Use digits in the place of letters to solve the above equation. Each letter represents different digit.
]]>Probability puzzle is available on google app store:
it comprises of 3 levels of difficulty:
3. Outrageous:
That is my progress so far.
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To present this to my Guides I discussed the Caesar Cipher and then had each girl make her own decoding ring using this pattern. The girls could then practice coding and decoding messages for each other.
]]>That is the question Fiona Presly faced last spring when she noticed a wingless queen bumblebee in her garden in Inverness, Scotland, one blustery afternoon.
“To bee,” as it happened, won out, and now Fiona’s photos and videos of her sweet relationship with Bee — as she named her — have gone viral, racking up more than 4 million views on social media since she published her observations in a story for the Royal Entomological Society earlier this year.
Entomologists told Fiona that the bee likely lost her wings to Deformed Wing Virus, one of several viruses that threaten bees globally.
“I love that the world is now talking about bees — they are extremely important, as are all of our pollinators,” Fiona, a 55-year-old library assistant, tells PEOPLE. “No bees, no us!”
On that chilly day in the Highlands, she coaxed the bee to climb onto her hand, then carefully carried her inside and made a nest of soft black cotton inside a plastic box covered with a domed mesh cover. She also added a feeding station filled with sugar water, a small pond for bathing and a miniature paper ladder so that Bee could climb out whenever she liked.
Then, a couple of weeks later, Fiona built a larger home inside a wooden crate and filled it with heather, anemone and sedum — flowers she’d noticed that other bees appreciated.
Truth be told, she says, “I didn’t think of it as bonding at first — bees are intelligent and I figured she knew that I was a source of food. But I did start to feel responsible for her. I noticed that Bee would physically perk up when I came into her presence if she’d been alone for a while. Basically, if a creature moves I will talk to it, so I talked quite a bit to Bee. I got the sense that she really was quite grateful for what I did.”
At the beginning of last May, Fiona put Bee back into the garden for a few hours and was devastated when the bumblebee disappeared for three days.
“I thought that was the end — I’d done my best,” she tells PEOPLE. “But then she emerged, looking a bit disheveled and thirsty, so I was delighted to give her a drink of sugary water from my finger.”
Bee was soon crawling onto Fiona’s nose to clean herself, and she enjoyed snuggling in the warmth of her hand. Because a bee’s life span is short but sweet, Fiona didn’t expect that her wee and fuzzy pet would live for more than a couple of weeks.
Can’t get enough of cats, dogs and other furry friends? Click here to get the cutest pet news and photos delivered directly to your inbox.
“I’ve always loved bees and I’ve rescued many in my lifetime, but I’m a realist,” she says. “She should have died in the height of summer when new queens emerge, typically at the end of July or early August. I was surprised to have her with me for five months.”
In September, Fiona noticed that Bee moved more slowly and didn’t eat or drink as much, “so I was prepared,” she says. “On September 15th, she was in my hand when she passed. I felt a tinge of sadness, but I knew it was a privilege to be able to have this special bond with her.”
Bee was buried in the garden close to a favorite patch of heather, which Fiona visits often. She says she feels humbled to have had a relationship with the queen bee.
“People have a bond with their dogs and cats,” she says, “and I’ve now proven that you can also have a relationship with an insect. I think the world has enjoyed a nice story amongst all of the doom and gloom. I spent a wonderful summer with a very special bee.”
]]>Papers 1 and 2 are made up of two sections and each paper is worth 300 points. Each section is worth 150 points and there are 9 questions in total you will need to answer. You must answer all the questions. Every question has multiple parts. Most of the questions are word problems.
Section A is Concepts and Skills. There are 6 questions in this section.
Section B is Contexts and Applications. There are 3 questions in this section.
Best of luck with your exams!
]]>Why isn’t there more research done on the kinetic power of shoes? Couldn’t we sneak a few dollars from the military budget?
