Calculate the Taylor Polynomials $T_3(x)$ and $T_4(x)$ for the function $f(x) = \tan{x}$ centered at $\frac{\pi}{4}$.
To do this, we need the value of the function and it's first four derivatives at $\frac{\pi}{4}$.
For this function $f(x)$, $f'(x) = \sec^2{x}$, $f''(x) = 2\sec^2{x}\tan{x}$ and you can continue to compute more derivatives.
Then plugging in $\frac{\pi}{4}$ gives that $f(\frac{\pi}{4}) = 1$, $f'(\frac{\pi}{4}) = 2$, $f''(\frac{\pi}{4}) = 4$ and so on.
Then, divide each term by $n!$ before writing out the polynomials.
$\displaystyle T_3(x) = 1 + 2\left( x - \frac{\pi}{4}\right) + 2\left(x - \frac{\pi}{4}\right)^2 + \frac{8}{3}\left(x - \frac{\pi}{4}\right)^3$ and $\displaystyle T_4(x) = 1 + 2\left( x - \frac{\pi}{4}\right) + 2\left(x - \frac{\pi}{4}\right)^2 + \frac{8}{3}\left(x - \frac{\pi}{4}\right)^3 + \frac{10}{3}\left(x - \frac{\pi}{4}\right)^4$.
Find the Taylor Polynomial $T_4(x)$ centered at $c=-2$ for the function $f(x) = e^x$.
We need to take derivatives of this function and evaluate at $-2$ to build this polynomial.
The derivatives of $f(x)$ are all $e^x$. Therefore the value of these derivatives is always $e^{-2}$.
Then you can write out the polynomial by dividing by $n!$ at each step.
$\displaystyle T_4(x) = e^{-2}\left(1 + (x-2) + \frac{1}{2}(x-2)^2 + \frac{1}{6}(x-2)^3 + \frac{1}{24}(x-2)^4 \right)$
Determine the maximum possible error is using $T_2\left(\frac{\pi}{12}\right)$ to approximate $\cos\left(\frac{\pi}{12}\right)$ where the Taylor Polynomial is centered at $c=0$. Evaluate both $T_2\left(\frac{\pi}{12}\right)$ and $\cos\left(\frac{\pi}{12}\right)$ and confirm that this meets the error bound.
To find $T_2$, we need to either take derivatives of the cosine function, or just use the Taylor Series and cut it off after 2 terms.
We get that $T_2(x) = 1 - \frac{x^2}{2}$.
Since $f'''(x) = \sin(x)$, we know that this is bounded by $1$, so we can use $K = 1$ in our error formula.
The error bound is then $\displaystyle \frac{1 (\frac{\pi}{12} - 0)^3}{3!}$.
The error bound we get is $0.00299$. $\cos(\pi/12) = 0.96593$ and $T_2(\pi/12) = .96573$, so we are good.
How many terms of the Maclaurin Series for $f(x) = \ln(1+x)$ are needed to approximate the value of $\ln(1.2)$ to within $0.0001$?
To do this, we need to find the Maclaurin Series of $\ln(1+x)$. We should do this by computing derivatives.
For this $f'(x) = \frac{1}{1+x}$, $f''(x) = \frac{-1}{(1+x)^2}$ and the pattern keeps going.
With the pattern, we get that $f^{(k)}(x) = \frac{(-1)^k (k-1)!}{(1+x)^{k}}$. Thus, if we want to bound the $k$th derivative on $[1, 1.1]$, we can bound it by $(k-1)!$
Thus, for the error bounds, we get that the error in using $T_k(x)$ is $\frac{(k-1)!(0.2)^{k+1}}{(k+1)!}$
4 terms
Use $T_6(x)$ to approximate $\displaystyle \int_0^1 e^{-x^2}\ dx$.
First, we need to find $T_6(x)$ as a function. We can do this using the Taylor Series for $e^{-x^2}$.
The series for $e^{-x^2}$ is $\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{n!}x^{2n}$ so $T_6(x)$ is this series up to degree 6.
So $T_6(x) = 1 - x^2 + \frac{x^4}{2} - \frac{x^6}{6}$. We can then integrate this function and plug in $0$ and $1$.
This will need the reduction formulas and the fact that the antiderivative of $\sec(\theta)$ is $\ln |\sec{\theta} + \tan{\theta}|$ to finish up the problem.
$\displaystyle 1 - \frac{1}{3} + \frac{1}{10} - \frac{1}{42} = \frac{26}{35} = 0.7429$. The actual integral is $0.7468$.
Find the Taylor Polynomial $T_5(x)$ for the function $\sqrt{x}$ centered at $a=9$.
Use the Error Bound to find a value $N$ for which the error in using $T_n(-0.1)$ to approximate $e^{-0.1}$ to within $10^{-9}$, where $T_n$ is the Taylor polynomial centered at $0$.