# Expand the Binomial

## Challenge

Given three numbers $a$, $b$, and $n$, you must expand $(a + bx)^n$ in ascending powers of $x$ up to $x^3$.

## Binomial Expansion

Note that this method is an approximation

The binomial expansion works like so:

(a + bx)^n = a^n(1 + bx/a)^n
= a^n(1 + n(bx/a) + n(n -1)(bx/a)^2/2! + n(n-1)(n-1)(bx/a)^3/3!)


Or, more readable:

\begin{align} (a+bx)^n &= a^n\left(1+\frac{b}{a}x\right)^n\\ &\approx a^n\left(1 + n\left(\frac{b}{a}x\right)+ \frac{n(n-1)}{2!}\left(\frac{b}{a}x\right)^2 + \frac{n(n-1)(n-2)}{3!}\left(\frac{b}{a}x\right)^3\right) \end{align}

Which simplifies down to an expression in the form

$$\alpha + \beta x + \gamma x^2 + \delta x^3$$

Where $\alpha$, $\beta$, $\gamma$ and $\delta$ are constants which you must calculate.

These constants should be given to at least 3 decimal places where appropriate (i.e. 1.2537 should be output instead of 1.25, but 1.2 can be output if the answer is exactly 1.2).

You must then output these four constants in a list (in order as they appear in the final equation). This list may be separated however you wish.

# Input

The inputs $a$ and $b$ will be integers in the range -100 to 100 inclusive where $a \neq 0$.

$n$ will be given to one decimal place and will be in the range $-2 \leq n \leq 2$

# Examples

## a = 1, b = 6, n = -1.0

1, -6, 36, -216


## a = 3, b = 2, n = 2.0

9, 12, 4, 0


## a = 5, b = 4, n = 0.3

1.621, 0.389, -0.109, 0.049


# Winning

The shortest code in bytes wins.

• Related – Beta Decay Apr 15 '17 at 9:10
• There is no way your equations can hold, as the left side has degree n in x, where as the right side has degree 3. – flawr Apr 15 '17 at 9:16
• @flawr This method is a way to approximate the value of the polynomial – Beta Decay Apr 15 '17 at 9:18
• is a guaranteed to be non-zero? – tsh Apr 15 '17 at 9:23
• Oh look! There goes Mathematica! – Matthew Roh Apr 15 '17 at 14:39

## Mathematica, 45 37 bytes

It turns out that Binomial threads over lists!

Binomial[#3,r={0,1,2,3}]#2^r#^(#3-r)&


Each of Binomial[#3,r={0,1,2,3}], #2^r and #^(#3-r) returns a length-4 list, which are implicitly multiplied together term-by-term, giving the binomial coefficients.

Old solution (45 bytes):

FoldList[#~D~x/#2&,(#+#2x)^#3,Range@3]/.x->0&


And a 43-byte solution that works except when n = 0:

Series[(#+#2x)^#3,{x,0,3}][[3]]~PadRight~4&

• ooo, I love the Series...[[3]] hack! – Greg Martin Apr 15 '17 at 18:22

# Haskell, 80 74 64 63 62 bytes

I've tried many more sophisticated approaches, but they all were longer than the straightforward approach. (In meantime golfed down a bit.)

Thanks for -1 byte @Laikoni!

(a#b)n|c<-b/a=(*a**n)<$>1:c*n:map(n*(n-1)*c*c*)[1/2,c*(n-2)/6]  Try it online! # Haskell, 46 bytes (a%b)n=scanl(*)(a**n)[(n-i+1)/i*b/a|i<-[1..3]]  Try it online! Expresses the ratios of consecutive terms, and uses them to recover the terms by scanl(*) starting with the first term a*n. The ratios have form (n-i+1)/i*b/a. # Casio BASIC, 144 bytes ?→A ?→B ?→N B÷A→B A^N→A A((N(N-1))÷2!×B²)→S A((N(N-1)(N-2)÷3!×B³)→T N×B×A→N ClrText Locate 1,1,A Locate 1,2,N Locate 1,3,S Locate 1,4,T  Output is in a line like so: 34.5 20 11 56.85  # JavaScript, 5553 50 bytes (a,b,n)=>[s=a**n,s*=n*(b/=a),s*=--n*b/2,s*--n*b/3]  (a,b,n)=>a?[s=a**n,s*=n*(b/=a),s*=--n*b/2,s*--n*b/3]:(t=[0,0,0,0])|n>3?t:(t[n]=b**n,t) # MATL, 30 bytes ^li1G/3:"t@^2G@:q-HM/p*w]x4$h*


Try it online!

Input in a, n, b order.

# Pari/GP, 29 bytes

(a,b,n)->Vec((a+b*x)^n)[1..4]


Try it online!

# Axiom, 80 bytes

h(a,b,n)==(v:=series((a+b*x)^(n::Fraction INT));[coefficient(v,i)for i in 0..3])


exercises

(91) -> h(1,6,-1.0)
...
(91)  [1,- 6,36,- 216]
Type: List Expression Integer
(92) -> h(3,2,2.0)
(92)  [9,12,4,0]
Type: List Expression Integer
(93) -> h(5,4,0.3)
10+-+3     10+-+3    10+-+3
10+-+3 6 \|5     42 \|5   476 \|5
(93)  [ \|5  ,-------,- --------,---------]
25        625     15625
Type: List Expression Integer
(94) -> map(numeric, %)
(94)
[1.6206565966 927624351, 0.3889575832 0626298443, - 0.1089081232 9775363564,
0.0493716825 6164831482 4]
Type: List Float