Tag: Taylor Polynomials

Find the limit as x goes to 0 of (sin x) / (arctan x)

by RoRi

December 21, 2015

Evaluate the limit.

$\lim_{x \to 0} \frac{\sin x}{\arctan x}.$

We know (p. 287) the following expansions as $x \to 0$ ,

$\sin x = x + o(x^2), \qquad \text{and} \qquad \arctan x = x + o(x^2).$

(Note that these are the same expansion when we use only the first order terms. This tells us that $\sin x$ and $\arctan x$ behave similarly near 0. We would need to take higher order terms to differentiate between the two. For instance, if we wanted to include cubic terms we would have $\sin x = x - \frac{1}{6}x^3 + o(x^4)$ , but $\arctan x = x - \frac{1}{3}x^3 + o(x^4)$ .) From here we compute the limit,

$\lim_{x \to 0} \frac{\sin x}{\arctan x} = \lim_{x \to 0} \frac{x + o(x^2)}{x+o(x^2)} = 1.$

Find the limit as x goes to 0 of (1 – cos² x) / (x tan x)

by RoRi

December 21, 2015

Evaluate the limit.

$\lim_{x \to 0} \frac{1 - \cos^2 x}{x \tan x}.$

To evaluate this we use the trig identity $1 - \cos^2 x = \sin^2 x$ to simplify

$\begin{align*} \lim_{x \to 0} \frac{1 - \cos^2 x}{x \tan x} &= \lim_{x \to 0} \frac{\sin^2 x}{x \tan x} \\[9pt] &= \lim_{x \to 0} \frac{\sin^2 x \cos x}{x \sin x} \\[9pt] &= \lim_{x \to 0} \left( \frac{\sin x}{x} \cdot \cos x \right) \\[9pt] &= 1 \cdot 1 = 1. \end{align*}$

We have proved the limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$ earlier (at least once), but let’s do it again using the techniques of this section and $o$ -notation.

$\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{x + o(x^2)}{x} = \lim_{x \to 0} (1 + o(x)) = 1.$

Find the limit as x goes to 0 of (log (1+x)) / (e^2x – 1)

by RoRi

December 21, 2015

Evaluate the limit.

$\lim_{x \to 0} \frac{\log (1+x)}{e^{2x} - 1}.$

Using the expansions as $x \to 0$ (p. 287)

$\log(1+x) = x + o(x), \qquad \text{and} \qquad e^x = 1 + x + o(x),$

we compute,

$\begin{align*} \lim_{x \to 0} \frac{\log (1+x)}{e^{2x} - 1} &= \lim_{x \to 0} \frac{x + o(x)}{1+2x+o(x) - 1} \\[9pt] &= \lim_{x \to 0} \frac{1 + \frac{o(x)}{x}}{2 + \frac{o(x)}{x}} \\[9pt] &= \frac{1}{2}. \end{align*}$

Find the limit as x goes to 0 of (sin x – x) / x³

by RoRi

December 21, 2015

Evaluate the limit.

$\lim_{x \to 0} \frac{\sin x - x}{x^3}.$

Using the expansion (p. 287)

$\sin x = x - \frac{x^3}{3!} + o(x^4)$

we compute,

$\begin{align*} \lim_{x \to 0} \frac{\sin x - x}{x^3} &= \lim_{x \to 0} \frac{x - \frac{x^3}{6} + o(x^4) - x}{x^3} \\[9pt] &= \lim_{x \to 0} \frac{-1 + o(x)}{6} \\[9pt] &= \lim_{x \to 0} \left(\frac{-1}{6} + o(x)\right) \\[9pt] &= -\frac{1}{6}. \end{align*}$

Prove some limits of cos x

by RoRi

December 21, 2015

Using the expansion

$\cos x = 1 - \frac{1}{2}x^2 + o(x^3) \qquad \text{as} \qquad x \to 0,$

prove that

$\frac{1-\cos x}{x^2} \to \frac{1}{2} \qquad \text{as} \qquad x \to 0.$

Using a similar method, find

$\lim_{x \to 0} \frac{1-\cos (2x) - 2x^2}{x^4}.$

Proof. Since

$\cos x = 1 - \frac{1}{2}x^2 + o(x^3)$

we have

$\begin{align*} &&1 - \cos x &= \frac{1}{2} x^2 + o(x^3) \\[9pt] \implies && \frac{1 - \cos x}{x^2} &= \frac{1}{2} + o(x) &(\text{Theorem 7.8(c)}) \\[9pt] \implies && \frac{1 - \cos x}{x^2} &\to \frac{1}{2} \quad \text{as} \quad x \to 0. \end{align*}$

(We use Theorem 7.8(c) since in the second line to bring the $\frac{1}{x^2}$ inside the little $o(x^3)$ .)

