17
Related: Calculate Power Series Coefficients
Given a positive integer \$X\$ and a max exponent (Also a positive integer too) \$N\$ calculate the result of a power series. Example:
$$X^0+X^1+X^2+\cdots +X^N$$
Test Cases
1 2 => 3
2 3 => 15
3 4 => 121
2 19 => 1048575
Standard code-golf rules apply.
Luis felipe De jesus Munoz
Posted 2018-07-30T20:19:16.113
Reputation: 9 639
9
ngm
Posted 2018-07-30T20:19:16.113
Reputation: 3 974
2I love the way vectors work in R – Beta Decay – 2018-07-30T22:04:31.973
@BetaDecay maybe we should nominate R for the language of the month! – Giuseppe – 2018-07-31T01:56:14.800
1@giuseppe I've been thinking about that for a while now... let's target September? I'll post an answer in Meta. – JayCe – 2018-08-01T18:52:19.720
@JayCe yeah, sounds good. We should open up the R chatroom again (for the third time; I've twice opened one and both times it has died due to inactivity); I'm sure I'd love to add to whatever description you end up going with on the submission :-) – Giuseppe – 2018-08-01T19:00:02.667
5
The straightforward approach:
x#n=sum$map(x^)[0..n]
For the case \$ x \neq 1 \$ we alternatively could use following function with the same length:
x#n=div(x*x^n-1)$x-1
This uses the fact that
\$ 1 + x + x^2 + \ldots + x^n = \frac{x^{n+1} -1}{x-1} \forall x \neq 1.\$
flawr
Posted 2018-07-30T20:19:16.113
Reputation: 40 560
1also 21 bytes: x#0=1;x#n=x^n+x#(n-1). – nimi – 2018-07-30T22:21:51.040
@nimi Oh that is clever:) – flawr – 2018-07-31T18:07:24.967
5
Saved 1 byte thanks to @tsh
x=>g=n=>~n&&x*g(n-1)+1
Using the direct formula mentioned by @flawr:
x=>n=>~-(x**++n)/~-x||n
Arnauld
Posted 2018-07-30T20:19:16.113
Reputation: 111 334
x=>g=n=>n?g(n-1)*x+1:1 – tsh – 2018-07-31T05:36:10.653
5
ÝmO
Explanation:
Input: 4, 3
Ý [0..input] - [0, 1, 2, 3, 4]
m Vectorized exponent - [1, 3, 9, 27, 81]
O Sum - 121
Okx
Posted 2018-07-30T20:19:16.113
Reputation: 15 025
4
Erik the Outgolfer
Posted 2018-07-30T20:19:16.113
Reputation: 38 134
Not sure it's worth another answer to post this other 4 byter: *ⱮS‘ – Jonathan Allan – 2018-07-30T21:09:06.403
@JonathanAllan I actually edited that away (note: I do this pretty often), but, eh, a grace period exists. :P (It was actually me forgetting about vectorization of *...) – Erik the Outgolfer – 2018-07-30T21:27:16.103
3
lambda x,n:sum(x**k for k in range(n+1))
lambda x,n:sum(map(x.__pow__,range(n+1))) is cool too but it's 1 byte longer lol.
HyperNeutrino
Posted 2018-07-30T20:19:16.113
Reputation: 26 575
3
param($x,$n)0..$n|%{$o+=[math]::pow($x,$_)};$o
Not bad for needing a .NET call for pow.
-7 bytes thanks to mazzy.
AdmBorkBork
Posted 2018-07-30T20:19:16.113
Reputation: 41 581
It's stupid, but this param($x,$n)$s=0;0..$n|%{$s+=[math]::pow($x,$_)};$s is shorter. I don't sure about $s=0;: this expression is needed on a second run. Perhaps, it better to append ;rv s to the end.
2 bytes less: param($x,$n,$s)0..$n|%{$s+=[math]::pow($x,$_)};$s. A caller still sends 2 parameters. – mazzy – 2018-08-01T10:16:47.673
1@mazzy Thanks! There's no need to re-initialize $s if it's run as a full program (e.g., how it's done on Try It Online), so that saves a few more bytes. – AdmBorkBork – 2018-08-01T12:33:07.470
3
Giuseppe
Posted 2018-07-30T20:19:16.113
Reputation: 21 077
3
I~xI(ax!i^+~ai)aP
I~x - Store the input in the variable x
I( ) - Loop until the top of the stack equals the input n
a - Push a
x - Push x
!i - Increment i
^ - Calculate the value of x^i
+~a - Add x^i to a and store in a
i - Push i
aP - Print the value of a
Beta Decay
Posted 2018-07-30T20:19:16.113
Reputation: 21 478
3
-1 Thanks to tsh (f(x,n-1)+x**n -> f(x,n-1)*x+1)
Port of Arnauld's Javascript answer - not sure if it is the first, so do shout if you know who deserves the credit!
f=lambda x,n:~n and f(x,n-1)+x**n
A recursive function which sums the terms right-to-left, with a base case of zero when n reaches -1 (since ~(-1) is -1 - (-1) which is 0 which is falsey).
