Informal Derivation of Euler’s Identity

“Who would have thought that \pi which enters as the ratio of circumference to diameter, e, as the natural base for logarithms, i, as the fundamental imaginary unit and 0 and 1 (which we know all about from infancy) would all be tied together in any way, not to mention such a simple and compact way? I hope I never stumble into anything like this formula, for nothing I do after that in life would have any significance.” – Physicist Ramamurti Shankar 1


(1)   \begin{equation*} e^{i \pi} + 1 = 0 \end{equation*}

The Nobel Prize winning physicist Richard Feynman called equation (1), known as Euler’s identity, “one of the most remarkable, almost astounding, formulas in all of mathematics” 2.

Euler’s identity can be derived easily from Taylor series…

(2)   \begin{equation*} f\left( x \right) = \sum\limits_{n = 0}^\infty {\frac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}{x^n}} \end{equation*}

…as well as defining e^{ax} as

(3)   \begin{equation*} {e^{ax}} = \sum\limits_{n = 0}^\infty {\frac{{{{\left( {ax} \right)}^n}}}{{n!}}} \end{equation*}

In equation (2), f^{(n)} represents the n-th derivative of the function f, with the zeroth derivative of f being equal to f.

The Taylor series of \cos{x} can be easily derived from equation (2).

    \[{f^0}\left( x \right) = f\left( x \right) = \cos x\]

    \[{f^0}\left( 0 \right) = f\left( x \right) = \cos 0 = 1\]

    \[{f^1}\left( x \right) = \frac{{df\left( x \right)}}{{dx}} = \frac{d}{{dx}}\cos x = - \sin x\]

    \[{f^1}\left( 0 \right) = - \sin 0 = 0\]

    \[{f^2}\left( x \right) = \frac{{{d^2}f\left( x \right)}}{{dx}} = - \frac{d}{{dx}}\sin x = - \cos x\]

    \[{f^2}\left( 0 \right) = - \cos 0 = - 1\]

    \[{f^3}\left( x \right) = - \frac{d}{{dx}}\cos x = \sin x\]

    \[{f^3}\left( 0 \right) = \sin 0 = 0\]

    \[{f^4}\left( x \right) = \frac{d}{{dx}}\sin x = \cos x\]

    \[{f^4}\left( 0 \right) = \cos 0 = 1\]

From the equations above, as well as equation (2), it follows that:

(4)   \begin{equation*} \cos x = 1 - \frac{{{x^2}}}{{2!}} + \frac{{{x^4}}}{{4!}} - ...\sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{\left( {2n} \right)!}}{x^{2n}}} \end{equation*}

The Taylor series of \sin{x} can also be derived easily from equation (2).

    \[{f^0}(x) = f\left( x \right) = \sin x\]

    \[{f^0}\left( 0 \right) = \sin 0 = 0\]

    \[{f^1}\left( x \right) = \frac{d}{{dx}}\sin x = \cos x\]

    \[{f^1}\left( 0 \right) = \cos 0 = 1\]

    \[{f^2}\left( x \right) = \frac{d}{{dx}}\cos x = - \sin x\]

    \[{f^2}\left( 0 \right) = - \sin 0 = 0\]

    \[{f^3}\left( x \right) = - \frac{d}{{dx}}\sin x = - \cos x\]

    \[f^3\left( 0 \right) = - \cos 0 = - 1\]

    \[{f^4}\left( x \right) = - \frac{d}{{dx}}\cos x = \sin x\]

    \[{f^5}\left( 0 \right) = \sin 0 = 0\]

    \[{f^5}\left( x \right) = \frac{d}{{dx}}\sin x = \cos x\]

    \[{f^5}\left( 0 \right) = \cos 0 = 1\]

From the equations above and equation (2), it follows that:

(5)   \begin{equation*} \sin x = x - \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} - \cdots = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{\left( {2n + 1} \right)!}}{x^{2n + 1}}} \end{equation*}

