Winter 2016 - Q.11 - Grad Plus

# Winter 2016 – Q.11

11. a) Fit a curve $latex y=ab^x$ to the following data. [7M]

 x 2 3 4 5 6 y 144 172.8 207.4 248.8 298.6

Given curve is $latex y=ab^x$ taking logarithm on both sides, we get

$latex \begin{array}{l}\log y=\log a+x\log b\\\\y=A+Bx\end{array}$

where y = logy, A = loga, B = logb

The Normal equation are

$latex \begin{array}{l}\Sigma\;y=nA+B\Sigma\;x\\\\\Sigma\;xy=A\Sigma\;x+B\Sigma\;x^2\end{array}$

so we have a table of the following form

 x y y = logy xy x2 2 144 2.1584 4.3168 4 3 172.8 2.2375 6.7125 9 4 207.4 2.3168 9.2672 16 5 248.8 2.3959 11.9795 25 6 298.5 2.4750 14.85 36 20 11.5836 47.126 90

∴ we have,

$latex \begin{array}{l}11.5836=5A+20B—–\left(i\right)\\\\47.126=20A+90B—–\left(ii\right)\end{array}$

(ii)-(i)×4

0.7916=10B

∴ B=0.07916

$latex \begin{array}{l}A=\frac{11.5836-20\left(0.07916\right)}5\\\\\;\;\;\;=\frac{10.0008}5\\\\\;\;\;\;=2\\\\\;\;\;\;=\log a\end{array}$

$latex \begin{array}{l}\therefore a=Anti\;\log\;A\\\\\;\;\;\;\;\;=Anit\;\log\;2\\\\\;\;\;\;\;\;=10^2\\\\\;\;\;\;\;\;=100\end{array}$

$latex \begin{array}{l}\therefore b=Anti\;\log\;B\\\\\;\;\;\;\;\;=Anit\;\log\;\left(0.07916\right)\\\\\;\;\;\;\;\;=1.199\end{array}$

∴ The required curve is y = 100 (1.199)

b) Find the function whose first order forward difference is $latex x^3-3x^2+9.$ [6M]

Let f(x) be the required function

we have,

$latex \triangle f\left(x\right)=x^3-3x^2+9$

we first express $latex \triangle f\left(x\right)$ in the factorial notation

$latex \triangle f\left(x\right)=x^{\left(3\right)}+0x^{\left(2\right)}-2x^{\left(1\right)}+9x^{\left(0\right)}$

Taking Antidifference, we get

$latex \begin{array}{l}f\left(x\right)=\frac1\triangle\left[x^3-2x^1+9\right]\\\\\;\;\;\;\;\;\;\;=\frac{x^4}4-\frac{2x^2}2+9x+c\\\\\;\;\;\;\;\;\;\;=\frac{x^4}4-x^2+9x+c\\\\\;\;\;\;\;\;\;\;=\frac14x\left(x-1\right)\left(x-2\right)\left(x-3\right)-x\left(x-1\right)+9x+c\\\\\;\;\;\;\;\;\;\;=\frac14\left(x^4-6x^3+11x^2-6x\right)-\left(x^2-x\right)+9x+c\\\\\;\;\;\;\;\;\;\;=\frac14x^4-\frac146x^3+\frac{11}4x^2-\frac64x-x^2+x+9x+c\\\\\;\;\;\;\;\;\;\;=\frac14x^4-\frac32x^3+\frac{11}4x^2-\frac32x-x^2+10x+c\\\\\;\;\;\;\;\;\;\;=\frac14x^4-\frac32x^3+\frac74x^2+\frac{17}2x+c\\\\\;\;\;\;\;\;\;\;=\frac14\left[x^4-6x^3+7x^2+34x+4c\right]\end{array}$

$latex \therefore f\left(x\right)=\frac14\left[x^4-6x^3+7x^2+34x+4c\right]$

Where c is a constant

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