Projects

Project 8. An Expansion on Project 5

So, during the summer, I've been updating Project 5 and generalizing it more. First, that formula that Project 5 has? IGNORE IT. It's correct, don't get me wrong, but it's incredibly too complicated. I found a much simpler one. I'll elaborate. Here is "Project 5 in Review":

Given x^n + x^(n-1) + x^(n-2) + ... = SUM{m = 0 , n }( x^m ).

Let the polynomial above under go the following transformation and become:

A_n*(x+L)^n + A_(n-1)*(x+L)^(n-1) + ... + A_2*(x+L)^2 + A_1*(x+L)^1 + A_0

= SUM{ m = 0 , n }( A_m*(x+L)^m ) = P.

For any nonzero integer L, n > 0, and k > 0, the coefficient of the k-th term for an n-degree polynomial P given by: Now, looking again at Project 5, I found a formula for the coefficient of the (n-k)-th term. As you can see, it's a lot harder. But this one above will be of importance. First, I'll show you it works. Let's let L = -2, n = 3, and k = 2.

Here is the task: Find A_2 in the following expression.

x^3 + x^2 + x + 1 = A_3*(x-2)^3 + A_2*(x-2)^2 + A_1*(x-2)^1 + A_0.

Using algebra:

It is obvious that A_3 = 1. Expanding out (x-2)^3 = x^3 - 6x^2 + 12x - 8. Well, we want the quadratic term to have the coefficient 1 (so it matches up with the left-hand side). So we must have A_2 = 7. That way, the quadratic term that comes out of (x-2)^2 (which will have a coefficient of 1) will be multiplied by 7 then added to the -6x^2. So A_2 = 7.

Using the formula:

A_2 = 1/2!*SUM{ i = 0, 3}( (2)^(i-2)*i*(i-1) ). ( i*(i-1) is the product expanded).

= 1/2*(2^(-2)*0*-1 + 2^(-1)*1*0 + 2^(0)*2*1 + 2^(1)*3*2)

= (1/2)*(0 + 0 + 2 + 12) = 14/2 = 7. Same answer.

Try it with other numbers! You'll also see that when k = n, you get 1 and when k > n, you get 0. Now for the continuation!

I proposed the following question at the end: "What if, instead of just a constant, there was a sequence? Instead of (x+1)^n + (x+1)^(n-1) + ... there was (x+n)^n + (x+n-1)^(n-1) + ..., how would the coefficients change?"

This is where Project 8 begins.

(I'm skipping the algebra...)

What if we have this transformation instead:

A_n*(x+L*n)^n + A_(n-1)*(x+L*(n-1))^(n-1) + ... + A_1*(x+L*1)^1 + A_0 = SUM{ m = 0, n }(A_m*(x+L*m)^m).

[random note: I apologize for the wacky background...I didn't make that and I'm not sure who did or how to fix it. It looks cool but it's very hard to read, I understand. Please copy and paste the text somewhere else if you can't read it! I do apologize!]

So, the coefficients do change, slightly. At first, when computing this formula, I didn't think it would look anything like the above formula but it does! With the same prerequisites above, here's the formula: We can see it's very similar (same 1/k!, same product inside a sum...).

So, the next task: A_n*(x + L*n^2)^n + A_(n-1)*(x + L*(n-1)^2)^(n-1) + ...

THEN A_n*(x + L*n^q)^n + A_(n-1)*(x + L*(n-1)^q)^(n-1) + ... for some q. If you think about it: Project 5 had q = 0 and Project 8 has q = 1.

Then later, maybe reversing it:

A_n*(x + L*1)^n + A_(n-1)*(x + L*2)^(n-1) + ... + A_2*(x + L*(n-1))^2 + A_1*(x + L*n)^1 + A_0.

Many many more sequences too after all these! Triangular numbers, primes (very tough I think), factorials, etc.