From equations (3), (4), and (5), if \theta is a real number:

    \[{e^{i\theta }} = \sum\limits_{n = 0}^\infty {\frac{{{{\left( {i\theta } \right)}^n}}}{{n!}}} = 1 + i\theta + \underbrace {{i^2}}_{ - 1}\frac{{{\theta ^2}}}{{2!}} + \underbrace {{i^3}}_{ - i}\frac{{{\theta ^3}}}{{3!}} + \underbrace {{i^4}}_1\frac{{{\theta ^4}}}{{4!}} + \underbrace {{i^5}}_i\frac{{{\theta ^5}}}{!} - ...\]

    \[{e^{i\theta }} = 1 + i\theta - \frac{{{\theta ^2}}}{{2!}} - \frac{{i{\theta ^3}}}{{3!}} + \frac{{{\theta ^4}}}{{4!}} + \frac{{i{\theta ^5}}}{{5!}}\]

(6)   \begin{equation*} {e^{i\theta }} = \underbrace {\left( {1 - \frac{{{\theta ^2}}}{{2!}} + \frac{{{\theta ^4}}}{{4!}} - ...} \right)}_{\cos \theta } + i\underbrace {\left( {\theta - \frac{{{\theta ^3}}}{{3!}} + \frac{{{\theta ^5}}}{{5!}} - ...} \right)}_{\sin \theta } = \cos \theta + i\sin \theta \end{equation*}

Finally, let \theta = \pi. Equation (6) becomes:

    \[{e^{i\pi }} = \cos \pi + i\sin \pi = - 1\]


    \[{e^{i\pi }} + 1 = 0\]


  1. “Basic Training in Mathematics: a Fitness Program for Science Students.” Basic Training in Mathematics: a Fitness Program for Science Students, by Ramamurti Shankar, Springer, 2006, p. 95.
  2. Feynman, Richard. “Algebra.” The Feynman Lectures on Physics,

Radioactive Decay And Isochron Dating of Moon Rocks

The rate of decay of a sample of a radioactive isotope is proportional to the number, N, of particles of that isotope. Mathematically, this can be represented as

(1)   \begin{equation*} \frac{{dN\left( t \right)}}{{dt}} = - \lambda N\left( t \right) \end{equation*}

Equation (1) is a simple first order differential equation. It can be solved as follows. First, divide both sides of (1) by N(t):

    \[\frac{1}{{N\left( t \right)}}\frac{{dN\left( t \right)}}{{dt}} = - \lambda\]


(2)   \begin{equation*} \frac{d}{{dt}}\left[ {\ln \left( {N\left( t \right)} \right)} \right] = \frac{d}{{dN}}\left[ {\ln \left( {N\left( t \right)} \right)} \right]\frac{{dN\left( t \right)}}{{dt}} = \frac{1}{{N\left( t \right)}}\frac{{dN\left( t \right)}}{{dt}} = - \lambda \end{equation*}

Next, integrate both sides of equation (2):

(3)   \begin{equation*} \int_{{t_0}}^t {\underbrace {\frac{d}{{dt}}\left[ {\ln \left( {N\left( t \right)} \right)} \right]}_{\frac{1}{{N\left( t \right)}}\frac{{dN\left( t \right)}}{{dt}}}dt = - } \int_{{t_0}}^t {\lambda dt} \end{equation*}

Apply the fundamental theorem of calculus to both sides of equation (3). The integral on the left side is equal to

(4)   \begin{equation*} \ln \left( {N\left( t \right)} \right) - \ln \left( {N\left( {{t_0}} \right)} \right) = \ln \left( {\frac{{N\left( t \right)}}{{N\left( {{t_0}} \right)}}} \right) \end{equation*}

And, since \lambda is a constant, the integral on the right side of equation (3) is equal to

(5)   \begin{equation*} - \int_{{t_0}}^t {\lambda dt} = - \lambda \left( {t - {t_0}} \right) \end{equation*}

