Announcements
UPDATE August 2nd 2024: All charts have been replaced in Lesson 4 and Lesson 5. You will now have all the information required to complete the assignments for these lessons.
Welcome to Arithmancy 601
Before you enrol in this course, there are just a few things the Arithmancy team would like you to know.
- This course will involve some mathematical theory. Do not let this deter you, as we're always happy to assist with any questions you may have, whether they're about the general mathematical concepts, or Arithmancy-specific.
- We grade your assignments as quickly as we can but please do be patient.
- If you believe that there has been an error in the grading of your assignment, please send an owl to Professor Buchanan. We make mistakes sometimes, too. Just make sure to explain why you think there is an error, and include your assignment's ID number.
- If you have achieved 70% or higher in an assignment, you will not be able to retake it. Therefore, make sure you are happy with your work before submitting. If you are uncertain of something in the lesson, please ask for help before you complete the assignment.
Lesson 6) Introduction to Advanced Operations
Professor Buchanan enters the classroom just in time for the bell, carrying a pile of marked midterm papers. “Alright, settle down, please. We’ve got a lot to cover today.” He puts the pile of papers down and then perches himself on the edge of the desk, facing the class. With a clap of his hands, he declares, “Wands can go on your desks today. We’ve got a spell to learn later!”
Introduction to Advanced Operations
I hope that your Midterms have treated you well last week! I see that many of you provided compelling analyses on the names of people and their meanings, and I am confident that you will do well in the future if a querent ever asks you to calculate their personality numbers. Having said that, this means it’s time to take another step forward and discuss a few operations that will be a bit more complicated than the basic additions and multiplications we’ve studied so far.
Before we begin, though, I’d like to discuss why we cover these operations. Many simple Arithmantical tools - such as the personality numbers we discussed last week - only require addition. Sometimes techniques, however, will require you to multiply sequential numbers or find the “origin” of a multiplication to identify the best number to be used. For that reason, it is important for us to understand these operations. Note that these calculations will particularly come in handy when dealing with more technical applications of Arithmancy next year.
However, I also bring you good news! Although calculators or other pieces of Muggle technology, such as the one you could find on this website, do not work within our school’s bounds, there’s a charm that performs more or less the same function. This spell, aptly named the Calculation Charm, produces the result of any mathematical expression you can think of, provided that you know how to calculate it. The final number, variable, or comment on the expression having no solution will be briefly displayed in crimson writing atop the surface of the object that your wand is pointing to.
At this point, I’d like to emphasize that you must know how you’d calculate the expression by hand. This is a corollary of the Gamp’s Principal Exception of Knowledge; you may cast a spell that simplifies the process of obtaining knowledge, but not a spell that will give you knowledge that you don’t have. Trying to use the Calculation Charm on an expression you have no idea how to solve will net you no results or, worse, a nasty backfire! Backfires may include temporary confusion and uncontrollable hair raising for long periods of time.
Calculation Charm
Incantation: Numerfinis (noo-mehr-FEE-nees)
Wand Movement: Three jabs in a triangular formation pointing upwards, in reading order. Jab in the direction of where you want the answer to be displayed, with one jab per non-stressed syllable.
Willpower: Medium, on producing the answer
Concentration: High to very high, depending on the complexity of the expression to be solved. Focus on what you want to calculate.
Exponentiations
Now that we know how to use the Calculation Charm, it’s time for us to cover some new operations! The first one we will study today is exponentiation - which I hope you will find easy, in spite of how long its name is.
In simple terms, exponentiation refers to a series of repeated multiplications, in the same way that multiplication refers to a series of repeated additions. We have covered this in Lesson Three, but this is probably a good time for a refresher: whenever we write an operation such as 8 * 3, we are essentially rewriting 8 + 8 + 8 (‘eight’ added ‘three’ times) in a simpler way. Exponentiation is the same - for example, 83 means the exact same as 8 * 8 * 8 (‘eight’ multiplied ‘three’ times).
Every exponentiation contains two terms - a base, the larger number that indicates what is being multiplied, and an exponent, the smaller number sitting on the upper-right side of the base indicating how many times we multiply it. You should note that, unlike multiplication, changing the order of the base and the exponent may give you different results! For instance, consider the expressions 32 and 23. The former is equivalent to 3 * 3, or 9, whereas the latter is equivalent to 2 * 2 * 2, or 8 - even though we used the same numbers.