UPDATE:

Apparently, the transformations with constant L terms have been discovered before. Behold, these are Laguerre polynomials! The Laguerre polynomials are for L = 1 (I believe). For other L values, there are "Generalized Laguerre Polynomials" which cover all other L-values. Perhaps the extensions (1*L, 2*L, ... n*L) of this haven't been looked into yet!

Project 7. Canceling Digits

Ever seen the joke sin(x)/n = six because you just cancel out the n's? Well, it turns out, it could be this simple! Not for sines and cosines, but for actual numbers. The most common example is 64/16. How can you simplify 64/16? Well, you look at the prime factors and you'll see that 16 goes into 64 evenly. 4 times. So 64/16=4. But, what if I said 64/16 = 4/1 because you can cancel the 6's. .... May not be the right reason but it works. Maybe you aren't convinced yet. Okay, 64/16 = 4, that was easy to divide and figure out. But, what if we have 1612 / 1144? Not too sure now... Well both have two 1's on the top and bottom. So, using this method, 1612 / 1144 = 62 / 44. And yes, that is correct.

Of course, this does not work for all numbers (my friend was telling me "What about 73/17?" Obviously, it is not 3...). But it works for more than I thought! I let the numerator go up to 1000, and I got 201 solutions! Even more surprising, I only have 116 different numbers on the numerator. Many numbers have more than one solution. 303 has two solutions: 303 / 202 = 33 / 22 and 303 / 101 = 33 / 11. 392 has three solutions: 392 / 49 = 32 / 4, 392 / 98 = 32 / 8, and 392 = 196 = 32 / 16.

I wrote a Python program for this (which was very hard to come up with, there are many things that can go wrong...) but if you want to see it, send me an email! (In my About Me section). You can fascinate your friends!

A continuation of this project might to figure out which numbers work and if there is a pattern. I did find a pattern for some numbers. 65/26 = 5/2, 665/266 = 5/2, 6665/2666 = 5/2 ...etc. Same thing with 95/19 (appending 9's) and 756/27 (appending 7's). So there are infinitely many solutions to this. But other numbers have other patterns that are waiting to be solved!

Project 6. English in Math

I got the idea from Alexander Bogomolny @CutTheKnotMath. He (well, his son) proposed the following:

40 - 32 / 2 = 4!

Now, at first glance, it may look right but it may not. Reading left-to-right, ignoring PEMDAS, we see that this statement is true if the exclamation is the English exclamation point. But, if we account for PEMDAS, we also see this statement is true if the exclamation is the Math exclamation point: that is, the factorial.

It's a nice equation that works for the math-impaired (4 = 4) and the math-educated (24 = 24). I decided to explore more solutions to A - B / C = D! Notice that there are restrictions to this. B < A, C < B, and D < 10 (probably...10! is quite large). Also, mathematically speaking A - (B/C) = D! and (A-B)/C = D. I didn't want any trivial solutions (i.e. D = 1 or D = 2). So I started D at 3. Letting A vary between 0 and 100, I only found 2 solutions! Going to 1000, there were 13 solutions:

30 - 18 / 3 = 4!

40 - 32 / 2 = 4!

138 - 108 / 6 = 5!

230 - 220 / 2 = 5!

728 - 416 / 52 = 6!

731 - 473 / 43 = 6!

735 - 525 / 35 = 6!

748 - 616 / 22 = 6!

756 - 648 / 18 = 6!

765 - 675 / 15 = 6!

816 - 768 / 8 = 6!

833 - 791 / 7 = 6!

952 - 928 / 4 = 6!

I don't know if it's just me, but I wanted more solutions. I didn't think it was this rare. I thought one A value would have multiple solutions. Maybe it was the minus sign? I changed it to a plus and saw what happened. For A < 1000, I didn't get anything. Maybe that made sense...by making it a plus sign, I'm stretching the gap between D and D! even further.

There has to be more solutions. I tried another format: A / B + C = D! and A / B - C = D!

I did the minus case first. For A < 1000, I got 9 solutions:

12 / 3 - 1 = 3!

