Finding Taylor Polynomial for tan(x) - Wondering

[mathmari]

Hey!

Let $f :\rightarrow \mathbb{R}$, $f(x) := tan(x)$.

I want to find a $N\in \mathbb{N}$ such that for the $N$-th Taylor polynomial $P_N$ at $0$, that is defined as follows
$P_N(x)=\sum_{n=0}^N\frac{f^{(n)}(0)}{n!}x^n$, it holds that
$$\left |f(x)-P_N(x)\right |\leq 10^{-5}, \ \ x\in \left [-\frac{1}{10}, \frac{1}{10}\right ]$$ For that do we have to write $f$ as a power series? But which is the formula? (Wondering)

Or do we have to do something else? (Wondering)

 

[Prove It]

Hey!

Let $f :\rightarrow \mathbb{R}$, $f(x) := tan(x)$.

I want to find a $N\in \mathbb{N}$ such that for the $N$-th Taylor polynomial $P_N$ at $0$, that is defined as follows
$P_N(x)=\sum_{n=0}^N\frac{f^{(n)}(0)}{n!}x^n$, it holds that
$$\left |f(x)-P_N(x)\right |\leq 10^{-5}, \ \ x\in \left [-\frac{1}{10}, \frac{1}{10}\right ]$$ For that do we have to write $f$ as a power series? But which is the formula? (Wondering)

Or do we have to do something else? (Wondering)

Yes write the power series.

 

[mathmari]

Yes write the power series.

We have that the power series of $\sin (x)$ is $$\sin (x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}$$ and the power series of $\cos (x)$ is $$\cos (x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}$$

Since $\tan (x)=\frac{\sin (x)}{\cos (x)}$, we get that $$\tan (x)=\frac{\displaystyle{\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}}}{\displaystyle{\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}}}$$

To what is that equal? (Wondering)

 

[I like Serena]

How about using the known expansion for $\tan$? (Wondering)

See e.g. here.

 

[mathmari]

How about using the know expansion for $\tan$? (Wondering)

See e.g. here.

So, we have that $$\tan (x)=\sum_{n=0}^{\infty}\frac{(-1)^n2^{2n+2}(2^{2n+2}-1)B_{2n+2}}{(2n+2)!}x^{2n+1}$$ where $B_n$ is a Bernoulli number.

$P_N$ is the power series of $\tan (x)$ that is stopped at $N$, or not? (Wondering)

Then we have that $$\left |f(x)-P_N(x)\right |=\left |\sum_{n=0}^{\infty}\frac{(-1)^n2^{2n+2}(2^{2n+2}-1)B_{2n+2}}{(2n+2)!}x^{2n+1}-\sum_{n=0}^N\frac{f^{(n)}(0)}{n!}x^n\right |=\left |\sum_{n=N+1}^{\infty}\frac{(-1)^n2^{2n+2}(2^{2n+2}-1)B_{2n+2}}{(2n+2)!}x^{2n+1}\right |\leq \sum_{n=N+1}^{\infty}\frac{2^{2n+2}(2^{2n+2}-1)B_{2n+2}}{(2n+2)!}\left (\frac{1}{10}\right )^{2n+1}$$

It holds that $$\frac{x}{e^x-1}=\sum_{n=0}^{\infty}\frac{B_nx^n}{n!}$$ Could we apply this in this case? But how? Now we have that the starts from $N+1$. (Wondering)

 

[I like Serena]

A Taylor series around 0 (aka MacLaurin series) up to degree $N$ has a remainder of:
$$R_N(x) = \frac{f^{(N+1)}(\theta x)}{(N+1)!}x^{N+1}$$
where $0 \le \theta \le 1$.

This is called the Lagrange form of the remainder. See e.g. here.

So:
$$|R_N(x)| \le \frac{\max_{0\le\theta\le 1} |f^{(N+1)}(\theta x)|}{(N+1)!}|x|^{N+1}$$

And perhaps we can pick the tangent expansion:
$$\tan x = x + \frac 13 x^3 + \frac 2{15} x^5 + \frac {17}{315} x^7 + \frac{62}{2835} x^9 + ...$$
(Thinking)

 

[mathmari]

A Taylor series around 0 (aka MacLaurin series) up to degree $N$ has a remainder of:
$$R_N(x) = \frac{f^{(N+1)}(\theta x)}{(N+1)!}x^{N+1}$$
where $0 \le \theta \le 1$.

This is called the Lagrange form of the remainder. See e.g. here.

