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)(x-1)^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=x-x^3/3!+x^5/5!-...
But the hyperbolic sine has all positive terms,
sinh x=x+x^3/3!+x^5/5!+...