MacLaurin/Taylor/Convergence Finals Help :(

[korth0221]

1. Find 2nd degree maclaurin polynomial that approximates f(x)=sec(x)

2. Find 3rd degree Taylor polynomial that approximates f(x)=(2/x) at c=1

3. Find radius of convergence of power series n=0 to infinity: ((2n)!*x^(2n))/(n!)

4. Find interval of convergence of power series n=1 to infinity: (-1^(n+1)*(x-4)^n)/(n*9^n)

My professor didn't have "time" to teach us this section so I'm very lost :( If you guys can please answer these with work that would help me a lot for this final. Thank you so much :)

 

[MarkFL]

Hello and welcome to MHB, korth0221! (Wave)

Before we get started, I do need to point out that our goal here at MHB is not to merely provide worked solutions to posted problems, but rather to engage the student to help them solve the problem(s). We also ask that no more than two question be initially posted in a thread. This helps prevent a thread from becoming convoluted and hard to follow.

So, that being said, let's look at the first question:

1. Find 2nd degree maclaurin polynomial that approximates f(x)=sec(x)

The Maclaurin series for $f$ is given by:

$$M(x)=\sum_{k=0}^{\infty}\left(\frac{f^{(k)}(0)}{k!}x^k\right)$$

We are asked for a partial sum...in order for $M$ to be a 2nd degree polynomial, what will our upper limit for $k$ be?

 

[MarkFL]

Just to follow up, we want the upper limit for $k$ to be 2 in order to get a second degree polynomial...that is:

$$M_2(x)=\sum_{k=0}^{2}\left(\frac{f^{(k)}(0)}{k!}x^k\right)=f(0)+f'(0)x+\frac{f''(0)}{2}x^2$$

Now, we are given:

$$f(x)=\sec(x)\implies f(0)=1$$

And so differentiating, we get:

$$f'(x)=\sec(x)\tan(x)\implies f'(0)=0$$

$$f''(x)=\sec(x)\left(\tan^2(x)+\sec^2(x)\right)\implies f''(0)=1$$

Hence:

$$M_2(x)=1+\frac{1}{2}x^2$$

Here's a graph of $f$ and $M_2$ for comparison:

[DESMOS=-2,2,0.5,2]y=\sec\left(x\right);y=\frac{x^2}{2}+1[/DESMOS]

2.) For this problem, we need to use:

The Taylor series for $f$ centered at $a$ is given by:

$$T(x)=\sum_{k=0}^{\infty}\left(\frac{f^{(k)}(a)}{k!}(x-a)^k\right)$$

Since we are asked for a third degree polynomial, we want the partial sum:

$$T_3(x)=\sum_{k=0}^{3}\left(\frac{f^{(k)}(a)}{k!}(x-a)^k\right)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2}(x-a)^2+\frac{f'''(a)}{6}(x-a)^3$$

We are given:

$$a=1$$

$$f(x)=\frac{2}{x}\implies f(1)=2$$

And so differentiating, we obtain:

$$f'(x)=-\frac{2}{x^2}\implies f'(1)=-2$$

$$f''(x)=\frac{4}{x^3}\implies f''(1)=4$$

$$f'''(x)=-\frac{12}{x^4}\implies f'''(1)=-12$$

And so we have:

$$T_3(x)=2-2(x-1)+2(x-1)^2-2(x-1)^3=-2x^3+8x^2-12x+8$$

Here's a a graph for comparison:

[DESMOS=0,2,0,4]y=\frac{2}{x};y=-2x^3+8x^2-12x+8[/DESMOS]

I will leave the other two problems for anyone else who is inclined to post solutions. :)

 

[Prove It]

2. Find 3rd degree Taylor polynomial that approximates f(x)=(2/x) at c=1

A geometric series $\displaystyle \begin{align*} \sum_{n = 0}^{\infty}{r^n} \end{align*}$ converges to $\displaystyle \begin{align*} \frac{1}{1 - r} \end{align*}$ provided that $\displaystyle \begin{align*} \left| r \right| < 1 \end{align*}$. Notice that we can write

$\displaystyle \begin{align*} f(x) &= \frac{2}{x} \\ &= 2 \left[ \frac{1}{1 - \left( 1-x \right) } \right] \end{align*}$

Now this is in the same form as the closed form of the geometric series, with $\displaystyle \begin{align*} r = 1 - x \end{align*}$ and so provided that $\displaystyle \begin{align*} \left| 1 - x \right| < 1 \end{align*}$, i.e. $\displaystyle \begin{align*} 0 < x < 2 \end{align*}$ we have

