Tag Archives: sum

December 16, 2024 · 9:19 am

Interesting Sum

S=\sum_{n=1}^\infty (tan^{-1}(\frac{2}{n^2})), find S.

I came across this sum in An Imaginary Tale by Nahin and I was fascinated.

Let tan(\alpha)=n+1 and tan(\beta)=n-1.

Remember
tan(\alpha-\beta)=\frac{tan(\alpha)-tan(\beta)}{1+tan(\alpha)tan(\beta)}
Hence,
tan(\alpha-\beta)=\frac{(n+1)-(n-1)}{1+(n+1)(n-1)}
tan(\alpha-\beta)=\frac{2}{1+n^2-1}
tan(\alpha-\beta)=\frac{2}{n^2}
Therefore,
\alpha-\beta=tan^{-1}(\frac{2}{n^2})
and
\alpha=tan^{-1}(n+1) and \beta=tan^{-1}(n-1)

tan^{-1}(n+1)-tan^{-1}(n-1)=tan^{-1}(\frac{2}{n^2})

Which means,

    \begin{equation*}S=\sum_{n=1}^\infty(tan^{-1}(n+1)-tan^{-1}(n-1))\end{equation}

Let’s try a few partial sums

S_4=tan^{-1}(2)-tan^{-1}(0)+tan^{-1}(3)-tan^{-1}(1)+tan^{-1}(4)-tan^{-1}(2)+tan^{-1}(5)-tan^{-1}(3)

S_4=-tan^{-1}(0)+-tan^{-1}(1)+tan^{-1}(4)+tan^{-1}(5)

S_6=tan^{-1}(2)-tan^{-1}(0)+tan^{-1}(3)-tan^{-1}(1)+tan^{-1}(4)-tan^{-1}(2)+tan^{-1}(5)-tan^{-1}(3)+tan^{-1}(6)-tan^{-1}(4)+tan^{-1}(7)-tan^{-1}(5)

S_6=-tan^{-1}(0)+-tan^{-1}(1)+tan^{-1}(6)+tan^{-1}(7)

Hence, S_N=-tan^{-1}(0)+-tan^{-1}(1)+tan^{-1}(N)+tan^{-1}(N+1)

S_N=-\frac{\pi}{4}-0+tan^{-1}(N)+tan^{-1}(N+1)

What happens as N\rightarrow \infty ?

\lim\limits_{N\to \infty}\ S_N=-\frac{\pi}{4}+\frac{\pi}{2}+\frac{\pi}{2}=\frac{3\pi}{4}

Because we know tan(\frac{\pi}{2}) is undefined.

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September 14, 2024 · 9:45 am

Sum and Product of the Roots of Polynomials

There is a relationship between the sum and product of the roots of a polynomial and the co-efficient of the polynomial.

Let’s start with a quadratic.

The general form for a quadratic (polynomial of degree 2) is

y=ax^2+bx+c

Use the quadratic equation formula to find the roots

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

Hence the roots are

x_1=\frac{-b+\sqrt{b^2-4ac}}{2a} and x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}

Sum of the roots:

\frac{-b+\sqrt{b^2-4ac}}{2a}+\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{-2b}{2a}=\frac{-b}{a}

Product of the roots:

\frac{-b+\sqrt{b^2-4ac}}{2a}\times\frac{-b-\sqrt{b^2-4ac}}{2a}

\frac{b^2}{4a^2}-\frac{b^2-4ac}{4a^2}

\frac{4ac}{4a^2}

\frac{a}{c}

Worked Example
The equation 4x^2+bx+c=0 has two distinct roots. The product of the roots is \frac{3}{4} and the sum is 2. Find b and c.
\frac{-b}{a}=2
a=4
\frac{-b}{4}=2
b=-8

\frac{c}{a}=\frac{3}{4}
\frac{c}{4}=\frac{3}{3}
c=3
The equations is 4x^2-8x+3

Solve the equation to prove the roots do in fact sum to 2 and multiply to \frac{3}{4}
4x^2-8x+3
4x^2-6x-2x+3
2x(2x-3)-1(2x-3)
(2x-1)(2x-3)
x_1=\frac{1}{2} and x_2=\frac{3}{2}
\frac{1}{2}+\frac{3}{2}=2 and \frac{1}{2}\times\frac{3}{2}=\frac{3}{4}

Let’s move to a cubic function.

