Question 9
(a) Expand $\left(1-8x^2\right)^{-\frac{1}{2}}$ in ascending powers of $x$, up to and including the term in $x^6$ giving each coefficient as an integer. (3)
$$ g(x)=\frac{a+bx}{\sqrt{1-8x^2}} \text{ where } a \text{ and } b \text{ are prime numbers} $$Given that the fourth and fifth terms, in ascending powers of $x$, in the series expansion of $g(x)$ are $20x^3$ and $48x^4$ respectively,
(b) find the value of $a$ and the value of $b$ (4)
Using the first five terms, in ascending powers of $x$, in the series expansion of $g(x)$
(c) obtain an estimate, to 4 significant figures, of $\int_0^{0.2} g(x)\,dx$ (4)
Solution
Part (a)
We aim to expand $\left(1-8x^2\right)^{-\frac{1}{2}}$ in ascending powers of $x$ up to and including the term in $x^6$. Using the binomial expansion for $(1 + u)^n$:
$$ (1 + u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \cdots, $$where $n = -\frac{1}{2}$ and $u = -8x^2$, we proceed as follows:
Step 1: Expand the series
Substitute $n = -\frac{1}{2}$ and $u = -8x^2$ into the formula:
$$ \left(1 - 8x^2\right)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)(-8x^2) + \frac{-\frac{1}{2}\left(-\frac{1}{2}-1\right)}{2!}(-8x^2)^2 $$ $$ \quad + \frac{-\frac{1}{2}\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right)}{3!}(-8x^2)^3 + \cdots. $$Step 2: Simplify each term
- Zeroth term: $1$
- First term: $\left(-\frac{1}{2}\right)(-8x^2) = 4x^2$
Second term:
$$ \frac{-\frac{1}{2}(-\frac{1}{2}-1)}{2!}(-8x^2)^2 = \frac{-\frac{1}{2}\left(-\frac{3}{2}\right)}{2}(64x^4) $$ $$ = \frac{3}{8}(64x^4) = 24x^4 $$Third term:
$$ \frac{-\frac{1}{2}\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right)}{3!}(-8x^2)^3 $$ $$ = \frac{-\frac{1}{2}\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{6}(512x^6) $$ $$ = \frac{1 \cdot 3 \cdot 5}{2 \cdot 2 \cdot 2 \cdot 6}(512x^6) = 160x^6 $$Step 3: Final expansion
Collecting terms, we have:
$$ \left(1 - 8x^2\right)^{-\frac{1}{2}} = 1 + 4x^2 + 24x^4 + 160x^6 + \cdots $$Part (b)
We are given $\mathrm{g}(x) = \frac{a + bx}{\sqrt{1 - 8x^2}}$, where $a$ and $b$ are prime numbers. The fourth and fifth terms in the expansion of $\mathrm{g}(x)$ are $20x^3$ and $48x^4$, respectively. We determine $a$ and $b$ step-by-step.
Step 1: Expand $\mathrm{g}(x)$
Substitute the expansion of $\left(1 - 8x^2\right)^{-\frac{1}{2}}$ into $\mathrm{g}(x)$:
$$ \mathrm{g}(x) = (a + bx)\left(1 + 4x^2 + 24x^4 + 160x^6 + \cdots\right) $$ $$ = a(1 + 4x^2 + 24x^4 + 160x^6) + bx(1 + 4x^2 + 24x^4 + 160x^6). $$Expand and collect terms:
$$ \mathrm{g}(x) = a + 4ax^2 + 24ax^4 + 160ax^6 + bx + 4bx^3 + 24bx^5 + 160bx^7 + \cdots. $$Step 2: Match coefficients
The fourth term in the expansion is $20x^3$, and the fifth term is $48x^4$. Match coefficients for $x^3$ and $x^4$:
- Coefficient of $x^3$: $4b = 20 \implies b = 5$
- Coefficient of $x^4$: $24a = 48 \implies a = 2$
Step 3: Verify the values of $a$ and $b$
Substitute $a = 2$ and $b = 5$ back into the expansion of $\mathrm{g}(x)$:
$$ \mathrm{g}(x) = 2 + 5x + 8x^2 + 20x^3 + 48x^4 + \cdots. $$The coefficients match the given conditions, confirming that $a = 2$ and $b = 5$.
Part (c) $\int_0^{0.2} \mathrm{g}(x) \, \mathrm{d}x$
We now estimate $\int_0^{0.2} \mathrm{g}(x) \, \mathrm{d}x$ using the first five terms in the expansion of $\mathrm{g}(x)$.
Substitute $\mathrm{g}(x) = 2 + 5x + 8x^2 + 20x^3 + 48x^4$ into the integral:
$$ \int_0^{0.2} \mathrm{g}(x) \, \mathrm{d}x = \int_0^{0.2} \left(2 + 5x + 8x^2 + 20x^3 + 48x^4\right) \, \mathrm{d}x $$First, find the indefinite integral:
$$ \int \left( 2 + 5x + 8x^2 + 20x^3 + 48x^4 \right) \, dx = 2x + \frac{5}{2}x^2 + \frac{8}{3}x^3 + 5x^4 + \frac{48}{5}x^5 + C $$Now, evaluate this expression at the limits $x = 0.2$ and $x = 0$:
$$ I = \left[ 2x + \frac{5}{2}x^2 + \frac{8}{3}x^3 + 5x^4 + \frac{48}{5}x^5 \right]_0^{0.2} $$Substitute the upper limit $x = 0.2$ into the expression:
$$ = \left( 2(0.2) + \frac{5}{2}(0.2)^2 + \frac{8}{3}(0.2)^3 + 5(0.2)^4 + \frac{48}{5}(0.2)^5 \right) $$Now, compute each term:
$$ = 0.4 + \frac{5}{2}(0.04) + \frac{8}{3}(0.008) + 5(0.0016) + \frac{48}{5}(0.00032) $$ $$ = 0.4 + 0.1 + 0.0213333 + 0.008 + 0.003072 $$ $$ = 0.5324053 $$Therefore, the value of the definite integral is approximately:
$$ I \approx 0.5324 $$
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