Taylor Swifts 34th Birthday

[lpetrich]

Having handled the Taylor part, I turn to the Swift part.

Let's look at infinite-series convergence. One of the series posted by SLD was the binomial series:

•  Binomial theorem and Binomial Theorem -- from Wolfram MathWorld
•  Binomial series and Binomial Series -- from Wolfram MathWorld

\( \displaystyle{ (1+x)^p = \sum_{k=0}^\infty \frac{p(p-1)(p-2)\cdots(p-k+1)}{k!} x^k } \)
Taking the Taylor series of the LHS around x = 0 gives the RHS.

For nonnegative-integer p, one gets the binomial theorem.

For p = -1, one gets
\( \displaystyle{ \frac{1}{1+x} = 1 - x + x^2 - x^3 + x^4 - x^5 + \cdots } \)

For x = -1 one gets 1/0 = 1 + 1 + 1 + 1 + 1 + ... which obviously diverges.

But for x = 1, one gets 1/2 = 1 - 1 + 1 - 1 + ... which is indeterminate.

For x = 2 one gets 1/3 = 1 - 2 + 4 - 8 + 16 - 32 + ... which diverges

For x = -2 one gets -1 = 1 + 2 + 4 + 8 + 16 + 32 + ... which diverges

So the binomial series only gives meaningful results inside its radius of convergence: |x| < 1.

 

[lpetrich]

 Radius of convergence -- for S(x), a series in variable x: \( \displaystyle{ S(x) = \sum_n a_n x^n } \) -- the radius of convergence r is the largest value where S(x) converges for all |x| < r.

It can be found from the  Ratio test \( \displaystyle{ \frac{1}{r} = \lim_{n\to\infty} \frac{|a_{n+1}|}{|a_n|} } \) or from the  Root test \( \displaystyle{ \frac{1}{r} = \limsup_{n\to\infty} |a_n|^{1/n} } \)

The triangle inequality is also useful: \( \displaystyle{ |\sum_n a_n| \le \sum_n |a_n| } \) It also works for integrals.

Also, if for each n, \( 0 \le a_n \le b_n \), \( \displaystyle{ \sum_n a_n \le \sum_n b_n } \)

Sums can be turned into integrals by turning a sum into a step function with a step at each summed value. One can then find some more easily integrated function that bounds this step function and then integrate it. For summed values given as a(n) for index value n, one can turn that sum into an integral of a(n) over n. One has to be careful about non-integer values of n when doing this, however.

As an example, consider \( \displaystyle{ \sum_{n=1}^\infty \frac{1}{n^2} } \)

It can be expressed as \( \displaystyle{ 1 + \int_1^\infty \frac{1}{(\text{ceil}(n))^2} dn } \)

Since \(n < \text{ceil}(n) \) for non-integer n, \(1/n^2 > 1/(\text{ceil}(n))^2 \) for non-integer n, and thus

\( \displaystyle{ \sum_{n=1}^\infty \frac{1}{n^2} = 1 + \int_1^\infty \frac{1}{(\text{ceil}(n))^2} dn < 1 + \int_1^\infty \frac{1}{n^2} dn = 2 } \)

thus

\( \displaystyle{ \sum_{n=1}^\infty \frac{1}{n^2} < 2 } \)

meaning that that sum converges. Its value is (pi)²/6 ~ 1.645 < 2.

 

[SLD]

 Radius of convergence -- for S(x), a series in variable x: \( \displaystyle{ S(x) = \sum_n a_n x^n } \) -- the radius of convergence r is the largest value where S(x) converges for all |x| < r.

It can be found from the  Ratio test \( \displaystyle{ \frac{1}{r} = \lim_{n\to\infty} \frac{|a_{n+1}|}{|a_n|} } \) or from the  Root test \( \displaystyle{ \frac{1}{r} = \limsup_{n\to\infty} |a_n|^{1/n} } \)

The triangle inequality is also useful: \( \displaystyle{ |\sum_n a_n| \le \sum_n |a_n| } \) It also works for integrals.

