# Weird problems

• Dec 2nd 2006, 11:13 AM
12MED34
Weird problems
I'm in an analysis class currently discussing differentiation. I'm pretty sure I need to use Taylor's Theorem on these but I'm not sure. Could someone work me through them?

First, if x is greater than zero, show that (1+x)^(1/3)-(1+(1/3)x-(1/9)x^2) is less than or equal to (5/81)x^3.

Second, if abs(x) is less than or equal to one, show that abs([sin(x)-(x-(x^3)/6+(x^5)/120]) is less than (1/5040).

Sorry. I know it's quite complicated, but I have no idea. Anyways, thank you for any help you can give.
• Dec 2nd 2006, 12:16 PM
Soroban
Hello, 12MED34!

You're right . . . you need the Taylor Series.
Here's the first one . . .

Quote:

If $x < 0$, show that: . $(1+x)^{\frac{1}{3}} -1 + \frac{1}{3}x - \frac{1}{9}x^2\;\leq\;\frac{5}{81}x^3$

First, we'll crank out the Taylor Series for $(1 + x)^{\frac{1}{3}}$

Formula: . $f(x) \;=\;f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \hdots$

We have: . $\begin{array}{ccccc}f(x)\;= \\ f'(x)\:= \\ f''(x)\,= \\ f'''(x) = \\ \vdots\end{array}
\begin{array}{ccccc}(1+x)^{\frac{1}{3}} \\ \frac{1}{3}(1+x)^{\text{-}\frac{2}{3}} \\ \text{-}\frac{2}{9}(1+x)^{\text{-}\frac{5}{3}} \\ \frac{10}{27}(1+x)^{\text{-}\frac{8}{3}} \\ \vdots\end{array}$

Then: . $\begin{array}{ccccc}f(0)\;= \\ f'(0)\:= \\ f''(0)\,= \\ f'''(0)= \\ \vdots\end{array}
\begin{array}{ccccc}1 \\ \frac{1}{3} \\ \text{-}\frac{2}{9} \\ \frac{10}{27} \\ \vdots\end{array}$

We have: . $(1+x)^{\frac{1}{3}}\;=\;1 + \frac{1}{3}x -\frac{\frac{2}{9}}{2!}x^2 + \frac{\frac{10}{27}}{3!}x^3 - \hdots$

Hence: . $(1 + x)^{\frac{1}{3}} \;\leq \;1 + \frac{1}{3}x - \frac{1}{9}x^2 + \frac{5}{81}x^2$

Therefore: . $(1 + x)^{\frac{1}{3}} - 1 + \frac{1}{3}x - \frac{1}{9}x^2\;\leq \;\frac{5}{81}x^3$ . . . . There!

• Dec 2nd 2006, 12:40 PM
CaptainBlack
Quote:

Originally Posted by Soroban
Hello, 12MED34!

You're right . . . you need the Taylor Series.
Here's the first one . . .

First, we'll crank out the Taylor Series for $(1 + x)^{\frac{1}{3}}$

Formula: . $f(x) \;=\;f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \hdots$

We have: . $\begin{array}{ccccc}f(x)\;= \\ f'(x)\:= \\ f''(x)\,= \\ f'''(x) = \\ \vdots\end{array}$ $
\begin{array}{ccccc}(1+x)^{\frac{1}{3}} \\ \frac{1}{3}(1+x)^{\text{-}\frac{2}{3}} \\ \text{-}\frac{2}{9}(1+x)^{\text{-}\frac{5}{3}} \\ \frac{10}{27}(1+x)^{\text{-}\frac{8}{3}} \\ \vdots\end{array}" alt="
\begin{array}{ccccc}(1+x)^{\frac{1}{3}} \\ \frac{1}{3}(1+x)^{\text{-}\frac{2}{3}} \\ \text{-}\frac{2}{9}(1+x)^{\text{-}\frac{5}{3}} \\ \frac{10}{27}(1+x)^{\text{-}\frac{8}{3}} \\ \vdots\end{array}" />

Then: . $\begin{array}{ccccc}f(0)\;= \\ f'(0)\:= \\ f''(0)\,= \\ f'''(0)= \\ \vdots\end{array}$ $
\begin{array}{ccccc}1 \\ \frac{1}{3} \\ \text{-}\frac{2}{9} \\ \frac{10}{27} \\ \vdots\end{array}" alt="
\begin{array}{ccccc}1 \\ \frac{1}{3} \\ \text{-}\frac{2}{9} \\ \frac{10}{27} \\ \vdots\end{array}" />

We have: . $(1+x)^{\frac{1}{3}}\;=\;1 + \frac{1}{3}x -\frac{\frac{2}{9}}{2!}x^2 + \frac{\frac{10}{27}}{3!}x^3 - \hdots$

Hence: . $(1 + x)^{\frac{1}{3}} \;\leq \;1 + \frac{1}{3}x - \frac{1}{9}x^2 + \frac{5}{81}x^2$

Therefore: . $(1 + x)^{\frac{1}{3}} - 1 + \frac{1}{3}x - \frac{1}{9}x^2\;\leq \;\frac{5}{81}x^3$ . . . . There!

What follows is a bit pedantic, but the OP does say that he is in an analysis
class, so I think I will assume he needs to treat the remainder in a Taylor series
explicitly.

In fact this is not a proof, you are making an assumption about the remainder
for a truncated Taylor series, expanding about $a$ :

$
f(x)=f(a)+f'(a)(x-a)+...+\frac{f^{(n-1)}(a)(x-a)^{n-1}}{(n-1)!} + R_n
$

where:

$
R_n=\frac{f^{(n)}(\xi)(x-a)^n}{n!},\ \ \xi \mbox{ is between a and x}
$

In this case $a=0$, $x>0$, and $f'''(x)$ is a decreasing function of $x$, which allows one to obtain a bound on the remainder:

$
R_3 < \frac{f'''(0) x^3}{3!}
$

and the conclusion follows.

RonL
• Dec 2nd 2006, 01:04 PM
12MED34
I really appreciate the help.

For the second one, what would the derivative of abs(sin x) be?
• Dec 2nd 2006, 01:15 PM
CaptainBlack
Quote:

Originally Posted by 12MED34
I really appreciate the help.

For the second one, what would the derivative of abs(sin x) be?

You don't need the derivative of $|\sin(x)|$, what you are asked to show is that the absolute value of the difference between $\sin(x)$, and its truncated Taylor series truncated at the $x^5$ term, about $0$ is less than $1/5040$ when $|x| \le 1$.

So the series you are interested in is that of $\sin(x)$ about $0$.

RonL
• Dec 2nd 2006, 01:28 PM
12MED34
Yeah I end up with at the end that

abs([sin(x)-(x-(x^3)/6+(x^5)/120]) is less than or equal to abs((x^7)/5040).
How would I show that this is just greater than the expression on the left (and not equal)?
• Dec 2nd 2006, 01:49 PM
CaptainBlack
Quote:

Originally Posted by 12MED34
Yeah I end up with at the end that

abs([sin(x)-(x-(x^3)/6+(x^5)/120]) is less than or equal to abs((x^7)/5040).
How would I show that this is just greater than the expression on the left (and not equal)?

As x^7<=1 for all x such that |x|<=1, so you have abs([sin(x)-(x-(x^3)/6+(x^5)/120])<=1/5040.

Equality can only occur when |x|=1, as if |x|<1, then the bound can be made tighter. So you need only check that equality does not apply at x=+/-1.

RonL