So:
$$|R_N(x)| \le \frac{\max_{0\le\theta\le 1} |f^{(N+1)}(\theta x)|}{(N+1)!}|x|^{N+1}$$

And perhaps we can pick the tangent expansion:
$$\tan x = x + \frac 13 x^3 + \frac 2{15} x^5 + \frac {17}{315} x^7 + \frac{62}{2835} x^9 + ...$$
(Thinking)

So, we have that $f-P_N$ is the remainder, right? (Wondering)

$$|R_N(x)| \le \frac{\max_{0\le\theta\le 1} |f^{(N+1)}(\theta x)|}{(N+1)!}|x|^{N+1}\leq \frac{\max_{0\le\theta\le 1} |f^{(N+1)}(\theta x)|}{(N+1)!}\left |\frac{1}{10}\right |^{N+1}$$

Do we use the tangent expansion to find the $\max_{0\le\theta\le 1} |f^{(N+1)}(\theta x)|$ ? (Wondering)

 

[I like Serena]

So, we have that $f-P_N$ is the remainder, right? (Wondering)

$$|R_N(x)| \le \frac{\max_{0\le\theta\le 1} |f^{(N+1)}(\theta x)|}{(N+1)!}|x|^{N+1}\leq \frac{\max_{0\le\theta\le 1} |f^{(N+1)}(\theta x)|}{(N+1)!}\left |\frac{1}{10}\right |^{N+1}$$

Do we use the tangent expansion to find the $\max_{0\le\theta\le 1} |f^{(N+1)}(\theta x)|$ ? (Wondering)

Yes.

Formally, we indeed need to find the higher order derivatives of $\tan$ to find that upper bound.
However, we usually make the assumption that the remainder is less than the last term of the series.
So for $N=3$ we would assume that:
$$|R_3(x)| = |f(x) - P_3(x)| \le \frac 13 |x|^3 \le \frac 13 \left(\frac 1{10}\right)^3 < 4 \cdot 10^{-4}$$
(Thinking)

 

[mathmari]

Yes.

Formally, we indeed need to find the higher order derivatives of $\tan$ to find that upper bound.
However, we usually make the assumption that the remainder is less than the last term of the series.
So for $N=3$ we would assume that:
$$|R_3(x)| = |f(x) - P_3(x)| \le \frac 13 |x|^3 \le \frac 13 \left(\frac 1{10}\right)^3 < 4 \cdot 10^{-4}$$
(Thinking)

And which value do we take in this case where we don't know the value of N? (Wondering)

 

[I like Serena]

And which value do we take in this case where we don't know the value of N? (Wondering)

We should try them.

Note that if we ask Wolfram|Alpha, we find that:
$$\max_{-1/10 \le x\le 1/10} \tan^{(4)} x < 2$$
so that:
$$|R_3(x)| < \frac {2}{4!} \left| \frac 1{10}\right|^4 < 10^{-5}$$
(Thinking)

Or alternatively we assume that:
$$|R_5(x)| \le \frac{2}{15} \left|\frac 1{10}\right|^5 < 10^{-5}$$

 

[mathmari]

Do we have to take all the N's till we get a result $>10^{-5}$ ? (Wondering)

 

[I like Serena]

Do we have to take all the N's till we get a result $>10^{-5}$ ? (Wondering)

Not really. The problem statement does not require us to find the lowest $N$.

So we can pick any $N$ that seems to be sufficiently high and check the remainder.
Actually, we can already predict that to get $R_N(x) < 10^{-5}$ it will suffice to pick $N=5$, since the 5th term has a fraction that is less than $1$. (Nerd)

 

[I like Serena]

Alternatively, we can analyze:
$$R_3(x) = \tan x - (x+\frac 13 x^3)$$
It's anti-symmetric and if we take the derivative we can see that it's an increasing function.
So it's maximum absolute value is at $x=\frac 1{10}$.
W|A tells us that:
$$R_3(\frac 1{10}) \approx 1.3 \cdot 10^{-6}$$

 

[mathmari]

So we can pick any $N$ that seems to be sufficiently high and check the remainder.
Actually, we can already predict that to get $R_N(x) < 10^{-5}$ it will suffice to pick $N=5$, since the 5th term has a fraction that is less than $1$. (Nerd)

For $N=5$ do we take the part of the tangent expansion of degree $5$ and that will we bigger that the $R_5$ ? (Wondering)

 

[Prove It]

An alternative (and probably easier) method is to remember that in an alternating series, the error in truncating to n terms is never any more than the absolute value of the "n+1"th term.

