tag:blogger.com,1999:blog-26379284728594538312020-02-28T11:25:49.784-05:00Accidental Desultory CogitationsRandom thoughts on a variety of topics, but mostly on science and technology.Unknownnoreply@blogger.comBlogger105125tag:blogger.com,1999:blog-2637928472859453831.post-54369266609612220402019-11-07T13:11:00.000-05:002019-11-07T13:11:20.809-05:00Prepending and appending a repunit with the same numberConsider a repunit R(n) of n digits in decimal, i,e, the number consisting of n 1's which is equal to $(10^n-1)/9$. If we append and prepend R(n) with the same number k, which we will denote as&nbsp; k|R(n)|k with | denoting concatenation of digits, will k|R(n)|k be prime? For instance, for k = 324 and n = 1111, $k|R(n)|k = 3241111324 = 2^2\times 11^2 \times 281\times 23831$. It turns out that if n is a power of 2, then k needs to have at least n digits for k|R(n)|k to be prime. In particular, we have the following result:<br /><br /><b>Theorem:</b> If $2^r$ is the largest power of 2 that divides n and k has at most $2^r-1$ digits, then k|R(n)|k is not prime. This is true even if some of the digits of k are leading zeros.<br /><br /><i>Proof:</i> If k has m digits, then $k|R(n)|k = k(10^{n+m}+1) + R(n)10^m$ which is a multiple of gcd$(R(n),10^{n+m}+1)$.<br /><br />Since $10^{2^r} - 1 = (10^{2^{r-1} -1})\times (10^{2^{r-1} + 1})$ and 9 does not divide $10^n+1$, by induction it is easy to show that $10^{2^w} + 1$ is a divisor of&nbsp; $R(2^r)$ for $1 \leq w &lt; r$. Since $R(2^r)$ is a divisor of $R(2^r s)$, $10^{2^w} + 1$ is also a divisor of $R(2^r s)$.<br /><br />For $1 \leq m &lt; 2^r$, let $t$ be the 2-adic valuation (or 2-adic order) of m, i.e. t is the largest integer such that $2^t$ divides $m$. This means that&nbsp; $0 \leq t&nbsp; &lt; r$.<br /><br />Then $10^{2^r s+m}+1 = 10^{2^t q}+1 = (10^{2^t})^q + 1$ for some odd number $q$.<br /><br />Since the sum of two odd powers $a^q+b^q$ is divisible by $a+b$, this implies that $10^{2^r s+m}+1$ is divisible by $10^{2^t}+1$.<br /><br />This implies that for $n = 2^r s$ and for all $1 \leq m &lt; 2^r$, gcd$(R(n),10^{n+m}+1) \geq 10^{2^t}+1 &gt; 1$, i.e. $k|R(n)|k$ is not prime. QED<br /><br />For example, for n = 32, the smallest $k$ such that $k|R(n)|k$ is prime is k =&nbsp;10000000000000000000000000000077, a number with 32 digits. The smallest such $k$ for different $n$'s can be found in OEIS sequence <a href="https://oeis.org/A263182">A263182</a>&nbsp;(see also OEIS sequence <a href="https://oeis.org/A329121">A329121</a>).<br /><br />Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-2637928472859453831.post-30177517698969322382019-11-06T20:48:00.000-05:002019-11-06T20:48:11.872-05:00Product of a number and its reversal in decimal<span style="background-color: white; font-size: 13px; text-indent: -16px;"><span style="font-family: inherit;">The number 10119126106147652568 is the smallest number such that it is the product of a number and its reversal in decimal notation in exactly 5 ways.&nbsp;</span></span><br /><span style="background-color: white; font-size: 13px; text-indent: -16px;"><span style="font-family: inherit;"><br /></span></span><span style="font-family: inherit;"><span style="background-color: white; font-size: 13px; text-indent: -16px;">10119126106147652568&nbsp;</span><span style="background-color: white; font-size: 13px; text-indent: -16px;">= 8848263411 * 1143628488 = 8044687521 * 1257864408 = 4884561702 * 2071654884 = 4440958722 * 2278590444 = 4082378742 * 2478732804.</span></span><br /><span style="font-family: inherit;"><span style="background-color: white; font-size: 13px; text-indent: -16px;"><br /></span></span><span style="font-family: inherit;"><span style="background-color: white; font-size: 13px; text-indent: -16px;">Multiply this number by 10 and it is the smallest number&nbsp;</span></span><span style="background-color: white; font-size: 13px; text-indent: -16px;">such that it is the product of a number and its reversal in decimal notation in exactly 10 ways.&nbsp;</span><br /><span style="background-color: white; font-size: 13px; text-indent: -16px;"><br /></span><span style="background-color: white; font-size: 13px; text-indent: -16px;">101191261061476525680 = 88482634110 * 1143628488 = 80446875210 * 1257864408 = 48845617020 * 2071654884 = 44409587220 * 2278590444 = 40823787420 * 2478732804 = 24787328040 * 4082378742 = 22785904440 * 4440958722 = 20716548840 * 4884561702 = 12578644080 * 8044687521 = 11436284880 * 8848263411.</span>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-2637928472859453831.post-22118952490475433182019-01-11T13:04:00.000-05:002019-01-11T13:04:26.418-05:00Numbers whose binary complement is larger than the binary complement of its square.<a href="https://oeis.org/">OEIS</a> sequence <a href="https://oeis.org/A323192">A323192</a> lists numbers whose binary complement is larger than the binary complement of its square. The binary complement is defined as flipping each bit in its binary representation. For instance, 29 is 11101 in binary, so its binary complement is 2 which is 00010 in binary. Note that the binary complement function is not one-to-one, for instance the binary complement of 61 (111101 in binary) is also 2.&nbsp; This function is not exactly the same as <a href="https://en.wikipedia.org/wiki/Ones%27_complement">1's complement</a> in computer arithmetic, since the number of bits flipped is not fixed but is equal to the number of bits in the binary representation of the number.<br /><br />Giovanni Resta noted that the first 13 terms of sequence A323192 are of the form $\lfloor\sqrt{2^k-1}\rfloor$. It turns out this is true in general. In fact we can prove more. In particular, we can show that the terms of this sequence corresponds to numbers of the form $\lfloor\sqrt{2^{2k-1}}\rfloor$ which are larger than $\frac{1}{2} + \sqrt{\frac{1}{4} + 2^{2k-1} - 2^k}$ with $k &gt; 0$.<br /><br />This result follows from the following theorem:<br /><br /><b>Theorem:</b> if $n$ is a term of OEIS sequence A323192, then $n$ is of the form $\lfloor\sqrt{2^{2k-1}}\rfloor$ for some $k &gt; 0$. In addition, $\lfloor\sqrt{2^{2k-1}}\rfloor$ is a term if and only if it is larger than $\frac{1}{2} + \sqrt{\frac{1}{4} + 2^{2k-1} - 2^k}$.<br /><br /><i>Proof:</i> suppose $n$ has $k$ bits in its binary representation, i.e. $2^{k-1} \leq n &lt; 2^k$. Then $n^2$ has either $2k-1$ or $2k$ bits.<br /><br />First we show that if $n$ is a term of the sequence, then $n^2$ has $2k-1$ bits. Suppose $n^2$ has $2k$ bits, i.e. $n^2 \geq 2^{2k-1}$. Then $2^{2k} - 1 - n^2 &lt; 2^k - 1 - n$. This is rearranged as $n^2 - n + 2^k - 2^{2k} &gt; 0$. Solving this quadratic inequality leads to $n &gt; 2^k$ which contradicts the fact that $n$ has $k$ bits.<br /><br />Thus $n^2 &lt; 2^{2k-1}$ and $2^{2k-1} - 1 - n^2 &lt; 2^k - 1 - n$. Solving this inequality and combining it with $n &lt; \sqrt{2^{2k-1}}$ shows that $n$ must satisfy $\sqrt{2^{2k-1}} &gt; n &gt;&nbsp; \frac{1}{2}+ \sqrt{\frac{1}{4} + 2^{2k-1} - 2^k}$.