Letting N\left( t \right) = N, t_0 = 0, and {N_0} = N\left( {{t_0}} \right) = N\left( {t = 0} \right), from (3), (4), and (5):

    \[\ln \left( {\frac{N}{{{N_0}}}} \right) = - \lambda t\]

    \[\underbrace {{e^{\ln \left( {\frac{N}{{{N_0}}}} \right)}}}_{N/{N_0}} = {e^{ - \lambda t}}\]

(6)   \begin{equation*} \boxed{N = {N_0}{e^{ - \lambda t}}} \end{equation*}

(7)   \begin{equation*} \boxed{t = - \frac{1}{\lambda }\ln \left( {\frac{N}{{{N_0}}}} \right)} \end{equation*}

If we define the half life of the radioactive isotope,the time it takes for half of the sample to decay, as t_{1/2}, we can solve for the decay constant \lambda in terms of the half life:

    \[N\left( {{t_{1/2}}} \right) = \frac{{{N_0}}}{2} = {N_0}{e^{ - \lambda {t_{1/2}}}}\]

    \[\frac{1}{2} = {e^{ - \lambda {t_{1/2}}}}\]

    \[{e^{\lambda {t_{1/2}}}} = 2\]

    \[\underbrace {\ln \left( {{e^{\lambda {t_{1/2}}}}} \right)}_{\lambda {t_{1/2}}} = \ln \left( 2 \right)\]

(8)   \begin{equation*} \boxed{\lambda = \frac{{\ln \left( 2 \right)}}{{{t_{1/2}}}}} \end{equation*}

The value of \lambda can be determined by experimentally measuring the half-life of the radioactive isotope. Assuming that the half life of a radioactive isotope is in fact constant (and therefore \lambda is a constant), we are still faced with a problem. Let’s assume that we have discovered rocks that we believe formed at the same time. If we want to date each of the rocks using equation (7), it appears that unless we know N_0, the number of particles a radioactive isotope present in the rock at its formation, even if \lambda is a constant, we’ll have no way to determine the age of the rock. Making matters worse, it appears that there is no way of determining whether or not the rocks in question are ‘closed’ systems. If the ‘system’ (the rock in question) has been a closed system since its formation, that means that no material has been added or removed from the rock since its formation.

So is there no way to determine the age of rocks from radioactive isotopes present in the rocks? No. There are multiple possible methods for determining the age of rocks. The method I will focus on is the rubidium-strontium isochron method.

There are two naturally occuring isotopes of rubidium. ^{87}Rb is a radioactive isotope of rubidium, and has half life of 48.8 billion years. ^{85} Rb is a stable isotope of rubidium 1. ^{87} Rb decays to ^{87} Sr via beta decay (one of ^{87} Rb's decays into a proton, an electron, and an antineutrino).

    \[\begin{array}{*{20}{c}} {87}\\ {37} \end{array}Rb \to \begin{array}{*{20}{c}} {87}\\ {37} \end{array}Sr + \begin{array}{*{20}{c}} 0\\ { - 1} \end{array}\beta\]

Strontium-87 and Strontium-86 are two stable isotopes of strontium. If a group of rocks are co-genetic, meaning they are formed at the same time and place and are derived from the same parent material, they will have the same ratio of ^{87}Sr/^{86}Sr when they are formed 2. Assuming that the rocks are closed systems, the amount of ^{86}Sr at the time of in the rocks should be the same today as when the rocks were formed, since ^{86}Sr is not produced by radioactive decay. The amount of ^{87}Sr in each rock at the time of formation should be different from the amount of ^{87}Sr in each rock today, since ^{87}Sr can form from the decay of ^{87}Rb.