 

[I like Serena]

For $N=5$ do we take the part of the tangent expansion of degree $5$ and that will we bigger that the $R_5$ ? (Wondering)

Yes. That's true by assumption, which is generally true for 'nice' series like tan.

An alternative (and probably easier) method is to remember that in an alternating series, the error in truncating to n terms is never any more than the absolute value of the "n+1"th term.

That's not true for just any alternating series, is it? There's an additional condition.
And anyway, the series at hand is not alternating.

 

[Prove It]

That's not true for just any alternating series, is it? There's an additional condition.
And anyway, the series at hand is not alternating.

As a matter of fact, it IS true for EVERY alternating series.

But I was mistaken about the series itself, it is not alternating, I forgot that the Bernoulli numbers do not have a pattern to the positives and negatives.

 

[I like Serena]

As a matter of fact, it IS true for EVERY alternating series.

Counter example: 1 - 2 + 10000 - 1/2 + 1/3 - 1/4 + ...

In particular the remainder is bigger than term 1 or term 2.
I believe we need the additional condition that the terms are absolutely decreasing (which is a sufficient condition to ensure the series converges as well).
It also means we need to prove that for any Taylor expansion we want to apply it to.

 

[mathmari]

Not really. The problem statement does not require us to find the lowest $N$.

We cannot find an interval of $N$ such that the inequality holds, can we? (Wondering)

 

[I like Serena]

We cannot find an interval of $N$ such that the inequality holds, can we? (Wondering)

To find the lowest value of N for which the inequality holds we need to evaluate $\tan(1/10) - P_N(1/10)$ for $N=0,1,...$ until we find it. (Thinking)

 

[mathmari]

To find the lowest value of N for which the inequality holds we need to evaluate $\tan(1/10) - P_N(1/10)$ for $N=0,1,...$ until we find it. (Thinking)

Ah ok. (Thinking)

we can analyze:
$$R_3(x) = \tan x - (x+\frac 13 x^3)$$
It's anti-symmetric and if we take the derivative we can see that it's an increasing function.

The derivative is $\frac{1}{\cos^2(x)}-1-x^2$. Is this always positive? (Wondering)

 

[mathmari]

One root of $\frac{1}{\cos^2(x)}-1-x^2$ is $x=0$.

This function is symmetric, right? (Wondering)

But can we find it the function is positive or negative? (Wondering)

 

[I like Serena]

One root of $\frac{1}{\cos^2(x)}-1-x^2$ is $x=0$.

This function is symmetric, right? (Wondering)

But can we find it the function is positive or negative? (Wondering)

The function is non-negative for $\frac \pi 2<x<\frac \pi 2$, as we can see here:
\begin{tikzpicture}[scale=2]
\draw[<->] (-1.2,0) -- (1.2,0);
\draw[->] (0,0) -- (0,1.7);
\foreach \i in {-1,-.5,...,1} {\draw (\i,.05) -- (\i,-.05) node[below] {$\i$}; }
\foreach \i in {.5,1,...,1.5} { \draw (.05,\i) -- (-.05,\i) node

{$\i$}; }
\draw[blue, ultra thick, domain=-1:1, variable=\x] plot ({\x}, {(sec(deg(\x))^2) - 1 - (\x)^2});
\end{tikzpicture}

However, it's a bit painful to formally prove it.
We can make a Taylor expansion, which is:
$$\frac{1}{\cos^2(x)}-1-x^2 = \frac 23 x^4 + \frac{17}{45} x^6 + ...$$
It confirms that it's indeed non-negative, although formally this is not a proof. (Doh)​

 

[mathmari]

Ah ok. (Thinking)

Is it maybe better to use the formula for the remainder term $R_N(x)=\frac{f^{(N+1)}(\xi )}{(N+1)!}x^{N+1}$ ? (Wondering) We have that $f^{(4)}=8\frac{ 2\sin (x)+\sin^3 (x)}{\cos^5(x)}$.