For the second limit, we use the Taylor expansion for $\cos x$ and replace $x$ with $2x$ to obtain,

$\begin{align*} &&\cos (2x) &= 1 - 2x^2 + \frac{2}{3}x^4 + o(x^5) \\[9pt] \implies && 1 - \cos (2x) - 2x^2 &= 1 - 1 + 2x^2 - \frac{2}{3}x^4 + 2x^2 + o(x^5) \\[9pt] \implies && 1 - \cos (2x) - 2x^2 &= -\frac{2}{3}x^4 + o(x^5) \\[9pt] \implies && \frac{1 - \cos (2x) - 2x^2}{x^4} &\to -\frac{2}{3} \quad \text{as} \quad x \to 0. \qquad \blacksquare \end{align*}$

Find an expression for log x as a quadratic polynomial in (x-1)

by RoRi

December 21, 2015

Find constants $a,b$ , and $c$ such that

$\log x = a + b(x-1) + c(x-1)^2 + o((x-1)^2) \qquad \text{as} \qquad x \to 1.$

From the Taylor formula for $\log (1+x)$ we have

$\log(1+x) = x - \frac{x^2}{2} + o(x^2) \quad \text{as} \quad x \to 0.$

Replacing $x$ by $x-1$ we then have

$\begin{align*} \log (1+x-1) = \log x &= (x-1) - \frac{(x-1)^2}{2} + o((x-1)^2) & \text{as } (x-1) \to 0 \\[9pt] &= (x-1) - \frac{1}{2}(x-1)^2 + o((x-1)^2) & \text{as } x \to 1. \end{align*}$

Therefore,

$a = 0, \qquad b = 1, \qquad c = -\frac{1}{2}.$

Find a polynomial of minimal degree such that sin (x – x²) = P(x) + o(x⁶)

by RoRi

December 21, 2015

Find the polynomial $P(x)$ of minimal degree such that

$\sin (x - x^2) = P(x) + o(x^6) \qquad \text{as} \qquad x \to 0.$

Using the Taylor expansion of $\sin x$ we know as $x \to 0$ we have

$\begin{align*} \sin (x-x^2) &= (x-x^2) - \frac{(x-x^2)^3}{3!} + \frac{(x-x^2)^5}{5!} + o(x^6) \\[9pt] &= x - x^2 - \frac{x^3}{6} (1-x)^3 + \frac{x^5}{120} (1-x)^5 + o(x^6) \\[9pt] &= x - x^2 - \frac{x^3}{6} (1 - 3x + 3x^2 - x^3) \\ & \qquad +\frac{x^5}{120} (1 - 5x + 10x^2 - 10x^3 + 5x^4 - x^5) + o(x^6) \\[9pt] &= x - x^2 - \frac{x^3}{6} + \frac{x^4}{2} + \left( -\frac{1}{2} + \frac{1}{120} \right)x^5 + o(x^6) \\[9pt] &= x - x^2 - \frac{x^3}{6} + \frac{x^4}{2} - \frac{59}{120} x^5 + \frac{x^6}{8} + o(x^6) \end{align*}$

(This is where we see how nice $o$ -notation can be. All of the terms in the polynomials larger than $x^6$ will get absorbed into the $o(x^6)$ . This simplifies computations tremendously when we don’t care about the higher order terms.) Therefore, we have

$P(x) = x - x^2 - \frac{x^3}{6} + \frac{x^4}{2} - \frac{59}{120} x^5} + \frac{x^6}{8} + o(x^6).$

Find a cubic polynomial such that x cos x = P(x) + o((x-1)³)

by RoRi

December 21, 2015

Find a cubic polynomial $P(x)$ such that

$x \cos x = P(x) + o ((x-1)^3) \qquad \text{as} \qquad x \to 1.$

Let $f(x) = x \cos x$ and take some derivatives,

$\begin{align*} f'(x) &= \cos x - x \sin x \\ f''(x) &= -2 \sin x - x \cos x \\ f'''(x) &= -3 \cos x + x \sin x. \end{align*}$

Therefore, the Taylor polynomial approximation to $x \cos x$ around 1 is given by

$\begin{align*} T_n (x \cos x) &= \sum_{k=0}^n \frac{f^{(k)}(a)}{k!} (x-a)^k \\[9pt] &= \cos 1 + (\cos 1 - \sin 1)(x-1) - \frac{1}{2}(2 \sin 1 + \cos 1)(x-1)^2 \\ &\qquad + \frac{1}{6}(\sin 1 - 3 \cos 1)(x-1)^3 + o((x-1)^3). \end{align*}$

Therefore, we can take

$P(x) = \cos 1 + (\cos 1 - \sin 1)(x-1) - \frac{1}{2}(2 \sin 1 + \cos 1)(x-1)^2 + \frac{1}{6}(\sin 1 - 3 \cos 1)(x-1)^3.$

Find a quadratic polynomial such that 2^x = P(x) + o(x²)

by RoRi

December 21, 2015

Find a quadratic polynomial $P(x)$ such that

$2^x = P(x) + o(x^2) \qquad \text{as} \qquad x \to 0.$

Since $2^x = e^{x \log 2}$ and

$e^x = 1 + x + \frac{x^2}{2} + o(x^2) \qquad \text{as} \qquad x \to 0,$

we have

$2^x = 1 + (x \log 2) + \left( \frac{\log^2 2}{2} \right) x^2 + o(x^2) \qquad \text{as} \qquad x \to 0.$