48 / 6 - 4 = 4!

60 / 12 - 2 = 3!

72 / 8 - 5 = 4!

192 / 16 - 8 = 4!

336 / 24 - 10 = 4!

360 / 18 - 15 = 5!

576 / 36 - 12 = 4!

840 / 30 - 23 = 5!

Even less than the first format. What about A / B + C = D! instead? Based on my find with A + B / C = D! , I kind of assumed that I would be doing the same thing. Turns out, I was right; I didn't get any solutions. Nonetheless, this was a clever find! I hope you find more and perhaps other combinations (A + B / C + D = E!) to trick some of your friends!

Project 5. Transformation of "Unit Polynomials"

(See Project 4 to understand the motivation and the process).

This is very similar to Project 4; however, I will consider the simplest polynomials:

1

x + 1

x^2 + x + 1

x^3 + x^2 + x + 1

....

I've called these polynomials "unit" polynomials, as all the coefficients are 1.

Again, we can transform these using (for example) (x+1) into the following:

A*(x+1)^0

B*(x+1)^1 + C*(x+1)^0

D*(x+1)^2 + E*(x+1)^1 + F*(x+1)^0

G*(x+1)^3 + H*(x+1)^2 + I*(x+1)^1 + J*(x+1)^0

....

And we can find out what these coefficients A,B,...I,J,... are. Skipping the algebra, I've arrived at a very general formula for any transformation (x+L). It is shown below:

For a unit polynomial of degree n, we can use the transformation x --> x + L. The new coefficient for the term with a degree of (n-k) (for k = 1,2,...,n) is given by the following formula: Unlike the formula below, it's much more complicated. It's an interesting thought and the next logical step here would be to find the general formula for a general sequence of initial coefficients. In Project 4, the coefficients were 1, n, n*(n-1), ... n! and these are 1,1,1....1. But what if we have a sequence given for the coefficients, say 1*C, 2*C, 3*C, ... n*C. What would the result coefficients be?

Update in Project 8!

Project 4. Derangements and Derivatives

Let's consider a simple polynomial: x^n (for some positive integer n).

Its derivative is nx^(n-1). What if we add its derivatives? With a term like x^n, we know it has n derivatives. If we add them all together, for some n, we get a different polynomial; here are a few of them:

1

x + 1

x^2 + 2x + 2

x^3 + 3x^2 + 6x + 6

x^4 + 4x^3 + 12x^2 + 24x + 24

x^5 + 5x^4 + 20x^3 + 60x^2 + 120x + 120

...

and so on.

The general formula for x^n is as follows:

x^n + nx^(n-1) + n*(n-1)*x^(n-2) + .... + n*(n-1)*(n-2)*(...)(3)(2)x^1 + n*(n-1)*(n-2)*(...)(3)(2)(1)

Let us write these coefficients as combinations "nCk" where nCk = n!/(k!*(n-k)!).

(nC0)*x^n + (nC1)*x^(n-1) + 2*(nC2)*x^(n-2) + 6*(nC3)*x^(n-3) + ... + (n-1)!*(nC(n-1))*x^1 + n!*(nCn).

And this can be expressed as a sum:

sum(k!*(nCk)*x^(n-k)) from k = 0 to k = n.

I was playing around with these and, it turns out, you can write these as the sum of A*(x+1)^n for different n-values and different A-values (A is some constant). I will demonstrate this with the x^4 polynomial then write them out for the others.

x^4 + 4x^3 + 12x^2 + 24x + 24.

We must start with (x+1)^4 with a coefficient of 1 on the outside. Once we have this, we have covered x^4 + 4x^3 + 6x^2 + 4x + 1. Thus, we have 6x^2 + 20x + 23 left.

Since the highest order of x is now 2, we need (x+1)^2. Since we have 6x^2, we must add 6(x+1)^2. So, by adding 6(x+1)^2, we account for 6x^2 + 12x + 6. Now, we have 8x + 17 left.

Applying the same rules above, we can write this as 8(x+1) + 9.

