# f(x)=ln(1-3x)   a. Find the first four derivatives and n^th derivative. b. Find the Maclaurin series using the derivative from part a c.Find the radius of convergence and its interval of...

f(x)=ln(1-3x)

a. Find the first four derivatives and n^th derivative.
b. Find the Maclaurin series using the derivative from part a
c.Find the radius of convergence and its interval of convergence

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tiburtius | High School Teacher | (Level 2) Educator

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a.

`f'(x)=-3/(1-3x)`

`f''(x)=-3^2/(1-3x)^2`

`f'''(x)=-(2*3^3)/(1-3x)^3,`

`f^((4))(x)=-(2*3*3^4)/(1-3x)^4`

Now to be able to write `n`-th derivative you need to see the pattern in the derivatives above. You see that the power of 3 in numerator as well as the power in the denominator is growing by 1, in other words it's equal to `n.` On the other hand in third and fourth derivative we see something else in the numerator, that is 2 and 2*3 and in fifth derivative we would have 2*3*4 (2*3 from before and 4 from differentiating the power of the denominator). So the `n`-th derivative is

`f^((n))(x)=-((n-1)!3^n)/(1-3x)^n, n geq 1`

b.

Maclaurin series is Taylor series about 0.

`f(x)=f(0)+sum_(n=0)^oof^((n))(0)x^n/(n!)=0-sum_(n=0)^oo((n-1)!3^n)/(1-3*0)x^n/(n!)=`

Now we cancel `(n-1)!` and `n!` and get only `n` in the denominator so finally we have

`f(x)=-sum_(n=0)^oo(3^nx^n)/n`

c.

Convergence radius of this kind of series is easily found. We just need to find such real number `r` that the series converges for all `x` such that `|x|<r.` Obviously `r=1` i.e. `-1<x<1.`

But we need to take into consideration the domain of our original function `f(x)=ln(1-3x)` which is defined only when `x<1/3` (we can take logarithm of only positive numbers) which means that that the series will be good for approximating our function `f` only on interval `(-1/3,1/3).` ` ` ````

On the image below blue is the graph of function `f` and red is the graph of it's Maclaurin series of 10th degree.

In this case lower degree series give better approximation outside the interval `(-1/3,1/3).`

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