Thread: approximate through infinite series

Use the series expansion for sinh(x) to evaluate, correct 4 d.p
a) sinh 0.5
b) sinh 2

sinh x = x + (x^3)/6 + (x^5)/120 + (x^7)/5040

put in the values of x into this

0.5 + (0.5^3)/6 + (0.5^5)/120 + (0.5^7)/5040 = 0.5210953

2 + (2^3)/6 + (2^5)/120 + (2^7)/5040 = 3.62539683

to check these are correct type into google

sinh(0.5)

or

sinh(2)

sorry and the formula for the power series of sinh is

sinh(x) = x + (x^3)/ 3! + (x^5)/5! + .... + (x^(2n+1)) / (2n+1)!

the maclaurin series which is a Taylor series expansion of a function about 0