Finally, we arrive at our result:

x^4 + 4x^3 + 12x^2 + 24x + 24 = (x+1)^4 + 6(x+1)^2 + 8(x+1) + 9.

We now have the following table:

(x+1)^0 + 0

(x+1)^1 + 0

(x+1)^2 + 1

(x+1)^3 + 3*(x+1)^1 + 2

(x+1)^4 + 6*(x+1)^2 + 8*(x+1)^1 + 9

(x+1)^5 + 10*(x+1)^3 + 20*(x+1)^2 + 45*(n+1) + 44

(x+1)^6 + 15*(x+1)^4 + 40*(x+1)^3 + 135*(n+1)^2 + 264*(n+1) + 265

...

and so on.

Again, we can use combinations to write this sum as well (for a polynomial degree n):

(x+1)^n + (nC2)*(x+1)^(n-2) + 2*(nC3)*(x+1)^(n-3) + 9*(nC4)*(x+1)^(n-4) +...+ A*(nC(n-1))*(x+1)^1 + B*(nCn)

where A and B are constants. But, what are those constants in front of the combinations? First, note that (x+1)^(n-1) is not mentioned up there. We can assume that the coefficient in front of the would-be combination nC1 is 0. So, let's write this as a sum as well:

sum((A_k)*(nCk)*(x+1)^(n-k)) from k = 0 to k = n. A_k should be read "A subscript k" and is a sequence of coefficients.

A_k = {1,0,1,2,9,44,265,...} (OEIS Sequence A000166).

These are known as subfactorial numbers or derangements. Factorials (1,1,2,6,24,...) describe the number of possibilities to arrange n objects. Subfactorials describe the number of possibilities to arrange n objects; however, it disregards any arrangement that does not move at least one object. Let's have an example:

A teacher wants her students to grade each other's papers. But, she doesn't want a student grading his or her own paper. Let's say there are four papers, ABCD. How many ways can she distribute the tests without someone getting back his/her own test? Let's see....

BADC, BCDA, BDAC, CADB, CDAB, CDBA, DABC, DCAB, and DCBA. There are 9 different ways. So, when n = 4, we have 9. Derangements are denoted by "!n". It can be represented a few ways:

!n = n!*sum((-1)^k/k!) from k = 0 to k = n.

!n = |_n!/e + 1/2 _| for n >= 1

!n = [n!/e] for n >= 1 where [ ] denotes the nearest integer function.

Curiously, lim(!n/n!) = 1/e (.3679...). This is the probability that if you select a random permutation, you pick a derangement.

Back to the polynomial, these polynomials are also seen in Calc 1 as a student uses integration by parts. To save time, let's call one of these polynomials P_n for some degree n.

Integral of (x^n*e^x) dx = e^x*(P_n) + C.

Possible next project:

What if instead of (x+1), we use (x+2)? (x+3)? etc.. Just more things to think about!

UPDATE!:

So I've been looking at different constants, namely (x+2). Let's talk about this generally. Let's say we want to use (x+L) (above, L = 1). Turns out, the constants in all of these polynomials can be represented as a sum:

sum((-L)^(n-k)*n!/(n-k)!) from k = 0 to k = n.

So, when converting to (x+1), like we did above, each polynomial has a constant term (0,0,1,2,9,44,265,...) and each can be represented as this sum with L = 1. It just happens that these are also known as subfactorials.

FURTHER UPDATE:

Last night I kept working on this. I was able to find a formula for the term in the polynomial of (n-k)th degree (for k between 0 and n). Again, this formula is generalized to include all constants L.

Consider the polynomial x^n + nx^(n-1) + n*(n-1)x^(n-2) + ... + n!

The transformation of the polynomial to the form A_0*(x+L)^n + A_1*(x+L)^(n-1) + A_2*(x+L)^(n-2) + ... + A_n.

