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Mathematics
Mathematics
Problem
The Maclaurin series for
arctan
x
\arctan x
arctan
x
evaluated at
x
=
1
x=1
x
=
1
gives
π
4
=
∑
n
=
0
∞
(
−
1
)
n
1
2
n
+
1
.
\displaystyle \frac{\pi}{4}= \sum_{n=0}^{\infty} (-1)^{n}\dfrac{1}{2n+1}.
4
π
=
n
=
0
∑
∞
(
−
1
)
n
2
n
+
1
1
.
Determine the smallest integer
N
N
N
so that the partial sum
S
N
=
∑
n
=
0
N
(
−
1
)
n
1
2
n
+
1
S_N=\displaystyle\sum_{n=0}^{N} (-1)^{n}\dfrac{1}{2n+1}
S
N
=
n
=
0
∑
N
(
−
1
)
n
2
n
+
1
1
approximates
π
4
\dfrac{\pi}{4}
4
π
with an absolute error less than
1
×
10
−
4
1\times10^{-4}
1
×
1
0
−
4
.
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