Tag: Taylor Polynomials

Determine the coefficients in the Taylor series of (2 + x²)^5/2

by RoRi

April 5, 2016

Consider the function $f(x) = (2+x^2)^{\frac{5}{2}}$ . Determine the first five coefficients ( $a_0, a_1, \ldots, a_4$ ) in the Taylor series expansion of $f(x)$ at $x = 0$ .

First, we take some derivatives,

$\begin{align*} && f(x) &= (2+x^2)^{\frac{5}{2}} \\[9pt] \implies && f'(x) &= 5x(2+x^2)^{\frac{3}{2}} \\[9pt] \implies && f''(x) &= 15x^2 (2+x^2)^{\frac{1}{2}} + 5 (2+x^2)^{\frac{3}{2}} \\[9pt] \implies && f'''(x) &= \frac{15x^3}{(2+x^2)^{\frac{1}{2}}} + 45x (2+x^2)^{\frac{1}{2}} \\[9pt] \implies && f^{(4)} (x) &= -\frac{15x^4}{(2+x^2)^{\frac{3}{2}}} + \frac{90x^2}{(2+x^2)^{\frac{1}{2}}} + 45 (2+x^2)^{\frac{1}{2}}. \end{align*}$

Then, since the coefficients in the Taylor series at $x = 0$ are of the form $a_n = \frac{f^{(n)}(0)}{n!}$ we can compute as follows:

$\begin{align*} a_0 &= \frac{f(0)}{0!} = 2^{\frac{5}{2}} = 4 \sqrt{2} \\[9pt] a_1 &= \frac{f'(0)}{1!} = \frac{0}{1} = 0 \\[9pt] a_2 &= \frac{f''(0)}{2!} = \frac{5 \cdot 2^{\frac{3}{2}}}{2} = 5 \sqrt{2} \\[9pt] a_3 &= \frac{f'''(0)}{6} = 0 \\[9pt] a_4 &= \frac{f^{(4)}(0)}{4!} = \frac{45 \sqrt{2}}{24} = \frac{15}{8} \sqrt{2}. \end{align*}$

Prove some properties of the function e^-1/x²

by RoRi

January 13, 2016

Consider the function

$f(x) = e^{-\frac{1}{x^2}} \qquad \text{when} \quad x \neq 0$

and $f(0) = 0$ .

Prove that for every positive number $m$ we have
$\lim_{x \to 0} \frac{f(x)}{x^m} = 0.$
Prove that if $x \neq 0$ then
$f^{(n)}(x) = f(x) P \left( \frac{1}{x} \right)$

where $P(t)$ is a polynomial in $t$ .
Prove that
$f^{(n)}(0) = 0 \qquad \text{for all } n \geq 1.$

Proof. (A specific case of this general theorem is actually the first problem of this section, here. Maybe it’s worth taking a look since this proof is just generalizing that particular case.) We make the substitution $t = \frac{1}{x^2}$ , so that $t \to +\infty$ as $x \to 0$ and we have
$\begin{align*} \lim_{x \to 0} \frac{e^{-\frac{1}{x^2}}}{x^m} &= \lim_{t \to +\infty} \frac{t^{\frac{m}{2}}}{e^t} \\[9pt] &= 0 \end{align*}$

by Theorem 7.11 (page 301 of Apostol) since $m > 0$ implies $\frac{m}{2} > 0$ as well $. \qquad \blacksquare$
Proof. The proof is by induction on $n$ . In the case $n = 1$ we have
$\begin{align*} f'(x) &= \left( \frac{2}{x^3} \right) e^{-\frac{1}{x^2}} \\[9pt] &= 2 \left( \frac{1}{x} \right)^3 e^{-\frac{1}{x^2}} \\[9pt] &= P \left(\frac{1}{x} \right) e^{-\frac{1}{x^2}}. \end{align*}$

So, indeed the formula is valid in the case $n = 1$ . Assume then that the formula holds for some positive integer $k$ . We want to show this implies the formula holds for the case $k + 1$ .