So, we have that $$R_3(x)=\frac{f^{(4)}(\xi )}{(4)!}x^{4}=\frac{8\frac{ 2\sin (\xi)+\sin^3 (\xi)}{\cos^5(\xi)}}{1\cdot 2\cdot 3\cdot 4}x^4=\frac{1}{3}\frac{ 2\sin (\xi)+\sin^3 (\xi)}{\cos^5(\xi)}x^4$$

Therefore $$|f(x)-P_N(x)|=|R_3|=\left |\frac{1}{3}\frac{ 2\sin (\xi)+\sin^3 (\xi)}{\cos^5(\xi)}x^4\right |=\frac{1}{3}\left |\frac{ 2\sin (\xi)+\sin^3 (\xi)}{\cos^5(\xi)}x^4\right |\leq \frac{1}{3}\left |\frac{ 2\sin (\xi)+\sin^3 (\xi)}{\cos^5(\xi)}\right |\cdot \left |x^4\right | \\ \leq \frac{1}{3}\left |\frac{ 2\sin (\xi)+\sin^3 (\xi)}{\cos^5(\xi)}\right |\cdot \left (\frac{1}{10}\right )^4$$

But how could we bound the absolute value $\left |\frac{ 2\sin (\xi)+\sin^3 (\xi)}{\cos^5(\xi)}\right |$ ? (Wondering)

 

[Prove It]

Counter example: 1 - 2 + 10000 - 1/2 + 1/3 - 1/4 + ...

In particular the remainder is bigger than term 1 or term 2.
I believe we need the additional condition that the terms are absolutely decreasing (which is a sufficient condition to ensure the series converges as well).
It also means we need to prove that for any Taylor expansion we want to apply it to.

I did mean every convergent alternating series (I thought that was implied)...

 

[I like Serena]

But how could we bound the absolute value $\left |\frac{ 2\sin (\xi)+\sin^3 (\xi)}{\cos^5(\xi)}\right |$ ? (Wondering)

For $0<\xi\le\frac 1{10}$ the function $f^{(4)}$ is positive since both $\sin$ and $\cos$ are positive then.
The numerator is largest when $\xi$ is largest, since $\sin$ is an increasing function.
And the denominator is smallest when $\xi$ is largest as well, since $\cos$ is a decreasing function.
So the function has the largest value when $\xi=\frac 1{10}$.

Since the function is anti-symmetric, the function has the same largest value for $\xi=-\frac 1{10}$, just negative.
(Thinking)

I did mean every convergent alternating series (I thought that was implied)...

Erm... the counter example I gave is a convergent alternating series.

 

[mathmari]

For $0<\xi\le\frac 1{10}$ the function $f^{(4)}$ is positive since both $\sin$ and $\cos$ are positive then.
The numerator is largest when $\xi$ is largest, since $\sin$ is an increasing function.
And the denominator is smallest when $\xi$ is largest as well, since $\cos$ is a decreasing function.
So the function has the largest value when $\xi=\frac 1{10}$.

Since the function is anti-symmetric, the function has the same largest value for $\xi=-\frac 1{10}$, just negative.
(Thinking)

So, we have that \begin{equation*}\left |\frac{ 2\sin (\xi)+\sin^3 (\xi)}{\cos^5(\xi)}\right |\leq \left |\frac{ 2\sin \left (\frac{1}{10}\right )+\sin^3 \left (\frac{1}{10}\right )}{\cos^5\left (\frac{1}{10}\right )}\right |\end{equation*} right? (Wondering)

Do we have to use here a calculator, or could we do here something without it? (Wondering)

It holds that $|\sin (x)|\leq x$, right? Do we have a similar inequality for $\frac{1}{\cos (x)}$ ? (Wondering)

 

[I like Serena]

How about $\frac{1}{\cos (x)} \le 1 + |x|$? (Wondering)

 

[mathmari]

How about $\frac{1}{\cos (x)} \le 1 + |x|$? (Wondering)

We get this inequality from the Taylor expansion of $\cos (x)$, right? (Wondering) So, we have that \begin{equation*}\left |\frac{ 2\sin (\xi)+\sin^3 (\xi)}{\cos^5(\xi)}\right |\leq \left |\frac{ 2\sin \left (\frac{1}{10}\right )+\sin^3 \left (\frac{1}{10}\right )}{\cos^5\left (\frac{1}{10}\right )}\right |= \left |2\sin \left (\frac{1}{10}\right )+\sin^3 \left (\frac{1}{10}\right )\right |\cdot \left |\frac{1}{\cos^5\left (\frac{1}{10}\right )}\right |\leq \left (2 \frac{1}{10}+\frac{1}{10^3}\right )\cdot \left (1+\frac{1}{10}\right )^5\end{equation*}

The last value is not smaller than $10^{-5}$ ( Wolfram ) (Wondering)

So, we have to do something else...

We have that $$|\cos (x)|=\left |\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!} \right |=\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}\geq 1$$

Therefore $\frac{1}{|\cos(x)|}\leq 1$.