My previous 35 byter:
lambda x,n:x^1and~-x**-~n/~-x or-~n
The ^ operator is bitwise-xor, so x^1 is zero when x is one and non-zero otherwise.
In Python non-zeros are truthy, so the right of the logical and is executed when x is not one, but not executed when x is one, whereupon the right of the logical or is executed instead.
So, when x is one we execute
-~n which is equivalent to
-1 * ~n which is equivalent to
-1 * (-1 - n) which is equivalent to
1 + n...
...and when x is not one we execute
~-x**-~n/~-x which, adding parentheses to signify precedence, is
(~-(x**(-~n)))/(~-x) which is equivalent to
(-1 - -1 * (x ** (-1 * (-1 - n))))/(-1 - -x) which is equivalent to
((x ** (n + 1)) - 1)/(x - 1)
Jonathan Allan
Posted 2018-07-30T20:19:16.113
Reputation: 67 804
The same to JS answer: +x**n -> *x+1 save 1 byte. – tsh – 2018-07-31T06:27:08.127
3
t⟦R&h;Rz^ᵐ+
Nothing particularly interesting, construct the range 0 to n, raise x to the power of each number in it.
And since I wanted to learn how ᵃccumulate works, here's a slightly longer version that uses that:
,1↺⟨t×{bh}⟩ᵃ⁾k+
Form array [1, x], and do this n times, accumulating the results into that array after each iteration: multiply the last element of the array, by the second element of the array (i.e. x). Since this calculates [1, x, x^2, ... x^n, x^(n+1)], knife off the last value and add the rest of them up as the output.
sundar - Reinstate Monica
Posted 2018-07-30T20:19:16.113
Reputation: 5 296
2
Kirill L.
Posted 2018-07-30T20:19:16.113
Reputation: 6 693
2
Daniel
Posted 2018-07-30T20:19:16.113
Reputation: 6 425
Since x is guaranteed to be positive, I believe you can do if x<2. – Mario Ishac – 2018-07-31T02:02:19.790
@MarDev, ah yes that’s right. Thanks. – Daniel – 2018-07-31T02:12:42.743
2
$0;A^$
Implicit inputs: A and n
$0 Zero indexing
; Output sum of first n terms in sequence
A^$ Each term is A to the power of the current index
Stephen
Posted 2018-07-30T20:19:16.113
Reputation: 12 293
2
#.1$~>:
#.1$~>: Left argument = X, Right argument = N
1$~>: Generate a list of N+1 ones
#. Interpret as base X digits and convert to single integer
Bubbler
Posted 2018-07-30T20:19:16.113
Reputation: 16 616
2
@(x,n)sum(x.^(0:n))
Just learnt Octave 15 minutes ago for this challenge... Hoping it is already optimized.
tsh
Posted 2018-07-30T20:19:16.113
Reputation: 13 072
2
Uriel
Posted 2018-07-30T20:19:16.113
Reputation: 11 708
1
wastl
Posted 2018-07-30T20:19:16.113
Reputation: 3 089
1
mbomb007
Posted 2018-07-30T20:19:16.113
Reputation: 21 944
r is redundant, so f=lambda x,n:n>=0and x**n+f(x,n-1) is 34, although golfing that with n>=0 -> ~n we get my port of 33 bytes. – Jonathan Allan – 2018-07-30T22:24:40.410
1
Οurous
Posted 2018-07-30T20:19:16.113
Reputation: 7 916
1
Digital Trauma
Posted 2018-07-30T20:19:16.113
Reputation: 64 644
1
Digital Trauma
Posted 2018-07-30T20:19:16.113
Reputation: 64 644
1
{sum $^a X**0..$^b}
{ } # Anonymous code block
sum # Get the sum of
X** # The cross product with the meta operator exponential
$^a # With the first parameter
0..$^b # And the range of 0 to the second parameter
Jo King
Posted 2018-07-30T20:19:16.113
Reputation: 38 234
1
l~),f#:+
Try it online! Or verify all test cases.
l e# Read a line from STDIN
~ e# Evaluate: pushes x, then n
) e# Add 1 to n
, e# Range: gives [0 1 ... n]
f# e# Map with extra parameter: gives [x^0 x^1 ... x^n]
:+ e# Fold addition over array: gives x^0 + x^1 + ... + x^n
e# Implicit display in STDOUT
Luis Mendo
Posted 2018-07-30T20:19:16.113
Reputation: 87 464
1CJam... Now that's a name I haven't heard in a while – Beta Decay – 2018-07-31T09:29:05.347
@BetaDecay Judge me by my age, do you? – Luis Mendo – 2018-07-31T09:32:04.817
0
alephalpha
Posted 2018-07-30T20:19:16.113
Reputation: 23 988
0
x->n->x>1?~-(int)Math.pow(x,n+1)/~-x:n+1
Uses the approach mentioned by @flawr.
