Nth term of a Leibniz Series

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Eric Brown
Eric Brown 2022-2-25
回答： Onimisi 2023-2-22
I'm trying to create a function that will take a signle input (N) and output the Nth term in the Leibniz Series but for some reason my code keeps failing to pass these test: First term check, Check random variable.
Here is my function and how i am calling it.
test = LeibnizTerm(7)
function y = LeibnizTerm(N)
y = (-1)^N/(2*N+1);
end

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采纳的回答

Voss
Voss 2022-2-25
Ran in:
It depends on if the first term should correspond to input N == 0 or input N == 1. If it's N == 0, I think it's ok how you have it.
LeibnizTerm(0)
ans = 1
LeibnizTerm(1)
ans = -0.3333
LeibnizTerm(2)
ans = 0.2000
If it should start with N == 1, then you have to make a couple of adjustments (see modified function LeibnizTerm1 below).
LeibnizTerm1(1)
ans = 1
LeibnizTerm1(2)
ans = -0.3333
LeibnizTerm1(3)
ans = 0.2000
function y = LeibnizTerm(N)
y = (-1)^N/(2*N+1);
end
function y = LeibnizTerm1(N)
y = (-1)^(N+1)/(2*N-1);
end

更多回答（2 个）

Onimisi
Onimisi 2023-2-22
n=0:19
x=(1)*1./((2*n)+1)
y=x.*(-1).^n
LeibnizTerms=y
Onimisi
Onimisi 2023-2-22
Ran in:
n=0:19
n = 1×20
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
x=(1)*1./((2*n)+1)
x = 1×20
1.0000 0.3333 0.2000 0.1429 0.1111 0.0909 0.0769 0.0667 0.0588 0.0526 0.0476 0.0435 0.0400 0.0370 0.0345 0.0323 0.0303 0.0286 0.0270 0.0256
y=x.*(-1).^n
y = 1×20
1.0000 -0.3333 0.2000 -0.1429 0.1111 -0.0909 0.0769 -0.0667 0.0588 -0.0526 0.0476 -0.0435 0.0400 -0.0370 0.0345 -0.0323 0.0303 -0.0286 0.0270 -0.0256
LeibnizTerms=y
LeibnizTerms = 1×20
1.0000 -0.3333 0.2000 -0.1429 0.1111 -0.0909 0.0769 -0.0667 0.0588 -0.0526 0.0476 -0.0435 0.0400 -0.0370 0.0345 -0.0323 0.0303 -0.0286 0.0270 -0.0256

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