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Thread: Analytical mathematics-please help.

  1. #1
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    Analytical mathematics-please help.

    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
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  2. #2
    Senior Member JaneBennet's Avatar
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    #1
    Let’s do a table. The following table shows production output at the end of the given week.

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

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

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

    (ii) Let the number of weeks be n. Then solve for n for the given inequalities: (A) $\displaystyle 40r^n>300$; (B) $\displaystyle 40+nd>300$.
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  3. #3
    Junior Member roy_zhang's Avatar
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    2.Using the appropriate power series, determine, using 4 terms

    (a)e 3

    (b) (1.02) 4
    2a) Recall that the function $\displaystyle e^x$ can be expanded by power series as: $\displaystyle e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}=1+x+\frac{x^ 2}{2!}+\frac{x^3}{3!}+\cdots$

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

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

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