Recall the natural logarithm, `ln x`. Recall also the graph of y = ln x:
Graph of `f(x)=ln(x)`.
Our aim is to find a good polynomal approximation to the curve in the region near x = 10.
We need to use the Taylor Series with a = 10.
The first term in the Taylor Series is f(a). In this example,
`f(a)` ` = f(10) = ln 10 = 2. 302\ 585\ 093.`
Now for the derivatives.
Recall the derivative of ln x, which is `1/x`. So
We need `f'(10)`, which is `1/10 = 0.1`.
Now for the second derivative:
At `x = 10`, this has value `-0.01`.
The third derivative at `x = 10` has value `0.002`.
At `x = 10`, this has value `-0.0006`.
You can see that we could continue forever. This function is infinitely differentiable.
Now to substitute these values into the Taylor Series:
`f(x)~~f(a)` `+f’(a)(x-a)` `+(f’’(a))/(2!)(x-a)^2` `+(f’’’(a))/(3!)(x-a)^3` `+(f^("iv")(a))/(4!)(x-a)^4+...`
`f(x)~~f(10)` `+f’(10)(x-10)` `+(f’’(10))/(2!)(x-10)^2` `+(f’’’(10))/(3!)(x-10)^3` `+(f^("iv")(10))/(4!)(x-10)^4+...`
`~~2.302585093+0.1(x-10)` `+(-0.01)/(2!)(x-10)^2` `+(2xx0.001)/(3!)(x-10)^3` `+(-6xx0.0001)/(4!)(x-10)^4+...`
Expanding this all out and collecting like terms, we obtain the polynomial which approximates `ln x`:
`ln x ~~ 0.21925 ` `+ 0.4x` ` − 0.03x^2 ` `+ 0.00133x^3` ` − 0.000025x^4+ ...`
This is the approximating polynomial that we were looking for.
We see from the graph that our polynomial (in grey) is a good approximation for the graph of the natural logarithm function (in green) in the region near `x = 10`.
Graph of the approximating polynomial, and `f(x)=ln(x)`.
Notice that the graph is not so good as we get further away from `x = 10`. The regions near `x = 0` and `x = 20` are showing some divergence.
Let's zoom out some more and observe what happens with the approximation:
Graph of the approximating polynomial, and `f(x)=ln(x)`, zoomed further out.
Clearly, it is no longer a good approximation for values of x less than 3 or greater than 20. How do we get a better approximation? We would need to take more terms of the polynomial.