$\begin{align*} f^{(k+1)}(x) = \left( f^{(k)}(x) \right)' &= \left( P\left( \frac{1}{x} \right) e^{-\frac{1}{x^2}} \right)' \\[9pt] &= P' \left( \frac{1}{x} \right) e^{-\frac{1}{x^2}} + \frac{2}{x^3} P \left( \frac{1}{x} \right) e^{-\frac{1}{x^2}} \\[9pt] &= \left( P' \left( \frac{1}{x} \right) + \frac{2}{x^3} P \left( \frac{1}{x} \right) \right)e^{-\frac{1}{x^2}}. \end{align*}$

But then the leading term

$P' \left( \frac{1}{x} \right) + 2 \left( \frac{1}{x} \right)^3 P \left( \frac{1}{x} \right)$

is still a polynomial in $\frac{1}{x}$ since the derivative of a polynomial in $\frac{1}{x}$ is still a polynomial in $\frac{1}{x}$ , and so is the sum of two polynomials in $\frac{1}{x}$ . Therefore, we have that the formula holds for the case $k+1$ ; hence, it holds for all positive integers $n. \qquad \blacksquare$
Proof. The proof is by induction on $n$ . If $n = 1$ then we use the limit definition of the derivative to compute the derivative at 0,
$\begin{align*} f'(0) &= \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} \\[9pt] &= \lim_{h \to 0} \frac{f(h)}{h} &(f(0) = 0 \text{ by def of } f)\\[9pt] &= 0 & \text{by part (a)}. \end{align*}$

So, indeed $f'(0) = 0$ and the statement is true for the case $n = 1$ . Assume then that $f^{(k)} (0) = 0$ for some positive integer $k$ . Then, we use the limit definition of the derivative again to compute the derivative $f^{(k+1)}(0)$ ,

$\begin{align*} f^{(k+1)}(0) &= \lim_{h \to 0} \frac{f^{(k)}(0+h) - f^{(k)}(0)}{h} \\[9pt] &= \lim_{h \to 0} \frac{f^{(k)}(h)}{h} &(f^{(k)}(0) = 0 \text{ by Ind. Hyp.}) \\[9pt] &= \lim_{h \to 0} f(h) P \left( \frac{1}{h} \right) \cdot \frac{1}{h} &(\text{part (b)}) \\[9pt] &= \lim_{h \to 0} P \left( \frac{1}{h} \right) e^{-\frac{1}{h^2}}. \end{align*}$

This follow since $\frac{1}{h} \cdot P \left( \frac{1}{h} \right)$ is still a polynomial in $\frac{1}{h}$ , and by the definition of $f(x)$ for $x \neq 0$ . But then, by part (a) we know

$\lim_{h \to 0} \frac{f(h)}{h^m} = 0 \qquad \text{for all } m \in \mathbb{Z}^+.$

Therefore,

$f^{(k+1)}(0) = \lim_{h \to 0} P \left( \frac{1}{h} \right) e^{-\frac{1}{h^2}} = 0.$

Thus, the formula holds for the case $k+1$ , and hence, for all positive integers $n. \qquad \blacksquare$

Compute derivatives of a function given a limit equation that it satisfies

by RoRi

December 22, 2015

Consider a function $f(x)$ that satisfies the limit relation

$\lim_{x \to 0} \left( 1 + x + \frac{f(x)}{x} \right)^{\frac{1}{x}} = e^3,$

and has a continuous third derivative everywhere. Determine the values

$f(0), \quad f'(0), \quad f''(0), \quad \text{and} \quad \lim_{x \to 0} \left( 1 + \frac{f(x)}{x} \right)^{\frac{1}{x}}.$

We do some simplification to the expression first.

$\begin{align*} &&\lim_{x \to 0} \left( 1 + x + \frac{f(x)}{x} \right)^{\frac{1}{x}} &= e^3 \\[9pt] \implies && \lim_{x \to 0} e^{\frac{1}{x} \log \left( 1 + x + \frac{f(x)}{x} \right)} &= e^3 \\[9pt] \implies && \exp \left( \lim_{x \to 0} \left( \frac{1}{x} \log \left( 1 +x + \frac{f(x)}{x} \right) \right) &= e^3 \\[9pt] \implies && \lim_{x \to 0} \left( \frac{1}{x} \log \left( 1 + x + \frac{f(x)}{x} \right) \right) &= 3. \end{align*}$

Now, we apply the hint (that if $\lim_{x \to 0} g(x) = A$ then $g(x) = A + o(1)$ ),

$\begin{align*} &&\frac{1}{x} \log \left( 1 + x + \frac{f(x)}{x} \right) &= 3 + o(1) \\[9pt] \implies && \log \left( 1 + x + \frac{f(x)}{x} \right) &= 3x + o(x) \\[9pt] \implies && 1 + x + \frac{f(x)}{x} &= e^{3x + o(x)} \\[9pt] \implies && x + x^2 + f(x) &= xe^{3x}e^{o(x)} \\[9pt] \implies && f(x) &= - x - x^2 + xe^{3x}e^{o(x)}. \end{align*}$

