Obtain an expression for the error in $ y = x^2 - 1/x^2$. Find the error bound for $y$ when $x = 0.76$

Question

Here's my working:

$y = \left(0.76\right)^2-\left(\frac{1}{\left(0.76\right)^2}\right) = -1.1537$

1) Finding Relative Error:

$R.E = 2\left(\frac{5.10^{-3}}{0.76}\right)+\frac{0.5}{1}+2\left(\frac{5.10^{-3}}{0.76}\right) = 0.5263$

2) Finding Absolute Error

$A.E. = |R.E.| * |y| = 0.5263 * 1.1537 = 0.6071$

3) Finding Error Bounds:

$Error Bound = y +/- A.E. = -1.7608$ to $-0.5466$

but in my book, the answer is: $-1.184$ to $-1.123$

where am i missing?

Answer 1

The number $1$ has no error associated with it. As such you should not include the $\frac{0.5}{1}$ term in your relative error.

Without it the working would go as follows:

Here's my working:

$y = \left(0.76\right)^2-\left(\frac{1}{\left(0.76\right)^2}\right) = -1.1537$

1) Finding Relative Error:

$R.E = 2\left(\frac{5.10^{-3}}{0.76}\right)+2\left(\frac{5.10^{-3}}{0.76}\right) = 0.0263$

2) Finding Absolute Error

$A.E. = |R.E.| * |y| = 0.0263 * 1.1537 = 0.0303$

3) Finding Error Bounds:

$Error Bound = y +/- A.E. = -1.1840$ to $-1.1234$

Answer 2

Hint

$$y=x^2-\frac 1{x^2}\implies \frac{dy}{dx}=2x+\frac 2{x^3}\implies \Delta y=\left(2x+\frac 2{x^3} \right)\Delta x$$ So, using $x=0.76$ and $\Delta x=5\times 10^{-3}$, $\Delta y= 0.0304$.