Update badges and use LaTeX for math

- We're using GHA + Codecov rather than Travis + Coveralls
- GitHub now renders LaTeX math in Markdown

README.md

@@ -2,12 +2,12 @@
 
 A general framework for fast Fourier transforms (FFTs) in Julia.
 
-[![Travis](https://travis-ci.org/JuliaMath/AbstractFFTs.jl.svg?branch=master)](https://travis-ci.org/JuliaMath/AbstractFFTs.jl)
-[![Coveralls](https://coveralls.io/repos/github/JuliaMath/AbstractFFTs.jl/badge.svg?branch=master)](https://coveralls.io/github/JuliaMath/AbstractFFTs.jl?branch=master)
+[![GHA](https://github.com/JuliaMath/AbstractFFTs.jl/workflows/CI/badge.svg)](https://github.com/JuliaMath/AbstractFFTs.jl/actions?query=workflow%3ACI+branch%3Amaster)
+[![Codecov](http://codecov.io/github/JuliaMath/AbstractFFTs.jl/coverage.svg?branch=master)](http://codecov.io/github/JuliaMath/AbstractFFTs.jl?branch=master)
 
 Documentation:
 [![](https://img.shields.io/badge/docs-stable-blue.svg)](https://JuliaMath.github.io/AbstractFFTs.jl/stable)
-[![](https://img.shields.io/badge/docs-latest-blue.svg)](https://JuliaMath.github.io/AbstractFFTs.jl/latest)
+[![](https://img.shields.io/badge/docs-latest-blue.svg)](https://JuliaMath.github.io/AbstractFFTs.jl/dev)
 
 This package is mainly not intended to be used directly.
 Instead, developers of packages that implement FFTs (such as [FFTW.jl](https://github.com/JuliaMath/FFTW.jl) or [FastTransforms.jl](https://github.com/JuliaApproximation/FastTransforms.jl))
@@ -36,5 +36,6 @@ To define a new FFT implementation in your own module, you should
 
 * You can also define similar methods of `plan_rfft` and `plan_brfft` for real-input FFTs.
 
-The normalization convention for your FFT should be that it computes yₖ = ∑ⱼ xⱼ exp(-2πi jk/n) for a transform of
-length n, and the "backwards" (unnormalized inverse) transform computes the same thing but with exp(+2πi jk/n).
+The normalization convention for your FFT should be that it computes $y_k = \sum_j \exp\(-2 \pi i \cdot \frac{j k}{n}\)$
+for a transform of length $n$, and the "backwards" (unnormalized inverse) transform computes the same thing but with
+$\exp\(+2 \pi i \cdot \frac{j k}{n}\)$.