Monthly Archives: September 2018

Meeting Mr. Bernoulli

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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)

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Fig. 1

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

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Pandora’s Box

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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}} = {{\sum\limits_{i=1}^{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.

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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)

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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.

It reads:

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

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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.