<br /><br />To complete the proof, we need to show that for each $k$, there is at most one integer satisfying this inequality. This is easily verified for $k = 1$. Assume that $k &gt; 1$. Let $a = \sqrt{2^{2k-1}}$ and $b = \frac{1}{2} + \sqrt{\frac{1}{4}+ 2^{2k-1} - 2^k}$.<br /><br />Using the identity $\sqrt{x} - \sqrt{y} = (x-y)/(\sqrt{x}+\sqrt{y})$ it follows that $a-b = -\frac{1}{2} + \frac{2^k - \frac{1}{4}}{\sqrt{2^{2k-1}}+\sqrt{\frac{1}{4}+ 2^{2k-1} - 2^k}} &lt; -\frac{1}{2} + \frac{2^k}{\sqrt{2^{2k-1}} + \sqrt{2^{2k-1}-2^k}}$.<br /><br />Since $k \geq 2$, $\sqrt{2^{2k-1}-2^k} = \sqrt{2^{2k-1}(1-2^{1-k})} \geq \sqrt{\frac{1}{2}}\ \sqrt{2^{2k-1}}$. This implies that $a-b &lt; -\frac{1}{2} + \frac{2^k}{\frac{\sqrt{2}+1}{\sqrt{2}}\sqrt{2^{2k-1}}} = \frac{2\sqrt{2}}{\sqrt{2}+1} - \frac{1}{2} &lt; 0.672$. Thus it follows that there is at most one integer in the range between $b$ and $a$.<br /><br />The above discussion also completely characterizes the sequence. $\lfloor\sqrt{2^{2k-1}}\rfloor$ is a term if and only if $\sqrt{2^{2k-1}}-\lfloor\sqrt{2^{2k-1)}}\rfloor &lt; a-b$ where $a-b$ is as defined above.<br /><br />The criterion can be simplified as:<br />$\lfloor\sqrt{2^{2k-1}}\rfloor$ is a term if and only if it is larger than $\frac{1}{2} + \sqrt{\frac{1}{4} + 2^{2k-1} - 2^k}$. This concludes the proof.$\blacksquare$<br /><br />Note that $a-b \rightarrow \frac{1}{2}$ as $k \rightarrow \infty$, i.e. for large $k$, the fractional part of $\sqrt{2^{2k-1}}$ should be less than about $\frac{1}{2}$ in order for the integer part to be a term.<br /><br />The numbers $k$ such that $\lfloor\sqrt{2^{2k-1}}\rfloor$ is a term of OEIS sequence&nbsp;A323192 are listed in OEIS sequence <a href="https://oeis.org/A323062">A323062</a>.Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-2637928472859453831.post-65179562887258302052019-01-07T14:25:00.000-05:002019-01-07T14:25:31.998-05:00560106 and 601065There is a remarkable property of the sequence of numbers 560106, 5606106, 56066106, 560666106, etc. Take any number of the sequence, say 56066106, reverse the digits: 60166560, multiply these 2 numbers, multiply the result by 10 and the result is a perfect square. Thus<br />$56066106 \times 60166065 \times 10 =&nbsp;33732769778928900 =&nbsp;183664830^2$.<br /><br />To see this, let us denote the decimal digits reversal of a number $n$ as $R(n)$.&nbsp; &nbsp;Let $a = 56\times 10^{4+k}+106 + 6000\times (10^k-1)/9$ for $k\geq 0$. Then $R(a) = 601\times 10^{3+k}+65 + 6000\times (10^k-1)/9$. The number $10\times a\times R(a)$ can be written as $30360100\times (10^{k + 3} - 1)^2/9$ whose square root is $5510\times (10^{k + 3} - 1)/3$.<br /><br />It is clear the the digit reversals of these numbers, i.e. 601065, 6016065, 60166065, 601666065, ..., satisfy the same property.<br /><br />Other numbers with this property can be found in <a href="https://oeis.org/">OEIS</a> sequence <a href="https://oeis.org/A323061">A323061</a>.<br /><br />Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-2637928472859453831.post-42700688712940755412018-12-30T12:49:00.000-05:002018-12-30T12:49:24.733-05:00Christmas lights ring oscillatorWe set up a string of Christmas lights wrapped around a pine tree outside and plug it into an outlet with a dusk to dawn sensor so it will only turn on at dusk. After a windy day, some of the lights fell from the tree so we adjusted it in the evening. After adjustment, my son noticed that the lights turn on and off at a regular interval, which was strange since these lights are not flashing lights, but are supposed to be on all the time. It turns out one of the light bulb is too close to the light sensor on the outlet, thus tricking it to turn off since the outlet assumed it was dawn. Since it is now dark without the lights, the sensor assumes it is dusk and turns the light back on and so on and so on. Since there is a delay in the sensor, this resulted in the lights being turned on and off periodically. In essence, we have constructed a ring oscillator. A <a href="https://en.wikipedia.org/wiki/Ring_oscillator">ring oscillator</a>&nbsp;consists of a loop of an odd number of NOT logic gates. The number of stages and the delay in each stage of the logic gates determine the frequency of the oscillator. In our case, we have a single NOT gate represented by the dawn-to-dusk light sensor/outlet combination.Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-2637928472859453831.post-84209372924962223022018-05-20T23:36:00.000-04:002018-05-21T00:30:56.322-04:00An equation involving radicalsWhile browsing the internet, the following brainteaser popped up on the screen: find $x$ such that<br /><br />$$\sqrt{x+15} + \sqrt{x} = 15$$<br /><br />A straight forward approach is to square both sides, move the single term with radicals to one side and square again, and reducing terms to obtain a linear equation in $x$ whose solution gives the answer.<br /><br />Here is another (somewhat simpler) way to solve this problem. First note that the left hand side is a increasing function of $x$ and at $x=0$ is equal to $\sqrt{15} &lt; 15$, so there is a single real solution to the equation above.<br />If we set $x = y^2$ and $x+15 = (y+z)^2$, we get $15 = 2yz + z^2 = z(2y+z)$. The left hand side of the original equation then becomes:<br /><br />$y+z +y = 2y+z = 15$. Combine this with the above it follows that $z = 1$ and $y = 7$. Thus the answer is $x = 49$.<br /><br />In general, this method shows that the equation<br /><br /><span style="background-color: transparent; color: black; display: inline; float: none; font-family: &quot;times new roman&quot;; font-size: 16px; font-style: normal; font-variant: normal; font-weight: 400; letter-spacing: normal; text-align: left; text-decoration: none; text-indent: 0px; text-transform: none; white-space: normal; word-spacing: 0px;">$$\sqrt{x+a} + \sqrt{x} = b$$ where $b \geq 0$ and $b^2 \geq a$ has as the only real solution:</span><br /><span style="background-color: transparent; color: black; display: inline; float: none; font-family: &quot;times new roman&quot;; font-size: 16px; font-style: normal; font-variant: normal; font-weight: 400; letter-spacing: normal; text-align: left; text-decoration: none; text-indent: 0px; text-transform: none; white-space: normal; word-spacing: 0px;"><br /></span><span style="background-color: transparent; color: black; display: inline; float: none; font-family: &quot;times new roman&quot;; font-size: 16px; font-style: normal; font-variant: normal; font-weight: 400; letter-spacing: normal; text-align: left; text-decoration: none; text-indent: 0px; text-transform: none; white-space: normal; word-spacing: 0px;">$$x = \left(\frac{b^2-a}{2b}\right)^2$$</span><br /><span style="background-color: transparent; color: black; display: inline; float: none; font-family: &quot;times new roman&quot;; font-size: 16px; font-style: normal; font-variant: normal; font-weight: 400; letter-spacing: normal; text-align: left; text-decoration: none; text-indent: 0px; text-transform: none; white-space: normal; word-spacing: 0px;"><br /></span><span style="background-color: transparent; color: black; display: inline; float: none; font-family: &quot;times new roman&quot;; font-size: 16px; font-style: normal; font-variant: normal; font-weight: 400; letter-spacing: normal; text-align: left; text-decoration: none; text-indent: 0px; text-transform: none; white-space: normal; word-spacing: 0px;">For the case $a = b \geq 0$, this reduces to</span><br /><span style="background-color: transparent; color: black; display: inline; float: none; font-family: &quot;times new roman&quot;; font-size: 16px; font-style: normal; font-variant: normal; font-weight: 400; letter-spacing: normal; text-align: left; text-decoration: none; text-indent: 0px; text-transform: none; white-space: normal; word-spacing: 0px;"><br /></span><span style="background-color: transparent; color: black; display: inline; float: none; font-family: &quot;times new roman&quot;; font-size: 16px; font-variant: normal; letter-spacing: normal; text-align: left; text-indent: 0px; text-transform: none; white-space: normal; word-spacing: 0px;"></span><br /><div style="-webkit-text-stroke-width: 0px; background-color: transparent; color: black; font-family: Times New Roman; font-size: 16px; font-style: normal; font-variant: normal; font-weight: 400; letter-spacing: normal; margin-bottom: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; orphans: 2; text-align: left; text-decoration: none; text-indent: 0px; text-transform: none; white-space: normal; word-spacing: 0px;"><span style="background-color: transparent; color: black; display: inline; float: none; font-family: &quot;times new roman&quot;; font-size: 16px; font-variant: normal; letter-spacing: normal; text-align: left; text-indent: 0px; text-transform: none; white-space: normal; word-spacing: 0px;"><span style="background-color: transparent; color: black; display: inline; float: none; font-family: &quot;times new roman&quot;; font-size: 16px; font-style: normal; font-variant: normal; font-weight: 400; letter-spacing: normal; text-align: left; text-decoration: none; text-indent: 0px; text-transform: none; white-space: normal; word-spacing: 0px;">$$x = \left(\frac{a-1}{2}\right)^2$$</span></span></div><span style="background-color: transparent; color: black; display: inline; float: none; font-family: &quot;times new roman&quot;; font-size: 16px; font-variant: normal; letter-spacing: normal; text-align: left; text-indent: 0px; text-transform: none; white-space: normal; word-spacing: 0px;"><div style="-webkit-text-stroke-width: 0px; background-color: transparent; color: black; font-family: Times New Roman; font-size: 16px; font-style: normal; font-variant: normal; font-weight: 400; letter-spacing: normal; margin-bottom: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; orphans: 2; text-align: left; text-decoration: none; text-indent: 0px; text-transform: none; white-space: normal; word-spacing: 0px;"><span style="background-color: transparent; color: black; display: inline; float: none; font-family: &quot;times new roman&quot;; font-size: 16px; font-style: normal; font-variant: normal; font-weight: 400; letter-spacing: normal; text-align: left; text-decoration: none; text-indent: 0px; text-transform: none; white-space: normal; word-spacing: 0px;"><br /></span></div><div style="-webkit-text-stroke-width: 0px; background-color: transparent; color: black; font-family: Times New Roman; font-size: 16px; font-style: normal; font-variant: normal; font-weight: 400; letter-spacing: normal; margin-bottom: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; orphans: 2; text-align: left; text-decoration: none; text-indent: 0px; text-transform: none; white-space: normal; word-spacing: 0px;">w<span style="background-color: transparent; color: black; display: inline; float: none; font-family: &quot;times new roman&quot;; font-size: 16px; font-style: normal; font-variant: normal; font-weight: 400; letter-spacing: normal; text-align: left; text-decoration: none; text-indent: 0px; text-transform: none; white-space: normal; word-spacing: 0px;">hich is an integer if $a$ is odd.</span></div><div style="-webkit-text-stroke-width: 0px; background-color: transparent; color: black; font-family: Times New Roman; font-size: 16px; font-style: normal; font-variant: normal; font-weight: 400; letter-spacing: normal; margin-bottom: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; orphans: 2; text-align: left; text-decoration: none; text-indent: 0px; text-transform: none; white-space: normal; word-spacing: 0px;"><span style="background-color: transparent; color: black; display: inline; float: none; font-family: &quot;times new roman&quot;; font-size: 16px; font-style: normal; font-variant: normal; font-weight: 400; letter-spacing: normal; text-align: left; text-decoration: none; text-indent: 0px; text-transform: none; white-space: normal; word-spacing: 0px;"></span></div><b></b><i></i><u></u><sub></sub><sup></sup><strike></strike></span><b></b><i></i><u></u><sub></sub><sup></sup><strike></strike>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-2637928472859453831.post-45716001334724447522018-04-16T17:49:00.001-04:002018-04-16T17:49:45.945-04:00What is a slide rule for?I was listening to Sam Cooke's classic song "Wonderful World" and thought of the line: "Don't know what the slide rule is for". When the song was released in 1960, the line was describing how disinterested the protagonist was about math and algebra, as the slide rule was a common tool for doing calculations. Today, that line would probably describe most young adults or younger, as the slide rule has not been used for calculations for many years with the rise of the calculator. When I was in high school, we used electronic calculators in our tests and exams, but out of curiosity my brother and I bought a slide rule anyway (they were still being sold in bookstores at the time, but disappeared soon after from the shelves). I was fascinated by how you can multiply two numbers merely by aligning the start of one ruler with the first number and reading off the result off the second number.&nbsp; This is due a property of logarithm: log(ab) = log(a) + log(b). Thus multiplication is reduced to addition. By printing the numbers in logarithmic scale and lining up segment end-to-end (corresponding to addition), we achieve the operation of multiplication using a slide rule. Similarly, division can be done with a slide rule as it corresponds to subtraction. And it can do a lot more, such as trigonometry and taking square and cube roots. Accuracy was a problem, and they were soon supplanted by calculators which can compute to many digits of precision. As the song is continuously being covered by many artists, I wonder if the current audience would find the lyrics strange?Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-2637928472859453831.post-9873457219646611592018-03-14T13:16:00.001-04:002018-03-14T13:16:55.324-04:00Primes of the form H(n,-k)-1OEIS sequence <a href="https://oeis.org/A299145">A299145</a>&nbsp;lists the primes of the form $Q(n,k) = \sum_{i=2}^n i^k$ for $n \geq 2$ and $k&gt; 0$. It is clear that except for the case $k =1$ and $n=2$ resulting in the prime $2$, we must have $n\geq 3$.