Use binomal theorem to evaluate ##(0.90)^{2.2}##

[chwala]

Homework Statement

Use binomal theorem to evaluate $(0.90)^{2.2}$ correct to four significant figures

Relevant Equations

series

My approach,

$(0.90)^{2.2}=(1-0.1)^{2.2}-1+\dfrac{2.2×-0.1}{1!}+\dfrac{1.2 ×2.2×(-0.1)^2}{2!}+\dfrac{0.2×1.2×2.2×(-0.1)^3}{3!}+...$
$=1-0.22+0.0132-0.000088=0.7931$

There may be other approach. Insight welcome.

 

[fresh_42]

This looks like Taylor, not the binomial formula.
\begin{align*}
(0.9)^{2.2}&=(0.9)^2\cdot (0.9)^{1/5}=(1-0.1)^2\cdot (0.9)^{1/5}=(1-0.2+0.01)=0.81\cdot \underbrace{(0.9)^{1/5}}_{=x}
\end{align*}
and it remains to calculate the second factor $x.$ We have the equation $x^5=0.9$ and want to achieve a precision $x=a_0+\dfrac{a_1}{10}+\dfrac{a_2}{100}+\dfrac{a_3}{1000}+\dfrac{a_4}{10000}+r.$ Now we can use the binomial formula for
$$
x^5 \left(a_0+\dfrac{a_1}{10}+\dfrac{a_2}{100}+\dfrac{a_3}{1000}+\dfrac{a_4}{10000}+r\right)^4=0.9
$$
However, this would be a nightmare. In this case, I would use Bernoulli's inequality
$$
(1+a)^r\leq 1+ra \, , \,(1-0.1)^{0.2}\leq 1-0.02 = 0.98
$$
which admittedly is Taylor in disguise again but the total error is $-0.000689826...$ as required. More precision can be achieved by taking the quadratic term of the Taylor series into account, too.

A solution in accordance to the problem statement would be to use
$$
x=(1-0.1)^{0.2}=\sum_{k=0}^\infty \binom{0.2}{k}(0.2)^k
$$
where you need the generalized binomial coefficients. Maybe that's what you used but you didn't tell.

 

[Mark44]

$(0.90)^{2.2}=(1-0.1)^{2.2}-1+\dfrac{2.2×-0.1}{1!}+\dfrac{1.2 ×2.2×(-0.1)^2}{2!}+\dfrac{0.2×1.2×2.2×(-0.1)^3}{3!}+...$
$=1-0.22+0.0132-0.000088=0.7931$

Typo at the beginning - the first minus sign on the left should be = .

As noted already, you are not using the binomial theorem. This wikipedia article discusses the binomial theorem - https://en.wikipedia.org/wiki/Binomial_theorem#Generalizations. This section shows how to deal with situations for which the exponent is not an integer (but is nonnegative).

 

[chwala]

Typo at the beginning - the first minus sign on the left should be = .

As noted already, you are not using the binomial theorem. This wikipedia article discusses the binomial theorem - https://en.wikipedia.org/wiki/Binomial_theorem#Generalizations. This section shows how to deal with situations for which the exponent is not an integer (but is nonnegative).

So was the textbook wrong to indicate using binomal theorem evaluate...

Thanks for redirecting me to the correct path...

 

[Mark44]

So was the textbook wrong to indicate using binomal theorem evaluate...

Not wrong, but they could have been clearer by specifying the generalized binomial theorem. Taking a closer look at your work, it seems that you have used this theorem, although I haven't taken the time to get your result.

 

[chwala]

Not wrong, but they could have been clearer by specifying the generalized binomial theorem. Taking a closer look at your work, it seems that you have used this theorem, although I haven't taken the time to get your result.

Yes i did use binomal theorem. There was no need for me to include $1$ and its exponents.

 

[Agent Smith]

I dunno but ...

$0.9^{2.2} = (1 - 0.1)^2 \times 0.9^{0.2} = \left(^2C_0 1 - ^2C_1 (1)(0.1) + ^2C_2 0.1^2 \right) \times 0.9^{0.2}$
$= \left(^2C_0 1 - ^2C_1 (1)(0.1) + ^2C_2 0.1^2 \right) \times \left(0.9^2 \right)^{\frac{1}{10}}$
$= \left(^2C_0 1 - ^2C_1 (1)(0.1) + ^2C_2 0.1^2 \right) \times \left(^2C_0 1 - ^2C_1 (1)(0.1) + ^2C_2 0.1^2 \right)^{\frac{1}{10}}$

 

[chwala]

Will check your approach later...now at rural africa doing some farming...big up to physicsforums...

 

[WWGD]

May all the roots you remove be perfectly square.