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.
Sep 25th 2006, 11:33 PM
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
Sep 26th 2006, 07:36 AM
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,
But the hyperbolic sine has all positive terms,