Thread: 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.

Hello, 12MED34!

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

If , show that: .

First, we'll crank out the Taylor Series for

Formula: .

We have: .

Then: .


We have: .

Hence: .

Therefore: . . . . . There!

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

Formula: .

We have: . \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}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: .

Hence: .

Therefore: . . . . . 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 :



where:



In this case , , and is a decreasing function of , which allows one to obtain a bound on the remainder:



and the conclusion follows.

RonL

I really appreciate the help.

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

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 , what you are asked to show is that the absolute value of the difference between , and its truncated Taylor series truncated at the term, about is less than when .

So the series you are interested in is that of about .

RonL

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)?

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