<br />The sum $Q(n,k)$ is equal to $H(n,-k) - 1$ where $H(n,m) = \sum_{i=1}^n \frac{1}{i^m}$ is the <a href="https://en.wikipedia.org/wiki/Harmonic_number#Generalized_harmonic_numbers">generalized harmonic number</a> of order $n$ of $m$. It is well known that $H(n,-k)$ is a polynomial of $n$ of degree $k+1$. In particular, <a href="https://en.wikipedia.org/wiki/Faulhaber%27s_formula">Faulhaber's formula</a> shows that<br />$$H(n,-k) = \frac{1}{2} n^k + \frac{1}{k+1}\sum_{j=0}^{\lfloor k/2 \rfloor} \left(\begin{array}{c} k+1\\2j\end{array}\right) B_{2j} n^{k+1-2j}$$<br />where $B_i$ is the $i$-th <a href="https://en.wikipedia.org/wiki/Bernoulli_number">Bernoulli number</a>.<br /><br />$H(n,k)$ and thus $Q(n,k)$ can be written as a degree $k+1$ polynomial of $n$ with rational coefficients. The smallest denominator of this polynomial is found in OEIS <a href="https://oeis.org/A064538">A064538</a>. In 2017, Kellner and Sondow showed that this is equal to $(k+1)\prod_{p\in S}$ where $S$ is the set of primes $p \leq \frac{k+2}{2+(k \mod 2)}$ such that the sum of the base $p$ digits of $k+1$ is $p$ or larger.<br /><br />Since $H(1,-k) = 1$, this implies that $Q(1,k) = 0$ and thus $n-1$ is a factor of $Q(n,k)$. Since $n\geq 3$, if $n-1$ does not get cancelled out by a factor of the denominator, we would have a nontrivial factor of $Q(n,k)$ and thus $Q(n,k)$ is not prime. This implies that $Q(n,k)$ is prime only if $n-1$ is a divisor of a(k) in sequence A064538. Thus for each $k$ there is only a finite number of values of $n$ to check. This provides an efficient algorithm to find terms of this sequence by looking only for primes in the numbers $Q(n,k)$ where $n-1$ is a divisor of A064538(k). There are <a href="https://oeis.org/A299145/b299145.txt">45</a> such numbers with $1000$ or less digits.<br /><br /><b>References</b><br />Bernd C. Kellner, Jonathan Sondow, <a href="https://doi.org/10.4169/amer.math.monthly.124.8.695">Power-Sum Denominators</a>, Amer. Math. Monthly 124 (2017), 695-709.Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-2637928472859453831.post-78792324651979135892018-02-20T21:51:00.000-05:002018-02-20T21:51:27.026-05:00Chaos in musicIn an earlier&nbsp;<a href="http://accidentaldesultorycogitations.blogspot.com/2014/03/fractals-in-fairy-tale.html">post</a>, I talked about fractals mentioned in the lyrics of a song. Today, I heard the song "Jurassic Park" by Al Yankovic, which mentions <a href="https://en.wikipedia.org/wiki/Chaos_theory">chaos theory</a>, a related branch of mathematics that studies the unpredictable and complex behavior of dynamical systems, often manifested as <a href="https://en.wikipedia.org/wiki/Butterfly_effect">sensitive dependence on initial conditions</a>. On the other hand, musical compositions themselves can exhibit fractal and chaotic behavior [1,2].<br /><br /><b>References:</b><br /><br /> K J Hsü and A Hsü, "Self-similarity of the "1/f noise" called music," PNAS, vol. 88, no. 8, pp. 3507-3509, 1991.<br /> J. Beran, "Music - Chaos, Fractals, and Information," CHANCE, vol. 17, no. 4, pp. 7-16, 2004.Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-2637928472859453831.post-4204552633183766862017-12-18T12:48:00.000-05:002017-12-18T12:48:52.416-05:00O, zero and eightWhen my parents bought us our first computer, a <a href="https://en.wikipedia.org/wiki/Commodore_VIC-20">Commodore VIC-20</a>, in my early teens, I learned that to distinguish capital O with the digit zero, a diagonal line is put through the digit zero (also known as a <a href="https://en.wikipedia.org/wiki/Slashed_zero">slashed zero</a>). This is due to the relatively low resolution computer fonts used in those days, where O is easily confused with 0 (but its use actually <a href="https://en.wikipedia.org/wiki/Slashed_zero">predates</a> computers). On the other hand, some Scandinavian languages has something similar to a slashed zero in their alphabet, and the slashed zero can cause more problems than it solves. The use of the slashed zero is also useful when writing computer code using pen and paper, an activity that is becoming rare. Recently, I have not seen the use of slashed zero in word processing since given the number of pixels they have at their disposal, many fonts do not use it to represent the digit 0. In some fonts such as DejaVu Sans Mono and the font used in the Windows command prompt, a dotted zero is used to represent 0. Incidentally, for the Default font used in this blog, zero (0) looks almost identical to the lowercase letter "o".<br /><br />The empty set (the set with zero elements) is denoted as {}, and is sometimes also denoted with a symbol similar to a slashed zero, but with the slash extending beyond the boundary of "0". This notation makes sense since in <a href="https://en.wikipedia.org/wiki/Set-theoretic_definition_of_natural_numbers">Zermelo-Fraenkel set theory</a> the natural numbers are defined as sets with 0 being the empty set and the number n+1 defined as $n \cup \{n\}$, i.e. n+1 is the set obtained by augmenting the set n with a single element: the set consisting of the set n. In other words, 0 = {}, 1 = {{}}, 2 = {{},{{}}}, etc.<br /><br />The use of a slashed zero is a good idea, but recently I found that it is causing problems for me. There is one place where I consistently see the use of a slashed zero and that is on credit card receipts. With advancing age and the onset of farsightedness, I have a hard time separating the slashed zero from the digit 8. When we go out to eat at a restaurant, the lighting is typically dim which exacerbates the problem and sometimes I have a hard time determining how much tip to leave and add it up correctly on the bill. I wish the receipt printer would not use the slash zero in their font (I don't think the dotted zero is much better either in this case).Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-2637928472859453831.post-61948037345600438392017-10-10T13:06:00.000-04:002017-10-10T13:06:11.715-04:00The triangle inequality and its converse<h2>The triangle inequality</h2>The triangle inequality is a workhorse in many branches of mathematics. It expresses the subadditivity of norms and is stated as:<br />$$\| A+B\| \leq \|A\| + \|B\|.$$<br />The name comes from its interpretation on the plane. If $A$ and $B$ are complex numbers, then we can draw a triangle with vertices at the points $0$, $A$ and $A+B$, with the resulting sides having norms equal to $|A|$, $|B|$ and $|A+B|$. Stated in another way, if $a$, $b$ and $c$ are the lengths of the $3$ sides of a triangle, then $a \leq b+c$. This is illustrated in Fig. 1.<br />A slightly different, but clearly equivalent statement is the following: if $a \geq b \geq c \geq 0$ are the lengths of the $3$ sides of a triangle, then $a \leq b+c$. We will use this alternative formulation as it is easier to state and prove the converse and its generalization.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-UUyo-lQaCmA/Wdwj3tfX7HI/AAAAAAAAAlc/LmIaNDApXIgYwmB5VPHURYf2zGliPTdhQCLcBGAs/s1600/triangle1.