Analytical mathematics-please help.

**Diane Charlton** · March 5th 2008, 11:54 AM

Can someone please help me with these questions, i'm so stuck!!

1.Due to an increase in orders, a firm increases its weekly production from 40 to 100 components in 5 weeks.
If the weekly productions form
(A)A Geometric progression (b)Arithmetic Progression
determine
(i) the percentage/fixed production increase
(ii)the number of full weeks before production demands exceeds the maximum capacity of 300 components assuming the progressions continue.

2.Using the appropriate power series, determine, using 4 terms

(a)e 3

(b) (1.02) 4

**JaneBennet** · March 5th 2008, 12:42 PM

#1
Let’s do a table. The following table shows production output at the end of the given week.

$\begin{tabular}{l|c|c|c|c|c|c|}<br /> {} & Week 0 & Week 1 & Week 2 & Week 3 & Week 4 & Week 5 \\\hline<br /> G.P. & $40$ & $40r$ & $40r^2$ & $40r^3$ & $40r^4$ & $40r^5$\\\hline<br /> A.P. & $40$ & $40+d$ & $40+2d$ & $40+3d$ & $40+4d$ & $40+5d$\\\hline<br /> \end{tabular}$

Set (A) $40r^5=100$ and (B) $40+5d=100$ and solve for r and d.

(i) The percentage increase at the end of week k from the previous week is (A) $\frac{40r^k-40r^{k-1}}{40r^{k-1}}\times100\%$ ; (B) $\frac{40+kd-(40+(k-1)d)}{40+(k-1)d}\times100\%$ . (Simplify.)

(ii) Let the number of weeks be n. Then solve for n for the given inequalities: (A) $40r^n>300$ ; (B) $40+nd>300$ .

**roy_zhang** · March 5th 2008, 01:10 PM

2.Using the appropriate power series, determine, using 4 terms

(a)e 3

(b) (1.02) 4

2a) Recall that the function $e^x$ can be expanded by power series as: $e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}=1+x+\frac{x^ 2}{2!}+\frac{x^3}{3!}+\cdots$

Thus, $e^3 \approx 1+3+\frac{9}{2}+\frac{27}{6}$

2b) Let's consider the Maclaurin series for the function $f(x)=(1+x)^k$ , which is actually a binomial series:

$(1+x)^k = \sum_{n=0}^{\infty} \binom{k}{n}x^n = 1+ kx + \frac{k(k-1)x^2}{2!} + \frac{k(k-1)(k-2)x^3}{3!} + \cdots$

Hence, $(1.02)^4 = (1+0.02)^4$ , let $x=0.02$ and $k=4$ , plug into the series above.

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