The coefficient A_k multiplied by the term with degree (n-k) for k = 0, 1, ... n is given by the following formula:

Coefficient = (nCk)*sum((-L)^(m)*(k!/m!)) from m = 0 to m = k.

Project 3. Algebra Mistake Actually Correct?

Teachers, I'm sure you've gotten this before:

(x+y)^2 = x^2 + y^2

and yes, it is incorrect. However, for certain x and y, this equation is actually correct. What x and y work for this?

(x+y)^2 = x^2 + 2xy + y^2 = x^2 + y^2

==> 2xy = 0

==> x = 0 or y = 0.

Nothing to interesting...If one of them is zero, the term drops and sure, the algebra error is still correct. What if you have this:

(x+y+z)^2 = x^2 + y^2 + z^2.

This is a different story; that is, if x,y,z < 0. If they are all positive, two must be 0 in order for this to be true. But, when x,y,z are less than 0, we have many different solutions. Let's compute the possible answers for these algebraically...

(x+y+z)^2 = x^2 + xy + xz + yx + y^2 + yz + zx + zy + z^2 = x^2 + y^2 + z^2

==> 2xy + 2yz + 2xz = 0

==> z = -xy/(x+y)

Here are the unique ordered points for (x,y,z) (the range for x,y,z is [-10,10] excluding 0):

(1,-2,-2)

(2,-6,-3)

(2,-4,-4)

(2,-1,2)

(3,-6,-6)

(3,-2,6)

(4,-8,-8)

(4,-2,4)

(5,-10,-10)

(6,-3,6)

(8,-4,8)

(10,-5,10)

Since addition is commutative, we can order these in any way we like. The different values for (x+y+z) are the following: {+/- 3, +/- 7, +/- 6, +/- 15, +/- 9, +/- 12}. It is interesting that all of the choices are multiples of 3 (excluding 7). Why is 7 there? Is it significant that they are multiples of 3? If we have (w+x+y+z)^2 = w^2 + x^2 + y^2 + z^2, will we get multiples of 4? Let's try it out. There's a pattern that we see already that I will skip to:

wx + wy + wz + xy + xz + yz = 0

w = -(xy+yz+xz)/(x+y+z)

There are over 40 unique solutions with this so I'll just state the list of possibilities for (w+x+y+z). They are the following: {+/- 11, +/- 6, +/- 10, +/- 5, +/- 2, +/- 12, +/- 3, +/- 9, +/- 4, +/- 13, +/- 8, +/- 17, +/- 14, +/- 16, +/- 18, +/- 20}. From this, it doesn't seem like there is a pattern with the sum. Both of these sets have one thing in common: no +/-1. This, however, makes sense. If the sum = 1, one side of the equation would be 1^2. The only way for 1 to equal a sum of squares is if one of them is 0 (which we have excluded). Eventually, I believe the sum can be any natural number > 1 (but someone may want to confirm that).

What if (x+y)^3 = x^3 + y^3. (x+y)^n? (a+b+c+...)^n? Just some more projects to think about.

UPDATE:

Today, I decided to work on this a bit more. I turns out, for (x+y)^n = x^n + y^n, this ONLY works if x or y are 0 or if x = -y (for odd values of n).

Additionally, for (x+y+z)^n = x^n + y^n + z^n, this will NOT work unless n = 2. If n is not 2, then it will only work if at least one is zero or at least one equals the opposite of another (ergo, x = -y, y = -z, or x = -z). Why does it only work for n = 2? Algebraic proof?

What about larger terms? (w+x+y+z)^2 = w^2 + x^2 + y^2 + z^2 works. So does  (w+x+y+z)^3 = w^3 + x^3 + y^3 + z^3. But not for the exponent equal to 4. Going off of induction, I had the following proposition:

Proposition 1: For integers a_1,...a_n, and natural numbers m and n, (a_1 + a_2 + ... + a_n)^m = (a_1)^m + (a_2)^m + ... + (a_n)^m has nontrivial solutions when m < n.

................