]]>This post is the third in four that will appear each Monday in May. The series walks through the main steps in hypothesis testing, which are:
In the first part, we identified the null hypothesis and alternative hypotheses of interest in our case study, “Meta-Analysis of Intellectual and Neuropsychological Test Performance in Attention-Deficit/Hyperactivity Disorder”. In our second part, identified the z-test statistics for the variables of interest, namely the mean scores of ADHD groups versus healthy groups on various cognitive and neuropsychological assessments. These test statistics told us how much the observed results deviated from the expected results as predicted by the null hypothesis.
Recall that the null hypothesis and alternative hypotheses in our review are as follows:
The first alternative hypothesis focuses on overall cognitive ability, while the second focuses on total neuropsychological performance. For the FSIQ [Full Scale IQ] scores of the ADHD groups relative to the control groups, z = 27.72, meaning that the mean scores of the ADHD group differed from the control group by 27.72 standard deviations. For the comparison of neuropsychological performance, z = .
Now in step 3 of hypothesis testing, we use a p-value to evaluate the significance of the results.
A p-value basically tells us that we have a x percent change of observing the results that we did given that the null hypothesis is true. In our case study, the p-value says that the probability of observing a difference of 27.72 standard deviations in the FSIQ and 2.5 or more standard deviations in the range of neuropsychological assessments between the ADHD group and the non-ADHD group is such-and-such amount if there really is no difference in the mean scores on a given cognitive or neuropsychological assessment between the two (the ).
To use a p-value, you must first set a confidence level. When a study uses, say, a 90% confidence level, the researchers are informing their colleagues, “We can say with 90% confidence that these results are true.” If their research successfully rejected the null hypothesis of their study, then the p-value must have been less than 1.0 – 0.9, or 0.1.
Most statistical studies use a confidence level of 95%, and our ADHD is no different. With a confidence level of 95%, the p-value must be less than 1.0 – 0.95, or 0.05, for the researchers to reject the null hypothesis. A p-value less than 0.05 says that it is very unlikely to observe data at least as extreme as those that we did if the null hypothesis was true. Thus, the difference must come not from study design or error, but rather from circumstances that the alternative hypotheses posit. Ergo, the null hypothesis is very likely wrong.
As the above graph shows, a small enough p-value falls outside the range of “Fail to reject “. (The -2 and +2 represent a difference of -2 and +2 standard deviations. Any points beyond this degree of deviation falls outside of a 95% confidence interval.)
Unlike a test statistic, p-values are not hand-calculated. To employ them in our most scientific and official ADHD research, we turn to statistical packages (Excel, R, Minitab, etc.), to z tables (obviously only applicable to z-statistics), or online calculators. We’ll use the latter, as z-tables commonly found online go only up to 3.49.
For z = 27.72, which concerns the first , p < 0.00001.
For z = 2.5, which concerns the second , p = 0.00621.
Both of these p-values fall below the confidence level 0.05. So, what now? In the next and last “Hypothesis Testing” post, we will wrap up this case study by interpreting the p-values in context.
Other posts in “Hypothesis Testing”
Let and define S to be the set of all bijections (one-one and onto) from every non-empty subset Y of X to Y. So essentially S is the set of all possible permutations of every non-empty subset of X.
Example: Consider , then
So , , , are a few of the bijections that we have shown.
Cycle: A cycle is an element of the set S (as is defined above) which cyclically permutes these integers of any subset of X. Hence a cycle is the mapping which sends to , such that and send to .
Example: From the previous example it is evident that and are cycles.
Disjoint Cycles: If and are cycles, then these cycles are said to be disjoint if , and .
Now we are ready to define “Stirling Numbers of the First Kind” .
Stirling Numbers of the First Kind: It is the number of permutations of n-elements comprising of exactly k-disjoint cycles.
These numbers are usually denoted in the following manner
Example: So how do you calculate ?
Here we are discussing the naive way of finding by simply checking all possible -cycles.
Hence clearly .
One thing one should keep in mind is that the permutations that we have written down are the only distinct permutations possible in -disjoint cycles. Like for example one might wonder why was not considered. We did not consider because we had already considered and is exactly the same permutation as .
So now we are going to discuss the easier and more general method to calculate .
