How to find approximate value of $\frac{((2.03)^4)}{((3.998)^2)}$?

Question

I am having trouble to find the approximate value for this questions..

The methods I have already tried are:

1) I have let them as, $f(x)=2+0.03$ and $f(y)=4-0.002$ then I've used the method,

$f(x+Δx) = f(x)+ f'(x).Δx$.

I couldn't go further with this method. (It's probably solving with this method too, but I have some mistakes I guess..) anyways,

2) then I have factorized them $(a^2 - b^2)$ and $(a^2 +b^2)$. I had something but it's not looking like the answer,

1.1) Lastly, I have tried the first method again (see image), and I had some answers but I'm not sure. So could someone please help me to solve this question??

Answer 1

First, $$ 2.03^2 = (2+0.03)^2 = 2^2+2\cdot 2\cdot 0.03 + 0.03^2 = 4.1209 $$ We could discard the last $0.0009$ as too small to care about, but we can still keep everything in our head, so just let's keep it for now.

We're now looking at $$ \frac{4.1209^2}{3.998^2} = \Bigl(\frac{4.1209}{3.998}\Bigr)^2$$ If we don't want just to round the denominator to $4$, we can multiply both sides of the fraction by $1+\frac{1}{2000}$ to get $$ \cdots \approx \Bigl(\frac{4.1209+0.00206}{3.998+0.002}\Bigr)^2 \approx \Bigl(\frac{4.122}{4.000}\Bigr)^2 \approx 1.0305^2 \approx 1.061$$ where the last $\approx$ is another application of $(1+x)^2=1+2x+x^2$ and we ignore $x^2$ because it is small.

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According to my calculator, the true value is $1.0624257107\ldots$ -- close enough.

(Though I should probably have rounded $4.12296$ to $4.123$ and then rounded one fourth of that up to $1.031$ instead -- that would gotten closer).

Answer 2

Hint: Use total differential formula for $f(x,y)=\frac{x^4}{y^2}$ at the point $(x,y)=(2,4)$.

Full solution: $$\Delta f=f_x \Delta x+f_y \Delta y=$$ $$\frac{4x^3}{y^2}\cdot 0.03-\frac{2x^4}{y^3}\cdot (-0.002)=$$ $$\frac{32}{16}\cdot 0.03-\frac{32}{64}\cdot (-0.002)=0.06+0.001=0.061.$$ Hence: $$f(2.03,3.998)=f(2,4)+\Delta f=1+0.061=1.061.$$ P.S. Yes, your solution is correct.

Answer 3

The expression can be written (after dividing numerator / denominator by 16) as $(1+0.015)^{4}(1-0.0005)^{-2}$ and then via binomial theorem the approximation $(1+4\times 0.015)(1+2\times 0.0005)$ is obtained. This is further approximated as $1+0.06+0.001=1.061$.