Once I kept going, however, with 5 terms in the parenthesis [Meaning, (v+w+x+y+z)^n], it still did not work for m = 4 or m = 5. It also didn't work for 6 terms...this made me come up with a revised proposition.

Proposition 1: For integers a_1,...a_n, and natural numbers m and n, (a_1 + a_2 + ... + a_n)^m = (a_1)^m + (a_2)^m + ... + (a_n)^m has nontrivial solutions when m < 4.

...............

Whoops! Apparently for seven terms, the exponent can happily be 4 and 5. Well, I'm stuck. Thoughts anyone?

Project 2. Calculus Mistake Actually Correct?

PRODUCT RULE

Once again, teachers, does this look familiar?

h(x) = f(x)*g(x)

h'(x) = (f(x)*g(x))' = f '(x)*g'(x)

and yes, this is wrong. But, what if your student rebutted and claimed that his/her answer is actually correct for certain functions (and the student's right). Let's say f(x) = x^2. What could g(x) be so this actually works?

For example, let f(x) = x^2. So...

f '(x)*g'(x) = [f(x)*g(x)]' = f(x)*g'(x) + f '(x)*g(x)

==> (2x)*g'(x) = (x^2)*g'(x) + (2x)*g(x)

==> (2x - x^2)*g'(x) = (2x)*g(x)

If g(x) = y, g'(x) = dy/dx

==> (2x - x^2)*(dy/dx) = y*(2x)

==> dy/dx = y*[(2x)/(2x - x^2)]

==> dy/y = [(2x)/(2x-x^2)]*dx = (2/(2-x))*dx

We can integrate both sides...

==> ln(y) = -2*ln(2-x) + C

==> y = e^(-2ln(2-x) + C) = A/(2-x)^2 where A is any number > 0.

Let's generalize this: f(x) = x^n for any number n. I'll skip a few steps.

dy/y = ([nx^(n-1)]/[nx^(n-1) - x^n])*dx = (n/(n-x))*dx

==> ln(y) = -n*ln(n-x) + C

==> y = e^(-n*ln(n-x) + C) = A/(n-x)^n where A is any number > 0. We can also see there is no restriction on n, it can be any real number. So, if n = 0 [f(x) is the constant function] then g(x) = y = A (another constant function).

What if f(x) = sin(x) or cos(x)? It's possible to do but a bit messy to show on a website. Try it for yourself and see what you get.

Now, generalize. Let's solve for f(x) and see what f(x) would have to be...

f '(x)*g'(x) = (f(x)*g(x))' = f '(x)*g(x) + f(x)*g'(x)

==> f '(x)*g'(x) - f '(x)*g(x) = f(x)*g'(x)

==> f '(x) = [f(x)*g'(x)]/[g'(x) - g(x)]

Let f(x) = y and f '(x) = dy/dx

==> dy/dx = [y*g'(x)]/[g'(x) - g(x)]

==> dy/y = [g'(x)]/[g'(x) - g(x)] * dx

We can integrate both sides...

==> ln(y) = integral(g'(x)/[g'(x) - g(x)] dx)

= integral of      g'(x)*dx

g'(x) - g(x)

And we see that y = e^(integral....). Hence, f(x) depends on g(x) (as it should). So, you can use this quick go-to equation and solve for any function you like!

QUOTIENT RULE

We did the product rule. What about the quotient rule? Consider this:

(f(x)/g(x))' = f '(x)/g'(x)

Let's let g(x) = x^2 and solve for f(x). You'll see that if we let f(x) = x^2, it's near impossible to solve for g(x). If g(x) = x^2...

f '(x)/g'(x) = (g(x)*f '(x) - f(x)*g'(x))/(g(x))^2

==> f '(x)/(2x) = ((x^2)*f'(x) - f(x)*(2x))/(x^2)^2

==> (f '(x))*(x^3) = 2*f '(x)*(x^2) - 2*f(x)*(2x)

==> f '(x)*[2x^2 - x^3] = (4x)*f(x)

Once again, let f(x) = y...

