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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.Originally Posted by mark1950
I have a problem with this particular question:
Expandin ascending powers of x up to and including the term in
and determine the set of values of x for this expansion to be valid. Calculate the value of
giving your answer correct to 3 decimal places.
This is my take on the question:
![=[ 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 + ...]](http://latex.codecogs.com/png.latex?=[ 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 + ...])
I got the above expression but when I try to substituteinto 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
). Where have I gone wrong? Thanks.
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, whose cube root is
. Powers of 1/33 go to zero really fast, and you will get a very accurate approximation
, which you only have to multiply by 3/2 to find
.