Also, if for each n, \( 0 \le a_n \le b_n \), \( \displaystyle{ \sum_n a_n \le \sum_n b_n } \)

Sums can be turned into integrals by turning a sum into a step function with a step at each summed value. One can then find some more easily integrated function that bounds this step function and then integrate it. For summed values given as a(n) for index value n, one can turn that sum into an integral of a(n) over n. One has to be careful about non-integer values of n when doing this, however.

As an example, consider \( \displaystyle{ \sum_{n=1}^\infty \frac{1}{n^2} } \)

It can be expressed as \( \displaystyle{ 1 + \int_1^\infty \frac{1}{(\text{ceil}(n))^2} dn } \)

Since \(n < \text{ceil}(n) \) for non-integer n, \(1/n^2 > 1/(\text{ceil}(n))^2 \) for non-integer n, and thus

\( \displaystyle{ \sum_{n=1}^\infty \frac{1}{n^2} = 1 + \int_1^\infty \frac{1}{(\text{ceil}(n))^2} dn < 1 + \int_1^\infty \frac{1}{n^2} dn = 2 } \)

thus

\( \displaystyle{ \sum_{n=1}^\infty \frac{1}{n^2} < 2 } \)

meaning that that sum converges. Its value is (pi)²/6 ~ 1.645 < 2.

You’re really enjoying this thread! I’m glad I started it.

 

[lpetrich]

A clarification. If for each n, \( 0 \le a_n \le b_n \), \( \displaystyle{ 0 \le \sum_n a_n \le \sum_n b_n } \)

So if the b series converges, then the a series will also converge. Likewise, if the a series diverges, then the b series will also diverge. This is useful if one series is much easier to evaluate than the other.

Thus, for x > 0, \( \displaystyle{ \arctan x < \text{arctanh } x < \frac{x}{1-x^2} } \)

Because of their series: \( \displaystyle{ \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1} < \sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1} < \sum_{n=0}^\infty x^{2n+1} } \)

Also of their integrals: \( \displaystyle{ \int_0^x \frac{dt}{1+t^2} < \int_0^x \frac{dt}{1-t^2} < \int_0^x \frac{dt}{1-x^2} } \)

 

[lpetrich]

Riemann Zeta Function -- from Wolfram MathWorld
\( \displaystyle{ \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} = \prod_{primes\ p}\frac{1}{1-p^{-s}} = \frac{1}{\Gamma(s)} \int_0^\infty \frac{x^{s-1} \, dx}{e^x-1} } \)

Dirichlet Eta Function -- from Wolfram MathWorld
\( \displaystyle{ \eta(s) = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} = (1 - 2^{1-s}) \zeta(s) = \frac{1}{\Gamma(s)} \int_0^\infty \frac{x^{s-1} \, dx}{e^x+1} } \)

Dirichlet Lambda Function -- from Wolfram MathWorld
\( \displaystyle{ \lambda(s) = \sum_{n=0}^\infty \frac{1}{(2n+1)^s} = (1 - 2^{-s}) \zeta(s) = \frac{1}{\Gamma(s)} \int_0^\infty \frac{x^{s-1} e^x \, dx}{e^{2x}-1} } \)

Dirichlet Beta Function -- from Wolfram MathWorld
\( \displaystyle{ \beta(s) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s} = \frac{1}{\Gamma(s)} \int_0^\infty \frac{x^{s-1} e^x \, dx}{e^{2x}+1} } \)

 

[lpetrich]

The zeta function has special values in terms of the Bernoulli numbers for positive even args:
\( \displaystyle{ \zeta(2n) = \frac{(-1)^{n-1} (2\pi)^{2n} B_{2n} }{2 (2n)!} } \)

\( \displaystyle{ \zeta(2) = \frac{\pi^2}{6} ,\ \zeta(2) = \frac{\pi^4}{90} ,\ \zeta(6) = \frac{\pi^6}{945},\ \cdots } \)

There are no comparable formulas for positive odd args, however.