The general equation is f(x)=ax^3+bx^2+cx+d

Let’s say the roots of this cubic are \alpha, \beta, \gamma

Then ax^3+bx^2+cx+d=a(x-\alpha)(x-\beta)(x-\gamma)

=a(x^2-\beta x - \alpha x+\alpha\beta)(x-\gamma)

=a(x^2-(\alpha+\beta)x+\alpha\beta)(x-\gamma)

=a(x^3-\gamma x^2-x^2(\alpha+\beta)+\gamma(\alpha+\beta)x-\alpha\beta\gamma)

=a(x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\alpha\gamma+\beta\gamma)x-\alpha\beta\gamma)

The sum of the roots

\alpha+\beta+\gamma=\frac{-b}{a}

The product of the roots

\alpha\beta\gamma=\frac{-d}{a}

Also, it can be handy to know

\alpha\beta+\alpha\gamma+\beta\gamma=\frac{c}{a}

Worked example
f(x)=x^3-6x^2+4x+12, the roots are \alpha, \beta and \gamma
Find
(a) \alpha+\beta+\gamma
(b) \alpha^2+\beta^2+\gamma^2
(c) \frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}

(a) \alpha+\beta+\gamma=\frac{-b}{a}
\alpha+\beta+\gamma=\frac{6}{1}
\alpha+\beta+\gamma=6

(b)\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2\alpha\beta-2\alpha\gamma-2\beta\gamma
=6^2-2(4)
=28

(c)\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=\frac{\beta\gamma}{\alpha\beta\gamma}+\frac{\alpha\gamma}{\alpha\beta\gamma}+\frac{\alpha\beta}{\alpha\beta\gamma}
=\frac{\beta\gamma+\alpha\gamma+\alpha\beta}{\alpha\beta\gamma}
=\frac{4}{-13}
=\frac{-1}{3}

We can extend the method we used for finding the sum and product of the roots of cubic to polynomials of greater degree.

If the four roots of a quartic are \alpha, \beta, \gamma and \delta, and the general equation is ax^4+bx^3+cx^2+dx+e, then

\alpha+\beta+\gamma+\delta=\frac{-b}{a}

\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta=\frac{c}{a}

\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta=\frac{-d}{a}

\alpha\beta\gamma\delta=\frac{e}{a}

Worked Example (just one more)
The roots of the cubic equation x^3-4x^2-3x-2 are \alpha, \beta and \gamma. Find the cubic equation whose roots are \alpha+\beta, \alpha+\gamma, and \beta+\gamma

\alpha+\beta+\gamma=4
\alpha\beta+\alpha\gamma+\beta\gamma=-3
\alpha\beta\gamma=2

\frac{-b}{a}=\alpha+\beta+\alpha+\gamma+\beta+\gamma
\frac{-b}{a}=2(\alpha+\beta+\gamma
\frac{-b}{a}=2(4)
\frac{-b}{a}=8

\frac{c}{a}=(\alpha+\beta)(\alpha+\gamma)+(\alpha+\beta)(\beta+\gamma)+(\alpha+\gamma)(\beta+\gamma)
=\alpha^2+\alpha\gamma+\beta\gamma+\gamma^2+\alpha\beta+\alpha\gamma+\beta^2+\beta\gamma+\alpha\beta+\alpha\gamma+\gamma\beta+\gamma^2
=\alpha^2+\beta^2+\gamma^2+3\alpha\gamma+3\beta\gamma+3\alpha\beta
=(\alpha+\beta+\gamma)^2-2\alpha\gamma-2\beta\gamma-2\alpha\beta+3\alpha\gamma+3\beta\gamma+3\alpha\beta
=(\alpha+\beta+\gamma)^2+\alpha\gamma+\alpha\beta+\beta\gamma
=4^2-3
\frac{c}{a}=13

\frac{-d}{a}=(\alpha+\beta)(\alpha+\gamma)(\beta+\gamma)
=(\alpha^2+\alpha\gamma+\beta\alpha+\beta\gamma)(\beta+\gamma)
=\alpha^2\beta+\alpha^2\gamma+\alpha\gamma\beta+\alpha\gamma^2+\beta^2\alpha+\beta\alpha\gamma+\beta^2\gamma+\beta\gamma^2
=2\alpha\beta\gamma+\alpha^2\beta+\beta^2\alpha+\alpha^2\gamma+\gamma^2\alpha+\beta^2\gamma+\gamma^2\beta
=2(2)+\alpha\beta(\alpha+\beta)+\alpha\gamma(\alpha+\gamma)+\beta\gamma(\beta+\gamma)
=-24+\alpha\beta(\alpha+\beta+\gamma)-\alpha\beta\gamma+\alpha\gamma(\alpha+\gamma+\beta)-\alpha\beta\gamma+\beta\gamma(\beta+\gamma+\alpha)-\beta\gamma\alpha
=4+(\alpha+\beta+\gamma)(\alpha\beta+\alpha\gamma+\beta\gamma)-3(2)
=-2+(4)(-3)
\frac{-d}{a}=-14

If a=1 then b=-8, c=13 and d=14
The cubic is x^3-8x^2+13x+14

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