Math Notes 2024
I enjoy tinkering with math.
Unfortunately, I often forget the little things I discover.
On this page you will find all the tidbits I found interesting enough to record.
I don't strive for academic perfection, however I also do not want to spread blatant errors.
If you notice any errors in these notes, please email me .
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Integrals of Step Functions
December 10th 2024
Single Page
The integral of the function
⌊ x ⌋ \lfloor x \rfloor ⌊ x ⌋ x x x
∫ ⌊ x ⌋ d x = = ∑ n = 0 ⌊ x ⌋ − 1 n + ( x − ⌊ x ⌋ ) ( ⌊ x ⌋ ) = 1 2 ( ⌊ x ⌋ − 1 ) ( ⌊ x ⌋ − 1 + 1 ) + ( x − ⌊ x ⌋ ) ⌊ x ⌋ = ( x − 1 2 ) ⌊ x ⌋ − 1 2 ⌊ x ⌋ 2
\begin{align*}
\int \lfloor x \rfloor \, dx =& \\
=& \sum_{n=0}^{\lfloor x \rfloor -1} n + (x-\lfloor x \rfloor )(\lfloor x \rfloor )\\
=& \frac{1 }{2} (\lfloor x \rfloor -1)(\lfloor x \rfloor -1 +1 ) + (x-\lfloor x \rfloor )\lfloor x \rfloor \\
=& \left( x-\frac{1}{2} \right) \lfloor x \rfloor - \frac{1}{2} \lfloor x \rfloor^2
\end{align*}
∫ ⌊ x ⌋ d x = = = = n = 0 ∑ ⌊ x ⌋ − 1 n + ( x − ⌊ x ⌋) (⌊ x ⌋) 2 1 (⌊ x ⌋ − 1 ) (⌊ x ⌋ − 1 + 1 ) + ( x − ⌊ x ⌋) ⌊ x ⌋ ( x − 2 1 ) ⌊ x ⌋ − 2 1 ⌊ x ⌋ 2 (Constant of integration omitted)
A similar process can be done for the ceiling function
⌈ x ⌉ \lceil x \rceil ⌈ x ⌉
∫ ⌈ x ⌉ d x = = ∑ n = 0 ⌈ x ⌉ n + ( x − ⌈ ( x ) ⌉ ) ⌈ x ⌉ = 1 2 ⌈ x ⌉ ( ⌈ x ⌉ + 1 ) + ( x − ⌈ x ⌉ ) ⌈ x ⌉ = ( x + 1 2 ) ⌈ x ⌉ − 1 2 ⌈ x ⌉ 2
\begin{align*}
\int \lceil x \rceil \, dx =& \\
=&\sum_{n=0}^{\lceil x \rceil } n +(x-\lceil (x) \rceil )\lceil x \rceil \\
=& \frac{1}{2} \lceil x \rceil(\lceil x \rceil +1 ) + (x-\lceil x \rceil )\lceil x \rceil \\
=& \left( x+\frac{1}{2} \right)\lceil x \rceil -\frac{1}{2} \lceil x \rceil^2
\end{align*}
∫ ⌈ x ⌉ d x = = = = n = 0 ∑ ⌈ x ⌉ n + ( x − ⌈( x )⌉) ⌈ x ⌉ 2 1 ⌈ x ⌉ (⌈ x ⌉ + 1 ) + ( x − ⌈ x ⌉) ⌈ x ⌉ ( x + 2 1 ) ⌈ x ⌉ − 2 1 ⌈ x ⌉ 2 I’ll spare the fine details on the final one. The ‘round to the nearest integer’ function
nint ( x ) \operatorname{nint}(x) nint ( x )
∫ nint ( x ) d x = x ( nint ( x ) ) − 1 2 nint ( x ) 2
\int \operatorname{nint}(x) \, dx = x(\operatorname{nint} (x)) - \frac{1}{2} \operatorname{nint}(x)^2
∫ nint ( x ) d x = x ( nint ( x )) − 2 1 nint ( x ) 2 The Fibonacci sequence is defined as
F 1 = F 2 = 1 F n + 2 = F n + 1 + F n
\begin{align*}
F_{1} =& F_{2} = 1\\
F_{n+2} =& F_{n+1} + F_{n}
\end{align*}
F 1 = F n + 2 = F 2 = 1 F n + 1 + F n Are there any Fibonacci numbers
F n F_{n} F n F n + 1 > 2 F n F_{n+1} > 2F_{n} F n + 1 > 2 F n
F n + 1 = F n + F n − 1
F_{n+1} = F_{n} + F_{n-1}
F n + 1 = F n + F n − 1 Substitute into inequality
F n + F n − 1 > 2 F n
F_{n} + F_{n-1} > 2F_{n}
F n + F n − 1 > 2 F n F n − 1 > F n
F_{n-1} > F_{n}
F n − 1 > F n Which is never true for the sequence.
