ok i really hate series, they are confusing and i almost have no idea how to do them. These questions are really driving me up the wall.
Any help would be greatly appreciated.
Lets look at the first few terms of the Taylor series for f(x)=ln(1+x) centred
so df/dx(0) = 1
So the series starts:
Now look at your candidate answers. The first two can be ruled out as
the absolute value of the coefficient of x^2 is 2.
That leaves us with the second two. The fourth gives a -ve coefficient for
x, so it is ruled out, which leaves us with (c), which agrees with the terms
of the series that we know.
Look at the Lagrange form of the remainder for a truncated Taylor series.
In this case it is R(x)=f''''(zeta)(x-1)^4/4!, for some zeta in [0.9,1.1].
|R(x)|< 120 (1.1) (0.1)^4 /4!=5 (1.1) (0.1)^4
And so one,
b is the answer.
Also, the hyperbolic sine gets its name because it is in ways similar to the sine function. For example,
But the hyperbolic sine has all positive terms,