# Meeting Mr. Bernoulli The differential equation ${d \over dx} y + f(x) y = g(x) y^{\alpha}\quad\quad\quad(1)$

where $\alpha \neq 0, 1$ and $g(x) \not \equiv 0$, is known as the Bernoulli’s equation.

When $\alpha$ is an integer, (1) has trivial solution $y(x) \equiv 0$.

To obtain nontrivial solution, we divide each term of (1) by $y^{\alpha}$ to get, $\boxed{y^{-\alpha}{d \over dx}y} + f(x) y^{1-\alpha} = g(x)\quad\quad\quad(2)$

Since ${d \over {dx}}({{1 \over {1-\alpha}}y^{1-\alpha}}) ={1 \over {1-\alpha}}\cdot (1-\alpha) y^{1-\alpha-1}{d \over dx}y=\boxed{y^{-\alpha}{d \over dx}y}$

(2) can be expressed as ${d \over dx} ({{1 \over {1-\alpha}} y^{1-\alpha}}) + f(x) y^{1-\alpha} = g(x)$

which is ${{1 \over {1-\alpha}} {d \over dx} y^{1-\alpha}} + f(x) y^{1-\alpha} = g(x)$ .

Multiply $1-\alpha$ throughout, ${d \over dx} y^{1-\alpha} + (1-\alpha) f(x) y^{1-\alpha} = (1-\alpha) g(x)\quad\quad\quad(3)$

Let $z = y^{1-\alpha}$, (3) is transformed to a first order linear equation ${d \over dx} z + (1-\alpha) f(x) z = (1-\alpha) g(x)$,

giving the general solution of a Bernoulli’s equation (see Fig. 1) Fig. 1

For a concrete example of Bernoulli’s equation, see “What moves fast, will slow down

# Pandora’s Box Summations arise regularly in mathematical analysis. For example, $\sum\limits_{i=1}^{n}{1 \over {i (i+1)}} = {n \over {n+1}}$

Having a simple closed form expression such as ${n \over {n+1}}$ makes the summation easier to understand and evaluate.

The summation we focus on in this post is $\sum\limits_{i=1}^{n}i 2^i\quad\quad\quad(1)$

We will find a closed form for it.

In a recent post, I derived the closed form of a simpler summation (see “Beer theorems and their proofs“) Namely, $\sum\limits_{i=0}^{n}x^i={{x^{n+1}-1} \over {x-1}}\quad\quad\quad(2)$

From (2) it follows that ${d \over {dx}}{\sum\limits_{i=0}^{n}x^i} = {d \over {dx}}({ {x^{n+1}-1} \over {x-1} })$

which gives us ${\sum\limits_{i=0}^{n}{{d \over dx}x^i}}={{(n+1)x^{n}(x-1)-(x^{n+1}-1)} \over {(x-1)^2}}$.

Or, ${\sum\limits_{i=0}^{n}{i x^{i-1}}}= {{\sum\limits_{i=0}^{n}{i x^{i}}} \over {x}}$ $= {{(n+1)x^{n}(x-1)-(x^{n+1}-1)} \over {(x-1)^2}}$.

Therefore, ${\sum\limits_{i=1}^{n}{i x^{i}}}={{(n+1)x^{n+1}(x-1)-x^{n+2}+x} \over {(x-1)^2}}$.

Let $x=2,$ we arrived at (1)’s closed form: ${\sum\limits_{i=1}^{n}i 2^i} = {{(n+1)2^{n+1} -2 ^{n+2} + 2} \over {2-1}} = 2^{n+1} (n-1) + 2$. I have a Computer Algebra aided solution too.

Let $s_n \triangleq \sum\limits_{i=1}^{n} i x^i$,

we have $s_1 = x, s_{n}-s_{n-1}=n x^n$

Therefore, the closed form of $s_n$ is the solution of initial-value problem $\begin{cases} {s_{n}-s_{n-1} }= {n x^n} \\ s_1=x\end{cases}$

It is solved by Omega CAS Explorer (see Fig. 1) Fig. 1

At ACA 2017 in Jerusalem, I gave a talk on “Generating Power Summation Formulas using a Computer Algebra System“.

I had a dream that night. In the dream, I was taking a test.

Derive the closed form for $\sum\limits_{i=1}^{n} {1 \over {(3i-2)(3i+1)}}$ $\sum\limits_{i=1}^{n} {1 \over {(2i+1)^2-1}}$ $\sum\limits_{i=1}^{n} {i \over {(4i^2-1)^2}}$ $\sum\limits_{i=1}^{n} {{i^2 4^i} \over {(i+1)(i+2)}}$ $\sum\limits_{i=1}^{n} { i \cdot i!}$

I woke up with a sweat.

# My shot at Harmonic Series To prove Beer Theorem 2 (see “Beer theorems and their proofs“) is to show that the Harmonic Series $1 + {1 \over 2} + {1 \over 3} + ...$ diverges.

Below is my shot at it.

Yaser S. Abu-Mostafa proved a theorem in an article titled “A differentiation test for absolute convergence” (see Mathematics Magazine 57(4), 228-231)

His theorem states that

Let $f$ be a real function such that ${d^2 f} \over {dx^2}$ exists at $x = 0$. Then $\sum\limits_{n=1}^{\infty} f({1 \over n})$ converges absolutely if and only if $f(0) = f'(0)=0$.

Let $f(x) = x$, we have $\sum\limits_{n=1}^{\infty}f({1 \over n}) = \sum\limits_{n=1}^{\infty}{1 \over n}$,

the Harmonic Series. And, $f'(x) = {d \over dx} x = 1 \implies f'(0) \neq 0$.

Therefore, by Abu-Mostafa’s theorem, the Harmonic Series diverges.