So as $x \to 0$ we have

$f(x) = -x - x^2 + x\left( 1 + 3x + o(x) \right) = 2x^2 + o(x^2).$

But then since $f$ has three derivatives at 0 we know its Taylor expansion at 0 is unique and is given by (Theorem 7.1, page 274 of Apostol)

$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2} x^2 + o(x^2).$

Hence, equating the coefficients of like powers of $x$ we have

$f(0) = 0, \quad f'(0) = 0, \quad f''(0) = 4.$

Next, to compute the limit

$\lim_{x \to 0} \left( 1 + \frac{f(x)}{x} \right)^{\frac{1}{x}}$

we write

$\begin{align*} f(x) = 2x^2 + o(x^2) && \implies && \frac{f(x)}{x} &= 2x + o(x) \\[9pt] && \implies && 1 + \frac{f(x)}{x} &= 1 + 2x + o(x) \\[9pt] && \implies && \log \left(1 + \frac{f(x)}{x} \right) &= \log (1+2x+o(x)). \end{align*}$

Then, using the Taylor expansion of $\log(1+x)$ (page 287 of Apostol) we know as $x \to 0$ we have

$\log (1+2x+o(x)) = 2x+o(x) + o(x) = 2x + o(x).$

Therefore we have

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

Find the limit as x goes to 1 of (1 / log x – 1 / (x-1))

by RoRi

December 22, 2015

Evaluate the limit.

$\lim_{x \to 1} \left( \frac{1}{\log x} - \frac{1}{x-1} \right).$

We use this exercise (Section 7.11, Exercise #4) to evaluate,

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

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

by RoRi

December 22, 2015

Evaluate the limit.

$\lim_{x \to 0} \left( \frac{1}{x} - \frac{1}{e^x - 1} \right).$

We have

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

Find the limit as x goes to 0 of (arcsin x / x)^1/x²

by RoRi

December 22, 2015

Evaluate the limit.

$\lim_{x \to 0} \left( \frac{\arcsin x}{x} \right)^{\frac{1}{x}}.$

First, we rewrite the expression using the definition of the exponential:

$\left( \frac{\arcsin x}{x} \right)^{\frac{1}{x}} = e^{\frac{1}{x^2} \log \left(\frac{\arcsin x}{x} \right)}.$

Next, we need to get a series expansion for $\arcsin x$ as $x \to 0$ . The most straightforward way is to take the first few derivatives (we’ll only need the $x^3$ term).

$\begin{align*} f(x) &= \arcsin x & \implies && f(0) &= 0 \\[9pt] f'(x) &= \frac{1}{\sqrt{1-x^2}} & \implies && f'(0) &= 1 \\[9pt] f''(x) &= \frac{-x}{(1-x^2)^{\frac{3}{2}}} & \implies && f''(0) &= 0 \\[9pt] f'''(x) &= \frac{1+2x^2}{(1-x^2)^{\frac{5}{2}}} & \implies && f'''(0) &= 1. \end{align*}$

Therefore, we have as $x \to 0$

$\arcsin x = x + \frac{x^3}{6} + o(x^4).$

Hence, as $x \to 0$

$\frac{\arcsin x}{x} = 1 + \frac{x^2}{6} + o(x^3).$

Since this is going to 1 as $x \to 0$ we may apply this exercise (Section 7.11, Exercise #4) to conclude

$\begin{align*} \log \left( \frac{\arcsin x}{x} \right) &= \left( \frac{\arcsin x}{x} - 1\right) - \frac{1}{2} \left( \frac{\arcsin x}{x} - 1 \right)^2 + o \left( \left( \frac{\arcsin x}{x} - 1 \right)^2 \right) \\[9pt] &= \left( \frac{x^2}{6} + o(x^3) \right) - \frac{1}{2} \left( \frac{x^2}{6} + o(x^3) \right)^2 + o \left( \left( \frac{x^2}{6} + o(x^3) \right)^2 \right) \\[9pt] &= \frac{x^2}{6} + o (x^3). \end{align*}$

Now, we have the expansion as $x \to 0$

$\frac{1}{x^2}\log \left( \frac{\arcsin x}{x} \right) = \frac{1}{6} + o (x).$

Therefore,

$\begin{align*} \lim_{x \to 0} \left( \frac{\arcsin x}{x} \right)^{\frac{1}{x}} &= \lim_{x \to 0} e^{ \frac{1}{x^2} \log \left( \frac{\arcsin x}{x} \right)} \\[9pt] &= e^{\frac{1}{6}} \lim_{x \to 0} e^{o(x)} \\ &= e^{\frac{1}{6}}. \end{align*}$

Find the limit as x to 0 of the given expression

by RoRi

December 22, 2015

Evaluate the limit.

