shows the total cash flows over the five-year period. Net cash flow over this five-year period is$25,000 on an initial investment of $100,000. What if this$25,000 was spread out equally over five years? The cash flow table would then look like this:

Note that the IRR, in this case, is now only 5.00%.

The substantial difference in the IRR between these two scenarios—despite the initial investment and total net cash flows being the same in both cases—has to do with the timing of the cash inflows. In the first case, substantially larger cash inflows are received in the first four years. Because of the time value of money, these larger inflows in the earlier years have a positive impact on IRR.

## Benefits of ROI

The biggest benefit of ROI is that it is an uncomplicated metric, easy to calculate and intuitively easy to understand. ROI's simplicity means that it is a standardized, universal measure of profitability with the same connotation anywhere in the world, and hence not liable to be misunderstood or misinterpreted. "This investment has an ROI of 20%" has the same meaning whether you hear it in Argentina or Zimbabwe.

Despite its simplicity, the ROI metric is versatile enough to be used to evaluate the efficiency of a single stand-alone investment or to compare returns from different investments.

## Limitations of ROI

ROI does not take into account the holding period of an investment, which can be an issue when comparing investment alternatives. For example, assume investment X generates an ROI of 25% while investment Y produces an ROI of 15%. One cannot assume that X is the superior investment unless the timeframe of investment is also known. What if the 25% ROI from X is generated over a period of five years, but the 15% ROI from Y only takes one year? Calculating annualized ROI can overcome this hurdle when comparing investment choices.

ROI does not adjust for risk. It is common knowledge that investment returns have a direct correlation with risk – the higher the potential returns, the greater the possible risk. This can be observed firsthand in the investment world, where small-cap stocks typically have higher returns than large-cap stocks but are accompanied by significantly greater risk. An investor who is targeting a portfolio return of 12%, for example, would have to assume a substantially higher degree of risk than an investor who wants a return of 4%. If one focuses only on the ROI number without evaluating the concomitant risk, the eventual outcome of the investment decision may be very different from the expected result.

ROI figures can be exaggerated if all the expected costs are not included in the calculation, whether deliberately or inadvertently. For example, in evaluating the ROI on a piece of real estate, associated expenses such as mortgage interest, property taxes, insurance and maintenance costs must be considered because they can take a hefty chunk out of ROI. Not including all these expenses in the ROI calculation can result in a grossly overstated return figure.

Like many profitability metrics, ROI only emphasizes financial gain and does not consider ancillary benefits such as social or environmental ones. A relatively new ROI metric known as "Social Return on Investment" (SROI) helps quantify some of these benefits.

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## The Bottom Line

Return on investment (ROI) is a simple and intuitive metric of profitability used to measure the return or gain from an investment. Despite its simplicity, it is versatile enough to be used to evaluate the efficiency of a single stand-alone investment or to compare returns from different investments. ROI's limitations are that it does not consider the holding period of an investment (which can be rectified by using the annualized ROI calculation) and is not adjusted for risk. Despite these limitations, ROI finds the widespread application and is one of the key metrics—along with other cash flow measures such as IRR and NPV—used in business analysis to evaluate and rank returns from competing for investment alternatives. (For related reading, see "How to Calculate the ROI on a Rental Property")


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## Harold Jeffreys’ default Bayes factor [for psychologists]

“One of Jeffreys’ goals was to create default Bayes factors by using prior distributions that obeyed a series of general desiderata.”

The paper Harold Jeffreys’s default Bayes factor hypothesis tests: explanation, extension, and application in Psychology by Alexander Ly, Josine Verhagen, and Eric-Jan Wagenmakers is both a survey and a reinterpretation cum explanation of Harold Jeffreys‘ views on testing. At about the same time, I received a copy from Alexander and a copy from the journal it had been submitted to! This work starts with a short historical entry on Jeffreys’ work and career, which includes four of his principles, quoted verbatim from the paper:

1. “scientific progress depends primarily on induction”;
2. “in order to formalize induction one requires a logic of partial belief” [enters the Bayesian paradigm];
3. “scientific hypotheses can be assigned prior plausibility in accordance with their complexity” [a.k.a., Occam’s razor];
4. “classical “Fisherian” p-values are inadequate for the purpose of hypothesis testing”.

“The choice of π(σ)  therefore irrelevant for the Bayes factor as long as we use the same weighting function in both models”

A very relevant point made by the authors is that Jeffreys only considered embedded or nested hypotheses, a fact that allows for having common parameters between models and hence some form of reference prior. Even though (a) I dislike the notion of “common” parameters and (b) I do not think it is entirely legit (I was going to write proper!) from a mathematical viewpoint to use the same (improper) prior on both sides, as discussed in our Statistical Science paper. And in our most recent alternative proposal. The most delicate issue however is to derive a reference prior on the parameter of interest, which is fixed under the null and unknown under the alternative. Hence preventing the use of improper priors. Jeffreys tried to calibrate the corresponding prior by imposing asymptotic consistency under the alternative. And exact indeterminacy under “completely uninformative” data. Unfortunately, this is not a well-defined notion. In the normal example, the authors recall and follow the proposal of Jeffreys to use an improper prior π(σ)∝1/σ on the nuisance parameter and argue in his defence the quote above. I find this argument quite weak because suddenly the prior on σ becomes a weighting function... A notion foreign to the Bayesian cosmology. If we use an improper prior for π(σ), the marginal likelihood on the data is no longer a probability density and I do not buy the argument that one should use the same measure with the same constant both on σ alone [for the nested hypothesis] and on the σ part of (μ,σ) [for the nesting hypothesis]. We are considering two spaces with different dimensions and hence orthogonal measures. This quote thus sounds more like wishful thinking than like a justification. Similarly, the assumption of independence between δ=μ/σ and σ does not make sense for σ-finite measures. Note that the authors later point out that (a) the posterior on σ varies between models despite using the same data [which shows that the parameter σ is far from common to both models!] and (b) the [testing] Cauchy prior on δ is only useful for the testing part and should be replaced with another [estimation] prior when the model has been selected. Which may end up as a backfiring argument about this default choice.

“Each updated weighting function should be interpreted as a posterior in estimating σ within their own context, the model.”

The re-derivation of Jeffreys’ conclusion that a Cauchy prior should be used on δ=μ/σ makes it clear that this choice only proceeds from an imperative of fat tails in the prior, without solving the calibration of the Cauchy scale. (Given the now-available modern computing tools, it would be nice to see the impact of this scale γ on the numerical value of the Bayes factor.) And maybe it also proceeds from a “hidden agenda” to achieve a Bayes factor that solely depends on the t statistic. Although this does not sound like a compelling reason to me, since the t statistic is not sufficient in this setting.

In a differently interesting way, the authors mention the Savage-Dickey ratio (p.16) as a way to represent the Bayes factor for nested models, without necessarily perceiving the mathematical difficulty with this ratio that we pointed out a few years ago. For instance, in the psychology example processed in the paper, the test is between δ=0 and δ≥0; however, if I set π(δ=0)=0 under the alternative prior, which should not matter [from a measure-theoretic perspective where the density is uniquely defined almost everywhere], the Savage-Dickey representation of the Bayes factor returns zero, instead of 9.18!