If a rock has been a closed system since its time of formation, each ^{87}Rb particle that decays forms a single ^{87}Sr particle. So the amount of ^{87}Sr particles measured long after the rock has formed is

(9)   \begin{equation*} {\left( {^{87}Sr} \right)_{final}} = {\left( {^{87}Sr} \right)_{initial}} + \left[ {{{\left( {^{87}Rb} \right)}_{initial}} - {{\left( {^{87}Rb} \right)}_{final}}} \right] \end{equation*}

Since the amount of ^{86}Sr is assumed to be constant, it follows that

(10)   \begin{equation*} {\left( {\frac{{^{87}Sr}}{{^{86}Sr}}} \right)_{final}} = {\left( {\frac{{^{87}Sr}}{{^{86}Sr}}} \right)_{initial}} + \frac{1}{{^{86}Sr}}\left[ {{{\left( {^{87}Rb} \right)}_{initial}} - {{\left( {^{87}Rb} \right)}_{final}}} \right] \end{equation*}

From equation (6), letting N = {\left( {^{87}Rb} \right)_{final}} and N_0 = {\left( {^{87}Rb} \right)_{initial}}:

    \[{\left( {^{87}Rb} \right)_{final}} = {\left( {^{87}Rb} \right)_{initial}}{e^{ - \lambda t}}\]

    \[{\left( {^{87}Rb} \right)_{final}}{e^{\lambda t}} = {\left( {^{87}Rb} \right)_{initial}}\]

(11)   \begin{equation*} {\left( {^{87}Rb} \right)_{initial}} - {\left( {^{87}Rb} \right)_{final}} = {\left( {^{87}Rb} \right)_{final}}\left( {{e^{\lambda t}} - 1} \right) \end{equation*}

From (10) and (11) it follows that:

(12)   \begin{equation*} \boxed{{\left( {\frac{{^{87}Sr}}{{^{86}Sr}}} \right)_{final}} = {\left( {\frac{{^{87}Sr}}{{^{86}Sr}}} \right)_{initial}} + \frac{{{{\left( {^{87}Rb} \right)}_{final}}}}{{^{86}Sr}}\left( {{e^{\lambda t}} - 1} \right)} \end{equation*}

We are assuming that all of the rocks collected formed at the same time and thus are the same age. We are assuming that {\left( {\frac{{^{87}Sr}}{{^{86}Sr}}} \right)_{initial}} is the same for all rocks in our sample (since we assume they formed at the same time). {\left( {\frac{{^{87}Sr}}{{^{86}Sr}}} \right)_{final}} depends on the initial amount of ^{87}Sr, ^{86}Sr, and the amount of ^{87}Rb present at the time of formation (which will vary from rock to rock). \frac{{{{\left( {^{87}Rb} \right)}_{final}}}}{{^{86}Sr}} depends on both the amount of ^{87}Rb at the time of formation and the initial amount of ^{86}Sr present at the time of formation, which will vary from rock. From those assumptions, including the assumption that the half life of ^{87}Rb is a constant that has not changed over time (and is independent of geological conditions), equation (12) has the form of a simple linear equation.

(13)   \begin{equation*} \boxed{\underbrace {{{\left( {\frac{{^{87}Sr}}{{^{86}Sr}}} \right)}_{final}}}_y = \underbrace {{{\left( {\frac{{^{87}Sr}}{{^{86}Sr}}} \right)}_{initial}}}_b + \underbrace {\frac{{{{\left( {^{87}Rb} \right)}_{final}}}}{{^{86}Sr}}}_x\underbrace {\left( {{e^{\lambda t}} - 1} \right)}_m} \end{equation*}

So if we collect a sample of rocks that we assume were formed at the same time and place, by the same parent material, measure the ratios of ^{87}Rb to ^{86}Sr and ^{87}Sr to ^{86}Sr present in each rock, a graph of {{{\left( {\frac{{^{87}Sr}}{{^{86}Sr}}} \right)}_{final}}} (the ratio of ^{87}Rb to ^{86}Sr versus ^{87}Sr to ^{86}Sr should form a straight line. The slope of that line will be \left( {{e^{\lambda t}} - 1} \right)} and the y-intercept will give us the initial ratio of ^{87}Sr to ^{86}Sr.

