NastyAccident - you're almost there! However, you're not computing the power term correctly.
The full term (1.4/14) should be the power term.
Let's review. You've expanded the denominator correctly - Let's put it into a more general form. The expansion of the
denominator has the following form:
<br />
g(1+\xi) = g(1) + \sum_{k=1}^{\infty} \beta_k \, \xi^k<br />
where
<br />
\beta_k = (-1)^k \, \frac{\Pi_{m=1}^{k} (4m-3)}{4^k \cdot k!}<br />
and
<br />
\xi = x /14<br />
Of course, the full function approximation to f(x) would be f(x) = 7 * g(1+\xi).
Now, if we compute the first few coefficients:
<br />
\beta_1 = -1/4 = -0.25<br />
<br />
\beta_2 = \beta_1 \cdot (-5/4) \cdot (1/2) = 5/32 = 0.15625<br />
<br />
\beta_3 = \beta_2 \cdot (-9/4) \cdot (1/3) = -45/384 = -0.1171875<br />
Now, for the desired approximatiopn, you've also correctly realized that, here, the value of \xi=1.4/14 = 0.100.
It is here that
lanedance was trying to give you a hint to look at the value of the various terms, considering the powers of \xi and coefficients, to determine the numbers of terms that we might need for 3 digit accuracy. Since \xi=0.1, each succesive power will decrease the significance of the term by approximately one decade.
For example, consider the magnitude of \beta_2 \xi^2 term: the contribution to the overall sum
for f(x) from this term will be approximately
<br />
|7 \cdot \beta_2 * \xi^2 |= 7 \cdot |\beta_2| \cdot (0.1)^2 = 7 \cdot (1.56 \, \cdot 10^{-1}) \cdot 10^{-2} = 7 \cdot (1.56 \, \cdot 10^{-3}) = 1.09 \, \cdot 10^{-2}<br />.
(Note the multiplication by 7). Thus we see that terms beyond this need to be taken into account. What about the next term?
<br />
|7 \cdot \beta_3* \xi^3|= 7 \cdot |\beta_3| \cdot (0.1)^3= 7 \cdot (1.17 \, \cdot 10^{-1}) \cdot 10^{-3} = 7 \cdot (1.17 \, \cdot 10^{-4}) = 8.20 \, \cdot 10^{-4}<br />.
We see that the k=3 term will be less than 0.001 (least significant digit for three place accuracy) and will not effect the third decimal place on truncation (it will, however, effect the result upon rounding).
This is sufficient for the result to be
correct to three decimal places. If we wanted to be assured that the rounded result would be accurate to three decimal places, than we would have more work to do (see below).
Hence, to be correct to three decimal places, two terms (k =1, 2) plus the "zero" term (g(1)) is needed in the expansion
Computing this requires the following:
<br />
y = 7 \cdot (1+\beta_1 \cdot \xi + \beta_2 \cdot \xi^2 )<br />
which yields
<br />
y = 7 \cdot (1 - \frac{1}{4} \xi + \frac{5}{32} \xi^2)<br />
= 7 \cdot (1 - 0.25 \cdot 10^{-1} + 0.15625 \cdot 10^{-2})<br />
which computes as
<br />
y = 7 \cdot (0.976563)<br />
<br />
y = 6.835938<br />
The actual computed value is
<br />
\frac{7}{\sqrt[4]{1.1}} = 7 \cdot (0.976454) = 6.835179<br />
Notice that the approximation is correct to three decimal places, however, the rounded approximation to three decimals places would be in error in the last digit.
As an aside, consider the rounded to three decimal places condition (not part of OP problem).
From prior analysis with three \beta terms, we would expect (hope!) that further terms are not needed since \xi will influence the next term by a factor of ten, but what about the coefficient? A quick calculation shows that the next coefficient is \beta_4 = \beta_3 \cdot (-13/4)*(1/4) and its magnitude is less than \beta_3.
Hence, a total of three terms (k = 1,2,3) plus the "zero" term (g(1) = 1) should be used in the expansion for proper approximation to three digits.
Computing this requires the following:
<br />
y = 7 \cdot (1+\beta_1 \cdot \xi + \beta_2 \cdot \xi^2 + \beta_3 \cdot \xi^3)<br />
which yields
<br />
y = 7 \cdot (1 - \frac{1}{4} \xi + \frac{5}{32} \xi^2 - \frac{45}{384} \xi^3)<br />
= 7 \cdot (1 - 0.25 \cdot 10^{-1} + 0.15625 \cdot 10^{-2} - 0.1171875 \cdot 10^{-3})<br />
which computes as
<br />
y = 7 \cdot (0.976445)<br />
<br />
y = 6.835117<br />
The actual computed value is
<br />
\frac{7}{\sqrt[4]{1.1}} = 7 \cdot (0.976454) = 6.835179<br />
The above approximation is now accurate to three decimal places.