Another question: which Fibonacci numbers
F n F_{n} F n F n + 2 > 2 F n F_{n+2} > 2F_{n} F n + 2 > 2 F n
Again substitute
F n + 1 + F n > 2 F n
F_{n+1} + F_{n} > 2F_{n}
F n + 1 + F n > 2 F n F n + 1 > F n
F_{n+1} > F_{n}
F n + 1 > F n Which is true for all Fibonacci numbers except for when
n = 1 n=1 n = 1
In a similar spirit, it is obvious that
F n = F n − 1 + F n − 2
F_n = F_{n-1} + F_{n-2}
F n = F n − 1 + F n − 2 But this form can be reduced further to be a sum of
F n − 2 F_{n-2} F n − 2 F n − 3 F_{n-3} F n − 3
F n = F n − 1 + F n − 2 = F n − 2 + F n 3 + F n − 2 = 2 F n − 2 + F n − 3 and again = 2 F n − 3 + 2 F n − 4 + F n − 3 = 3 F n − 3 + 2 F n − 4
\begin{align*}
F_n =& F_{n-1} + F_{n-2}\\
=& F_{n-2} + F_{n_3} + F_{n-2}\\
=& 2F_{n-2} + F_{n-3}\\
& \text{and again}\\
=& 2F_{n-3} + 2F_{n-4} + F_{n-3}\\
=& 3F_{n-3} + 2F_{n-4}
\end{align*}
F n = = = = = F n − 1 + F n − 2 F n − 2 + F n 3 + F n − 2 2 F n − 2 + F n − 3 and again 2 F n − 3 + 2 F n − 4 + F n − 3 3 F n − 3 + 2 F n − 4 The coefficients on the reduced forms are interesting to me. Given coefficients
a k , b k a_{k},b_{k} a k , b k a k + 1 , b k + 1 a_{k+1},b_{k+1} a k + 1 , b k + 1
a k F n − 1 + b k F n − 2 = a k ( F n − 2 + F n − 3 ) + b k F n − 2 = ( a k + b k ) F n − 2 + a k F n − 3
a_k F_{n-1} + b_k F_{n-2} = a_k (F_{n-2} + F_{n-3}) + b_k F_{n-2} = (a_k +b_k)F_{n-2} + a_k F_{n-3}
a k F n − 1 + b k F n − 2 = a k ( F n − 2 + F n − 3 ) + b k F n − 2 = ( a k + b k ) F n − 2 + a k F n − 3 We define
a 1 = b 1 = 1 a_1 =b_1 = 1 a 1 = b 1 = 1
a k + 1 = a k + b k b k + 1 = a k
\begin{align*}
a_{k+1} =& a_k + b_k \\
b_{k+1} =& a_k
\end{align*}
a k + 1 = b k + 1 = a k + b k a k a k + 2 = a k + 1 + b k + 1 = a k + 1 + a k
a_{k+2} = a_{k+1} + b_{k+1} = a_{k+1} + a_{k}
a k + 2 = a k + 1 + b k + 1 = a k + 1 + a k Notice something peculiar about the above identity: it’s also the Fibonacci sequence!