Explanation:
x->n-> // Method with two integer parameters and integer return-type
x>1? // If `x` is larger than 1:
~-(int)Math.pow(x,n+1) // Return `x` to the power of `n+1` - 1
/~-x // integer-divided by `x-1`
: // Else:
n+1 // Return `n+1` as result instead
Kevin Cruijssen
Posted 2018-07-30T20:19:16.113
Reputation: 67 575
0
Galen Ivanov
Posted 2018-07-30T20:19:16.113
Reputation: 13 815
0
Throws errors, but still works I guess.
f(_,0,1).
f(X,N,Y):-M is N-1,f(X,M,Z),Y is Z+X^N.
qwertxzy
Posted 2018-07-30T20:19:16.113
Reputation: 101
0
Your usual, everyday recursive solution.
f(b,e,t,i){for(t=1,i=e;i--;t*=b);t=e--?t+f(b,e):1;}
ErikF
Posted 2018-07-30T20:19:16.113
Reputation: 2 149
f(x,n,v){for(v=1;n--;v=v*x+1);x=v;} – tsh – 2018-07-31T07:20:33.587
f(b,e){b=e?pow(b,e--)+f(b,e):1;} – ceilingcat – 2018-07-31T14:27:12.820
0
Ｉ↨ＮＥ⊕Ｎ¹
Try it online! Link is to verbose version of code. Uses @Bubbler's algorithm. Explanation:
Ｎ Second input as a number
⊕ Increment
Ｅ ¹ Make list of `1`s of that length
Ｎ First input as a number
↨ Base conversion
Ｉ Cast to string
Implicitly print
Neil
Posted 2018-07-30T20:19:16.113
Reputation: 95 035
0
AQVhH=+Z^GN)Z
How it works
assign('Q',eval_input())
assign('[G,H]',Q)
for N in num_to_range(head(H)):
assign('Z',plus(Z,Ppow(G,N)))
imp_print(Z)
ElPedro
Posted 2018-07-30T20:19:16.113
Reputation: 5 301
0
@set s=0
@for /l %%i in (0,1,%2)do @set/as=s*%1+1
@echo %s%
Using @tsh's algorithm.
Neil
Posted 2018-07-30T20:19:16.113
Reputation: 95 035
0
func a(x:Int,n:Int)->Int{return(0...n).reduce(0,{p,q in p+(q>=1 ?(1...q).reduce(1,{a,_ in a*x}):1)})}
func a(x:Int, n:Int) -> Int {
return (0...n).reduce(0, // For 0 to n
{ p, q in // p is result of last cycle, q the current element
p + // Add p to
( q > 0 ? // If current element is not 0
(1...q).reduce(1, { a, _ in a*x } ) // pow(x, n) *
: 1 // Otherwise, 1
)
}
)
}
* The reduce is shorter than:
import Foundation
[...] Int(pow(Double(x),Double(q)))
First golf I do, so could be much more golfable, but finally the super-verbose Swift has a chance to be used!
Simone Chelo
Posted 2018-07-30T20:19:16.113
Reputation: 543
4It would open up some more possibilities if we could assume that $ x\neq 1$, then we could use $ 1 + x + x^2 + \ldots + x^n = \frac{x^{n+1}-1}{x-1}$, but it is probably too late for that =) – flawr – 2018-07-30T20:35:44.450
I definitely remember this challenge appearing previously but I don't have time to look for a dupe now. – xnor – 2018-07-30T20:38:45.473
3"Assume 0 ≤ (X + N) ..." - but X & N are positive integers, so should that read "Assume 0 < (X + N) ..." or should X & N be non-negative integers? – Jonathan Allan – 2018-07-30T21:25:44.797
@JonathanAllan Most non-mathematicians would probably class zero as a positive integer – Beta Decay – 2018-07-30T21:58:14.173
1@BetaDecay really? Most non-mathematicians would probably class it as a "number" :p – Jonathan Allan – 2018-07-30T22:04:28.153
4This is the potential dupe I was thinking of, with a difference being that it goes to
n*n-1rather thann. Since my vote hammers, I'll wait for others to say if this is dupe-worthy. – xnor – 2018-07-30T22:54:56.8603@BetaDecay Most of the world considers
0to be neither positive nor negative. A couple of places (like France) don't consider positive to mean strictly positive, and treat0as both positive and negative. – Jo King – 2018-07-31T02:36:12.320This is a simply stated challenge with a clear title. Perhaps the other challenge with the irrelevant backstory and the confusing title should get the dupe label retroactively! – ngm – 2018-07-31T13:31:15.803
2@ngm, I don't find the title of this question clear, whereas the other title references a classic question that I've seen in printed puzzle books. However, if you want to propose flipping the duplicate closure relationship the place to do it is a [meta-tag:specific-question] [meta-tag:discussion] question on meta. – Peter Taylor – 2018-07-31T13:50:50.993