We have made multiple assumptions, and these can all be tested by plotting the ratio of ^{87}Rb/^{86}Sr versus ^{87}Sr to ^{86}Sr. If we find a linear relationship between ^{87}Rb/^{86}Sr and ^{87}Sr/^{86}Sr, we can be confident that our assumptions were correct.

The following data set is of moon rocks collected during the Apollo 17 mission. The relationship between ^{87}Rb/^{86}Sr and ^{87}Sr/^{86}Sr forms nearly a perfect straight line 3. The dataset can be downloaded here

The slope of the best-fit line is m = 6.56 \times 10^{-2}. From equation (13):

    \[\begin{array}{l} m = {e^{\lambda t}} - 1\\ {e^{\lambda t}} = m + 1\\ \lambda t = \ln \left( {m + 1} \right) \end{array}\]

(14)   \begin{equation*} \boxed{t = \frac{1}{\lambda }\ln \left( {m + 1} \right)} \end{equation*}

From equation (8), equation (14), and the fact that the measured half life of ^{87}Rb is 48 billion years:

    \[t = \frac{{{t_{1/2}}}}{{\ln \left( 2 \right)}}\ln \left( {m + 1} \right) = \frac{{48.8 \times {{10}^9}\ln \left( {6.56 \times {{10}^{ - 1}} + 1} \right)}}{{\ln \left( 2 \right)}}\]

(15)   \begin{equation*} \boxed{t \cong 4.47{\text{ billion years}}} \end{equation*}

The moon rocks that the Apollo 17 collected are approximately 4.5 billion years old.


  1. Dating Rocks with the Rb-Sr “Isochron” Method
  2. The Rubidium-Strontium System
  3. “Rb-Sr study of a lunar dunite and evidence for early lunar differentiates”, Papanastassiou, D. A. & Wasserburg, G. J..….6.1467P/0001467.000.html

Interesting Geometric Series and Applications of Geometric Series

A geometric series is a series that can be written as

(1)   \begin{equation*}a + ar + a{r^2} + a{r^3} +  \cdots  = \sum\limits_{n = 1}^\infty  {a{r^{n - 1}}}\end{equation*}

A geometric series will converge if \left| r \right| < 1. For a proof that that the only way a geometric series converges is if \left| r \right| < 1, click here.

Proving that a geometric series converges if \left| r \right| < 1 is simple. The sum of the first n terms of a geometric series is

(2)   \begin{equation*} {S_n} = \frac{{a\left( {1 - {r^n}} \right)}}{{1 - r}} \end{equation*}

Equation (2) is derived as followers.

    \[{S_n} = a + a{r^1} + ... + a{r^{n - 1}} = \sum\limits_{i = 1}^n {a{r^{i - 1}}}\]

Multiply the left side of the above equation and the far right side by r.

    \[r{S_n} = r\left( {a + ar + a{r^2} + ... + a{r^{n - 1}}} \right) = ar + a{r^2} + a{r^3} + ... + a{r^n}\]

Next, subtract the equation above from S_n and solve for S_n:

    \[{S_n} - r{S_n} = \left( {a + ar + a{r^2} + ... + a{r^{n - 1}}} \right) - \left( {ar + a{r^2} + a{r^3} + ... + a{r^n}} \right)\]

    \[S_n =  a - ar^n\]

    \[{S_n}\left( {1 - r} \right) = a\left( {1 - {r^n}} \right)\]

    \[{S_n} = \frac{{a - a{r^n}}}{{1 - r}}\]

(3)   \begin{equation*} {S_n} = \frac{a}{{1 - r}} - \frac{{a{r^n}}}{{1 - r}} \end{equation*}

