Math Notes Spring 25
Fibonacci Moduli
For the Fibonacci sequence, defined by
I believe the following is true (Haven’t found any counterexamples)
Calculus of inverses
Let be functions that are defined and differentiable over some interval. Further, let
Such that are inverses.
Differentiate by chain rule.
This already has some neat results
These aren’t the standard forms you’re familiar with, but they are equivalent.
By this we can create a very annoying looking integral (for fun of course).
Textbook derivatives
I know, it’s in the textbook, but I never really learned the derivative of
It’s not too complicated though using the logarithm change-of-base identity.
The exponential form isn’t that bad either.
Of course, given that and are inverses, we could find also find the exponential through the identity found in the previous section.
Inverse Inverse Trigonometry
In the section on derivatives of inverses, these came up
These forms can be simplified to remove any notion of trigonometry.
Draw a right triangle with angle , adjacent side length , opposite side length , and hypotenuse side length .
Of course, , so within bounds of course.
And by definition , so
Some more identities found by similar method
Arctangent is a bit more complex to deal with. Let’s do
Draw a right triangle with angle , adjacent side length , opposite side length of , and hypotenuse side length .
We know that so set . Also known is
So
And so, the triangle has been completed.
Finally,
This leads to another set of identities
As a sidenote, it seems weird to me that
Complete Factorization
Quick! what are all the factors of ?
Instead of giving you an answer, here’s an algorithm to find it.
First, by repeated division, find the prime factors.
Find the all unique factors and their cartesian products in to create .
Remove one of each unique factor from to make .
Find the all unique factors and their cartesian products in to create .
Multiply each number in by each number in to find Remove any duplicate values and remove any values which had been found in previous .
Remove one of each unique factor from to make .
Find the all unique factors and their cartesian products in to create .
Multiply each number in by each number in to find Remove any duplicate values and remove any values which had been found in previous .
Remove one of each unique factor from to make .
Find the all unique factors and their cartesian products in to create .
Multiply each number in by each number in to find Remove any duplicate values and remove any values which had been found in previous .
Remove one of each unique factor from to make .
Seeing as there are no more prime factors left in , Halt. The factors of are found in the union and of course .
I don’t know the name of this algorithm. I figured it out myself, but the simplicity of the algorithm leads me to believe it has been known for quite some time. There are likely more efficient algorithms, but compared to the naive algorithm of dividing by all numbers less than, it is appreciably efficient. An advantage of this algorithm is that it can be done on pen-and-paper and also can be done using a computer.
Squares mod 100
Two neat little identities for squares mod 100.
And
These can be algebraically shown to be true
What about the generic case of a modulus ? What is the specific to be added to to make the squares congruent?
Expand
The trivial case is for and . Let’s just ignore the possibility that both of those aren’t true and yet still
The smallest natural number such that for any integer is clearly if is even, and if is odd.
The smallest natural number such that is a bit trickier.
Obviously works, but the case of only works if is divisible by
So for
If is divisible by
Otherwise:
Partial Geometric sums
So, the geometric series: if you’ve taken some Calculus, you know of it.
Solve for
And there’s the formula the textbook gave ya'
But what about the partial sums?
Decompose into difference of sums
We know the value of the left sum, and the right sum can be found similarly.
So the partial sum can be found
Some alternate forms are common
Also note that, now that the upper bound of the sum is not infinity, the sum is defined for values
Partial Extended-Geometric Sums
This also applies to series like
or in general terms where
Finding a closed form for this is not too much more difficult. First find the closed for to the simpler
Now the more complex form can be found
Following similar reasoning as the regular geometric series, the partial form is found.
Pause solving , introduce new equation,
Pause solving , introduce new equation
Resume solving
Resume solving
Thus
So
For convenience, increment the value of by one.
When
I’m not sure if any of that made any sense, but from my bit of verification, I believe it’s true.
As far as verification goes, it wouldn’t hurt to do so through Induction.
First the base case of
The base case is satisfied
The equation is valid if the following holds
And so it has been proven.