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="207" data-original-width="500" height="132" src="https://3.bp.blogspot.com/-UUyo-lQaCmA/Wdwj3tfX7HI/AAAAAAAAAlc/LmIaNDApXIgYwmB5VPHURYf2zGliPTdhQCLcBGAs/s320/triangle1.png" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;">Fig. 1:&nbsp;Illustration of the triangle inequality and its converse.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><h2>Converse of the triangle inequality on the plane</h2>The converse of the triangle inequality is also true: If $a \geq b \geq c \geq 0$ and $a \leq b+c$, then there exists a triangle whose sides have lengths $a$, $b$ and $c$ respectively. Equivalently this can be restated as: If $a \geq b \geq c \geq 0$ and $a \leq b+c$, then there exists complex numbers $A$, $B$ and $C$ with $|A|=a$, $|B| = b$, $|C| = c$ such that $A+B+C = 0$.<br /><br />The following is a simple construction of this triangle.<br />First we show that $d = \frac{b^2+c^2 - a^2}{2bc}$ satisfies $|d| \leq 1$. This follows from the fact that $d = \frac{(b+c)^2 -a^2-2bc}{2bc} = -1 + \frac{(b+c)^2 - a^2}{2bc} \geq -1$<br />and $d= \frac{(b-c)^2 -a^2+2bc}{2bc} = \frac{(b-c)^2 -a^2}{2bc} + 1\leq 1$.<br />Thus we can choose $\phi = \cos^{-1}(d)$ which satisfies $a^2 = b^2+c^2-2bc\cos (\phi)$ and the <a href="http://mathworld.wolfram.com/LawofCosines.html">law of cosines</a> shows that the triangle with sides of length $b$ and $c$ and angle $\phi$ with have the third side of length $a$. (see Fig. 1).<br /><br /><h2>The generalized triangle inequality and its converse</h2>A simple induction argument generalizes the triangle inequality to the summation of more than 2 quantities:<br />$$\left\| \sum_i^n A_i\right\| \leq \sum_i^n \|A_i\|$$<br />The geometrical interpretation is that if $r_i$ are the lengths of edges of a polygon, then $r_i \leq \sum_{j\neq i} r_j$ or equivalently, if $r_i$ are the lengths of edges of a polygon, and $r_1\geq r_i$, then $r_1\leq \sum_{i &gt; 1}r_i$ (Fig. 2).<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-Q079PAYpCCc/WdwkcDuh5-I/AAAAAAAAAlg/-sAk_pNRpiAAhEPCZ2XwX1_hfqa1u_kNgCLcBGAs/s1600/polygon1.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="345" data-original-width="625" height="176" src="https://1.bp.blogspot.com/-Q079PAYpCCc/WdwkcDuh5-I/AAAAAAAAAlg/-sAk_pNRpiAAhEPCZ2XwX1_hfqa1u_kNgCLcBGAs/s320/polygon1.png" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;">Fig. 2:&nbsp;Illustration of the generalized triangle inequality and its converse.</div><div style="text-align: left;"><br /></div>The converse of the generalized triangle inequality is true as well. If $r_1\geq r_i \geq 0$ and $r_1\leq \sum_{i = 2}^n r_i$, then there is a $n$-polygon with $r_i$ as the lengths of its sides (Fig. 2). Having one vertex of the polygon at the origin of the complex plane this can be reformulated as:<br /><br /><b>Lemma 1: </b>(Ref. )<br />Let $n\geq 2$ and $r_1\geq r_2\geq \cdots r_n \geq 0$ be real sumbers such that $r_1 \leq \sum_{i=2}^n r_i$, then there exists complex numbers $c_i$ such that $|c_i| = r_i$ and $\sum_i c_i = 0$.<br /><br /><b>Proof:&nbsp;</b><br />As stated in  this is easily proved by induction. For $n=2$ this implies that $|r_1| = |r_2|$ and thus we can set $c_1 = r_1 = -c_2$. For $n=3$ this is the converse of the triangle inequality described above. Suppose the Lemma is true for $n = k \geq 3$. Let $n = k+1$, and $s_k = r_k+r_{k+1}$. If $s_k \leq r_1$, then applying the Lemma to $r_1,\cdots r_{k-1}, s_k$ and then splitting $c_k$ into $\frac{r_k}{s_k}c_k$ and $\frac{r_{k+1}}{s_k}c_k$ would prove it for $n=k+1$. If $s_k &gt; r_1$, then $s_k \leq r_1 + r_2 \leq \sum_{i=1}^{k-1} r_k$ and again we can apply the Lemma for $n=k$.<br /><br /> <br />The degenerate case occurs when $r_1 = \sum_{i=2}^n r_i$ in which the resulting polygon must have zero area (Fig. 3). The proof of Lemma 1 also shows that the polygon can be arranged to look like a triangle, i.e. there is a partition of $\{r_i\}$ into $3$ sets $R_1$, $R_2$ and $R_3$ such that $\sum_{r_i\in R_1}r_i \geq \sum_{r_j\in R_2}r_j \geq \sum_{r_k\in R_3}r_k$ and $\sum_{r_i\in R_1}r_i \leq \sum_{r_j\in R_2\cup R_3}r_j$. This is illustrated in Fig. 4.<br /> <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-jtRhPydn-L0/WdwxP_MzMqI/AAAAAAAAAl0/UAyZJeBnpYQ4Zg2g15QpRXkkYI0JqXy5QCLcBGAs/s1600/line1.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="325" data-original-width="616" height="168" src="https://2.bp.blogspot.com/-jtRhPydn-L0/WdwxP_MzMqI/AAAAAAAAAl0/UAyZJeBnpYQ4Zg2g15QpRXkkYI0JqXy5QCLcBGAs/s320/line1.png" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;">Fig. 3:&nbsp;The degenerate polygon when $r_1 = \sum_{i=2}^n r_i$.</div><div class="separator" style="clear: both; text-align: center;"><br /></div><div style="text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-bDq-l2LR8pA/WdwxdlM0B2I/AAAAAAAAAl4/qQqPwETiaYAUuN7Y5NC7fqNEmdLikpvUgCLcBGAs/s1600/triangle2.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="293" data-original-width="603" height="155" src="https://3.bp.blogspot.com/-bDq-l2LR8pA/WdwxdlM0B2I/AAAAAAAAAl4/qQqPwETiaYAUuN7Y5NC7fqNEmdLikpvUgCLcBGAs/s320/triangle2.png" width="320" /></a></div><div style="text-align: center;">Fig. 4:&nbsp;The edges of the polygon can be reordered to form a triangle.</div><div style="text-align: center;"><br /></div><h4>overlapping edges</h4>We have implicitly allowed the possibility that the edges of the polygon may overlap, i.e. the angles of the vertices are allowed to be $0$ (see for example the degenerate case in Fig. 3). The next result shows that we can have up to $n-3$ angles to be either $0$ or $\pi$. <br /><br /><br /><b>Lemma 2: </b>(Ref. )<br />Let $r_i$ be real numbers such that $r_1\geq r_2 \geq \cdots \geq r_n \geq 0$ and $r_1 \leq \sum_{i=2}^n r_i$, then there exists $c_i$ such that $|c_i| = r_i$ and $\sum_i c_i = 0$ and either all $c_i$'s are real or $n-2$ of the $c_i$'s are real.<br /><br /><b>Proof:</b><br />We follow the proof in . Select $j\geq 3$ to be the smallest number such that $\sum_{i=3}^j r_i \geq r_1-r_2$. Such an $j$ is possible since $\sum_{i=3}^n r_i \geq r_1-r_2$. Since $r_i\leq r_2$ for $i\geq 3$, this implies that $\sum_{i=3}^j r_i &lt; r_1$ as otherwise $\sum_{i=3}^{j-1} r_i \geq r_1-r_2$. For $k = j+1,\dots n$, if $\sum_{i=3}^{k-1} r_i \leq r_1$, then set $c_k = r_k$, otherwise set $c_k = -r_k$. Ths ensures that $r_1-r_2\leq \sum_{i=3}^k c_i \leq r_1+r_2$ for each $k$ and by the converse of the triangular inequality there exists $c_1$ and $c_2$ with $|c_1| = r_1$ and $|c_2| = r_2$ such that $\sum_i c_i = 0$. Note that in the degenerate case both $c_1$ and $c_2$ are real.<br /><br /><br />The geometric interpretation of this result is that the polygon can be folded into a triangle (possibly with some overlapping edges and angles equals to $0$) as illustrated in Fig. 5 with the edges with lengths $a$ and $b$ being the 2 longest edges.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-0DsoNWluXz8/WdwyXLrI4XI/AAAAAAAAAl8/9s5QvgDWX9MCok_9802RRjyoLqkKDyPpQCLcBGAs/s1600/triangle3.