Therefore we may take

$P(x) = 1 + (\log 2)x + \frac{1}{2} x^2 \log^2 x .$

Compute π using the Taylor polynomial of arctan x

by RoRi

December 21, 2015

For this exercise define

$\alpha = \arctan \frac{1}{5}, \qquad \beta = 4 \alpha - \frac{1}{4} \pi.$

Using the trig identity
$\tan (A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

twice, once with $A = B = \alpha$ , and then the second time with $A = B = 2 \alpha$ , show that

$\tan (2 \alpha) = \frac{5}{12}, \qquad \tan (4 \alpha) = \frac{120}{119}.$

Then use the same identity again with $A = 4 \alpha$ and $B = -\frac{1}{4} \pi$ to show

$\tan \beta = \frac{1}{239}.$

This establishes the identity

$\pi = 16 \arctan \frac{1}{5} - 4 \arctan \frac{1}{239}.$
Using the Taylor polynomial approximation $T_{11} (\arctan x)$ at $x =\frac{1}{2}$ prove that
$3.158328934 < 16 \arctan \frac{1}{5} < 3.158328972.$
Using the Taylor polynomial approximation $T_3 (\arctan x)$ at $x = \frac{1}{239}$ prove that
$-0.016736309 < -4\arctan \frac{1}{239} < -0.016736300.$
Using the above parts show that the value of $\pi$ to seven decimal places is 3.1415926.

Proof. Letting $A = B = \alpha = \arctan \frac{1}{5}$ we have
$\tan (A+B) = \tan (2 \alpha) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{(2/5)}{1 - (1/25)} = \frac{5}{12}.$

Letting $A = B = 2 \alpha = 2 \arctan \frac{1}{5}$ we have

$\tan (A+B) = \tan (4 \alpha) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{(10/12)}{1 - (25/144)} = \frac{120}{119}.$

Letting $A = 4 \alpha$ and $B = -\frac{\pi}{4}$ we have (recalling that $\beta = 4 \alpha - \frac{\pi}{4}$ )

$\tan (\beta) = \frac{\tan (4\alpha) + \tan\left(-\frac{\pi}{4}\right)}{1 - \tan (4\alpha) \tan \left( -\frac{\pi}{4} \right)} = \frac{(120/119) - 1}{1 - (120/119)(-1)} = \frac{1}{239}.$

But then

$\begin{align*} && \tan \left( 4 \alpha - \frac{\pi}{4} \right) &= \frac{1}{239}\\[9pt] \implies && 4 \alpha - \frac{\pi}{4} &= \arctan \left( \frac{1}{239} \right) \\[9pt] \implies && \frac{\pi}{4} &= 4 \arctan \frac{1}{5} - \arctan \frac{1}{239} \\[9pt] \implies && \pi &= 16 \arctan \frac{1}{5} - 4 \arctan \frac{1}{239}. \qquad \blacksquare \end{align*}$
Proof. We know the Taylor polynomial approximation for $\arctan x$ from this exercise (Section 7.8, Exercise #3):
$T_{11} (\arctan x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \frac{x^{11}}{11}.$

Therefore, we can compute an approximation to $\arctan x$ ,

$\begin{align*} &&\arctan x &= \left( x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \frac{x^{11}}{11} \right) + E_{10}(x) \\[9pt] \implies && 16 \arctan \left(\frac{1}{5} \right) &= 16 \left( \frac{1}{5} - \frac{1}{375} + \frac{1}{15625} - \frac{1}{546875} + \frac{1}{17578125} - \frac{1}{537109375} \right) + E_{10}\left( \frac{1}{5} \right)\\[9pt] \implies && 16 \arctan \left( \frac{1}{5} \right) &= 3.158328957 + E_{10} \left( \frac{1}{5} \right), \end{align*}$

where

$\left|E_{10}\left(\frac{1}{5} \right)\right| \leq \frac{(1/5)^{11}}{11} = .000000018.$

Therefore,

$3.158328934 < 16 \arctan \left( \frac{1}{5} \right) < 3.158328972. \qquad \blacksquare$
Proof. Again using the Taylor polynomial approximation to $\arctan x$ we have
$\begin{align*} &&T_3 (\arctan x) &= x - \frac{x^3}{3} \\[9pt] \implies && \arctan x &= x - \frac{x^3}{3} + E_2 (x), \qquad |E_2 (x)| \leq \frac{x^3}{3} \\[9pt] \implies && \arctan \left( \frac{1}{239} \right) &= \left( \frac{1}{239} \right) - \frac{(1/239)^3}{3} + E_2 \left( \frac{1}{239} \right) \\[9pt] \implies && -4 \arctan \left( \frac{1}{239} \right) &= -0.01673630401 + E_2 \left( \frac{1}{239} \right) \\[9pt] \implies && -0.016736309 &< -4 \arctan \left( \frac{1}{239} \right) < -0.016736300. \qquad \blacksquare \end{align*}$
Finally,
$\begin{align*} \pi &= 16 \arctan \frac{1}{5} - 4 \arctan \frac{1}{239} \\ &\approx 3.158328957 - 0.01673630401 \\ &\approx 3.1415926. \end{align*}$