Problem with this question

**mark1950** · Aug 29th 2009, 01:55 AM

I have a problem with this particular question:

Expand $(\frac{1 + x}{1 - 3x})^{\frac{1}{3}}$ in ascending powers of x up to and including the term in $x^3$ ad determine the set values of x for this expansion to be valid. Calculate the value of $\sqrt[3] 3$ giving your answer correct to 3 decimal places.

This is my take on the question:

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

$=[ 1 + \frac{1}{3} + \frac{(\frac{1}{3})(\frac{1}{3} - 1)}{2!}x^2 + \frac{(\frac{1}{3})(\frac{1}{3} - 1)(\frac{1}{3} - 2)}{3!}x^3 + ...]$ $[1 + (-\frac{1}{3})(-3x) + \frac{(-\frac{1}{3})(-\frac{1}{3} - 1)}{2!}(-3x)^2<br /> + \frac{(-\frac{1}{3})(-\frac{1}{3} - 1)(-\frac{1}{3} - 2)}{3!}(-3x)^3 + ...]$

$=[ 1 + \frac{1}{3}x - \frac{1}{9}x^2 + \frac{5}{81}x^3 + ... ][1 + x + 2x^2 + \frac{14}{3}x^3]$

$=[1 + x + 2x^2 + \frac{14}{3}x^3 + \frac{1}{3}x + \frac{1}{3}x^2 + \frac{2}{3}x^3 - \frac{1}{9}x^2 + \frac{5}{81}x^3 + ...]$

$=[1 + \frac{4}{3}x + \frac{20}{9}x^2 + \frac{428}{81}x^3]$

I got the above expression but when I try to substitute $x = \frac{1}{5}$ into the expression to solve the last part of the question, the value aren't the same and the difference is quite large (mine was 1.39) compared with the standard value of 1.442 (value of $\sqrt[3] 3$ ). Where have I gone wrong? Thanks.

**Opalg** · Aug 29th 2009, 02:28 AM

Originally Posted by mark1950

I have a problem with this particular question:

Expand $(\frac{1 + x}{1 - 3x})^{\frac{1}{3}}$ in ascending powers of x up to and including the term in $x^3$ and determine the set of values of x for this expansion to be valid. Calculate the value of $\sqrt[3] 3$ giving your answer correct to 3 decimal places.

This is my take on the question:

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

$=[ 1 + \frac{1}{3} + \frac{(\frac{1}{3})(\frac{1}{3} - 1)}{2!}x^2 + \frac{(\frac{1}{3})(\frac{1}{3} - 1)(\frac{1}{3} - 2)}{3!}x^3 + ...]$ $[1 + (-\frac{1}{3})(-3x) + \frac{(-\frac{1}{3})(-\frac{1}{3} - 1)}{2!} + (-\frac{1}{3})(-3x) <br /> + \frac{(-\frac{1}{3})(-\frac{1}{3} - 1)(-\frac{1}{3} - 2)}{3!} + ...]$

$=[ 1 + \frac{1}{3}x - \frac{1}{9}x^2 + \frac{5}{81}x^3 + ... ][1 + x + 2x^2 + \frac{14}{3}x^3]$

$=[1 + x + 2x^2 + \frac{14}{3}x^3 + \frac{1}{3}x + \frac{1}{3}x^2 + \frac{2}{3}x^3 - \frac{1}{9}x^2 + \frac{5}{81}x^3 + ...]$

$=[1 + \frac{4}{3}x + \frac{20}{9}x^2 + \frac{428}{81}x^3]$

I got the above expression but when I try to substitute $x = \frac{1}{5}$ into the expression to solve the last part of the question, the value aren't the same and the difference is quite large (mine was 1.39) compared with the standard value of 1.442 (value of $\sqrt[3] 3$ ). Where have I gone wrong? Thanks.

You have avoided the instruction in green. The expansion is only valid for |x| < 1/3. The trouble with your (inaccurate) approximation is related to this. In fact, the closer x is to 1/3, the less reliable the approximation will be. To get a good approximation, you want x to be as small as possible, so that its powers tend to 0 rapidly. That will mean that the first few terms of the series will be all that is needed to get a good approximation to the correct answer.

By choosing x = 1/5, you gave yourself a value of x that is dangerously close to 1/3, so the terms of the series do not decrease sufficiently fast to get a good approximation. Suppose that instead you tried x = –1/33. Then $\frac{1+x}{1-3x} = \frac89$ , whose cube root is $\tfrac23\sqrt[3]3$ . Powers of 1/33 go to zero really fast, and you will get a very accurate approximation $\tfrac23\sqrt[3]3\approx0.96149$ , which you only have to multiply by 3/2 to find $\sqrt[3]3\approx1.4422$ .

**mark1950** · Aug 29th 2009, 06:47 AM

Hm...so my methods are right, right? Its just about the x = 1/5 thing...

Actually, I got x = 1/5 from here:

$\sqrt[3]{\frac{1+ x}{1-3x}} = \sqrt[3] 3$

${\frac{1+ x}{1-3x}} = 3$

$1 + x = 3 - 9x$

$10x = 2$

$x = \frac{1}{5}$

Is there anything wrong there? Thanks.

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