However, there are still quite a few interesting properties when it comes to exponents, particularly when we multiply or divide them. The first of them surfaces when we multiply exponentiations with the same base - we can simplify them into a single exponentiation by preserving the base and adding the exponents. For example, 33 * 32 is the same as 33+2, or 35 - and the reason why is quite simple if you think about it. We know that 33 is the same as 3 * 3 * 3, and that 32 is the same as 3 * 3. Multiply them together and you’ll get 3 * 3 * 3 * 3 * 3, which is the same as three multiplied five times by itself; since this is also a repeated multiplication, we can represent it in exponential form as well.
Likewise, the division between exponentiations can also be simplified by subtracting the first exponent from the second - as an example, 47 / 45 is the same as 47-5, or 42. This becomes intuitive if we remember that divisions are the opposite of multiplication: 6 * 2 = 12 (if six wizards have two Chocolate Frogs each, there are twelve Chocolate Frogs total in the room) entails that 12 / 6 = 2 (if there are twelve Chocolate Frogs equally divided between six wizards, each wizard has two Chocolate Frogs each).
Let’s talk a little bit about zero, since that number is particularly interesting in exponentiation. One interesting property is that raising zero to (nearly) any number means the final result is also zero - for example, 04, 012 or 0674 are all equivalent to zero! This is also very intuitive when we remember that exponentiations are repeated multiplications, and that anything multiplied by zero is also zero - it’s no surprise that zero multiplied by itself many times does not break that rule.
Conversely, (nearly) any number raised to zero as an exponent is equal to one. This might seem odd at first, but remembering what we discussed previously about dividing exponentiations with the same base may come in handy. For example, imagine that we have a division such as 66 / 66 - in this case, based on the properties we discussed above, the final result will be 66-6, or 60. However, both the dividend and the divisor are equal to one another, meaning that the divisor can fit only one time inside the dividend! In other words, dividing a number by itself usually gives one as the final result. Therefore, we have that 60 is equal to 1.
You might be asking yourself - if zero raised to anything is zero, and anything raised to zero is one, then what’s the value of 00? This is why I was careful enough to use the word ‘nearly’ so many times, as 00 is actually undefined. If you’ll recall our third lesson, divisions by zero do not produce a real number, and for that reason raising zero to the zero-eth power (which is also a division by zero if we use the technique discussed in the last paragraph) must also be undefined.
Lastly comes the exponentiation of an exponentiation! Nothing prevents you from stacking multiple exponents, such as (32)4 - in that case, what is the final result? If you think closely about it, you’ll realize that we should have the exponent 32 multiplied by itself four times, which is yet again a multiplication of exponentiations with the same base. 32 * 32 * 32 * 32 is the same as 32+2+2+2 or, more simply, 32*4 - and here’s where the number four appears again! In other words, raising an exponentiation to an exponent is the same as multiplying the exponents by one another.
Radicals and Logarithms
Alright, now that we are done with exponentiations, let’s talk about radicals - which are represented by the symbol √ above and to the left of the value we want to break down. Radicals may have a root, or a number above its V section (for example, ā) indicating into how many parts the number must be broken, or it may be left empty. If empty, this means we are dealing with a square root, or finding a number that, when multiplied by itself, gives that final result.
In simple terms, radicals undo the operations done by exponentiations - the latter gets one number to be used for a repeated multiplication and one number to indicate how many times it must be repeated, giving the result of this successive multiplication, whereas the former gets one number to be tested and another number to indicate in how many “multiplying parts” it must be broken into.
To give a more concrete example, let us think about the number 64 (which can be broken into the multiplication 4 * 4 * 4). If we were to calculate ā64, we need to identify which number, if multiplied by itself three times, gives the final result 64. The answer to that question is four.
Calculating radicals by hand can be quite difficult and time-consuming, as it’s difficult to start from the final result and try to find the original pieces. If you ever have to calculate a radical, feel free to use the Calculation Charm!
Finally, there is yet another form of calculation that is quite similar to exponentiations and radicals: logarithms. While radicals start from the final number and figure out what number can be multiplied by itself a certain number of times, logarithms start from the final number and figure out how many times a base must be multiplied by itself to get there. This is perhaps best illustrated with yet another comparative example.
Take a look at the calculations above. As a reference, we know that three multiplied by itself four times is equal to 81. The first operation, exponentiation, asks how much is three multiplied by itself four times, whereas the second operation, the radical, asks what number, when multiplied by itself four times, will give 81 as a result. Last but not least, the final operation, the logarithm, asks how many times three must be multiplied by itself to reach a result of 81.