This allows us to define a specific level of reduced Fibonacci symbolically. Let
k > 1 k > 1 k > 1
F n = a k F n − k + 1 + b k F n − k
F_n = a_k F_{n-k+1} + b_k F_{n-k}
F n = a k F n − k + 1 + b k F n − k F n = F k F n − k + 1 + F k − 1 F n − k
F_n = F_k F_{n-k+1} + F_{k-1} F_{n-k}
F n = F k F n − k + 1 + F k − 1 F n − k As an example
n = 10 n=10 n = 10 k = 4 k=4 k = 4
F 10 = F 4 F 7 + F 3 F 6
F_{10} = F_{4}F_{7} + F_{3} F_{6}
F 10 = F 4 F 7 + F 3 F 6 55 = 3 ⋅ 13 + 2 ⋅ 8
55 = 3\cdot 13 + 2 \cdot 8
55 = 3 ⋅ 13 + 2 ⋅ 8
A friend entertained the idea that the closed form of the nth Fibonacci number could be used to further prove this.
ϕ = 1 + 5 2 , ψ = 1 − 5 2 , F n = 1 5 ( ϕ n − ψ n )
\phi = \frac{1+\sqrt{5}}{2},\quad \psi = \frac{1-\sqrt{5}}{2},\quad F_{n} = \frac{1}{\sqrt{5}} \left(\phi^n - \psi^n \right)
ϕ = 2 1 + 5 , ψ = 2 1 − 5 , F n = 5 1 ( ϕ n − ψ n ) A quick look shows that there’s a lot of algebra down this road.
Generalized Geometric Series
December 31st 2024
Single Page
The geometric series is defined as
∑ n = 1 ∞ r n
\sum_{n=1}^{\infty} r^n
n = 1 ∑ ∞ r n Where
− 1 < r < 1 -1 < r < 1 − 1 < r < 1 x x x
x = r + r 2 + r 3 + … + r n x = r ( 1 + r + r 2 + … + r n − 1 ) x = r ( 1 + x ) x = r 1 − r
\begin{align*}
x &= r + r^2+r^3 + \ldots + r^n\\
x &= r(1 +r + r^2 + \ldots + r^{n-1})\\
x &= r(1+x)\\
x &= \frac{r}{1-r}
\end{align*}
x x x x = r + r 2 + r 3 + … + r n = r ( 1 + r + r 2 + … + r n − 1 ) = r ( 1 + x ) = 1 − r r How about some other values for the exponent of
r r r 2 n − 1 2n-1 2 n − 1
x = r + r 3 + r 5 + … + r 2 n − 1 x = r ( 1 + r 2 + r 4 + … + r 2 n − 2 ) x = r ( 1 + r ( r + r 3 + … + r 2 n − 3 ) ) x = r ( 1 + r x ) x = r 1 − r 2
\begin{align*}
x &= r+r^3+r^5+\ldots + r^{2n-1}\\
x &= r(1+r^2+r^4+\ldots + r^{2n-2})\\
x &= r(1+r(r+r^3+\ldots + r^{2n-3}))\\
x &= r(1+rx)\\
x&= \frac{r}{1-r^2}
\end{align*}
x x x x x = r + r 3 + r 5 + … + r 2 n − 1 = r ( 1 + r 2 + r 4 + … + r 2 n − 2 ) = r ( 1 + r ( r + r 3 + … + r 2 n − 3 )) = r ( 1 + r x ) = 1 − r 2 r ∑ n = 1 ∞ r 2 n − 1 = r 1 − r 2
\sum_{n=1}^{\infty} r^{2n-1} = \frac{r}{1-r^2}
n = 1 ∑ ∞ r 2 n − 1 = 1 − r 2 r How about for the multiples of three
3 n 3n 3 n
x = r 3 + r 6 + r 9 + … + r 3 n x = r 3 ( 1 + r 3 + r 6 + … + r 3 n − 3 ) x = r 3 ( 1 + x ) x = r 3 1 − r 3