Recurrence Relation:
Proof: Every permutation of n objects into k cycles can be done either by adding the object as a new identity cycle (maps the object to itself) or by adding the object in one of the pre-existing cycles.
So one part of the counting that deals with counting the number of permutations of (n-1) objects into (k-1)-disjoint cycles and then adding the object as an identity cycle can be done in ways.
For the other part it is clear that we have constructed k-disjoint cycles from (n-1) objects and we just need to insert the object in any one of those cycles. This can be done if we can simply insert the object ahead of any one of those (n-1) objects. So for construction of k-disjoint cycles from (n-1) objects we need ways and to insert the object we have (n-1) options. Hence we can do the second part of the counting in ways.
Hence by combining the two cases we can easily say .
It is quite clear that by using the Recurrence Relation it will be easier to find the value of .
]]>除了摺紙的方式之外，網路上可以到不只一種證明方法，但是證明的套路都是一樣的，大抵都是做輔助線和使用「等邊與等角的關係」以及「三角形外角大於其任一內對角」性質，都可以輕鬆漂亮的完成證明。以下列出三種「大邊對大角」性質的證明方法。
第一種方式，就是課本使用的摺紙法，下圖使用「更」數學的語言，描述摺紙的過程，然後簡述證明的邏輯。如上說明，證明的關鍵在角AHE 是 △HBE 的外角，因為三角形角大於其任一內對角，所以 ∠AHE = ∠ACB ≧ ∠ABC。
網路上還可以找到另外一個證明方法， 在△FGI 中做輔助線，在線段GI上取一點J，使得GF=GJ，所以△GFJ是個等腰三角形。因為等角對等邊，以及上面已經提過的三角形外角與內對角的關係，同樣可以輕鬆得證。
之前燒腦系列文章提過在知乎網站上，有篇問答解答了如何以希爾伯特的公設體系證明「三角形任兩邊之和必大於第三邊」，證明過程中提到了一個證明「大邊對大角」的方法。如下圖描述，畫輔助線，使得 成為一個等腰三角形，然後如同上面證明一樣，運用「等角對等邊」，以及一再提到的「三角形外角與內對角的關係」，簡單得證。
簡單歸納，上述三個證明，都使用「等角對等邊」，以及「三角形外角與內對角的關係」兩個性質，作為證明的基礎。恰巧都和希爾伯特幾何公設體系的邏輯是相一致的，所以使用起來，格外放心（這是冷笑話，呵呵）。
]]>We again thank every problem proposer.
Let be real numbers and . Let be a differentiable function with and let . Show that . For the equality is it necessary that the function is constant ? (Proposed by the Mathematics Scouts team).
For more information see the problem of the week discussion and archive page of the website.
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For all positive integers show that the sum converges to an irrational number.
An additional thought on the problem (we have it still unsolved).
The solver of this problem will receive a reward or a special mention from the Mathematics Scouts team.
For all positive integers is it true that the sum converges to a transcendental number.
Here is the solution to the originally asked problem.
First note that this sum converges.
Let the sum be . We need to show that is irrational.
Assume the contrary,
Then for some integers $p,q$ with no common factor other than among the positive integers. And so $p>0,q>0$.
Now, that means must be an integer.
Substituting this in the sum gives us :
And if be the above sum then it can be easily shown that and this is .
So, this is a positive integer , which directly gives us a contradiction thus concluding the solution .
This was the official solution we had in mind while proposing the problem. A huge thanks to the entire team (with a special mention to Dorlir Ahmeti from Albania).
We also received the same solution on AOPS from user Tuzo. Congratulations to him !
You are still welcome to send thoughts and / or solutions to this problem.
For more details check out problem of the week discussion and archive page of the website.
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Up above are four copies of a translational fundamental piece. There are horizontal straight lines meeting in a square pattern, vertical symmetry planes intersecting the squares diagonally, vertical lines through edge midpoints of the squares and horizontal symmetry planes half way between squares at different heights. What more could one want? Well, CLP has genus 3, and we wouldn’t mind another handle.