Just like exponentiation, there are a few properties that are associated with logarithms - for those that have been paying attention to the class, you’ll notice they are quite similar to the ones we discussed when talking about exponentiation. Firstly, we know that, for any number a, loga1 is always zero, and that logaa is always one. This is directly correlated to the fact that any number to the power of zero is one, and that any number to the power of one (i.e., multiplied by itself only one time) is the very starting number.
We also know, as a callback to multiplying and dividing exponentiations with the same base, that loga(m * n) is the same as logam + logan and that loga(m/n) = logam - logan. Likewise, from the rule that raises a power to another power, we may derive that logamn is the same as n * logam. Do note that these are just rewritten forms of the properties we discussed above when talking about exponentiations: for the sake of brevity, I will not go over its proofs again.
Although there are a few more properties for logarithms, I believe this is quite enough for our first foray into the topic! In case you are interested about learning more, make sure to reach out to me or one of my PAs.
Factorials
It’s time for something I believe most of you will find much easier - factorials. A factorial (represented by the symbol !) is nothing more than the successive multiplication of all natural numbers up to the value listed next to the factorial symbol. For example, 7! is equivalent to 1 * 2 * 3 * 4 * 5 * 6 * 7 = 5040.
Factorials are particularly useful when dealing with the reorganization of groups, and for that reason they see a lot of application when discussing the Arithmantical properties of any type of human assembly. There are in fact three types of classical calculations that use factorials: permutations, combinations and arrangements.
Let’s discuss permutations first. In a nutshell, permutations are calculations relating to the reordering of elements in a group - for example, the number of different ways I can cast three spells in sequence or the number of different paths in a tour that visits four departments in the Ministry of Magic. In order to calculate the total number of permutations for a given number of elements, you simply have to calculate n!, where n is the number of elements involved.
Take a look at the example above regarding spellcasting. Imagine that I want to cast the Stunning Spell (S), the Disarming Charm (D) and the Impediment Jinx (I) during a duel - in that case, how many different orders are possible for my spellcast? We can do this by hand and take note of every different spell order (SDI / SID / DSI / DIS / ISD / IDS), but it’s easier to calculate 3!, or 1 * 2 * 3. There are six possible ways for you to cast all of these spells.
Next comes combinations, which calculate the number of different groups that can be formed from a pool of people. Imagine, for example, that five Aurors (out of nine trainees) will be selected to join the Auror Office - if you were to calculate how many different groups of Aurors could be selected, you’d use the combination formula.
To calculate the total number of combinations, we need two numbers - namely, the size of the pool from which to draw our options, and the size of the smaller selected group. Let’s call the pool size m and the size of the smaller group n. In that case, the total number of combinations we can form is equivalent to m! / [n! * (m - n)!]. I know this might seem like a handful, but it all gets easier when we put it into specific numbers!
Let’s go back to the Auror Office example. If we want to pick five Aurors out of nine trainees, we want to calculate 9! / (5! * 4!); note that 9 - 5 is equal to 4 and, as such, occupies the last number in our factorial calculation. By multiplying all factorials, we get a final result of 362880 / (120 * 24) = 126. That means we have quite a few different options to think about!
The last type of calculation that uses factorials is arrangements, which pertain to the creation of groups where their order matters; some examples of this type of group can be seen in sports competitions (being part of the Top 3 as first place is different from being third) or creating a study schedule for your NEWTs (studying Astronomy first thing in the morning or during the night can give you very different results if you want to observe the planets).
In order to calculate an arrangement, we can just multiply the formulas for combination and permutation - after all, what we’re doing is simply (1) finding how many different groups we can form from a larger pool, and then (2) finding how many different ways this smaller group can be reordered. If we define the pool size as m and the smaller group size as n, we can calculate the total number of arrangements as (n! * m!) / [n! * (m - n)!]. Since both sides of the division have an n!, this can be further simplified into m! / (m - n)! instead.
As usual, it is time for a practical example. Imagine that you are currently taking seven NEWTs, and want to review the contents of four of these classes on Sunday to prepare for the next week - how many different schedules can you create? By plugging the numbers in our formula, we determine that we can create a total of 7! / 3! schedules, which is equivalent to 840 ways to organize your weekend studies.
Closing
That should be quite enough for today! Although there are a few other techniques that weren’t mentioned so far, such as differential calculus and Bayesian inference. These exceed the scope of our course and would be best suited for those who want to pursue a specialization in Arithmancy. Next week, we will take a breather from all these calculations and embark on a historical journey.
Dismissed.
Original lesson written by Professor Vaylen Draekon
Improvements made by Professor Calum Buchanan
Image credits here, here, here and here
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