Taylor series

## Taylor series
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 ## Taylor series
 #### Definition
 If $f$ is infinitely differentiable at $a$, then the *Taylor series for $f(x)$ at $x=a$* is
 $\displaystyle \sum\_{n=0}^\infty \frac{f\^{(n)}(a)}{n!}(x-a)^n$.
 
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 ## Taylor series
 ### (Maclaurin's version)
 #### Definition
 If $f$ is infinitely differentiable at $a$, then the *Taylor series for $f(x)$ at $x=a$* is
 $\displaystyle \sum\_{n=0}^\infty \frac{f\^{(n)}(a)}{n!}(x-a)^n$.
 #### Definition
 If $f$ is infinitely differentiable at $0$, then the *Maclaurin series for $f$* is the Taylor series for $f(x)$ at $x=0$: the function
 $\displaystyle \sum\_{n=0}^\infty \frac{f\^{(n)}(0)}{n!}x^n$.
 
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 ## Taylor series
 #### Definition
 If $f$ is infinitely differentiable at $a$, then the *Taylor series for $f(x)$ at $x=a$* is
 $\displaystyle \sum\_{n=0}^\infty \frac{f\^{(n)}(a)}{n!}(x-a)^n$.
 #### Example 1
 If $\displaystyle f(x) = \mathrm{e}^x$, then $f'(x) = f(x)$, so $f\^{(n)}(0)=1$ for all $n\in\mathbb N\cup\\{0\\}$.
 Therefore, the Taylor series for $f$ at $0$ converges on $(-\infty,\infty)$, and it converges to $f$ there.

[Excel file demonstrating Taylor series of exp](https://nc.dasmithmaths.com/index.php/s/xiGSBG99CWgzW9G)
 
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 ## Taylor series
 #### Definition
 If $f$ is infinitely differentiable at $a$, then the *Taylor series for $f(x)$ at $x=a$* is
 $\displaystyle \sum\_{n=0}^\infty \frac{f\^{(n)}(a)}{n!}(x-a)^n$.
 #### Example 1
 If $\displaystyle f(x) = \mathrm{e}^x$, then $f'(x) = f(x)$, so $f\^{(n)}(0)=1$ for all $n\in\mathbb N\cup\\{0\\}$.
 Therefore, the Taylor series for $f$ at $0$ converges on $(-\infty,\infty)$, and it converges to $f$ there.

The Taylor/Maclaurin series may not converge at all, and may not converge to $f(x)$.
 #### Example 2
 If
 $\displaystyle f(x) = \begin{cases} \mathrm{e}^x & \text{if } x\leqslant1 \\\\ \mathrm{e}(2-x) & \text{if } x\gt1 \end{cases}$,
 then the Taylor series for $f$ converges everywhere, but not to $f$; see spreadsheet.
 
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 ## Taylor series
 #### Definition
 If $f$ is infinitely differentiable at $a$, then the *Taylor series for $f(x)$ at $x=a$* is
 $\displaystyle \sum\_{n=0}^\infty \frac{f\^{(n)}(a)}{n!}(x-a)^n$.
 #### Example 3
 Let $\displaystyle f(x)=e\^{-1/x^2}$.
 This is defined and infinitely differentiable everywhere except at $x=0$.
 We can fix this by redefining
 $\displaystyle f(x) = \begin{cases} 0 & \text{if } x=0, \\\\ e\^{-1/x^2} & \text{otherwise.} \end{cases}$

[Let's look at the derivatives of this function.](https://nc.dasmithmaths.com/index.php/s/zEAGSfR9z3os2gg)
 
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 ## Taylor series
 #### Definition
 If $f$ is infinitely differentiable at $a$, then the *Taylor series for $f(x)$ at $x=a$* is
 $\displaystyle \sum\_{n=0}^\infty \frac{f\^{(n)}(a)}{n!}(x-a)^n$.
 #### Example 3
 Let $\displaystyle f(x)=e\^{-1/x^2}$.
 This is defined and infinitely differentiable everywhere except at $x=0$.
 We can fix this by redefining
 $\displaystyle f(x) = \begin{cases} 0 & \text{if } x=0, \\\\ e\^{-1/x^2} & \text{otherwise.} \end{cases}$

It can be shown that $f\^{(n)}(0)=0$ for all $n\in\mathbb N\cup\\{0\\}$.

Proof: by induction,
 $\displaystyle f\^{(n)}(x) = \begin{cases} 0 & \text{if } x=0, \\\\ e\^{-1/x^2}P_n(x) & \text{otherwise,} \end{cases}$
 
 for $P_0(x)=1$ and $P_n(x) = 2x\^{-3}P\_{n-1}(x)+P'\_{n-1}(x)$.
 
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 ## Taylor series
 #### Definition
 If $f$ is infinitely differentiable at $a$, then the *Taylor series for $f(x)$ at $x=a$* is
 $\displaystyle \sum\_{n=0}^\infty \frac{f\^{(n)}(a)}{n!}(x-a)^n$.
 #### Example 3
 Let $\displaystyle f(x)=e\^{-1/x^2}$.
 This is defined and infinitely differentiable everywhere except at $x=0$.
 We can fix this by redefining
 $\displaystyle f(x) = \begin{cases} 0 & \text{if } x=0, \\\\ e\^{-1/x^2} & \text{otherwise.} \end{cases}$

It can be shown that $f\^{(n)}(0)=0$ for all $n\in\mathbb N\cup\\{0\\}$.

