9.4 Still better approximations

$T_3''(x)$ and $f''(x)$ agree when $x=1$.

$f''(1)=-4$

Since $T_3(x)$ is the third-degree Taylor polynomial for $f(x)$ about $x=1$:

In particular, $f''(1)=T_3''(1)$.

So, $f''(1)=-4$.

The $n$th degree Taylor polynomial for $f(x)$ about $x=5$ is

Match up the terms.

$f^{(10)}(5)=10!$

In Question 1, we differentiated the Taylor polynomial to find its derivative. We don't really want to differentiate this ten times, though, so let's look for another way. Unlike Question 1, our Taylor polynomial is given to us in a form very similar to its definition. The $n$th degree Taylor polynomial for $f(x)$ about $x=5$ is

So,

For any $k$ from 0 to $n$,

In particular, when $k=10$,

The fourth-degree Maclaurin polynomial for $f(x)$ is

while the third-degree Maclaurin polynomial for $f(x)$ is

$T_3(x)=-x^3+x^2-x+1$

The fourth-degree Maclaurin polynomial for $f(x)$ is

while the third-degree Maclaurin polynomial for $f(x)$ is

So, we simply “chop off" the part of $T_4(x)$ that includes $x^4$:

The third-degree Taylor polynomial for $f(x)$ about $x=1$ is

How can you recover $f(1)$, $f'(1)$, $f''(1)$, and $f'''(1)$ from $T_4(x)$?

$T_3(x)=-7+7(x-1)+9(x-1)^2+5(x-1)^3$, or equivalently, $T_3(x)=5x^3-6x^2+4x-10$

We saw this kind of problem in Question 3. The fourth-degree Taylor polynomial for $f(x)$ about $x=1$ is

while the third-degree Taylor polynomial for $f(x)$ about $x=1$ is

In Question 3 we “chopped off" the term of degree 4 to get $T_3(x)$. However, our polynomial is not in this form. It's not clear, right away, what the term $f^{(4)}(x-1)^4$ is in our given $T_4(x)$. So, we will use a different method from Question 3.

One option is to do some fancy algebra to get $T_4(x)$ into the standard form of a Taylor polynomial. Another option (which we will use) is to recover $f(1)$, $f'(1)$, $f''(1)$, and $f'''(1)$ from $T_4(x)$.

Recall that $T_4(x)$ and $f(x)$ have the same values at $x=1$ (although maybe not anywhere else!), and they also have the same first, second, third, and fourth derivatives at $x=1$ (but again, maybe not anywhere else, and maybe their fifth derivatives don't agree). This tells us the following:

Now, we can write the third-degree Taylor polynomial for $f(x)$ about $x=1$:

Remark: expanding the expression above, we get the equivalent polynomial
$T_3(x)=5x^3-6x^2+4x-10$. From this, it is clear that we can't just “chop off" the term with $x^4$ to change $T_4(x)$ into $T_3(x)$ when the Taylor polynomial is not centred about $x=0$.

Compare the given polynomial to the more standard form of the $n$th degree Taylor polynomial,

and notice that the term you want (containing $f^{(10)}(5)$) corresponds to $k=10$ in the standard form, but is not the term corresponding to $k=10$ in the polynomial given in the question.

$f^{(10)}(5)=\dfrac{11\cdot 10!}{6}$

The $n$th degree Taylor polynomial for $f(x)$ about $x=5$ is

We expand this somewhat:

So, the coefficient of $(x-5)^{10}$ is $\dfrac{1}{10!}f^{(10)}(5)$. Expanding the given form of the Taylor polynomial:

Equating the coefficients of $(x-5)^{10}$ in the two expressions:

$T_3'''(a)=f'''(a)$

$a=\sqrt{e}$

Since $T_3(x)$ is the third-degree Taylor polynomial for $f(x)$ about $x=a$, we know the following things to be true:

But, some of these don't look super useful. For instance, if we try to use the first bullet, we get this equation:

Solving this would be terrible. Instead, let's think about how the equations look when we move further down the list. Since $T_3(x)$ is a cubic equation, $T_3'''(x)$ is a constant (and so $T_3'''(a)$ does not depend on $a$). That sounds like it's probably the simplest option. Let's start differentiating. We'll need to know both $f'''(a)$ and $T_3'''(a)$.

Now, let's move to the Taylor polynomial. Remember that $e$ is a constant.

The final bullet point gives us the equation:

So, $a=\sqrt{e}$.