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="352" data-original-width="1005" height="112" src="https://1.bp.blogspot.com/-0DsoNWluXz8/WdwyXLrI4XI/AAAAAAAAAl8/9s5QvgDWX9MCok_9802RRjyoLqkKDyPpQCLcBGAs/s320/triangle3.png" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;">Fig. 5:&nbsp;The edges of the polygon can be reordered and folded to form a triangle with the edges with lengths $a$ and $b$ being the longest 2 edges. Some of the edges may be overlapping.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><h2><br />Singularity of matrices</h2>The Levy-Desplanques theorem [3,4] (which is equivalent to Gershgorin's circle criterion ) gives a sufficient condition for a complex matrix to be nonsingular:<br /><br /><b>Theorem 1</b><br />The matrix $A = \{a_{ij}\}$ is nonsingular if $|a_{ii}| &gt; \sum_{j\neq i}|a_{ij}|$ for all $i$.<br /><br />This is easily shown by using the generalized triangle inequality. Suppoe $A$ is singular, i.e. $Ax = 0$ for some nonzero vector $x$.<br />Let $i$ be such that $|x_i|\geq |x_j|$ for all $j$. Since $x\neq 0$, this implies that $|x_i| &gt; 0$. Then applying the generalized triangle inequality to $|\sum_{j\neq i}a_{ij}x_j| = |a_{ii}x_i|$ results in:<br />$$|a_{ii}||x_i| \leq \sum_{j\neq i}|a_{ij}||x_i|$$<br />Dividing both sides by $|x_i|$ shows that it violates the condition that $|a_{ii}| &gt; \sum_{j\neq i}|a_{ij}|$ for all $i$.<br /><br />Similarly, the converse of the triangle inequality can be used to prove the following statement:<br /><br /><b>Theorem 2:</b> (Ref. )<br />Let $A$ be a real nonnegative matrix and let $D$ is a nonzero diagonal matrix with nonnegative diagonal elements. If $B=DA$ satisfies $b_{ij}\leq \sum_{k\neq i} b_{kj}$ for all $i$, $j$, there there exists a complex singular matrix $C = \{c_{ij}\}$ with $|c_{ij}| = a_{ij}$.<br /><br />If $d_ia_{ij}\leq \sum_{k\neq i} d_ka_{kj}$ for all $i$, $j$, then Lemma 1 implies that there are $g_{ij}$ with $|g_{ij}| = d_ia_{ij}$ such that $\sum_i{g_{ij}} = 0$.<br />By choosing $c_{ij} = \frac{g_{ij}}{d_i}$ if $d_i \neq 0$ and $c_{ij} = a_{ij}$ if $d_i = 0$, we get a matrix such that $\sum_i d_ic_{ij} = 0$ and $|c_{ij}| = a_{ij}$. Since $D$ is not the zero matrix, this implies that the rows of $\{c_{ij}\}$ are linear dependent, hence $C$ is singular.<br /><br /><br />These results were extended by the Camion-Hoffman theorem  which gives necessary and sufficient conditions for a real matrix $A$ such that any complex matrix whose elements have the same norm as the corresponding elements in $A$ is nonsingular. More precisely it is stated as:<br /><br /><b>Theorem 3:</b> (Camion-Hoffman)<br />Let $A = \{a_{ij}\}$ be a real matrix of nonnegative numbers. The following conditions are equivalent:<br /><br /><ol><li>If $C = \{c_{ij}\}$ is a complex matrix with $|c_{ij}| = a_{ij}$ then $C$ is nonsingular.</li><li>If $D$ is a nonzero diagonal matrix with nonnegative diagonal elements, then $B = DA$ contains an entry $b_{ij}$ such that $b_{ij} &gt;\sum_{k\neq i} b_{kj}$.</li><li>There exists a permutation matrix $P$ and a diagonal matrix $D$ with positive diagonal elements such that $B = PAD$ is strictly row sum dominant, i.e. $b_{ii} &gt;\sum_{i\neq j} b_{ij}$.</li></ol><br /><br />If the conditions in Theorem 3 are not satisfied, then there is a candidate matrix $C$ with $|c_{ij}| = a_{ij}$ such that $C$ is singular.<br />Lemma 2 can be used to show that this candidate can be chosen such that each row is real except for possibly two elements.<br /><br /><b>Lemma 3: </b>(Ref. )<br />If $B$ is a singular complex matrix, then there exists a singular complex matrix $C$ such that each row has either $0$ or $2$ complex elements and $|b_{ij}| = |c_{ij}|$.<br /><br /><b>Proof:</b><br />Suppose $Bz = 0$. As before, we select $c_{ij} = \frac{b_{ij}z_j}{|z_j|}$ if $z_j\neq 0$ and $c_{ij} = |b_{ij}|$ otherwise, i.e. $\sum_i c_i = 0$. This means that $\sum_i c_{ij}|z_j| = 0$, i.e. $C$ is singular. By Lemma 2, we can replace up to $n-2$ of $c_{ij}$ with real numbers of the same norm. <br /><br />This result is extended in  to show that this candidate can be chosen with at most $2$ complex elements:<br /><br /><b>Theorem 4:</b> (Ref. )<br />If $B$ is a singular complex matrix, then there exists a singular complex matrix $C$ with either $0$ or $2$ complex elements such that $|b_{ij}| = |c_{ij}|$.<br /><br /><b>References</b><br /> P. Camion and A. J. Hoffman, "<a href="https://msp.org/pjm/1966/17-2/pjm-v17-n2-p03-p.pdf">On the nonsingularity of complex matrices</a>," <i>Pacifi c Journal of Mathematics</i>, vol. 17, no. 2, pp. 211-214, 1966.<br /> D. Coppersmith and A. J. Hoffman, "<a href="https://doi.org/10.1016/j.laa.2004.03.029">On the singularity of matrices</a>," <i>Linear Algebra and its Applications</i>, vol. 411, pp. 277-280, 2005.<br /> &nbsp;L. Lévy, "Sur la possibilité du l'équilibre électrique," <i>C. R. Acad. Sci. Paris</i>, vol. 93, pp. 706-708, 1881.<br /> J. Desplanques, "Théorème d'algèbre," <i>J. de Math. Spec.</i>, vol. 9, pp. 12-13, 1887.<br /> S. A. Geršgorin, "Über die Abgrenzung der Eigenwerte einer Matrix," <i>Izv. Akad. Nauk SSSR Otd. Fiz.-Mat. Nauk</i>, vol. 7, pp. 749-754, 1931.<br /><br /><br /> <br /><br /> <br /><br /> <br /><br />Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-2637928472859453831.post-8521707709229286372017-09-08T23:28:00.000-04:002017-09-08T23:28:26.740-04:00Numbers such that 7 is the smallest decimal digit of their squares.9949370777987917 is the smallest number whose square has as its smallest decimal digit 7.<br />Note that $9949370777987917^2 =&nbsp;98989978877879888789778997998889$. What is the next such number?Unknownnoreply@blogger.com2tag:blogger.com,1999:blog-2637928472859453831.post-40611437103828599982017-07-23T23:02:00.001-04:002017-07-23T23:02:38.726-04:00Solar eclipse 2017On August 21, 2017, there will be a <a href="https://www.space.com/33797-total-solar-eclipse-2017-guide.html">solar eclipse</a> occurring coast to coast across the continental United States. The last time this happened was almost a century ago in 1918. This has two effects on states that generate a lot of solar energy: the solar energy generated decreases during an Eclipse and people turn on their lights as the sky darkens. Both of these events stress the utility grid. &nbsp;California's utilities <a href="http://www.sfgate.com/business/article/Eclipse-plea-from-regulators-Stay-in-the-dark-11307981.php">are asking people not to turn on the lights during this event</a>. I am surprised such a short event (and California is not even in the path of totality) can cause concerns for power utilities operators. I wonder if something similar happens when there is no wind in the summer and more people turn on their fans to cool off?Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-2637928472859453831.post-83337018620909083602017-07-17T18:41:00.001-04:002017-07-17T19:33:21.742-04:00Happy Yellow Pig Day!Today, July 17 is <a href="http://holidayinsights.com/other/pigday.htm">Yellow Pig Day</a>. This "holiday" was created by Princeton math students&nbsp;Michael Spivak and David C. Kelly when they were analyzing the number 17 and wanted to have a day to celebrate this number. This year (2017) is extra special since the year also ends in the digits 17.