“In general, the fact that different priors result in different Bayes factors should not come as a surprise.”

The second example detailed in the paper is the test for a zero Gaussian correlation. This is a sort of “ideal case” in that the parameter of interest is between -1 and 1, hence makes the choice of a uniform U(-1,1) easy or easier to argue. Furthermore, the setting is also “ideal” in that the Bayes factor simplifies down into a marginal over the sample correlation only, under the usual Jeffreys priors on means and variances. So we have a second case where the frequentist statistic behind the frequentist test[ing procedure] is also the single (and insufficient) part of the data used in the Bayesian test[ing procedure]. Once again, we are in a setting where Bayesian and frequentist answers are in one-to-one correspondence (at least for a fixed sample size). And where the Bayes factor allows for a closed form through hypergeometric functions. Even in the one-sided case. (This is a result obtained by the authors, not by Jeffreys who, as the proper physicist he was, obtained approximations that are remarkably accurate!)

“The fact that the Bayes factor is independent of the intention with which the data have been collected is of considerable practical importance.”

The authors have a side argument in this section in favour of the Bayes factor against the p-value, namely that the “Bayes factor does not depend on the sampling plan” (p.29), but I find this fairly weak (or tongue in cheek) as the Bayes factor does depend on the sampling distribution imposed on top of the data. It appears that the argument is mostly used to defend sequential testing.

“The Bayes factor (…) balances the tension between parsimony and goodness of fit, (…) against overfitting the data.”

In fine, I liked very much this re-reading of Jeffreys’ approach to testing, maybe the more because I now think we should get away from it! I am not certain it will help in convincing psychologists to adopt Bayes factors for assessing their experiments as it may instead frighten them away. And it does not bring an answer to the vexing issue of the relevance of point null hypotheses. But it constitutes a lucid and innovative of the major advance represented by Jeffreys’ formalisation of Bayesian testing.

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### Section 1-4 : Partial Fractions

In this section we are going to take a look at integrals of rational expressions of polynomials and once again let’s start this section out with an integral that we can already do so we can contrast it with the integrals that we’ll be doing in this section.

\begin{align*}\int{{\frac{{2x - 1}}{{{x^2} - x - 6}}\,dx}} & = \int{{\frac{1}{u}\,du}}\hspace{0.25in}{\mbox{using }}u = {x^2} - x - 6\,\,{\mbox{ and }}du = \left( {2x - 1} \right)dx\\ & = \ln \left| {{x^2} - x - 6} \right| + c\end{align*}

So, if the numerator is the derivative of the denominator (or a constant multiple of the derivative of the denominator) doing this kind of integral is fairly simple. However, often the numerator isn’t the derivative of the denominator (or a constant multiple). For example, consider the following integral.

$\int{{\frac{{3x + 11}}{{{x^2} - x - 6}}\,dx}}$

In this case the numerator is definitely not the derivative of the denominator nor is it a constant multiple of the derivative of the denominator. Therefore, the simple substitution that we used above won’t work. However, if we notice that the integrand can be broken up as follows,

$\frac{{3x + 11}}{{{x^2} - x - 6}} = \frac{4}{{x - 3}} - \frac{1}{{x + 2}}$

then the integral is actually quite simple.

\begin{align*}\int{{\frac{{3x + 11}}{{{x^2} - x - 6}}\,dx}} & = \int{{\frac{4}{{x - 3}}\, - \frac{1}{{x + 2}}dx}}\\ & = 4\ln \left| {x - 3} \right| - \ln \left| {x + 2} \right| + c\end{align*}

This process of taking a rational expression and decomposing it into simpler rational expressions that we can add or subtract to get the original rational expression is called partial fraction decomposition. Many integrals involving rational expressions can be done if we first do partial fractions on the integrand.

So, let’s do a quick review of partial fractions. We’ll start with a rational expression in the form,

$f\left( x \right) = \frac{{P\left( x \right)}}{{Q\left( x \right)}}$

where both $$P\left( x \right)$$ and $$Q\left( x \right)$$ are polynomials and the degree of $$P\left( x \right)$$ is smaller than the degree of $$Q\left( x \right)$$. Recall that the degree of a polynomial is the largest exponent in the polynomial. Partial fractions can only be done if the degree of the numerator is strictly less than the degree of the denominator. That is important to remember.

So, once we’ve determined that partial fractions can be done we factor the denominator as completely as possible. Then for each factor in the denominator we can use the following table to determine the term(s) we pick up in the partial fraction decomposition.

Factor in
denominator
Term in partial
fraction decomposition
$$\displaystyle ax + b$$ $$\displaystyle \frac{A}{{ax + b}}$$
$$\displaystyle {\left( {ax + b} \right)^k}$$ $$\displaystyle \frac{{{A_1}}}{{ax + b}} + \frac{{{A_2}}}{{{{\left( {ax + b} \right)}^2}}} + \cdots + \frac{{{A_k}}}{{{{\left( {ax + b} \right)}^k}}}$$, $$k = 1,2,3, \ldots$$

$$\displaystyle a{x^2} + bx + c$$ $$\displaystyle \frac{{Ax + B}}{{a{x^2} + bx + c}}$$
$$\displaystyle {\left( {a{x^2} + bx + c} \right)^k}$$ $$\displaystyle \frac{{{A_1}x + {B_1}}}{{a{x^2} + bx + c}} + \frac{{{A_2}x + {B_2}}}{{{{\left( {a{x^2} + bx + c} \right)}^2}}} + \cdots + \frac{{{A_k}x + {B_k}}}{{{{\left( {a{x^2} + bx + c} \right)}^k}}}$$, $$k = 1,2,3, \ldots$$

Notice that the first and third cases are really special cases of the second and fourth cases respectively.

There are several methods for determining the coefficients for each term and we will go over each of those in the following examples.

Let’s start the examples by doing the integral above.

Example 1 Evaluate the following integral. $\int{{\frac{{3x + 11}}{{{x^2} - x - 6}}\,dx}}$
Show Solution

The first step is to factor the denominator as much as possible and get the form of the partial fraction decomposition. Doing this gives,

$\frac{{3x + 11}}{{\left( {x - 3} \right)\left( {x + 2} \right)}}\, = \frac{A}{{x - 3}} + \frac{B}{{x + 2}}$

The next step is to actually add the right side back up.

$\frac{{3x + 11}}{{\left( {x - 3} \right)\left( {x + 2} \right)}}\, = \frac{{A\left( {x + 2} \right) + B\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 2} \right)}}$

Now, we need to choose $$A$$ and $$B$$ so that the numerators of these two are equal for every $$x$$. To do this we’ll need to set the numerators equal.

$3x + 11 = A\left( {x + 2} \right) + B\left( {x - 3} \right)$

Note that in most problems we will go straight from the general form of the decomposition to this step and not bother with actually adding the terms back up. The only point to adding the terms is to get the numerator and we can get that without actually writing down the results of the addition.

At this point we have one of two ways to proceed. One way will always work but is often more work. The other, while it won’t always work, is often quicker when it does work. In this case both will work and so we’ll use the quicker way for this example. We’ll take a look at the other method in a later example.

