This talk got re-worked as The $\sin(x)$ Button.
This talk uses a bit of Python code.
You can download the relevant code here.
First, we consider the following mysterious function in Python.
def mystery(x, N):
"""This mystery function outputs some value based on x and N.
We assume that x is a real number, and N is a positive integer."""
output = 0;
for n in range(0, N + 1):
output += ((-1)**n)*(x**(2*n+1))/(math.factorial(2*n+1));
return output;
Our big question today is:
We will try various strategies to get a handle on it.
Calculate some specific values of the function.
Write a function table(C,R)
which produces a similar table of values with columns $x = 0, 1, 2, \dots, C$ and rows $N = 1, 2, \dots, R.$
Your function should return a two dimensional array of values.
$x = 0$ | $x = 1$ | $x = 2$ | $x = 3$ | $x = 4$ | $x = 5$ | $x = 6$ | $x = 7$ | $x = 8$ | $x = 9$ | $x = 10$ | |
---|---|---|---|---|---|---|---|---|---|---|---|
$N = 1$ | $0.0000$ | $0.8333$ | $0.6667$ | $-1.5000$ | $-6.6667$ | $-15.8333$ | $-30.0000$ | $-50.1667$ | $-77.3333$ | $-112.5000$ | $-156.6667$ |
$N = 2$ | $0.0000$ | $0.8417$ | $0.9333$ | $0.5250$ | $1.8667$ | $10.2083$ | $34.8000$ | $89.8917$ | $195.7333$ | $379.5750$ | $676.6667$ |
$N = 3$ | $0.0000$ | $0.8415$ | $0.9079$ | $0.0911$ | $-1.3841$ | $-5.2927$ | $-20.7429$ | $-73.5097$ | $-220.3683$ | $-569.4268$ | $-1307.4603$ |
$N = 4$ | $0.0000$ | $0.8415$ | $0.9093$ | $0.1453$ | $-0.6617$ | $0.0896$ | $7.0286$ | $37.6940$ | $149.4998$ | $498.2002$ | $1448.2716$ |
$N = 5$ | $0.0000$ | $0.8415$ | $0.9093$ | $0.1409$ | $-0.7668$ | $-1.1336$ | $-2.0603$ | $-11.8422$ | $-65.6961$ | $-287.9615$ | $-1056.9392$ |
$N = 6$ | $0.0000$ | $0.8415$ | $0.9093$ | $0.1411$ | $-0.7560$ | $-0.9376$ | $0.0372$ | $3.7172$ | $22.5894$ | $120.2379$ | $548.9652$ |
$N = 7$ | $0.0000$ | $0.8415$ | $0.9093$ | $0.1411$ | $-0.7568$ | $-0.9609$ | $-0.3224$ | $0.0867$ | $-4.3167$ | $-37.2105$ | $-215.7512$ |
$N = 8$ | $0.0000$ | $0.8415$ | $0.9093$ | $0.1411$ | $-0.7568$ | $-0.9588$ | $-0.2748$ | $0.7407$ | $2.0142$ | $9.6767$ | $65.3945$ |
$N = 9$ | $0.0000$ | $0.8415$ | $0.9093$ | $0.1411$ | $-0.7568$ | $-0.9589$ | $-0.2798$ | $0.6470$ | $0.8294$ | $-1.4281$ | $-16.8119$ |
$N = 10$ | $0.0000$ | $0.8415$ | $0.9093$ | $0.1411$ | $-0.7568$ | $-0.9589$ | $-0.2794$ | $0.6580$ | $1.0100$ | $0.7135$ | $2.7611$ |
If you examine the table of values, then you’ll notice that mystery(x,N)
eventually stops changing as a function of $N$.
That is, the value of mystery(x,N)
is essentially the same for all large values of $N$.
We call this the limiting value of the function.
To be concrete, we say that two numbers $a$ and $b$ are essentially the same if they differ by less than $10^{-10}$.
$$
|a - b| < 10^{-10} = 0. 000 000 000 1
$$
Create a new Python function limit(x)
which calculates the limiting value of mystery(x,N)
for each $x$.
You will want to use the built-in Python function
abs()
to compute the absolute value.
Use Python to approximate the smallest and biggest values of mystery(x,N)
on the domain $-3 \leq x \leq 3$ when $N$ is big.
To do so, calculate the value of limit(x)
at a large number of points in the domain $-3 \leq x \leq 3$.
Your final answers for the biggest and smallest values should specify both: the value of limit(x)
and the x
where this value is obtained.
On paper:
If you look carefully at the code, you’ll notice that mystery(x,N)
is a sum of powers of $x$.
Write it out as a polynomial, for $N = 3$.
To do so, you are going to need to use the factorial function.
The function math.factorial(n)
computes the product of the first $n$ whole numbers.
For example,
$$
\texttt{math.factorial(4)} = 1 \cdot 2 \cdot 3 \cdot 4 = 24.
$$
In mathematics, we usually write the factorial function as $n!$ because it grows so quickly.
It’s a very exciting function!
In Desmos: Graph the function mystery(x,N)
with a slider for $N$.
Notice that mystery(x,N)
is a summation from $n = 0$ to $n = N$.
In Desmos, typing sum
produces a summation sign $\displaystyle \sum_{n=}$.
What is the mystery function calculating?
Thanks to Mr. Nanthivarman for inviting me to speak at Lester B Pearson Collegiate Institute. Thanks to the following people for feedback on early drafts of this talk.
Published: Feb 5, 2024
Last Modified: Mar 26, 2024
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