So the Maclaurin series for $f(x)$ converges and is the constant function $0$, which is not $f(x)$.
 
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 ## Taylor series
 #### Definition
 If $f$ is infinitely differentiable at $a$, then the *Taylor series for $f(x)$ at $x=a$* is
 $\displaystyle \sum\_{n=0}^\infty \frac{f\^{(n)}(a)}{n!}(x-a)^n$.

So we do have to be careful with Taylor series.

For a Taylor series at $a$ to exist and be useful, we have to know:

* $f\^{(n)}(a)$ exists for all $n\in\mathbb N\cup\\{0\\}$. Calculus, MATH2340
 * The Taylor series converges on $(a-R,a+R)$ for some $R$ large enough for our application. MATH2340
 * The Taylor series converges to $f(x)$ on that interval. MATH2350

We may also want:

* The Taylor series converges fast enough. Asymptotic analysis
 
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 ## Taylor series
 #### Definition
 If $f$ is infinitely differentiable at $a$, then the *Taylor series for $f(x)$ at $x=a$* is
 $\displaystyle \sum\_{n=0}^\infty \frac{f\^{(n)}(a)}{n!}(x-a)^n$.

So we do have to be careful with Taylor series.

For a Taylor series at $a$ to exist and be useful, we have to know:

* $f\^{(n)}(a)$ exists for all $n\in\mathbb N\cup\\{0\\}$. Calculus, MATH2340
 * The Taylor series converges on $(a-R,a+R)$ for some $R$ large enough for our application. MATH2340
 * The Taylor series converges to $f(x)$ on that interval. MATH2350

We may also want:

* The Taylor series converges fast enough. Asymptotic analysis
 
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 ## Taylor series
 For a Taylor series at $a$ to exist and be useful, we have to know:

We may also want:

* The Taylor series converges fast enough. Asymptotic analysis

#### Exercise
 Assess how fast the Maclaurin series for sin converges using [this spreadsheet](https://nc.dasmithmaths.com/index.php/s/xiGSBG99CWgzW9G).
 
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 ## Taylor series
 For a Taylor series at $a$ to exist and be useful, we have to know:

We may also want:

* The Taylor series converges fast enough. Asymptotic analysis

#### Exercise
            Assess how fast the Maclaurin series for sin converges using [this spreadsheet](https://nc.dasmithmaths.com/index.php/s/xiGSBG99CWgzW9G).

#### Conclusion
 For $x\in(-0.1,0.1)$ very fast.
 But for $x\in(-10,10)$ quite slowly.
 
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 ## Application of Taylor series
 The Maclaurin series for $\sin(x)$ is
 $\displaystyle \sum\_{\substack{n=0 \\\\ n \text{ odd}}}^\infty \frac{(-1)\^{\frac{n-1}2}}{n!}x^n$.
 It converges to $\sin(x)$ on $(-\infty,\infty)$.

#### Exercise
            Evaluate
            $\displaystyle \lim\_{x\to0} \frac{\sin(x)}x$.

Evaluate
            $\displaystyle \lim\_{x\to0} \frac{\mathrm e^x - 1 - x}{x^2}$.

Think carefully about what we need to know about the relevant Taylor series to make this argument.
 
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 ## Application of Taylor series
 The Maclaurin series for $\sin(x)$ is
 $\displaystyle \sum\_{\substack{n=0 \\\\ n \text{ odd}}}^\infty \frac{(-1)\^{\frac{n-1}2}}{n!}x^n$.
 It converges to $\sin(x)$ on $(-\infty,\infty)$.

#### Exercise
            Evaluate
            $\displaystyle \lim\_{x\to0} \frac{\sin(x)}x$.

Evaluate
            $\displaystyle \lim\_{x\to0} \frac{\mathrm e^x - 1 - x}{x^2}$.

Think carefully about what we need to know about the relevant Taylor series to make this argument.

The limits are $1$ and $\frac12$.
 You only need to know that there is some (possibly very small) $\varepsilon\gt0$ such that the Maclaurin series converge on $(-\varepsilon,\varepsilon)$.
 It does not matter how fast.
 
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 ## Application of Taylor series
 We want to cut a curve of shape approximately $f(x)=\frac1{1-x}$ for $x\in[-3,\frac12]$, but the cutting machine only accepts polynomial functions.

Can we use a polynomial truncation of the Maclaurin series for $f$?

#### Exercise
            Calculate the first few coefficients of the Maclaurin series.

Hypothesise the general formula for the coefficients.

Then either:
            * prove the formula by induction, or
            * use [the spreadsheet](https://nc.dasmithmaths.com/index.php/s/zsn82EADec9rjf5) to evaluate the first few polynomial truncations of the Maclaurin series

Conclude.
 
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 ## Application of Taylor series
 We want to cut a curve of shape approximately $f(x)=\frac1{1-x}$ for $x\in[-3,\frac12]$, but the cutting machine only accepts polynomial functions.

Can we use a polynomial truncation of the Maclaurin series for $f$?

#### Exercise
            Calculate the first few coefficients of the Maclaurin series.

Hypothesise the general formula for the coefficients.

Conclude.

The coefficients are all $1$.
 The Maclaurin polynomials are not a good approximation for $x\lt-1$.
 In fact, the Maclaurin series diverges for $x\lt-1$.