The sum of a geometric series, if it exists, is defined as

(4)   \begin{equation*} \boxed{S = \mathop {\lim }\limits_{n \to \infty } {S_n} = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{a}{{1 - r}} - \frac{{a{r^n}}}{{1 - r}}} \right)}  \end{equation*}

Since a and r are constants, equation 4 can be written as

(5)   \begin{equation*} S = \frac{a}{{1 - r}} - \frac{a}{{1 - r}}\mathop {\lim }\limits_{n \to \infty } {r^n} \end{equation*}

If \left|r\right| <1:

    \[\mathop {\lim }\limits_{n \to \infty } {r^n} = 0\]

So if \left|r\right| <1, from equation (5), it follows that the geometric series converges, and is equal to

(6)   \begin{equation*} S = \frac{a}{{1 - r}} \end{equation*}

Here are a few interesting consequences of the definition of the sum of a geometric series (equation (4)).

    \[99999.... = .9 + .9\frac{1}{{10}} + .9{\left( {\frac{1}{{10}}} \right)^2} + .9{\left( {\frac{1}{{10}}} \right)^3} + ...\]

In the example above, a = .9 and r = 1/10. So,

    \[.99999... = \frac{a}{{1 - r}} = \frac{{.9}}{{1 - \frac{1}{{10}}}} = \frac{{.9}}{{.9}}\]


    \[\boxed{.99999... = 1}\]

    \[\sqrt {2\sqrt {2\sqrt {2\sqrt {2\sqrt {2\sqrt {2...} } } } } }  = 2\]


(7)   \begin{equation*}\sqrt {2\sqrt {2\sqrt 2 }... }  = {2^{\frac{1}{2}}} \cdot {2^{\frac{1}{4}}} \cdot {2^{\frac{1}{8}}} + ... = {2^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ....}} \end{equation*}

Focusing on the exponent on the right side, ^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ....}. This is a geometric series with a = \frac{1}{2} and r = \frac{1}{2}.

(8)   \begin{equation*}S = \frac{a}{{1 - r}} =\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... =  \frac{{\frac{1}{2}}}{{1 - \frac{1}{2}}} = \frac{{1/2}}{{1/2}} = 1 \end{equation*}

From (7) and (8):

    \[\sqrt {2\sqrt {2\sqrt 2 ...} }  = {2^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ....}} = {2^1}\]

    \[\boxed{\sqrt {2\sqrt {2\sqrt {2\sqrt {2\sqrt {2\sqrt {2...} } } } } }  = 2}\]

In the example above, I used 2. The choice of 2 was arbitrary. If c is a positive number:

    \[\boxed{\sqrt {c\sqrt {c\sqrt {c\sqrt {c\sqrt c ...} } } }  = c}}\]

Derivation of Generalization of the Gamma Function

The following definite integral problem was posted on stackexchange today.

(1)   \begin{equation*} 1.35 \times {10^{ - 7}}\int {{e^{ - 0.03x}}{x^4}dx} \end{equation*}


The version of the gamma function, a useful function for solving improper integrals, that I learned as an undergraduate physics major is

(2)   \begin{equation*}\int_0^\infty {{e^{ - x}}{x^n}dx = n!} \end{equation*}

At first glance, it appears that the gamma function (equation (2)) can be used to solve equation (1), since a constant times a definite integral is equal to the constant times the definite integral. However, since -0.03x \neq x, equation (2) can’t be used to solve equation (1); the integrands are not of the same form. But it is possible to derive a formula for solving improper integrals of the same form as equation (2) using u-substitution and the gamma function.

Equation (1) has the form

(3)   \begin{equation*} c \int_0^\infty {{e^{ - ax}}{x^n}dx} \end{equation*}

(with c and a being constants being any constant).