What we’re going to do here is to notice that the numerators must be equal for any x that we would choose to use. In particular the numerators must be equal for $$x = - 2$$ and $$x = 3$$. So, let’s plug these in and see what we get.

\begin{align*}x & = - 2 : & \hspace{0.5in}5 & = A\left( 0 \right) + B\left( { - 5} \right) & \hspace{0.25in} & \Rightarrow & \hspace{0.25in}B & = - 1\\ x & = 3 \,\,\,\,: & \hspace{0.5in}20 & = A\left( 5 \right) + B\left( 0 \right) & \hspace{0.25in} & \Rightarrow & \hspace{0.25in}A & = 4\end{align*}

So, by carefully picking the $$x$$’s we got the unknown constants to quickly drop out. Note that these are the values we claimed they would be above.

At this point there really isn’t a whole lot to do other than the integral.

\begin{align*}\int{{\frac{{3x + 11}}{{{x^2} - x - 6}}\,dx}} & = \int{{\frac{4}{{x - 3}}\, - \frac{1}{{x + 2}}dx}}\\ & = \int{{\frac{4}{{x - 3}}\,dx}} - \int{{\frac{1}{{x + 2}}dx}}\\ & = 4\ln \left| {x - 3} \right| - \ln \left| {x + 2} \right| + c\end{align*}

Recall that to do this integral we first split it up into two integrals and then used the substitutions,

$u = x - 3\hspace{0.5in}v = x + 2$

on the integrals to get the final answer.

Before moving onto the next example a couple of quick notes are in order here. First, many of the integrals in partial fractions problems come down to the type of integral seen above. Make sure that you can do those integrals.

There is also another integral that often shows up in these kinds of problems so we may as well give the formula for it here since we are already on the subject.

$\int{{\frac{1}{{{x^2} + {a^2}}}\,dx}} = \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{x}{a}} \right) + c$

It will be an example or two before we use this so don’t forget about it.

Now, let’s work some more examples.

Example 2 Evaluate the following integral. $\int{{\frac{{{x^2} + 4}}{{3{x^3} + 4{x^2} - 4x}}\,dx}}$
Show Solution

We won’t be putting as much detail into this solution as we did in the previous example. The first thing is to factor the denominator and get the form of the partial fraction decomposition.

$\frac{{{x^2} + 4}}{{x\left( {x + 2} \right)\left( {3x - 2} \right)}}\, = \frac{A}{x} + \frac{B}{{x + 2}} + \frac{C}{{3x - 2}}$

The next step is to set numerators equal. If you need to actually add the right side together to get the numerator for that side then you should do so, however, it will definitely make the problem quicker if you can do the addition in your head to get,

${x^2} + 4 = A\left( {x + 2} \right)\left( {3x - 2} \right) + Bx\left( {3x - 2} \right) + Cx\left( {x + 2} \right)$

As with the previous example it looks like we can just pick a few values of $$x$$ and find the constants so let’s do that.

\begin{align*}x & = 0 \,\,\,\,\, : & \hspace{0.5in}4 & = A\left( 2 \right)\left( { - 2} \right) & \hspace{0.5in} & \Rightarrow & \hspace{0.25in}A & = - 1\\ x & = - 2 : & \hspace{0.5in}8 & = B\left( { - 2} \right)\left( { - 8} \right) & \hspace{0.25in}&\Rightarrow & \hspace{0.25in}B & = \frac{1}{2}\\ x & = \frac{2}{3}\,\, : & \hspace{0.5in}\frac{{40}}{9} & = C\left( {\frac{2}{3}} \right)\left( {\frac{8}{3}} \right) & \hspace{0.25in} & \Rightarrow & \hspace{0.25in}C & = \frac{{40}}{{16}} = \frac{5}{2}\end{align*}

Note that unlike the first example most of the coefficients here are fractions. That is not unusual so don’t get excited about it when it happens.

Now, let’s do the integral.

\begin{align*}\int{{\frac{{{x^2} + 4}}{{3{x^3} + 4{x^2} - 4x}}\,dx}} & = \int{{ - \frac{1}{x} + \frac{{\frac{1}{2}}}{{x + 2}} + \frac{{\frac{5}{2}}}{{3x - 2}}\,dx}}\\ & = - \ln \left| x \right| + \frac{1}{2}\ln \left| {x + 2} \right| + \frac{5}{6}\ln \left| {3x - 2} \right| + c\end{align*}

Again, as noted above, integrals that generate natural logarithms are very common in these problems so make sure you can do them. Also, you were able to correctly do the last integral right? The coefficient of $$\frac{5}{6}$$ is correct. Make sure that you do the substitution required for the term properly.

Example 3 Evaluate the following integral. $\int{{\frac{{{x^2} - 29x + 5}}{{{{\left( {x - 4} \right)}^2}\left( {{x^2} + 3} \right)}}\,dx}}$
Show Solution

This time the denominator is already factored so let’s just jump right to the partial fraction decomposition.

$\frac{{{x^2} - 29x + 5}}{{{{\left( {x - 4} \right)}^2}\left( {{x^2} + 3} \right)}}\, = \frac{A}{{x - 4}} + \frac{B}{{{{\left( {x - 4} \right)}^2}}} + \frac{{Cx + D}}{{{x^2} + 3}}$

Setting numerators gives,

${x^2} - 29x + 5 = A\left( {x - 4} \right)\left( {{x^2} + 3} \right) + B\left( {{x^2} + 3} \right) + \left( {Cx + D} \right){\left( {x - 4} \right)^2}$

In this case we aren’t going to be able to just pick values of $$x$$ that will give us all the constants. Therefore, we will need to work this the second (and often longer) way. The first step is to multiply out the right side and collect all the like terms together. Doing this gives,

${x^2} - 29x + 5 = \left( {A + C} \right){x^3} + \left( { - 4A + B - 8C + D} \right){x^2} + \left( {3A + 16C - 8D} \right)x - 12A + 3B + 16D$

Now we need to choose $$A$$, $$B$$, $$C$$, and $$D$$ so that these two are equal. In other words, we will need to set the coefficients of like powers of $$x$$ equal. This will give a system of equations that can be solved.

\left. \begin{align*}{x^3} & :\hspace{0.25in} & A + C & = 0\\ {x^2} & :\hspace{0.25in} & - 4A + B - 8C + D & = 1\\ {x^1} & :\hspace{0.25in} & 3A + 16C - 8D & = - 29\\ {x^0} & :\hspace{0.25in} & - 12A + 3B + 16D & = 5\end{align*} \right\}\hspace{0.25in} \Rightarrow \hspace{0.25in}A = 1,\,B = - 5,\,C = - 1,\,D = 2

Note that we used $${x^0}$$ to represent the constants. Also note that these systems can often be quite large and have a fair amount of work involved in solving them. The best way to deal with these is to use some form of computer aided solving techniques.

