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Use the Maclaurin series for sinx,

cosx, and exto find the values of the following series.

1-π222·2!+π424·4!-π626·6!+⋯

Short Answer

Expert verified

The values of the series1-π222·2!+π424·4!-π626·6!+⋯is1-π222·2!+π424·4!-π626·6!+⋯=0

Step by step solution

01

Given information

The series is1-π222·2!+π424·4!-π626·6!+⋯

02

Find the Maclaurin series for the function f(x)=sinx 

The Maclaurin series for the function f(x)=sinxis

cosx=1-x22!+x33!-x44!+⋯

The series 1-π222·2!+π424·4!-π626·6!+⋯is the Maclaurin series for

cosxatx=Ï€2

03

Find the values of the series

1-x22!+x33!-x44!+⋯=cosx

Therefore,

1-π222·2!+π424·4!-π626·6!+⋯=cosπ21-π222·2!+π424·4!-π626·6!+⋯=0

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