MHB Can You Solve This Complex Real Number System Equation?

[anemone]

Here is this week's POTW:

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Given $x,\,y,\,u,\,v$ are real numbers that satisfy the following system:

$$x+y+u+v=\frac{1}{2}$$

$$8x+4y+2u+v=\frac{1}{3}$$

$$27x+9y+3u+v=\frac{1}{4}$$

$$64x+16y+4u+v=\frac{1}{5}$$

Evaluate $$343x+49y+7u+v.$$

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solution::)

1. Theia
2. Opalg
3. kaliprasad
4. MarkFL

Solution from Theia:

Spoiler

By noticing that the solution to the simultaneous equations is the same as coefficients of an interpolating polynomial for points $$(1, \tfrac{1}{2},\ (2, \tfrac{1}{3}),\ (3, \tfrac{1}{4}),\ (4, \tfrac{1}{5})$$, one can directly obtain the solution e.g. using Lagrangian method. Let the interpolating polynomial be $$f(t)$$, and so

$$\begin{align*}f(t) &= \frac{1}{2} \frac{(t-2)(t-3)(t-4)}{(1-2)(1-3)(1-4)} \\
&\ \ + \frac{1}{3} \frac{(t-1)(t-3)(t-4)}{(2-1)(2-3)(2-4)} \\
&\ \ + \frac{1}{4} \frac{(t-1)(t-2)(t-4)}{(3-1)(3-2)(3-4)} \\
&\ \ + \frac{1}{5} \frac{(t-1)(t-2)(t-3)}{(4-1)(4-3)(4-3)} \\
&= -\frac{t^3}{120} + \frac{11t^2}{120} - \frac{23t}{60} + \frac{4}{5}.\end{align*}$$

Now one can also note that the asked quantity $$l = 343x + 49y + 7u + v$$ is equal to $$f(7)$$. Thus the result

$$l = 343x + 49y + 7u + v = f(7) = \frac{-1}{4}.$$

Alternate solution from Opalg:

Spoiler

Let $f(n) = xn^3 + yn^2 + un + v$. We are told that $f(1) = \frac12$, $f(2) = \frac13$, $f(3) = \frac14$ and $f(4) = \frac15$. Make a table showing these values in the top row, the differences $f(n+1) - f(n)$ in the next row, then the second differences (that is, the differences of the differences) in the following row, and so on:
$$\begin{array}{ccccccc} 1&&2&&3&&4 \\ \hline \tfrac12 && \tfrac13 && \tfrac14 && \tfrac15 \\ &-\tfrac16 && -\tfrac1{12} && -\tfrac1{20} \\ && \tfrac1{12} && \tfrac1{30} \\ &&& -\tfrac1{20} \end{array}$$ Since $f(n)$ is a cubic polynomial, its third differences must be constant. So the third difference must always be $-\frac1{20}$. We can therefore add some extra elements $-\frac1{20}$ to the bottom row of the table, and then work our way back up to the top row, like this:
$$\begin{array}{ccccccccccccc} 1&&2&&3&&4 &&5&&6&&7 \\ \hline \tfrac12 && \tfrac13 && \tfrac14 && \tfrac15 &&\tfrac2{15} && 0 && -\tfrac14 \\ &-\tfrac16 && -\tfrac1{12} && -\tfrac1{20} && -\tfrac1{15} && -\tfrac2{15} && -\tfrac14 \\ && \tfrac1{12} && \tfrac1{30}&& -\tfrac1{60} && -\tfrac1{15} && -\tfrac7{60} \\ &&& -\tfrac1{20} && -\tfrac1{20} && -\tfrac1{20} && -\tfrac1{20}\end{array}$$
Conclusion: $343x + 49y + 7u + v = f(7) = -\tfrac14.$