$\displaystyle \begin{align*} f( x) &= 2\left[ \frac{1}{ 1 - \left( 1 - x \right) } \right] \\ &= 2\sum_{n =0}^{\infty}{ \left( 1 - x \right) ^n } \textrm{ where } \left| 1 - x \right| < 1 \\ &= 2\sum_{n = 0}^{\infty}{ \left[ - \left( x - 1 \right) \right] ^n } \\ &= \sum_{n = 0}^{\infty}{ 2 \left( - 1 \right) ^n \left( x - 1\right) ^n } \end{align*}$

 

[Prove It]

3. Find radius of convergence of power series n=0 to infinity: ((2n)!*x^(2n))/(n!)

By the ratio test, we know that a positive termed series $\displaystyle \begin{align*} \sum_{n = 0}^{\infty}{a_n} \end{align*}$ converges when $\displaystyle \begin{align*} \lim_{n \to \infty}{ \frac{a_{n+1}}{a_n} } < 1 \end{align*}$.

This series might not be positive termed, as it depends what value we have for x, so we will need to use the ratio test on the absolute value series to determine where the series is absolutely convergent. So

$\displaystyle \begin{align*} \lim_{n \to \infty}{ \frac{\left| a_{n +1} \right|}{\left| a_n \right|} } &< 1 \\ \lim_{n \to \infty}{ \frac{\left| \frac{\left[ 2\left( n + 1 \right) \right] ! \, x^{2\left( n + 1 \right) }}{\left( n + 1 \right) !} \right| }{ \left| \frac{\left( 2\,n \right) ! \, x^{2\,n}}{n!} \right| } } &< 1 \\ \lim_{n \to \infty}{ \frac{\frac{\left( 2\,n + 2 \right) ! \,\left| x \right| ^{2\,n + 2}}{\left( n + 1 \right) !}}{\frac{\left( 2\,n \right) ! \,\left| x \right| ^{2\,n}}{n!}} } &< 1 \\ \lim_{n \to \infty}{ \frac{n!\,\left( 2\,n + 2 \right) ! \,\left| x \right| ^{2\,n + 2}}{\left( n + 1 \right) !\,\left( 2\,n \right) ! \,\left| x \right| ^{2\,n}} } &< 1 \\ \lim_{n \to \infty}{ \frac{n!\,\left( 2\,n + 2 \right) \left( 2\,n + 1 \right) \left( 2\,n \right) ! \,\left| x \right| ^{2\,n} \,\left| x \right| ^2}{\left( n + 1 \right) \, n! \, \left( 2\,n \right) !\,\left| x \right| ^{2\,n}} } &< 1 \\ \lim_{n \to \infty}{ \frac{\left( 2\,n + 2 \right) \left( 2\,n + 1 \right) \left| x \right| ^2}{n + 1} } &< 1 \\ \lim_{n \to \infty} 2\left( 2\,n + 1 \right) \,\left|x\right| ^2 < 1 \end{align*}$

But as $\displaystyle \begin{align*} 2\left( 2\,n + 1 \right) \to \infty \end{align*}$ as $\displaystyle \begin{align*} n \to \infty \end{align*}$, that means that there are no values of x which will make this series absolutely convergent. Thus the radius of convergence is 0.

 

[MarkFL]

Yes, the Taylor polynomial in question 2.) can be simplified...let's begin with:

$$T(x)=\sum_{k=0}^{\infty}\left(\frac{f^{(k)}(a)}{k!}(x-a)^k\right)$$

Now, with $$f(x)=\frac{2}{x}$$, we have observed that:

$$f'(x)=-\frac{2}{x^2}$$

$$f''(x)=\frac{4}{x^3}$$

$$f'''(x)=-\frac{12}{x^4}$$

At this point, we can state the induction hypothesis $P_n$:

$$f^{(n)}(x)=\frac{2(-1)^{n}n!}{x^{n+1}}$$

Having already show the base case(s) to be true, our induction step can be to differentiate both sides:

$$f^{(n+1)}(x)=\frac{2(-1)^{n+1}(n+1)!}{x^{(n+1)+1}}$$

We have obtained $P_{n+1}$ from $P_n$ thereby completing the proof by induction. And so we may now state:

$$T(x)=\sum_{k=0}^{\infty}\left(\frac{2(-1)^{k}k!}{a^{k+1}}\cdot\frac{1}{k!}(x-a)^k\right)=2\sum_{k=0}^{\infty}\left(\frac{(a-x)^k}{a^{k+1}}\right)$$