Bernoulli Number -- from Wolfram MathWorld and  Bernoulli number
\( \displaystyle{ \frac{x}{e^x - 1} = \sum_{n=0}^\infty \frac{B_n x^n}{n!} } \)

The reflection formula:
\( \displaystyle{ \zeta(1-s) = 2 (2\pi)^{-s} \cos((1/2)\pi s) \Gamma(s) \zeta(s) } \)

\( \displaystyle{ \zeta(1-n) = \frac{(-1)^{n-1} B_n}{n} } \)

The defining series for the Riemann zeta function will only converge if the real part of the arg is greater than 1. So why is it possible to define that function over more of the complex plane?

Our first hint is that the integral that gives the Dirichlet eta function converges for Re(arg) > 0. So one can use that integral to fill in those missing values. One can then go the rest of the way with the reflection formula, covering the entire complex plane.

This is what gives us \( \displaystyle{ 1 + 2 + 3 + 4 + 5 + \cdots = \zeta(-1) = - \frac{1}{12} } \)

The behavior of the zeta function near 1 can be found from a series:

Stieltjes Constants -- from Wolfram MathWorld
\( \displaystyle{ \zeta(1+s) = \frac{1}{s} + \sum_{n=0}^\infty \frac{(-1)^n}{n!} \gamma_n s^n } \)
where
\( \displaystyle{ \gamma_n = \lim_{m \to \infty} \left[ \sum_{k=1}^m \frac{(\log k)^n}{k} - \frac{(\log m)^{n+1}}{n+1} \right] } \)

gamma(0) = Euler-Mascheroni Constant -- from Wolfram MathWorld

With this form of the zeta function, it is evident that one can go over to Re(arg) <= 1.

 

[lpetrich]

Infinite products can be tested for convergence in much the same way as infinite sums:

\( \displaystyle{ A = \prod_n a_n \ \to \ \log A = \sum_n \log a_n } \)

The sine and cosine functions have these infinite-product expansions:

\( \displaystyle{ \sin x = x \prod_{n=1}^\infty \left( 1 - \frac{x^2}{(n \pi)^2} \right) } \)

\( \displaystyle{ \cos x = \prod_{n=0}^\infty \left( 1 - \frac{x^2}{((n + 1/2) \pi)^2} \right) } \)

These give infinite sums for the tangent and cotangent functions:

\( \displaystyle{ - \frac{d}{dx} \log \cos x = \tan x = 2 \sum_{n=0}^\infty \frac{x}{x^2 + ((n + 1/2) \pi)^2} } \)

\( \displaystyle{ \frac{d}{dx} \log \sin x = \cot x = \frac{1}{x} + 2 \sum_{n=1}^\infty \frac{x}{x^2 + (n \pi)^2} } \)

From the Bernoulli-number generating function, one finds

\( \displaystyle{ \cot x = \sum_{n=0}^\infty \frac{(-1)^n 2^{2n} B_{2n} x^{2n-1}}{(2n)!} } \)

From the earlier series expanson, one finds

\( \displaystyle{ \cot x = \frac{1}{x} + 2 \sum_{n=1}^\infty \frac{(-1)^{n-1} x^{2n-1}}{\pi^{2n}} \zeta(2n) } \)

and that gives us the earlier expression for even positive arg:

\( \displaystyle{ \zeta(2n) = \frac{(-1)^{n-1} (2\pi)^{2n} B_{2n}}{2 (2n)!} } \)

 

[lpetrich]

I confess that my LaTeX math typesetting looks nicer than in previous years because I've been using \displaystyle a lot -- LaTeX's default is \textstyle -- Extending Factorials | Internet Infidels Discussion Board has a lot of \textstyle math formulas that ought to be \displaystyle for better readability.

 

[Elixir]

I confess that my LaTeX math typesetting looks nicer than in previous years because I've been using \displaystyle a lot -- LaTeX's default is \textstyle -- Extending Factorials | Internet Infidels Discussion Board has a lot of \textstyle math formulas that ought to be \displaystyle for better readability.

Taylor should be so proud!