Getting Started

We will quickly go through the setup and use our CLI to solve the previous problems.

Installing winkalc

I assume you have had Rust and Git installed on your device.

Open the terminal, navigate to your favorite working directory, and clone the source code:

git clone https://github.com/0xwink/winkalc

cd winkalc

Afterwards, use Cargo to build the binary and run it:

Bash/Zsh

cargo build --release

cd target/release

chmod +x winkalc

./winkalc

Powershell

cargo build --release

cd target/release

./winkalc.exe

Optional

Because the binary can be executed independently, you may now move the binary anywhere you see fit, for example a PATH directory like ~/.local/bin/. By doing this you successfully install the CLI as one of your user apps, so that you may run it through a single spell in Bash:

winkalc

Resolving the Problem 1

Run winkalc and enter an instruction. (Don't copy the leading >!)

> [Z] bezout {45499231} {793338813}

You will see an output like this:

Result: U * F + V * G = H, where
U = -372586328,
F = 45499231,
V = 21368413,
G = 793338813,
[GCD] H = 1.

The GCD of the numbers is one. Thus they are coprime.

Resolving the Problem 2

We don’t support automatic expansion of polynomials yet. So you need to first mannually expand (x+2)^2 and (x+3)^3 using the binomial theorem.

> [QPol] bezout {x^2+4x+4} {x^3+9x^2+27x+27}
Result: U * F + V * G = H, where
U = 3x^2 + 20x + 34,
F = x^2 + 4x + 4,
V = - 3x - 5,
G = x^3 + 9x^2 + 27x + 27,
[GCD] H = 1.

We basically have U * F + V * G = 1. Dividing it by F * G yields V / F + U / G = 1 / (F * G).

In formulae, we obtain

$$\frac{1}{(x+2{)}^2(x+3{)}^3}=\frac{-3x-5}{(x+2{)}^2}+\frac{3x^2+20x+34}{(x+3{)}^3}.$$

If you like, we can even further decompose the fractions:

> [QPol] divmod {-3x-5} {x+2}
Result: F / G = Q ... R, where
F = - 3x - 5,
G = x + 2,
Q = - 3,
R = 1.

> [QPol] divmod {3x^2+20x+34} {x^2+6x+9}
Result: F / G = Q ... R, where
F = 3x^2 + 20x + 34,
G = x^2 + 6x + 9,
Q = 3,
R = 2x + 7.

> [QPol] divmod {2x+7} {x+3}
Result: F / G = Q ... R, where
F = 2x + 7,
G = x + 3,
Q = 2,
R = 1.

These calculations means

$$-3x-5=-3(x+2)+1,$$$$3x^2+20x+34=3(x+3{)}^2+2(x+3)+1.$$

In conclusion, the ultimate partial fraction decomposition is

$$\frac{1}{(x+2{)}^2(x+3{)}^3}=\frac{-3}{x+2}+\frac{1}{(x+2{)}^2}+\frac{3}{x+3}+\frac{2}{(x+3{)}^2}+\frac{1}{(x+3{)}^3}.$$

Beautiful, isn’t it? Admittedly, current process is not fully automated. In the future, a completely automatic PFD algorithm may be implemented.