
Series
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.
http://users.bigpond.net.au/sterps/pictures/2.JPG
http://users.bigpond.net.au/sterps/pictures/3.JPG

First problem:
Lets look at the first few terms of the Taylor series for f(x)=ln(1+x) centred
at 0.
f(0)=0
df/dx=1/(1+x),
so df/dx(0) = 1
d^2f/dx^2=1/(1+x)^2
so d^2f/dx^2(0)=1.
So the series starts:
f(x)=0+x(1/2)x^2+...
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.
RonL

Second Problem:
Look at the Lagrange form of the remainder for a truncated Taylor series.
In this case it is R(x)=f''''(zeta)(x1)^4/4!, for some zeta in [0.9,1.1].
f''''(x)=120x,
so:
R(x)< 120 (1.1) (0.1)^4 /4!=5 (1.1) (0.1)^4
RonL

He have,
f(0)=0
f'(0)x/1!=x
f''(0)x^2/2!=0
f'''(0)x^3/3!=x^3/3!
And so one,
Thus,
b is the answer.
Also, the hyperbolic sine gets its name because it is in ways similar to the sine function. For example,
sin x=xx^3/3!+x^5/5!...
But the hyperbolic sine has all positive terms,
sinh x=x+x^3/3!+x^5/5!+...