(4)   \begin{equation*} u = ax \end{equation*}

It follows from equation (4) that

(5)   \begin{equation*} x = \frac{u}{a} \end{equation*}


(6)   \begin{equation*} dx = \frac{du}{a} \end{equation*}

Applying equations (4), (5), and (6) to equation (3):

(7)   \begin{equation*}c\int_0^\infty {{e^{ - ax}}{x^n}dx} = c\int_0^\infty {{e^{ - u}}{{\left( {\frac{u}{a}} \right)}^n}\frac{{du}}{a}} \end{equation*}

(8)   \begin{equation*} c\int_0^\infty {{e^{ - ax}}{x^n}dx} = \frac{c}{{{a^{n + 1}}}}\int_0^\infty {{e^{ - u}}{u^n}du}\end{equation*}

In equation (7), \int_0^\infty {{e^{ - u}}{u^n}du} is just the gamma function, and is thus equal to n!.

Note: the bounds of integration have not changed, since when x goes to zero, from equation (4), u goes to zero as well. And when x goes to infinity, u goes to infinity for any non-zero constant a.

Finally, from equations(7) and (2) (the gamma function):

(9)   \begin{equation*} \boxed{c\int_0^\infty {{e^{ - ax}}{x^n}dx} = \frac{c}{{{a^{n + 1}}}}n!}\end{equation*}

From (1), c = 1.35 \times 10^{-7} and a = 0.03. So the solution to (1) is:

    \[\boxed{1.35 \times {10^{ - 7}}\int_0^\infty {{e^{ - 0.03x}}{x^n}dx} = \frac{{1.35 \times {{10}^{ - 7}}}}{{{{0.03}^5}}}4! = \frac{{400}}{3}}\]

Without using equation (8), the integral could still be solved, but it would require using integration by parts multiple times (unless there is another method that I am unaware of), which is an unnecessarily long and arduous way to evaluate the integral.


Using Python To Find Square-Triangular Numbers

A number is triangular if it satisfies the formula \frac{n(n+1)}{2} for some positive integer n. The square-triangular numbers are numbers that are both triangular and the square of a positive integer.

Letting the n-th triangular number equal

\frac{{n\left( {n + 1} \right)}}{2}

…and the square of the m-th positive integer equal m^2… it follows that a positive integer is both triangular and a square if

\frac{{n\left( {n + 1} \right)}}{2} = m^2

for some value of n and some value of m. Finding pairs of values of m and n that satisfy this equation is not easy by hand. The following manipulations of the equations above allow for the use of a simple python algorithm to find square-triangular numbers.

\frac{1}{2}n\left( {n + 1} \right) = {m^2}

\frac{1}{2}{\left( {n + \frac{1}{2}} \right)^2} = \frac{1}{2}\left( {{n^2} + n + \frac{1}{4}} \right) = \frac{1}{2}\left( {{n^2} + n} \right) + \frac{1}{8}

\frac{1}{2}n\left( {n + 1} \right) = \frac{1}{2}\left( {{n^2} + n} \right) = \frac{1}{2}{\left( {n + \frac{1}{2}} \right)^2} - \frac{1}{8} = {m^2}

\frac{1}{2}{\left( {n + \frac{1}{2}} \right)^2} - {m^2} = \frac{1}{8}

4{\left( {n + \frac{1}{2}} \right)^2} - 8{m^2} = 1

2 \cdot \left( {n + \frac{1}{2}} \right) \cdot 2\left( {n + \frac{1}{2}} \right) - 8{m^2} = 1

{\left( {2n + 1} \right)^2} - 8{m^2} = 1

{\left( {2n + 1} \right)^2} - 2 \cdot {\left( {2m} \right)^2} = 1

{\left( {2n + 1} \right)^2} - 2 \cdot {\left( {2m} \right)^2} - 1 = 0

The following two definitions will help simplify the equation above:

w \equiv 2n + 1

z \equiv 2m

With the definitions of w and z above, {\left( {2n + 1} \right)^2} - 2 \cdot {\left( {2m} \right)^2} = 1 can be re-written as

{w^2} - 2{z^2} - 1 = 0

Finding values of w and z that satisfy the equation above is not simple, but solutions can be found easily in Python. The following short algorithm finds the first 6 square-triangular numbers.