<br />Michael Spivak is the author of the wonderful differential geometry texts (published by&nbsp;Publish-or-Perish Press which he founded) which I fondly remember reading in graduate school. He also wrote a popular typesetting book on TeX which I did not read since Leslie Lamport's book on LaTeX makes typesetting papers using LaTeX so easy that I have skipped learning TeX and went straight to using LaTeX (which sits on top of TeX).<br />If I remember correctly, the typesetting software I used prior to LaTeX was ChiWriter, which is quite easy to use as it is WYSIWYG, so moving to a typesetting paradigm where you relinquish control on the formatting, page breaks, etc. to the software is quite unnerving, but I soon learn to appreciate it as I can focus on the content of the document, rather than spend time improving its appearance.Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-2637928472859453831.post-70563981728433430522017-06-04T11:22:00.002-04:002017-06-04T11:40:59.505-04:00Moon and Jupiter on June 3, 2017The Moon and Jupiter are in close proximity (not physically, but as viewed from Earth) on the night of June 3, 2017.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-oLd0NjsdKtg/WTQlleG4I_I/AAAAAAAAAkg/nA3yG6rkpEwD-QdFAuEi_Qwk3FphDdvjQCLcB/s1600/Moon%2Band%2BJupiter.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1600" data-original-width="911" height="640" src="https://1.bp.blogspot.com/-oLd0NjsdKtg/WTQlleG4I_I/AAAAAAAAAkg/nA3yG6rkpEwD-QdFAuEi_Qwk3FphDdvjQCLcB/s640/Moon%2Band%2BJupiter.JPG" width="363" /></a></div><br />Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-2637928472859453831.post-86717597040703302262017-05-21T10:28:00.002-04:002017-05-21T10:28:41.988-04:00The race of Cloud ComputingI wrote <a href="http://accidentaldesultorycogitations.blogspot.com/2016/11/bots-creating-news-digests-about-bots.html">earlier</a> about a news digest app returning supplemental information which is not directly related to the article. Today, I read on the app that the horse "Cloud Computing" has won the Preakness race. And sure enough, there is a box underneath the article with a definition of cloud computing from Wikipedia. What is ironic is that this news summary is probably enabled by leveraging cloud computing. "Cloud Computing" won over "Classic Empire" by coming in from behind and won the race by a head. Perhaps this is an indication how cloud computing will overtake the classical empire of on premise computing😊.Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-2637928472859453831.post-61084621597873111042017-05-11T12:22:00.001-04:002017-05-11T12:22:22.014-04:00Find interesting properties of your favorite numberI created a simple web application where you can enter a positive integer &lt; 10<sup>12&nbsp;</sup>, wait a few moments and it will return a list of interesting properties about this number. Please give it a try and let me know what you think!<br /><br />To start, please visit:&nbsp;<a href="http://postvakje.pythonanywhere.com/">http://postvakje.pythonanywhere.com/</a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-2637928472859453831.post-65153458445777302502017-05-07T17:05:00.000-04:002017-05-07T17:35:58.236-04:00Spring is here, pt. 2<div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-yuhesL5GL1U/WQ-LJi0T9xI/AAAAAAAAAkA/xuYzTjU7-rEN1IXeKEBlN6Z1ldA-HLI3wCLcB/s1600/housefinch.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="376" src="https://1.bp.blogspot.com/-yuhesL5GL1U/WQ-LJi0T9xI/AAAAAAAAAkA/xuYzTjU7-rEN1IXeKEBlN6Z1ldA-HLI3wCLcB/s400/housefinch.JPG" width="400" /></a></div><div style="text-align: center;"><br /></div><div style="text-align: center;">House Finch</div><div style="text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-r2WMmYq3XPY/WQ-LhS9OV8I/AAAAAAAAAkE/KupeRX-htRwWnajF6mrs65y5Ik2NQNC3ACLcB/s1600/chippingsparrow.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="376" src="https://2.bp.blogspot.com/-r2WMmYq3XPY/WQ-LhS9OV8I/AAAAAAAAAkE/KupeRX-htRwWnajF6mrs65y5Ik2NQNC3ACLcB/s400/chippingsparrow.JPG" width="400" /></a></div><div style="text-align: center;"><br /></div><div style="text-align: center;">Chipping Sparrow</div><div style="text-align: center;">My wife says it reminds her of Red from Angry Birds</div>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-2637928472859453831.post-30704029233172138212017-05-02T21:37:00.000-04:002017-05-02T21:37:09.549-04:0012345The number 12345 is interesting as it is a sphenic number (product of 3 distinct primes) and its digit reversal 54321 is also a sphenic number.Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-2637928472859453831.post-69731025901704763722017-04-18T19:52:00.000-04:002017-04-18T21:19:32.497-04:00Palindromic primes which are sums of 3 consecutive primes<a href="https://oeis.org/">OEIS</a> sequence <a href="https://oeis.org/A113846">A113846</a> lists palindromic primes $q$ which are the sums of 3 consecutive primes $p_1$, $p_2$, $p_3$ such that $p_2$ is also a palindromic prime.<br /><br /><b>Theorem</b>: All terms of A113846 have an odd number of digits and the first digit (and last digit) is either 3 or 9. In addition, $p_2$ and $p_3$ have the same first digit and $p_2$ and $p_3$ have the same number of digits as $q$. The first digit of $p_2$ (and $p_3$) is equal to the first digit of $q$ divided by 3, i.e. if the first digit of $q$ is $3$, then the first digit of $p_2$ and $p_3$ is $1$.<br /><br /><b>Proof</b>: First note that all palindromes with an even number of digits is divisible by 11. Let $p_1$, $p_2$, $p_3$ be 3 consecutive primes such that $p_2$ is a palindrome and $q = p_1+p_2+p_3$ is a palindromic prime. &nbsp;It is clear that $q, p_2 &gt; 11$ and have a odd number of digits since the first term of A113846 is 31513.<br />The first digit of $p_2$ cannot be 2 since $p_2$ is odd. Next we show that the first digit of $p_2$ is either 1 or 3.<br />Let $k$ be an even integer such that $p_2$ has $k+1$ digits. Suppose the first digit of $p_2$ is 4 or larger, i.e. $4\cdot 10^k \leq p_2 &lt; 10^{k+1}$. Then by <a href="https://en.wikipedia.org/wiki/Bertrand%27s_postulate">Bertrand's postulate</a> (or Bertrand-Chebyshev theorem), $p_1\geq \frac{p_2}{2}$ and $p_3 \leq 2p_2$. This implies that $10^{k+1} &gt; p_1 \geq 2\cdot 10^k$, $4\cdot 10^k &lt; p3 \leq 2\cdot 10^{k+1}$ and thus $10^{k+1} &lt; q &lt; 4\cdot 10^{k+1}$, i.e. $q$ has an even number of digits, a contradiction.<br />Next we show that the first digit of $q$ is 3 or 9. To do this, we need a stronger result than Bertrand's postulate for the prime gap. Since $p_2 &gt; 647$, Rohrback and Weis's 1964 result  shows that $p_1 \geq \frac{12}{13}p_2$ and $p_3 \leq \frac{14}{13}p_2$.<br />This shows that $2\frac{12}{13} p_2 \leq q \leq 3\frac{1}{13} p_2$. Since $q$ is prime, it cannot start with the digit 5. If $p_2$ start with the digit 1, then $2\cdot 10^k \leq q \leq 6 \frac{2}{13} 10^k$ and thus $q$ must start with the digit 3.