Now, let’s take a look at the integral.

\begin{align*}\int{{\frac{{{x^2} - 29x + 5}}{{{{\left( {x - 4} \right)}^2}\left( {{x^2} + 3} \right)}}\,dx}} & = \int{{\frac{1}{{x - 4}} - \frac{5}{{{{\left( {x - 4} \right)}^2}}} + \frac{{ - x + 2}}{{{x^2} + 3}}\,dx}}\\ & = \int{{\frac{1}{{x - 4}} - \frac{5}{{{{\left( {x - 4} \right)}^2}}} - \frac{x}{{{x^2} + 3}}\, + \frac{2}{{{x^2} + 3}}\,dx}}\\ & = \ln \left| {x - 4} \right| + \frac{5}{{x - 4}} - \frac{1}{2}\ln \left| {{x^2} + 3} \right| + \frac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{x}{{\sqrt 3 }}} \right) + c\end{align*}

In order to take care of the third term we needed to split it up into two separate terms. Once we’ve done this we can do all the integrals in the problem. The first two use the substitution $$u = x - 4$$, the third uses the substitution $$v = {x^2} + 3$$ and the fourth term uses the formula given above for inverse tangents.

Example 4 Evaluate the following integral. $\int{{\frac{{{x^3} + 10{x^2} + 3x + 36}}{{\left( {x - 1} \right){{\left( {{x^2} + 4} \right)}^2}}}\,dx}}$
Show Solution

Let’s first get the general form of the partial fraction decomposition.

$\frac{{{x^3} + 10{x^2} + 3x + 36}}{{\left( {x - 1} \right){{\left( {{x^2} + 4} \right)}^2}}}\, = \frac{A}{{x - 1}} + \frac{{Bx + C}}{{{x^2} + 4}} + \frac{{Dx + E}}{{{{\left( {{x^2} + 4} \right)}^2}}}$

Now, set numerators equal, expand the right side and collect like terms.

\begin{align*}{x^3} + 10{x^2} + 3x + 36 & = A{\left( {{x^2} + 4} \right)^2} + \left( {Bx + C} \right)\left( {x - 1} \right)\left( {{x^2} + 4} \right) + \left( {Dx + E} \right)\left( {x - 1} \right)\\ & = \left( {A + B} \right){x^4} + \left( {C - B} \right){x^3} + \left( {8A + 4B - C + D} \right){x^2} + \\ & \hspace{0.5in}\hspace{0.25in}\left( { - 4B + 4C - D + E} \right)x + 16A - 4C - E\end{align*}

Setting coefficient equal gives the following system.

\left. \begin{align*}{x^4} & :\hspace{0.25in} & A + B & = 0\\ {x^3} & :\hspace{0.25in} & C - B & = 1\\ {x^2} & : \hspace{0.25in} & 8A + 4B - C + D & = 10\\ {x^1} & : \hspace{0.25in} & - 4B + 4C - D + E & = 3\\ {x^0} & :\hspace{0.25in} & 16A - 4C - E & = 36\end{align*} \right\}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,A = 2,\,B = - 2,\,C = - 1,\,D = 1,\,E = 0

Don’t get excited if some of the coefficients end up being zero. It happens on occasion.

Here’s the integral.

\begin{align*}\int{{\frac{{{x^3} + 10{x^2} + 3x + 36}}{{\left( {x - 1} \right){{\left( {{x^2} + 4} \right)}^2}}}\,dx}} & = \int{{\frac{2}{{x - 1}} + \frac{{ - 2x - 1}}{{{x^2} + 4}} + \frac{x}{{{{\left( {{x^2} + 4} \right)}^2}}}\,dx}}\\ & = \int{{\frac{2}{{x - 1}} - \frac{{2x}}{{{x^2} + 4}} - \frac{1}{{{x^2} + 4}} + \frac{x}{{{{\left( {{x^2} + 4} \right)}^2}}}\,dx}}\\ & = 2\ln \left| {x - 1} \right| - \ln \left| {{x^2} + 4} \right| - \frac{1}{2}{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) - \frac{1}{2}\frac{1}{{{x^2} + 4}} + c\end{align*}

To this point we’ve only looked at rational expressions where the degree of the numerator was strictly less that the degree of the denominator. Of course, not all rational expressions will fit into this form and so we need to take a look at a couple of examples where this isn’t the case.

Example 5 Evaluate the following integral. $\int{{\frac{{{x^4} - 5{x^3} + 6{x^2} - 18}}{{{x^3} - 3{x^2}}}\,dx}}$
Show Solution

So, in this case the degree of the numerator is 4 and the degree of the denominator is 3. Therefore, partial fractions can’t be done on this rational expression.

To fix this up we’ll need to do long division on this to get it into a form that we can deal with. Here is the work for that.

\require{enclose} \begin{align*} &\,\,\, x - 2\\ {x^3} - 3{x^2} & \enclose{longdiv}{{x^4} - 5{x^3} + 6{x^2} - 18} \\ - & \underline {\left( {{x^4} - 3{x^3}} \right)\hspace{0.9in}} \\ & \hspace{0.3in} - 2{x^3} + 6{x^2} - 18\\ & \underline { \hspace{0.08in}-\left( { - 2{x^3} + 6{x^2}} \right) \hspace{0.4in}} \\ & \hspace{1.35in} - 18\end{align*}

So, from the long division we see that,

$\frac{{{x^4} - 5{x^3} + 6{x^2} - 18}}{{{x^3} - 3{x^2}}}\, = x - 2 - \frac{{18}}{{{x^3} - 3{x^2}}}$

and the integral becomes,

\begin{align*}\int{{\frac{{{x^4} - 5{x^3} + 6{x^2} - 18}}{{{x^3} - 3{x^2}}}\,dx}} & = \int{{x - 2 - \frac{{18}}{{{x^3} - 3{x^2}}}\,dx}}\\ & = \int{{x - 2\,dx}} - \int{{\frac{{18}}{{{x^3} - 3{x^2}}}\,dx}}\end{align*}

The first integral we can do easily enough and the second integral is now in a form that allows us to do partial fractions. So, let’s get the general form of the partial fractions for the second integrand.

$\frac{{18}}{{{x^2}\left( {x - 3} \right)}} = \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x - 3}}$

Setting numerators equal gives us,

$18 = Ax\left( {x - 3} \right) + B\left( {x - 3} \right) + C{x^2}$

Now, there is a variation of the method we used in the first couple of examples that will work here. There are a couple of values of $$x$$ that will allow us to quickly get two of the three constants, but there is no value of $$x$$ that will just hand us the third.

