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If a is small, then show that $\dfrac{1}{(1+a)^3}$ is nearly equal to $(1-3a)$. Also show that if $0<a<0.01$, then the error$<0.0007$.

Can we solve this without using calculus?

If so can anyone help me with the solution?

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We know that $$(1+a)^3(1-a)^3=(1-a^2)^3$$ Therefore $$\frac{1}{(1+a)^3}=(1-a)^3/(1-a^2)^3\approx (1-3a+3a^2-a^3)\approx1-3a$$ Now for small a we can neglect $a^2$ and$a^3$ that's because higher powers of a get very small when compared to 1 or a when a is small Example $$a=0.001$$ then$$ a^2=0.00001$$ and$$ a^3=0.0000001$$ which are very small . $Q.E.D.$
Just plug in the value you need to calculate error.

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Suppose $a$ is small. Then $(1+a)^x \approx 1+xa$ holds and is known as the Binomial approximation. Thus $\frac{1}{(1+a)^3} = (1+a)^{-3} \approx 1+(-3)a = (1-3a)$.

To find the error, note that as $a \rightarrow 0$, $\frac{1}{(1+a)^3} \rightarrow 1$ and $(1-3a) \rightarrow 1$. When $a=0.01$, the error between the two expressions is $|\frac{1}{(1+a)^3} - (1-3a)| = |\frac{1}{[1+(0.01)]^3} - [1-3(0.01)]| \approx (0.9706) - (0.9700) = 0.0006$, which is less than $0.0007$. Noting that the difference in the expressions is monotonically increasing in the interval $0<a<0.01$, we can say that the error is thus less than $0.0007$.

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$$ \begin{align} \frac1{(1+a)^3}-(1-3a) &=\frac{6+8a+3a^2}{(1+a)^3}\cdot a^2\\[6pt] &\le\left\{\begin{array}{} 22a^2&\text{if }a\ge-\frac12\\ 6a^2&\text{if }a\ge0 \end{array}\right. \end{align} $$ since $\frac{6+8a+3a^2}{(1+a)^3}=\frac3{1+a}+\frac2{(1+a)^2}+\frac1{(1+a)^3}$ is decreasing for $a\gt-1$.

If $0\le a\le0.01$, then the error is at most $0.0006$.

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