 

[Prove It]

4. Find interval of convergence of power series n=1 to infinity: (-1^(n+1)*(x-4)^n)/(n*9^n)

$\displaystyle \begin{align*} \sum_{n = 1}^{\infty}{ \frac{\left( -1 \right) ^{n + 1}\left( x - 4 \right) ^n}{n\,9^n} } \end{align*}$

We can say for sure that the series is absolutely convergent where $\displaystyle \begin{align*} \lim_{n \to \infty}{ \frac{\left| a_{n+1} \right| }{\left| a_n \right| } } < 1 \end{align*}$, so

$\displaystyle \begin{align*} \lim_{n \to \infty}{ \frac{\left| \frac{\left( -1 \right) ^{n + 2} \left( x - 4 \right) ^{n + 1}}{\left( n + 1 \right) \, 9^{n+1}} \right| }{ \left| \frac{\left( -1 \right) ^{n + 1}\left( x - 4 \right) ^n}{n\,9^n} \right| } } &< 1 \\ \lim_{n \to \infty}{ \frac{\frac{\left| x - 4 \right| ^{n + 1}}{\left( n + 1 \right) \, 9^{n + 1}}}{\frac{\left| x - 4 \right| ^n}{n\,9^n}} } &< 1 \\ \lim_{n \to \infty}{ \frac{n\,9^n\,\left| x - 4 \right| ^{n+1}}{\left( n + 1 \right) \, 9^{n+1}\,\left| x - 4 \right| ^n} } &< 1 \\ \lim_{n \to \infty}{ \left( \frac{n}{n + 1} \right)\,\frac{\left| x - 4 \right|}{9} } &< 1 \\ \lim_{n \to \infty}{ \left( 1 - \frac{1}{n + 1} \right) \, \frac{ \left| x - 4 \right| }{9} } &< 1 \\ \left( 1 - 0 \right) \,\frac{\left| x - 4 \right| }{9} &< 1 \\ \frac{\left| x - 4 \right| }{9} &< 1 \\ \left| x - 4 \right| &< 9 \\ -9 < x - 4 &< 9 \\ -5 < x &< 13 \end{align*}$

We know for certain that the series is absolutely convergent where $\displaystyle \begin{align*} -5 < x < 13 \end{align*}$, but the ratio test tells us nothing about the convergence at the endpoints. So we will check what happens there... $\displaystyle \begin{align*} x = -5 \end{align*}$ gives

$\displaystyle \begin{align*} \sum_{n = 1}^{\infty}{ \frac{\left( -1 \right) ^{n+1} \left( -5 - 4 \right) ^n }{ n\,9^n } } &= \sum_{n = 1}^{\infty}{ \frac{\left( - 1 \right) ^{n + 1}\left( -9 \right) ^n }{ n\,9^n } } \\ &= \sum_{n = 1}^{\infty}{ \frac{\left( -1 \right) ^{n + 1} \left( -1 \right) ^n \, 9^n}{n\,9^n} } \\ &= \sum_{n = 1}^{\infty}{ \frac{\left( -1 \right) ^{n + 2}}{n} } \\ &= \sum_{n = 1}^{\infty}{ \frac{\left( -1 \right) ^n}{n} } \end{align*}$

The Alternating Harmonic Series is well known to be convergent, as $\displaystyle \begin{align*} \frac{1}{n} \end{align*}$ is decreasing. So the series converges at $\displaystyle \begin{align*} x = -5 \end{align*}$.

When $\displaystyle \begin{align*} x = 13 \end{align*}$ we have

$\displaystyle \begin{align*} \sum_{n =1}^{\infty}{ \frac{\left( -1 \right)^{n + 1}\left( 13 - 4 \right) ^n}{n\,9^n} } &= \sum_{ n = 1}^{\infty}{ \frac{\left( -1 \right) ^{n +1} \,9^n}{n\,9^n} } \\ &= -\sum_{n = 1}^{\infty}{ \frac{\left( -1 \right) ^n}{n} } \end{align*}$

again, this has been reduced to a multiple of the Alternating Harmonic Series, so we know the series is convergent at $\displaystyle \begin{align*} x = 13 \end{align*}$.

Thus the interval of convergence is $\displaystyle \begin{align*} -5 \leq x \leq 13 \end{align*}$.