Informal Proof of the Associative Law of Matrix Multiplication

In general, if A is an m \times n matrix (meaning it has m rows and n columns), the matrix product AB will exist if and only if the matrix B has n rows. If B is an n \times p matrix, AB will be an m \times p matrix. In general, if A is an m \times n matrix (meaning it has m rows and n columns), the matrix product AB will exist if and only if the matrix B has n rows. If B is an n \times p matrix, AB will be an m \times p matrix.

When multiplying one matrix by another, the number of columns of the matrix on the left must be equal to the number of rows of the matrix on the right. The number of rows of the resulting matrix will be equal to the number of rows of the matrix on the left, and the number of columns will be equal to the number of columns of the matrix on the right.

For (AB)C to exist, C must have p rows (the same number of rows as columns of AB). If C is a p \times r matrix, (AB)C will be an m \times r matrix.

If (AB)C exists, A(BC) will also exist and have the same shape as (AB)C. For these matrices to exist, A must have the same number of columns as B has rows, and C must have the same number of rows as B has columns. If B is an n \times p matrix, for BC to exist, C must have p rows. If C is a p \times r matrix, BC will be an n \times r matrix. If (AB)C exists, that means that AB exists. If BC is an n \times r matrix, then A must have n columns. If A is an m \times n matrix, A(BC) will be an m \times r matrix. So if A is an m \times n matrix, B is an n \times p matrix, and C is a p \times r matrix, it follows that (AB)C and A(BC) both exist and are m \times r matrices.

When (AB)C and A(BC) both exist, (AB)C = A(BC). To prove this, all we must do is prove that an arbitrary column of AB(C) will be equal to the same arbitrary column in A(BC).

If K_i represents the i-th column of any matrix K, n \times p matrix B can be represented as

B = \left[ {\begin{array}{*{20}{c}} {{B_i}}& \cdots &{{B_p}} \end{array}} \right]

B_i is the i-th column of B, and is a vector with n rows. The i-th column of AB is equal to A times the i-th column of B.

AB = \left[ {\begin{array}{*{20}{c}} {\underbrace {A{B_1}}_{{{\left( {AB} \right)}_1}}}& \cdots &{\underbrace {A{B_r}}_{{{\left( {AB} \right)}_r}}} \end{array}} \right]

The j-th column of the matrix C can be represented as

\left[ {\begin{array}{*{20}{c}} {{c_{1j}}} \\ \vdots \\ {{c_{pj}}} \end{array}} \right]

c_{ij} is the value in row i, column j of C.

Column j of (AB)C is equal to AB times column j of C. This is equal to the linear combination of the i-th columns of (AB) with the i-th values in C_j.

{\left( {\left( {AB} \right)C} \right)_j} = \left( {AB} \right){C_j} = \left[ {\begin{array}{*{20}{c}} {A{B_1}}& \cdots &{A{B_p}} \end{array}} \right]

The j-th column of BC is equal to BC_j, which is equal to the linear combination of the i-th columns of B with the i-th values in C_j:

{\left( {BC} \right)_j} = B{C_j} = {c_{1j}}{B_1} + {\kern 1pt} \cdots + {c_{pj}}{B_p}

The j-th column of A(BC) is equal to the jth column of BC, (BC)_j times A.

{\left( {A\left( {BC} \right)} \right)_j} = A{\left( {BC} \right)_j} = A\left( {{c_{1j}}{B_1} + \cdots + {c_{pj}}{B_p}} \right) = {c_1}A{B_1} + \cdots + {c_{pj}}{B_p} = {\left( {\left( {AB} \right)C} \right)_j}

Since the j-th column of (AB)C is equal to the j-th column of A(BC), it follows that (AB)C = A(BC).