\begin{align*}
x &= r^3+r^6+r^9+\ldots +r^{3n}\\
x &= r^3(1 + r^3+r^6+\ldots +r^{3n-3})\\
x &= r^3(1+x)\\
x &= \frac{r^3}{1-r^3}
\end{align*}
x x x x = r 3 + r 6 + r 9 + … + r 3 n = r 3 ( 1 + r 3 + r 6 + … + r 3 n − 3 ) = r 3 ( 1 + x ) = 1 − r 3 r 3 ∑ n = 1 ∞ r 3 n = r 3 1 − r 3
\sum_{n=1}^{\infty} r^{3n} =\frac{r^3}{1-r^3}
n = 1 ∑ ∞ r 3 n = 1 − r 3 r 3 Generally, it can be assumed that for any natural number
k > 0 k > 0 k > 0 r r r − 1 < r < 1 -1 < r < 1 − 1 < r < 1
∑ n = 1 ∞ r k n = r k 1 − r k
\sum_{n=1}^{\infty} r^{kn} = \frac{r^k}{1-r^k}
n = 1 ∑ ∞ r kn = 1 − r k r k To show that this is true, consider the following
x = r k + r 2 k + r 3 k + … + r n k x = r k ( 1 + r k + r 2 k + … + r n k − k ) x = r k ( 1 + x ) x = r k 1 − r k
\begin{align*}
x &= r^{k}+ r^{2k} + r^{3k} + \ldots + r^{nk}\\
x &= r^{k}(1+r^{k}+ r^{2k} + \ldots + r^{nk-k})\\
x &= r^{k}(1+x)\\
x &= \frac{r^{k}}{1-r^{k}}
\end{align*}
x x x x = r k + r 2 k + r 3 k + … + r nk = r k ( 1 + r k + r 2 k + … + r nk − k ) = r k ( 1 + x ) = 1 − r k r k How about for numbers which leave a remainder of two when divided by three:
3 n − 1 3n-1 3 n − 1
x = r 2 + r 5 + r 8 + … + r 3 n − 1 x = r 2 ( 1 + r 3 + r 6 + … + r 3 n − 3 )
\begin{align*}
x &= r^{2} + r^{5} + r^{8} + \ldots + r^{3n-1} \\
x &= r^2 (1+ r^3 + r^6 +\ldots + r^{3n-3})
\end{align*}
x x = r 2 + r 5 + r 8 + … + r 3 n − 1 = r 2 ( 1 + r 3 + r 6 + … + r 3 n − 3 ) The earlier found identity allows us to solve this
x = r 2 ( 1 + ∑ n = 1 ∞ r 3 n ) x = r 2 ( 1 + r 3 1 − r 3 ) x = r 2 + r 5 1 − r 3 x = r 2 1 − r 3
\begin{align*}
x &= r^2 \left(1+\sum_{n=1}^{\infty} r^{3n}\right)\\
x &= r^2\left(1+ \frac{r^3}{1-r^3}\right)\\
x &= r^2+ \frac{r^5}{1-r^3}\\
x &= \frac{r^2}{1-r^3}
\end{align*}
x x x x = r 2 ( 1 + n = 1 ∑ ∞ r 3 n ) = r 2 ( 1 + 1 − r 3 r 3 ) = r 2 + 1 − r 3 r 5 = 1 − r 3 r 2 ∑ n = 1 ∞ r 3 n − 1 = r 2 1 − r 3
\sum_{n=1}^{\infty} r^{3n-1} = \frac{r^2}{1-r^3}
n = 1 ∑ ∞ r 3 n − 1 = 1 − r 3 r 2 That got me thinking, is there a general closed form of
∑ n = 1 ∞ r a n − b = x
\sum_{n=1}^{\infty}r^{an-b} = x
n = 1 ∑ ∞ r an − b = x where
a , b ∈ N a,b \in \mathbb{N} a , b ∈ N 0 ≤ b < a 0 \leq b \lt a 0 ≤ b < a − 1 < r < 1 -1<r<1 − 1 < r < 1