What we’ll do in this example is pick $$x$$’s to get the two constants that we can easily get and then we’ll just pick another value of $$x$$ that will be easy to work with (i.e. it won’t give large/messy numbers anywhere) and then we’ll use the fact that we also know the other two constants to find the third.

\begin{align*}x & = 0 : & \hspace{0.25in} 18 & = B\left( { - 3} \right) & \hspace{0.15in}\Rightarrow \hspace{0.25in}B & = - 6\\ x & = 3 : & \hspace{0.25in} 18 & = C\left( 9 \right) & \hspace{0.15in} \Rightarrow \hspace{0.25in}C & = 2\\ x & = 1 : & 18 & = A\left( { - 2} \right) + B\left( { - 2} \right) + C = - 2A + 14 & \hspace{0.15in} \Rightarrow \hspace{0.25in}A & = - 2\end{align*}

The integral is then,

\begin{align*}\int{{\frac{{{x^4} - 5{x^3} + 6{x^2} - 18}}{{{x^3} - 3{x^2}}}\,dx}} & = \int{{x - 2\,dx}} - \int{{ - \frac{2}{x} - \frac{6}{{{x^2}}} + \frac{2}{{x - 3}}\,dx}}\\ & = \frac{1}{2}{x^2} - 2x + 2\ln \left| x \right| - \frac{6}{x} - 2\ln \left| {x - 3} \right| + c\end{align*}

In the previous example there were actually two different ways of dealing with the $${x^2}$$ in the denominator. One is to treat it as a quadratic which would give the following term in the decomposition

$\frac{{Ax + B}}{{{x^2}}}$

and the other is to treat it as a linear term in the following way,

${x^2} = {\left( {x - 0} \right)^2}$

which gives the following two terms in the decomposition,

$\frac{A}{x} + \frac{B}{{{x^2}}}$

We used the second way of thinking about it in our example. Notice however that the two will give identical partial fraction decompositions. So, why talk about this? Simple. This will work for $${x^2}$$, but what about $${x^3}$$ or $${x^4}$$? In these cases, we really will need to use the second way of thinking about these kinds of terms.

${x^3}\,\,\, \Rightarrow \,\,\,\,\,\frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{{x^3}}}\hspace{0.5in}{x^4}\,\,\, \Rightarrow \,\,\,\,\,\frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{{x^3}}} + \frac{D}{{{x^4}}}$

Let’s take a look at one more example.

Example 6 Evaluate the following integral. $\int{{\frac{{{x^2}}}{{{x^2} - 1}}\,dx}}$
Show Solution

In this case the numerator and denominator have the same degree. As with the last example we’ll need to do long division to get this into the correct form. We’ll leave the details of that to you to check.

$\int{{\frac{{{x^2}}}{{{x^2} - 1}}\,dx}} = \int{{1 + \frac{1}{{{x^2} - 1}}\,dx}} = \int{{dx}} + \int{{\frac{1}{{{x^2} - 1}}\,dx}}$

So, we’ll need to partial fraction the second integral. Here’s the decomposition.

$\frac{1}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{A}{{x - 1}} + \frac{B}{{x + 1}}$

Setting numerator equal gives,

$1 = A\left( {x + 1} \right) + B\left( {x - 1} \right)$

Picking value of $$x$$ gives us the following coefficients.

\begin{align*}x & = - 1 : & \hspace{0.25in} 1 & = B\left( { - 2} \right) & \hspace{0.25in} \Rightarrow \hspace{0.5in}B & = - \frac{1}{2}\\ x & = 1 \,\,\,\, : & \hspace{0.25in}1 & = A\left( 2 \right) & \hspace{0.25in} \Rightarrow \hspace{0.5in}A & = \frac{1}{2}\end{align*}

The integral is then,

\begin{align*}\int{{\frac{{{x^2}}}{{{x^2} - 1}}\,dx}} & = \int{{dx}} + \int{{\frac{{\frac{1}{2}}}{{x - 1}} - \frac{{\frac{1}{2}}}{{x + 1}}\,dx}}\\ & = x + \frac{1}{2}\ln \left| {x - 1} \right| - \frac{1}{2}\ln \left| {x + 1} \right| + c\end{align*}


text:
## If A Picture Paints A Thousand Words

And so begins the words of the a song by the Bread. The complexity we see in the world is a product of the interactions of simple building blocks. Everything we see is just a combination of a few elements in the periodic table. In turn every element is a combination of electrons, protons and neutrons.

A picture paints a thousands words but what paints a thousand pictures? Just like a Lego truck that is composed of a few Lego shapes, a “black and white” picture is also composed of a few levels of gray. The levels is just a set of 256 numbers starting from 0 up to 255. The number 0 represents black and the number 255 represents white. The numbers in between represent colors between extreme black and extreme white. The diagram below shows the levels of gray in a 16 by 16 matrix. The gradient goes from left to right, top to bottom.

Using just these 256 numbers, we can already draw interesting figures. Take for example this 8 by 8 matrix of numbers:

197   4  46  18 155 186 144 190
111  30   6 131 235  99 112 227
10   26 141 147 120  28  92 187
241 100 118 149 178   5 158 213
105  43 104  63 172 222  64 174
130 194 243  56  59 166 134 133
247 145  36  82 238 251  53 129
171  29  51 163 199 214 113 195

When you plot this on an 8×8 grid, you get the following pattern:

Not all pictures need to be boring. Here’s an example of a picture that is not that boring:

17 23 26 20 13 20 20 22 21 17 17 25 20 11 15 15 9 10 18 13 14 12 13 39 61 63 71 72 69 71 74 72 74 75 72 76 75 75 75 80 77 78 79 79 80 80 80 80 81 81 82 82 83 83 83 83 84 84 84 84 85 86 86 86 87 87 87 88 88 88 89 89 89 89 90 90 90 90 90 91 90 90 91 91 92 92 92 92 94 94 95 95 95 95 95 95 96 96 96 96 97 97 97 97 98 98 98 99 99 99 100 100 101 101 101 101 102 102 103 103 104 104 104 105 105 105 105 105 106 106 106 107 107 108 108 108 110 110 110 110 111 111 112 112 113 113 113 114 114 115 115 115 116 116 117 117 117 118 118 119 119 119 120 120 121 121 122 122 123 124 124 125 125 125 126 127 127 127 128 128 129 129 129 129 131 131 132 132 133 134 ...

That is just the start of a 310×204 matrix saved in the file “clifton.txt”. If you plot that using the R command

clifton = read.table("clifton.txt")

#convert this data frame to a vector
vclifton = as.numeric(clifton)

#create the 310x204 matrix
mclifton = matrix(vclifton,nrow=310,ncol=204)

#plot the image
image(mclifton[,ncol(mclifton):1],col=gray((0:255)/255))

you’ll get the following image

Histogram

If you look at the file “clifton.txt” and count the occurrence of each number, you will see that there are 9 occurrences of zero and 4 occurrences of 240. If you do this for all other numbers, you will get the following counts:

Level Frequency
0     9
1     16
2     31
3     34
4     49
:
:
237   115
238   113
239   18
240   4
:
:
254   0
255   0

If you plot this data as a bar chart, you will get what is known as the histogram of the picture:

The histogram above was obtained using the following R command:

hist(vclifton,breaks=256)

The histogram tells us the how much of each gray level contribute to the actual image.

Here’s another example of a picture that is relatively dark. In photography, the image below is called a “Low Key” image because it is mostly dark.

Low Key images tend to have the bulk of the colors at the right side of the scale. The next image is a “High Key” image because it is mostly white:

Histograms of High Key images tend to have the bulk of the colors at the right side of the scale. The histogram of the “clifton.txt” image tends to have a much better balance in that it contains most of the colors at the middle of the range.

It’s surprising how we can come up with complex images using just 256 levels of gray. We haven’t even considered colors yet! We also learned how histograms can be constructed by counting the frequency of each level and what histograms of High Key, Low Key and Optimum exposure look like.