Is this correct? (Wondering)

 

[I like Serena]

We get this inequality from the Taylor expansion of $\cos (x)$, right? (Wondering)

No, the Taylor expansion is $\cos x = 1-\frac 12 x^2 + \frac 1{4!}x^4 - ...$.
It's just that $\cos$ has a horizontal tangent at $x=0$.
We can set a bound with a line that has a non-horizontal tangent.
We can see it in this picture:
\begin{tikzpicture}[scale=2]
\draw[<->] (-1.2,0) -- (1.2,0);
\draw[->] (0,0) -- (0,2.2);
\foreach \i in {-1,-0.5,...,1} {\draw (\i,.05) -- (\i,-.05) node[below] {$\i$}; }
\foreach \i in {0.5,1,...,2} { \draw (.05,\i) -- (-.05,\i) node

{$\i$}; }
\draw[domain=-1:1, variable=\x, red, ultra thick] plot ({\x}, {(cos(deg(\x)))}) node

{$\cos x$};
\draw[domain=-1:1, variable=\x, red, ultra thick] plot ({\x}, {1-abs(\x)}) node[above right] {$1-|x|$};
\draw[domain=-1:1, variable=\x, blue, ultra thick] plot ({\x}, {(sec(deg(\x)))}) node

{$\frac{1}{\cos x}$};
\draw[domain=-1:1, variable=\x, blue, ultra thick] plot ({\x}, {1+abs(\x)}) node

{$1+|x|$};
\end{tikzpicture}

For the record, based on the $\cos$ expansion, we can write $\cos(x) \ge 1-\frac 12 x^2$.

So, we have that \begin{equation*}\left |\frac{ 2\sin (\xi)+\sin^3 (\xi)}{\cos^5(\xi)}\right |\leq \left |\frac{ 2\sin \left (\frac{1}{10}\right )+\sin^3 \left (\frac{1}{10}\right )}{\cos^5\left (\frac{1}{10}\right )}\right |= \left |2\sin \left (\frac{1}{10}\right )+\sin^3 \left (\frac{1}{10}\right )\right |\cdot \left |\frac{1}{\cos^5\left (\frac{1}{10}\right )}\right |\leq \left (2 \frac{1}{10}+\frac{1}{10^3}\right )\cdot \left (1+\frac{1}{10}\right )^5\end{equation*}

The last value is not smaller than $10^{-5}$ ( Wolfram ) (Wondering)

So, we have to do something else...

The remainder term is actually:
$$R_3(x) = \frac{1}{4!} f^{(4)}\,(\xi) x^4$$
I think we need a couple more factors... (Thinking)

We have that $$|\cos (x)|=\left |\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!} \right |=\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}\geq 1$$

Therefore $\frac{1}{|\cos(x)|}\leq 1$.

Is this correct? (Wondering)

Not quite. $|\cos x|\le 1$. Therefore $\frac{1}{|\cos(x)|}\ge 1$. (Nerd)​

 

[mathmari]

No, the Taylor expansion is $\cos x = 1-\frac 12 x^2 + \frac 1{4!}x^4 - ...$.
It's just that $\cos$ has a horizontal tangent at $x=0$.
We can set a bound with a line that has a non-horizontal tangent.$|\cos x|\le 1$. Therefore $\frac{1}{|\cos(x)|}\ge 1$. (Nerd)

Ah ok... (Thinking)

The remainder term is actually:
$$R_3(x) = \frac{1}{4!} f^{(4)}\,(\xi) x^4$$
I think we need a couple more factors... (Thinking)

What do you mean? (Wondering)

 

[I like Serena]

What do you mean? (Wondering)

We've evaluated an upper bound for $f^{(4)}(\xi)$, which is approximately $2.6$.
Btw, I think there's a factor $8$ missing in the expression. (Worried)

But we need an upper bound for $R_3(x) = \frac 1{4!}f^{(4)}(\xi)\left(\frac 1{10}\right)^4 = \frac{1}{24}\cdot 10^{-4}f^{(4)}(\xi) < 4\cdot 10^{-6}\cdot f^{(4)}(\xi)$. (Thinking)

 

[mathmari]

We've evaluated an upper bound for $f^{(4)}(\xi)$, which is approximately $2.6$.
Btw, I think there's a factor $8$ missing in the expression. (Worried)

But we need an upper bound for $R_3(x) = \frac 1{4!}f^{(4)}(\xi)\left(\frac 1{10}\right)^4 = \frac{1}{24}\cdot 10^{-4}f^{(4)}(\xi) < 4\cdot 10^{-6}\cdot f^{(4)}(\xi)$. (Thinking)

I see! Thanks a lot! (Mmm)