x = r a − b + r 2 a − b + r 3 a − b + … + r a n − b x = r a ( r − b + r a − b + r 2 a − b + … + r a ( n − 1 ) − b ) x = r a ( r − b + x ) x = r a − b + r a x x = r a − b 1 − r a
\begin{align*}
x&= r^{a-b} + r^{2a-b} + r^{3a-b} + \ldots + r^{an-b}\\
x&= r^{a}(r^{-b} + r^{a-b} + r^{2a-b} + \ldots + r^{a(n-1)-b}) \\
x&= r^{a}(r^{-b} +x)\\
x&= r^{a-b} +r^{a}x\\
x&= \frac{r^{a-b}}{1-r^{a}}
\end{align*}
x x x x x = r a − b + r 2 a − b + r 3 a − b + … + r an − b = r a ( r − b + r a − b + r 2 a − b + … + r a ( n − 1 ) − b ) = r a ( r − b + x ) = r a − b + r a x = 1 − r a r a − b Which seems to agree with the previous result of
a = 3 , b = 1 a=3,\,\, b=1 a = 3 , b = 1
r 2 1 − r 3
\frac{r^{2}}{1-r^{3}}
1 − r 3 r 2 And the even earlier result of odd numbers
a = 2 , b = 1 a=2,\,\, b=1 a = 2 , b = 1
r 1 − r 2
\frac{r}{1-r^2}
1 − r 2 r Thus I think it’s fair to say
∑ n = 1 ∞ r a n − b = r a − b 1 − r a
\sum_{n=1}^{\infty}r^{an-b} =\frac{r^{a-b}}{1-r^{a}}
n = 1 ∑ ∞ r an − b = 1 − r a r a − b There’s one final case that has not been covered
a n + b an +b an + b a , b ∈ N a,b \in \mathbb{N} a , b ∈ N a , b > 0 a,b > 0 a , b > 0
∑ n = 1 ∞ r a n + b
\sum_{n=1}^{\infty} r^{an+b}
n = 1 ∑ ∞ r an + b x = r a + b + r 2 a + b + r 3 a + b + … + r a n + b x = r a + b ( 1 + r a + r 2 a + … + r a ( n − 1 ) ) x = r a + b ( 1 + r a 1 − r a ) x = r a + b + r 2 a + b 1 − r a x = r a + b − r 2 a + b + r 2 a + b 1 − r a
\begin{align*}
x&=r^{a+b} + r^{2a+b} + r^{3a+b} + \ldots +r^{an+b}\\
x&=r^{a+b} (1+ r^{a} + r^{2a} + \ldots +r^{a(n-1)})\\
x&=r^{a+b} \left(1+ \frac{r^{a}}{1-r^{a}}\right)\\
x&= r^{a+b} + \frac{r^{2a+b}}{1-r^{a}}\\
x&= \frac{r^{a+b} - \cancel{r^{2a+b}} + \cancel{r^{2a+b}}}{1-r^a}\\
\end{align*}
x x x x x = r a + b + r 2 a + b + r 3 a + b + … + r an + b = r a + b ( 1 + r a + r 2 a + … + r a ( n − 1 ) ) = r a + b ( 1 + 1 − r a r a ) = r a + b + 1 − r a r 2 a + b = 1 − r a r a + b − r 2 a + b + r 2 a + b Thus
∑ n = 1 ∞ r a n + b = r a + b 1 − r a
\sum_{n=1}^{\infty} r^{an+b} = \frac{r^{a+b}}{1-r^a}
n = 1 ∑ ∞ r an + b = 1 − r a r a + b All this would seem to imply that if
f ( n ) f(n) f ( n ) N → N \mathbb{N} \to \mathbb{N} N → N
∑ n = 1 ∞ r f ( n )
\sum_{n=1}^{\infty} r^{f(n)}
n = 1 ∑ ∞ r f ( n ) I’d like to see this extended to other number systems than natural numbers, but I’m a bit exhausted of series at this point.