<br />If $p_2$ start with the digit 3, then $8\frac{10}{13}10^k \leq q \leq 12\frac{4}{13}10^k$. Since $q &lt; 10^{k+1}$, this means that $q$ must start with the digit 9.<br />Clearly $p_3$ must have the same number of digits as $p_2$ and as $q$.<br />Finally, we show that $p_3$ has the same starting digit as $p_2$. Suppose $p_2$ starts with the digit 1. If $p_3$ has a different starting digit, this must mean that $p_3 \geq 2\cdot 10^k$, i.e.<br />$p_2 \geq 1\frac{11}{13}10^k$, $p_1\geq 1\frac{119}{169} 10^k$ and $q \geq 4\cdot 10^k$, a contradiction.<br />Suppose $p_2$ start with the digit 3. If $p_3$ has a different starting digit, this must mean that $p_3 \geq 4\cdot 10^k$, i.e.<br />$p_2 \geq 3\frac{9}{13} 10^k$, $p_1 \geq 3\frac{69}{169}10^k$ and $q\geq 10^{k+1}$, a contradiction. QED<br /><br />This argument also shows that to search for terms, for each $k$, one only needs to consider for $p_2$ the palindromic primes in the ranges $(10^k, 1\frac{7}{19}10^k)$ and $(3\cdot 10^k, 3 \frac{8}{19}\cdot 10^k)$ which amount to approximately $0.79\cdot 10^{\frac{k}{2}}$ palindromes to consider rather than a full search of $10\cdot 10^{\frac{k}{2}}$ palindromes of length $k+1$.<br /><br /><b>Conjecture</b>: For all terms in A113846, the corresponding primes $p_1$ has the same number of digits and the same first digit as $p_2$ and $p_3$.<br /><br /><b>References:</b><br />1. H. Rohrback and J. Weis, "Zum finiten Fall des Bertrandschen Postulats", J. Reine Angew. Math., 214/215 (1964), 432–440.<br /><br />Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-2637928472859453831.post-39472009756355315752017-03-30T20:50:00.001-04:002017-03-30T20:50:24.877-04:00Spring is hereIt is amazing how a couple of days of sunshine and warmer weather erased most of the evidence that we had a major winter storm 2 weeks ago.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-GR1LfKdlXAI/WN2nozspfzI/AAAAAAAAAjk/kR3H46NBluwWjJMmhGEvhFqta7VfdalBACLcB/s1600/mourning%2Bdove.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="258" src="https://2.bp.blogspot.com/-GR1LfKdlXAI/WN2nozspfzI/AAAAAAAAAjk/kR3H46NBluwWjJMmhGEvhFqta7VfdalBACLcB/s400/mourning%2Bdove.JPG" width="400" /></a></div><br />Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-2637928472859453831.post-32122408108750276142017-03-15T19:29:00.000-04:002017-03-15T19:32:57.894-04:00Winter storm Stella<div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-x6FUNcvXv5s/WMnPHZl5dlI/AAAAAAAAAjI/MVhsdRbCsWEH7HK0gE98Pe2dUYevG0w-ACLcB/s1600/winterstormstella.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="276" src="https://1.bp.blogspot.com/-x6FUNcvXv5s/WMnPHZl5dlI/AAAAAAAAAjI/MVhsdRbCsWEH7HK0gE98Pe2dUYevG0w-ACLcB/s640/winterstormstella.JPG" width="640" /></a></div><br />Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-2637928472859453831.post-42726732514805362017-03-14T00:30:00.000-04:002017-03-14T09:53:16.039-04:00Pi Day 2017Happy Pi day (and a snowy one in the Northeastern US)! One of the early childhood mathematical memories I had was learning that&nbsp;<span style="background-color: white;">π</span>&nbsp;can be approximated by 22/7 and accurate up to 2 decimal places, a fact that was known since antiquity. In particular, Archimedes gave the first <a href="https://en.wikipedia.org/wiki/Proof_that_22/7_exceeds_%CF%80">proof</a> that 22/7 is strictly larger than&nbsp;<span style="background-color: white;">π. The fifth century Chinese mathematician and astronomer&nbsp;</span><a href="https://en.wikipedia.org/wiki/Zu_Chongzhi">Zu Chongzhi</a>&nbsp;gave the approximation <a href="https://en.wikipedia.org/wiki/Mil%C3%BC">355/113</a> that is accurate up to 6 decimal places. Today, sophisticated algorithms using Ramanujan's formula and its variants can compute<a href="http://www.numberworld.org/misc_runs/pi-12t/"> <span style="background-color: white;">trillions of digits of&nbsp;</span></a><span style="background-color: white;"><a href="http://www.numberworld.org/y-cruncher/">π</a>.</span>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-2637928472859453831.post-58410116509383636222017-03-09T20:46:00.000-05:002017-03-09T20:46:46.909-05:00Binomial coefficientsWhen studying binomial coefficients and how they form Pascal's Triangle, we learn that they are the coefficients of $(x+y)^n$, i.e.<br /><div>$$(x+y)^n = \sum_{k=0}^n \left(\begin{array}{c}n\\k\end{array}\right)x^ky^{n-k}$$</div><div>Replacing $y$ with $-y$ we get:</div><div>$$(x-y)^n = \sum_{k=0}^n -1^k\left(\begin{array}{c}n\\k\end{array}\right)x^{n-k}y^{k}$$</div><div><br /></div><div>For example, the coefficients of $(x+y)^4 =&nbsp;x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$ are 1,4,6,4,1 which form the 5th row of Pascal's triangle.</div><div><br /></div><div>What happens if the coefficients of the polynomial are repeated binomial coefficients such as 1,1,4,4,6,6,4,4,1,1? What is $$\sum_{k=0}^{2n+1} \left(\begin{array}{c}n\\\lfloor k/2\rfloor\end{array}\right)x^ky^{n-k}?$$</div><div><br /></div><div>It is easy to show that this is equal to $y(x^2+y^2)^n + x(x^2+y^2)^n = (x+y)(x^2+y^2)^n$.</div><div><br /></div><div>What about repeating coefficients with sign changes such as 1,1,-4,-4,6,6,-4,-4,1,1?</div><div>The same argument shows that&nbsp;</div><div><br /></div><div>$$\sum_{k=0}^{2n+1} -1^{\lfloor k/2\rfloor}\left(\begin{array}{c}n\\\lfloor k/2\rfloor\end{array}\right)x^{n-k}y^{k}$$</div><div><br /></div><div>is equal to $y(x^2-y^2)^n + x(x^2-y^2)^n = (x+y)(x^2-y^2)^n$ which can be further simplified as $(x+y)^{n+1}(x-y)^n$.</div><div><br /></div><div>If we now change the sign of every other coefficient as well, i.e. 1,-1,-4,4,6,-6,-4,4,1, -1,</div><div>we get&nbsp;</div><div><br /></div><div>$$\sum_{k=0}^{2n+1} -1^{\lfloor k/2\rfloor+n-k}\left(\begin{array}{c}n\\\lfloor k/2\rfloor\end{array}\right)x^{n-k}y^{k}$$</div><div><br /></div><div>This can be shown to be equal to $x(x^2-y^2)^n -y(x^2-y^2)^n = (x-y)(x^2-y^2)^n$ which is equal to $(x-y)^{n+1}(x+y)^n$.</div><div><br /></div><div>We can repeat the coefficients more than 2 times as well and show that</div><div><br /></div><div>&nbsp;$$\sum_{k=0}^{m(n+1)-1} \left(\begin{array}{c}n\\\lfloor k/m\rfloor\end{array}\right)x^ky^{n-k}$$</div><div><br /></div><div>is equal to $\sum_{t=0}^{t=m-1}x^ty^{m-1-t}(x^m+y^m)^n$.</div><div><br /></div><div><br /></div>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-2637928472859453831.post-61029137528646240052017-01-20T11:12:00.000-05:002017-01-20T12:12:35.823-05:00Interesting number: 1440000The number 1440000 satisfies the following equation:<br /><br />$$1440000^3 - 1 = 1439940^3 + 71999^3 = 1440060^3 - 72001^3$$<br /><br />By <a href="https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem">Fermat's Last Theorem</a> (for which the <a href="https://en.wikipedia.org/wiki/Proof_of_Fermat's_Last_Theorem_for_specific_exponents">case $n=3$</a> was proven by Euler), $1440000^3$ cannot be the sum or difference of 2 positive integer cubes. The above equation shows that $1440000^3-1$ is the sum and the difference of 2 positive integer cubes.<br /><br />Unknownnoreply@blogger.com0