$\lim_{x \to 0} \left( \frac{(1+x)^{\frac{1}{x}}}{e} \right)^{\frac{1}{x}}.$

First, we write the expression using the definition of the exponential,

$\begin{align*} \left( \frac{(1+x)^{\frac{1}{x}}}{e} \right)^{\frac{1}{x}} &= e^{ \frac{1}{x} \log \left( \frac{(1+x)^{\frac{1}{x}}}{e} \right)} \\[10pt] &= e^{\frac{1}{x} \left( \log (1+x)^{\frac{1}{x}} - \log e \right)} \\[10pt] &= e^{\frac{1}{x} \left( \frac{1}{x} \log(1+x) - 1 \right)} \\[10pt] &= e^{\frac{1}{x^2} \log(1+x) - \frac{1}{x}}. \end{align*}$

Now, considering the expression in the exponent and using the expansion of $\log(1+x)$ as $x \to 0$ (page 287 of Apostol) we have as $x \to 0$ ,

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

Therefore, we have

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

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

by RoRi

December 22, 2015

Evalue the limit.

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

First, we have

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

As $x \to 0$ we use the expansion for $\log (1+x)$ (page 287 of Apostol) to write,

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

From this we see that $\frac{1}{x} \log(1+x) \to 1$ as $x \to 0$ ; and so,

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

Hence, using the expansion for $e^x$ as $x \to 0$ (page 287 of Apostol) we have

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

Therefore, we have

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

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

by RoRi

December 22, 2015

Evaluate the limit.

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

From the definition of the exponential we have

$\left( x + e^{2x} \right)^{\frac{1}{x}} = e^{ \frac{1}{x} \log \left( x + e^{2x} \right)}.$

So, first we use the expansion of $e^x$ as $x \to 0$ (page 287 of Apostol) to write

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

Therefore, as $x \to 0$ we have

$\log (x + e^{2x}) = \log (x + 1 + 2x + o(x)) = \log (1 + 3x + o(x)).$

Now, since $3x + o(x) \to 0$ as $x \to 0$ we can use the expansion (again, page 287) of $\log (1+x)$ as $x \to 0$ to write

$\log (1 + 3x + o(x)) = 3x+o(x) + o(3x+o(x)) = 3x + o(x) \qquad \text{as} \quad x \to 0.$

Therefore, as $x \to 0$ we have

$\frac{1}{x} \log (x+e^{2x}) = \frac{1}{x}(3x + o(x)) = 3 + \frac{o(x)}{x}.$

So, getting back to the expression we started with,

$\lim_{x \to 0} \left( x + e^{2x} \right)^{\frac{1}{x}} = \lim_{x \to 0} e^{3 + \frac{o(x)}{x}} = e^3 \lim_{x \to 0} e^{\frac{o(x)}{x}}.$

But, as in the previous exercise (Section 7.11, Exercise #23) we know $\lim_{x \to 0} e^{\frac{o(x)}{x}} = 1$ . Hence,

$\lim_{x \to 0} \left( x + e^{2x} \right)^{\frac{1}{x}} = e^3.$

Find the limit as x goes to 1 of x^{1 / (1-x)}

by RoRi

December 22, 2015

Evaluate the limit.

$\lim_{x \to 1} x^{\frac{1}{1-x}}.$

First, we write

$x^{\frac{1}{1-x}} = e^{\frac{1}{1-x} \cdot \log x}.$

From this exercise (Section 7.11, Exercise #4) we know that as $x \to$ we have

$\log x = (x-1) + o((x-1))$

Therefore, as $x \to 1$ ,

$\frac{\log x}{1-x} = -1 + \frac{o(x-1)}{x-1}.$

So, we then have

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

(Here we could say that since the exponential is a continuous function we can bring the limit inside and so this becomes $e^0 = 1$ . I’m not sure we know we can pass limits through continuous functions like that, so we continue on with expanding the exponential as in previous exercises.)
Since $\frac{o(x-1)}{x-1} \to 0$ as $x \to 1$ we take the expansion of $e^x$ as $x \to 0$ ,

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

Therefore,

$\lim_{x \to 1} x^{\frac{1}{1-x}} = \frac{1}{e} \lim_{x \to 1} e^{\frac{o(x-1)}{x-1}} = \frac{1}{e}.$