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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`
Maclaurin series is Taylor series about 0.
Now we cancel `(n-1)!` and `n!` and get only `n` in the denominator so finally we have
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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