==> (dy/dx)*(2x - x^2) = 4*y

==> dy/y = (4/(2x-x^2))*dx     (yes, this is a nasty integral....I'll save you the trouble)

==> ln(y) = 2*(ln(x) - ln(2-x)) + C

==> y = e^(2*(ln(x) - ln(2-x)) + C) = (Ax^2)/(2-x)^2 where A is a number > 0.

For g(x) = x^n (I'll save you the algebra)...

y = (Ax^n)/(n-x)^n) where A is a number > 0.

Exercise: find out what f(x) is if g(x) = sin(x) or cos(x).

This can then be computed generally, just as before.

f '(x)/g'(x) = [g(x)*f '(x) - f(x)*g'(x)]/(g(x))^2

==> f '(x)*(g(x))^2 = g(x)*f '(x)*g'(x) - f(x)*g'(x)*g'(x)

We can see that it's near impossible to solve for g(x) in terms of f(x). At least, I don't know how to. If anyone has a solution to it, please let me know. Thus, let's have f(x) = y.

==> (dy/dx)*(g(x))^2 = g(x)*g'(x)*(dy/dx) - y*(g'(x))^2

==> (dy/dx)*[g(x)*g'(x) - (g(x))^2] = y*(g'(x))^2

==> dy/y = ([(g'(x))^2]/[g(x)*g'(x) - (g(x))^2])*dx

==> ln(y) = integral(([(g'(x))^2]/[g(x)*g'(x) - (g(x))^2])*dx)

=          (g'(x))^2          dx =   g'(x)     g'(x)      dx

g(x)*g'(x) - (g(x))^2            g(x)   g'(x) - g(x)

This general solution is very similar to the product rule solution. The only difference is we have the ratio of a function's derivative to the function itself. This is called the "relative derivative". See, even when your students are wrong, they can be right for certain functions!

Project 1. The Birthday Problem Generalized

We all know the problem: How many students do you need in a classroom for there to be a 50% chance that two of the students have the same birthday? It's 23. You can blow high school student's minds with this by asking everyone his or her birthday and (hopefully) two will be the same.

However, what if you wanted to know how many people you need in a class to have a 50% chance that 3 students have the same birthday. Turns out, it's 84.

The image on the left gives the generalized probability. The variables mean the following:

e: Euler's number-- 2.718281828...

n: number of students in the room that you need

m: number of students you want to have the same birthday

x: number of possibilities that your "subject" can be The "subject" in this case is birthdays. So, x = 365 because there are 365 days in the year.

So, if we want to find out how many people we need in class for a 50% chance that 4 students have the same birthday, let x = 365, m = 4, and the entire formula equal 1/2. We see, with computation, that we need 171.

Not only does this work for birthday but this works with other problems, too. This is the beauty of this equation. Consider the following:

"Think of a number between 1 and 10, inclusively. How many people do you need so you have a 50% chance that two people are thinking of the same number?"

We can go about this the same way. We're looking for n. Our m in this case is 2 (since it's 2 people). Our x is 10 because there are 10 different options here {1,2,3,4,5,6,7,8,9,10}. So, if we say

exp(-nC2/(10^(2-1))) = 1/2 (where nCm is said 'n choose m'), we see that n is 5. (Note that I'm rounding all of these numbers up.) Well, a student could've guessed that by saying "For two students to have a 50% chance, just cut 10 in half!"

Well, just try a bigger number.

Think of a number between 1 and 100, inclusively. Now, how many people do we need? (Naively, the student will say 50...). To the student's surprise, you will arrive at the low answer of 13! The student will be blown away. With this example, I like to think of the applicable problem below:

"You are picking a certain amount of people randomly. How many people do you need to pick to ensure that there is a 50% chance that two of the people you picked are the same age?"

With this, I'm assuming that the age limit is 100, but the same point is seen. I encourage you to think of other day-to-day problems (think of a letter between A and Z, here x = 26....) and tell your family/friends/students about this!

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