There are various ways of doing that, and one of them leads to today’s surface, shown above. For adding a handle we had to sacrifice the vertical straight lines, but all other symmetries are retained. These are, in fact, essentially the same symmetries we had in last week’s surface, except that there, the squares in consecutive layers were shifted against each other. The similarities go further.
Again we can ask how things look at the boundary. Pushing the one free parameter the the other limit, gives us again doubly periodic Scherk surfaces and Karcher-Scherk surfaces. There is a subtle difference (called a Dehn twist), however, how the two types of Scherk surfaces are attached to each other in both cases.
Finally, as usual, the cryptic rainbow polygons that encode everything. Today, the two fit together along their fractured edges, which has to do with the period condition these surfaces have to satisfy.
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Identify effectiveness of kd Tree solution by testing implementation on live data
CONDUCT:
RESULTS:
Comments:
An interesting point is the kd Tree depth is only going to between 8-15 levels (depending on max nodes per leaf). This means that it is not considering the full 35D data set. This may be tying in somewhat to the curse of dimensionality; the data that is being used to segregate the points is only a subset of the full data. It doesnt effect the endstate as the distance calculations consider all dimensions however the tree organisation doesnt consider all.
Another interesting note that caused some exceptions initially is that not all nodes are getting data points; even when there was a maximum of 10 points per leaf. In the initial testing errors were thrown when looking for points in a leaf node that didnt exist. This is due to the points passed to the parent node having the same value for the dimension the node was trying to split. The logic is if the data pont value at the relevant dimension is greater than or equal to the split value, the point is assigned to the right child, otherwise the left. A workaround was implemented to return from a query instantly if there are no data points on the leaf.
It occurs to me that as the number of datapoints gets smaller (coarse subset, specific subset) the curse of dimensionality will have a larger effect. An indicative guideline as to the number of points vs dimensionality for a kd Tree is N >> 2 ^k, where k is the number of dimensions. Thus as N gets smaller, the relevance of the curse of dimensionality becomes more significant.
There is one additional option for a candidate solution for the complicated weighting function; trading time for space by creating a graph to link all dimensions. Basically using a large database system or search engine form of solution. I still have some pretty big reservations that this will work however it is a potential avenue of exploration.
TIME SPENT:
7 hours (97 hours running total)
FROM HERE:
ONGOING:
Theorem 1.1:
Proof 1.1.1:
Let
Let
follows the Fibonacci recurrence.
Also , so and are identical sequences.
The purpose of this post is to derive the finite-difference equations. Specifically, I will be deriving the forward, backward, centered first order equations. We start with the Taylor expansion about the points :
and
Let . Therefore, if we consider the following differences…
and
and
and if we keep only linear terms, we get
and
where the first is the forward difference, the second is the backward difference, the last is the centered difference, and represents the quadratic, cubic, quartic,quintic,etc. terms. One can use similar logic to derive the second-order finite-difference equations.
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Scott and Laurie – with Ukhanyo School mathematics teaching assistant Thula.
Scott explained to us how the project has revolutionised the way Maths is being taught at the school. “Instead of children listening to the teacher talking for an hour, which is the normal practice, in their respective classes they do practical maths on the tables; the teacher does very little talking and goes around assisting groups.”
The project has helped the Ukhanyo teachers to incorporate novel teaching methods into their classes. It has also shown how to accommodate learners that progress at different rates so that both high achieving and struggling children can be stimulated in the same class.
The project attracted the attention of the local media and was featured in GroundUp in its early stages. It was also welcomed by Jessica Shelver, of the Western Cape Education Department, who said “This is an outstanding initiative”.
Scott and Laurie, who were primary school teachers in the USA prior to joining Masicorp, have run the project for three years and are confident that the teachers they have worked with will continue to deliver high quality Maths lessons to the pupils. The Maths Lab remains a priority among the many educational projects Masicorp offers the community.
We gathered at the Masicorp office to say a final thank you to Scott and Laurie and look forward to staying in touch with them to discuss the legacy they have left to mathematics teaching at Ukhanyo School.
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