## A Sound Correction

As I was taking my coffee over breakfast today, I realized that in the previous post, I did not incorporate the time it took for the sound waves to reach my ear when I pressed the stop button of the stopwatch. Sound waves travel at 340.29 meters per second. At a height of 25 meters, it would take 0.073 seconds for the sound of the explosion to reach the ground. This should be negligible for the purposes of our computation. However, what is really the effect if we incorporate this minor correction?

As seen in the figure below, the recorded time is composed of the time it took for the skyrocket to reach the maximum height plus the time it took for the sound waves to reach my ear, that is,

$t_r = t_a + t_s$

where $t_r$ is the recorded time, $t_a$ is the actual time the skyrocket reached its maximum height and $t_s$ is the time it took for the sound waves to travel from the maximum height to the ground.

We can rewrite the height as

$\displaystyle h_{\text{max}} = - \frac{gt_a^2}{2} + v_i t_a$

Since $v_i = gt_a$,

$\displaystyle h_{\text{max}} = - \frac{gt_a^2}{2} + gt_a^2$
$\displaystyle = \frac{gt_a^2}{2}$

At the maximum height, $h_{\text{max}}$, it would take the sound waves $t_s$ seconds to travel, that is,

$\displaystyle h_{\text{max}} = v_s t_s$

where $v_s$ is the speed of sound. Therefore,

$\displaystyle t_sv_s = \frac{gt_a^2}{2}$
$\displaystyle (t_r- t_a)v_s = \frac{gt_a^2}{2}$
$\displaystyle v_st_r - v_st_a = \frac{gt_a^2}{2}$
$\displaystyle \frac{gt_a^2}{2} + v_st_a - v_st_r = 0$

The last equation above is a quadratic equation which we can solve for $t_a$ using the quadratic formula:

$\displaystyle t_a = \frac{-v_s \pm \sqrt{v_s^2 - 4(1/2)gv_st_r}}{2(1/2)g}$

Substituting the values $g = 9.8$, $v_s=340.29$, and $t_r = 2.3$, we get

$\displaystyle t_a = 2.22$

Using this result, the maximum height is computed to be

$\displaystyle h_{\text{max}} = \frac{gt_a^2}{2} = 24.33$

Comparing this with our previous computation of $h_{\text{max}} = 25.829$, we find that we have overestimated the maximum height by about 1.49 meters. It’s not really that bad so we can just drop the effect of sound.

## New Year, New Heights and the Projectile

I spent my new year celebration at our friend’s place with 2 other colleagues. It was started with a dinner at about 9:30 pm and we bought some bottles of wine in addition to the spirits we already brought. At about 10:30, a street party was setup in the vicinity and about 5 other families joined in a potluck. Marijo’s brother started the fireworks at about that time with David and Gerard (my colleagues) taking turns setting them off. I did not attempt to set them off because I did a lot those things way back as a child and still lucky enough to have my fingers intact.

This is a video of our 2012 New Year’s Celebration:

As Gerard was flying off skyrockets (or kwitis as they are called in my language) David asked me “How high do you think these things go up?”. That was a really good question which got me to thinking. I said “Let’s compute!” So I took my iphone and launched its built-in stopwatch. Gerard flew about 3 skyrockets and I took the time it takes for them to fly until they blow up in the air. The average was found to be 2.3 seconds.

Independent of Mass

Legend has it that Galileo Galilei performed an experiment at the Leaning Tower of Pisa. He dropped objects of different masses to see which of them falls first. Starting from the heaviest to the lightest object, what he found was that, in a vacuum, they reach the ground at the same length of time when dropped from the same height. Using this result, we can compute for the distance travelled by any object under the influence of gravity. At heights near the earth’s surface, the acceleration of objects in the presence of gravity alone is given by

$\displaystyle \frac{d^2h}{dt^2}= \frac{dv}{dt} = -g$

where $h$ is the vertical distance (height), $v$ is the velocity and $g$ is the acceleration due to gravity given by $g = 9.8 \text{m/s}^2$.

The equation above is a simple differential equation which we can solve for the velocity at any time $t$ by integrating both sides:

$\displaystyle \int^{v}_{v_i} dv = \int^{t}_{t_i} -g dt$

which by the Fundamental Theorem of Calculus gives us

$\displaystyle v - v_i = -g(t - t_i)$
$\displaystyle v = -gt + v_i + t_i$

If we take initial time of the rocket launch to be zero, that is, $t_i=0$, we have

$\displaystyle v = \frac{dh}{dt}= -gt + v_i$

We don’t know the initial velocity $v_i$ of the rocket. However, at the maximum height attained by the rocket, the velocity is zero. Using this fact, we can solve for the initial velocity:

$\displaystyle 0 = -gt_{\text{max}} + v_i$
$\displaystyle v_i = gt_{\text{max}}$

where $t_{\text{max}}$ is the time the rocket reached maximum height.

At $t_{\text{max}}=2.3$ seconds, the initial velocity is therefore $v_i = 9.8 * 2.3 = 22.5$ m/s.

Maximum height

We can solve for the height as a function of time by integrating both sides of the velocity equation:

$\displaystyle \int^{h}_{h_i} dh = \int^{t}_{t_i} -gt dt + v_i dt$
$\displaystyle h - h_i = -\frac{gt^2}{2} + v_i t \big|^t_{t_i}$

Since the initial height is zero , that is , $h_i = 0$ and the initial time is zero $t_i = 0$, we have

$\displaystyle h = -\frac{gt^2}{2} + v_i t$

Since at t=2.3 seconds, the rocket reaches the maximum height, we have

$\displaystyle h_{\text{max}} = -\frac{9.8 \times (2.3)^2}{2} + 22.5 * 2.3 = 25.829 \text{ meters}$

Does this make sense? The average height of a building storey is about 3 meters. This is about 8 or 9 storeys high which I think makes sense.

It’s good that my new year not only started with a warm welcome from friends but it also appealed to my physics curiosity.

Thank you to Marijo Condes and family for letting us spend a memorable New Year’s Eve.

## Mathematically Romancing Mt Mayon

The name Mayon is short for Daragang Magayon, or “Beautiful Lady”, which is the legend associated with Mt Mayon. No wonder, I still find myself captivated by its charms after visiting it three weeks ago.

While playing with google maps this sunday, I decided to take a look at Mt Mayon and I was hoping I could get a three dimensional view of the beauty. I was disappointed. Although there is a 3D view, but it was not beautiful at all. Luckily, google map provides a topographical information of Mt Mayon. Suddenly, I was inspired to recreate the view of Mt Mayon using that information and using the tool called R.

First thing I did was to capture the image from the google maps and saved it as JPEG. Using the ReadImages library of R, I loaded the image and plotted it.

# load the library

# read the file and plot
plot(x)

grid(nx=20,ny=20,col="red")

The result of the code above is a plot of the image with red grid lines, as shown below.

You will notice that the image contains contour lines with a height information associated with it. I used the grid lines to manually interpolate the height of Mt Mayon at each intersection. I started at the bottom line and worked myself upward from left to right. However, I got lazy interpolating from point to point so I just stopped when I reached the middle line. I decided that the upper half of the image is just a mirror image of the bottom so I might as well use the same numbers. Here’s a dump of the interpolated height from the bottom half of the image:

0,        0,    0,     0,     0,    0,400,400,400,450,450,420,400,350,300,280,250,220,200,
0,     100,200,300,400,420,450,500,550,570,550,500,480,420,400,380,350,250,200,
200,300,350,400,420,500,550,580,600,650,600,600,580,550,500,400,350,300,320,
280,350,380,400,500,580,610,650,780,800,750,730,650,600,500,450,380,310,300,
300,400,430,500,580,650,780,800,950,990,970,900,800,700,580,500,520,350,300,
310,400,480,580,610,750,840,1000,1100,1200,1150,1100,1000,800,650,550,450,380,350,
320,450,500,600,700,800,1000,1200,1350,1500,1500,1380,1180,900,800,650,510,400,350,
350,450,500,600,750,900,1100,1350,1600,1800,1800,1600,1350,1100,900,700,550,450,350,
370,450,500,600,780,900,1200,1500,1800,2100,2000,1800,1500,1180,900,700,550,500,350,
380,450,500,620,780,970,1200,1500,1900,2250,2200,1800,1400,1150,900,700,550,520,400,

Reflecting this information about the middle line and inputting this to R, we get

z=c(0,        0,    0,     0,     0,    0,400,400,400,450,450,420,400,350,300,280,250,220,200,
0,     100,200,300,400,420,450,500,550,570,550,500,480,420,400,380,350,250,200,
200,300,350,400,420,500,550,580,600,650,600,600,580,550,500,400,350,300,320,
280,350,380,400,500,580,610,650,780,800,750,730,650,600,500,450,380,310,300,
300,400,430,500,580,650,780,800,950,990,970,900,800,700,580,500,520,350,300,
310,400,480,580,610,750,840,1000,1100,1200,1150,1100,1000,800,650,550,450,380,350,
320,450,500,600,700,800,1000,1200,1350,1500,1500,1380,1180,900,800,650,510,400,350,
350,450,500,600,750,900,1100,1350,1600,1800,1800,1600,1350,1100,900,700,550,450,350,
370,450,500,600,780,900,1200,1500,1800,2100,2000,1800,1500,1180,900,700,550,500,350,
380,450,500,620,780,970,1200,1500,1900,2250,2200,1800,1400,1150,900,700,550,520,400,
370,450,500,600,780,900,1200,1500,1800,2100,2000,1800,1500,1180,900,700,550,500,350,
350,450,500,600,750,900,1100,1350,1600,1800,1800,1600,1350,1100,900,700,550,450,350,
320,450,500,600,700,800,1000,1200,1350,1500,1500,1380,1180,900,800,650,510,400,350,
310,400,480,580,610,750,840,1000,1100,1200,1150,1100,1000,800,650,550,450,380,350,
300,400,430,500,580,650,780,800,950,990,970,900,800,700,580,500,520,350,300,
280,350,380,400,500,580,610,650,780,800,750,730,650,600,500,450,380,310,300,
200,300,350,400,420,500,550,580,600,650,600,600,580,550,500,400,350,300,320,
0,     100,200,300,400,420,450,500,550,570,550,500,480,420,400,380,350,250,200,
0,        0,    0,     0,     0,    0,400,400,400,450,450,420,400,350,300,280,250,220,200)

Next, I generated the coordinates of the x and y plane:

myx=seq(-9,9,1)
myy=seq(-9,9,1)

To generate the 3D view, I used the persp3d function of the rgl library and animated the 3D plot so that we can see the plot 360 degrees.

library(rgl)
persp3d(myx,myy,z,col="darkseagreen", aspect=c(4,4,1))
play3d(spin3d(axis=c(0,0,1), rpm=8), duration=20)

The result of the above code is an animated plot of Mt Mayon using partial topographical information. Shown below is the animation:

This is indeed a great waste of time. But I guess, that’s what you do when you fall in love…mathematically!

## Chances Are

I just came from a great weekend escapade, with my wife and two colleagues, in one of the beautiful regions of the Philippines. Bicol is home to Mayon Volcano, one of the seven perfect coned volcanoes on Earth. The scenery is so green and the air so refreshing. It was one of the best trips I ever made in the Philippines.

The trip was also full of coincidences. We met a colleague in one of the restaurants in Naga City where we had our dinner. In the same restaurant we also met one of our contacts in Ateneo de Naga. On the way home, we met another colleague at the airport. But the first coincidence happened on our way to Legazpi City where one of my companions (i’ll hide the name to protect the innocent) met a good friend in the airplane seated next to him. It was a pleasant surprise for my companion and he asked “What are the odds of that happening”? That’s a good question which got me thinking later that night.

To simplify the problem, let us not compute the odds of my companion and his friend to be in the same plane going to Bicol. Rather, let us compute the odds of him being seated next to his friend. The plane was an airbus A320 and about 180 seats. The seats are laid out in such a way that there is a middle isle and each row is composed of three seats. Let us draw a row of seats and assume we have persons A and B. How many ways can they be seated in that row in such a way that they sit next to each other? Below is a diagram on the ways two persons can be seated next to each other.

As you can see, there are four ways for them to be seated next to each other for any given row. In the first arrangement, we have A, B seated in that order. The third seat can be occupied by anyone. Having chosen the seats for A and B, how many choices do we have for the third person? Since there are 180 seats minus two, the third person has 178 other seats to choose from. Once person number 3 chooses one seat, the fourth person can then choose from the remaining 177 seats. Doing this for all other passengers, you will come up with 178! ways of assigning seating arrangements for the 180 passengers such that persons A and B are seated next to each other. Now, that is just for the case where A and B are in the first two seats. For the other three cases, you will see that it also takes 178! ways for each case. Since there are 4 cases per row and 60 rows, we have

$\displaystyle 4\cdot 60\cdot 178!$

ways to have two persons sit next to each other on a 180 seater plane. Since the total number of ways to assign passengers to 180 seats is 180!, then the probability of having two persons seat next to each other is

$\displaystyle \frac{4\cdot 60\cdot 178!}{180!}$
$\displaystyle = \frac{4\cdot 60\cdot 178!}{180\cdot 179\cdot 178!}$
$\displaystyle = \frac{4\cdot 60}{180\cdot 179}$
$\displaystyle = 0.00744879$

This is quite a low probability and it assumes that the seating arrangement is totally random. By totally random, I mean that my wife and I could be seated far apart from each other. In reality, this is not the case since many passengers come in groups and they prefer to be seated next to each other. If we take this into account, then the probability of my companion and his friend to be seated next to each other increases.

## Reflecting on Convexity

Last Friday, I was told of a puzzle by my colleague that was given to him on an interview. The puzzle goes like this: You and an opponent sit opposite a round table. You are to place coins (of the same size) on the table, one at a time. No coin should sit on top of another coin, even partially. No part of any coin should go beyond the table’s boundaries. You take turns putting coins on the table until you can no longer find any space to fit a coin. The last person to put a coin wins. If you had the first turn, what would be your strategy? This is a puzzle i’m not be able to answer on an interview and over 2 bottles of beer. So I told my colleague i’ll work on it on the weekend.

My first impulse was to treat this as a packing of circles inside a bigger circle. To start, let us have the circle small enough to fit only a 1 layer of circles along the circumference of the middle circle, as shown below.

In this configuration, the circles in dark blue are mine while the lighter ones are of the opponents. In this game, I win because I was the last one to put the coin. However, since my opponent is also rational, he will not allow me to win. He will put his lighter colored coin in a position shown below. This game started with me putting my coin in the middle. He then put his coin on top of the middle coin. Next I put my coin next to his coin, touching each other. But my opponent now puts his second coin in such a way that it has a space between my 2nd coin and his 2nd coin for which no coin can fit. Then I put my 3rd coin next to his 2nd coin. When he put his third coin, there was no more space for me to insert another coin. So I lost.

There must be a way for me to influence his next move. Looking at the geometry of circles, there must be something that can be made use of the circular symmetry. For one thing, all points on a circle have the same distance from the center. Another thing is if you pass a line through the center it will intersect the circle in 2 points. These points are antipodes of each other. This gives a very good way to control the outcome of the game. We let our opponent select a spot to put the coin, then we get the antipod of that location and put our coin there. We do this until no more coins can be put on the table.

Let us walkthrough the sequence of steps as shown above. After I put my coin in the middle, my opponent puts his coin above it. Then what I do is to put my 2nd coin on the opposite side (antipode). My opponent then employs his usual strategy which is to put his next coin so that it will have a small space in between. However, i put my 3rd coin opposite his 2nd coin. When he puts his 3rd coin, he will realize that there is no more space for him to squeeze it in, as shown by the red coin which already goes beyond the bounder of the table. So I win!

The good news is that this is not a coincidence. We will always be the last person to put a coin on the table. To see this, assume the contrary: our opponent was able to put a coin $C_o$ on the table but for which the antipode has no more space to accommodate the coin. This means that if we put our coin, it will intersect at least one coin $C\prime$. Let $p\prime$ be a point inside this intersection. By the way this was constructed, there is another coin $C$ which contains the antipode $p$ of point $p\prime$. This means that $p \in C_o \cap C$, that is, $C_o$ intersects $C$, which is a contradiction.

Convex Regions and Point Reflections

Will this work only with circular tables? Will it work also with rectangular tables? Will it work only with coins? Will any other shape also work? Consider a table of some arbitrary shape but with the property that for any two points $p$ and $q$ on this table, we can draw a line connecting them in such a way that the line is contained with in the table. We call this region a convex region. The property of a convex region is that the centroid of this region is contained within it. Shown below is an example of a convex region and a non-convex region.

A table shaped like a horseshoe is not convex and its centroid is outside of the table.

Suppose we have found the centroid of our convex region. Instead of using a coin, let’s use a star to put into the table. If our opponent puts his star anywhere on the table at a position $r_1$ from the centroid, then the antipode of this position must also be inside the region, by virtue of its being convex. Let us put our star in such a way that it is the reflection through the centroid of our opponent’s star. If we set up a coordinate system on the table centered at the centroid (surprise!), then point reflection through the origin transforms in the following manner:

$\displaystyle (x_1,y_1) \mapsto (-x_1, -y_1)$

Since this mapping is isometric, i.e, preserves distances, any two points $p_1$ and $p_2$ will retain the same distance after reflection. This means that if two stars don’t intersect, they will not suddenly intersect under this mapping.

The figure below shows a star $S_1$ defined by the following coordinates.

           x         y
1  0.8754257 0.4100271
2  0.8292804 0.4380167
3  0.8248856 0.4700049
4  0.7941220 0.4360175
5  0.7501740 0.4400160
6  0.7809376 0.3980315
7  0.7567662 0.3620448
8  0.8029116 0.3680426
9  0.8226882 0.3380536
10 0.8358726 0.3780389
11 0.8732283 0.4080278

This star is the one shown in blue in the first quadrant. Another star $S_2$ was defined by translating $S_1$ by 0.1 in both directions. This is the star shown in red in the first quadrant.

Applying the point reflection through the origin, we get the two stars shown in the third quadrant. The blue colored star is the reflection of the blue colored star in the first quadrant. The same thing applies to the red star. What we have just shown is that you can modify the problem to tables like squares or polygons by virtue of being convex regions. Also, by the isometry of the point reflection through the origin, we can use other shapes and not just circles.


text:
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text:
# American Institute of Mathematical Sciences

June  2015, 2(2): 193-225. doi: 10.3934/jcd.2015003

## Variational integrators for mechanical control systems with symmetries

 1 Department of Mathematics, University of Michigan, 530 Church Street, Ann Arbor, MI 48109 2 Zentrum Mathematik der Technische Universität München, D-85747 Garching bei Munchen 3 Instituto de Ciencias Matemáticas (CSIC-UAM-UC3M-UCM), Campus de Cantoblanco, UAM C/ Nicolas Cabrera, 15 - 28049 Madrid

Received  June 2013 Revised  May 2015 Published  May 2016

Optimal control problems for underactuated mechanical systems can be seen as a higher-order variational problem subject to higher-order constraints (that is, when the Lagrangian function and the constraints depend on higher-order derivatives such as the acceleration, jerk or jounces). In this paper we discuss the variational formalism for the class of underactuated mechanical control systems when the configuration space is a trivial principal bundle and the construction of variational integrators for such mechanical control systems.
An interesting family of geometric integrators can be defined using discretizations of the Hamilton's principle of critical action. This family of geometric integrators is called variational integrators, being one of their main properties the preservation of geometric features as the symplecticity, momentum preservation and good behavior of the energy. We construct variational integrators for higher-order mechanical systems on trivial principal bundles and their extension for higher-order constrained systems, paying particular attention to the case of underactuated mechanical systems.
Citation: Leonardo Colombo, Fernando Jiménez, David Martín de Diego. Variational integrators for mechanical control systems with symmetries. Journal of Computational Dynamics, 2015, 2 (2) : 193-225. doi: 10.3934/jcd.2015003
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show all references

##### References:
 [1] J. Baillieul, The geometry of controlled mechanical systems, Mathematical control theory, Springer, New York, (1999), 322-354.  Google Scholar [2] R. Benito, M. de León and D. Martín de Diego, Higher-order discrete lagrangian mechanics, Int. Journal of Geometric Methods in Modern Physics, 3 (2006), 421-436. doi: 10.1142/S0219887806001235.  Google Scholar [3] A. Bloch, Nonholonomic Mechanics and Control, Interdisciplinary Applied Mathematics Series, Springer-Verlag, New-York, 24, 2003. doi: 10.1007/b97376.  Google Scholar [4] A. M. Bloch and P. E. Crouch, Reduction of Euler Lagrange problems for constrained variational problems and relation with optimal control problems, Proceedings of 33rd IEEE Conference on Decision and Control, 3 (1994), 2584-2590. doi: 10.1109/CDC.1994.411534.  Google Scholar [5] A. M. Bloch and P. E. 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