MHB Express the value in a single fraction

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The discussion revolves around solving the equation involving real numbers \(x_1, x_2, x_3, x_4, x_5\) that satisfy a specific condition for \(m = 1, 2, 3, 4, 5\). The solution yields the value of \(\dfrac{x_1}{37}+\dfrac{x_2}{38}+\dfrac{x_3}{39}+ \dfrac{x_4}{40}+\dfrac{x_5}{41}\) as the fraction \(\frac{187465}{6744582}\). Additionally, a continued fraction representation and a unit fraction expansion are provided, showcasing the complexity of the solution. Participants are encouraged to share their methods for solving the problem, emphasizing collaborative learning. The thread highlights the intricate relationship between algebraic expressions and their fractional representations.
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Let $x_1, x_2, x_3, x_4, x_5$ be real numbers satisfying the following equation:

$\dfrac{x_1}{m^2+1}+\dfrac{x_2}{m^2+2}+\dfrac{x_3}{m^2+3}+\dfrac{x_4}{m^2+4}+\dfrac{x_5}{m^2+5}= \dfrac{1}{m^2}$ for $m=1, 2, 3, 4, 5$.

Find the value of
$\dfrac{x_1}{37}+\dfrac{x_2}{38}+\dfrac{x_3}{39}+ \dfrac{x_4}{40}+\dfrac{x_5}{41}$
 
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[sp]
Fraction:$\frac{187465}{6744582}$

Continued fraction: $[0,35,1,44,111,2,2,12,4]$

Unit fraction expansion: $\frac{1}{36}+\frac{1}{58395}+\frac{1}{9724688047}+\frac{1}{283708672824669334580}$
[/sp]
 
eddybob123 said:
[sp]
Fraction:$\frac{187465}{6744582}$

Continued fraction: $[0,35,1,44,111,2,2,12,4]$

Unit fraction expansion: $\frac{1}{36}+\frac{1}{58395}+\frac{1}{9724688047}+\frac{1}{283708672824669334580}$
[/sp]

Thank you so much for participating, eddybob123.

Your answer is correct, but I hope you can share with me the method you used to solve this problem. :o

By sharing that means a brief explanation on the concept that you employed will suffice.
 
anemone said:
Let $x_1, x_2, x_3, x_4, x_5$ be real numbers satisfying the following equation:

$\dfrac{x_1}{m^2+1}+\dfrac{x_2}{m^2+2}+\dfrac{x_3}{m^2+3}+\dfrac{x_4}{m^2+4}+\dfrac{x_5}{m^2+5}= \dfrac{1}{m^2}$ for $m=1, 2, 3, 4, 5$.

Find the value of
$\dfrac{x_1}{37}+\dfrac{x_2}{38}+\dfrac{x_3}{39}+ \dfrac{x_4}{40}+\dfrac{x_5}{41}$

Solution provided by other:

Let $f(x)=\dfrac{x_1}{m^2+1}+\dfrac{x_2}{m^2+2}+\dfrac{x_3}{m^2+3}+\dfrac{x_4}{m^2+4}+\dfrac{x_5}{m^2+5}$, then

$f(\pm 1)=1$, $f(\pm 2)=\dfrac{1}{4}$, $f(\pm 3)=\dfrac{1}{9}$, $f(\pm 4)=\dfrac{1}{16}$, $f(\pm 5)=\dfrac{1}{25}$, and $f(6)$ is the value to be found.

Next, we let $g(x)=(x^2+1)(x^2+2)(x^2+3)(x^2+4)(x^2+5)$ and $h(x)=f(x)g(x)$.

Then for $m=\pm1, \pm2, \pm3, \pm4, \pm5$, we get $h(x)=f(x)g(x)=\dfrac{g(x)}{m^2}$, i.e. $g(x)-m^2h(x)=0$.

Since $g(x)-x^2h(x)$ is a polynomial of degree 10 with roots $\pm1, \pm2, \pm3, \pm4, \pm5$, we get

$g(x)-x^2h(x)=A(x^2-1)(x^2-4)(x^2-9)(x^2-16)(x^2-25)$ (*)

Putting $x=0$ we get $A=\dfrac{g(0)}{(-1)(-4)(-9)(-16)(-25)}=-\dfrac{1}{120}$.

Finally, dividing both sides of (*) by $g(x)$ gives

$\dfrac{g(x)-x^2h(x)}{g(x)}=-\dfrac{(x^2-1)(x^2-4)(x^2-9)(x^2-16)(x^2-25)}{120g(x)}$ (*)

$1-x^2\dfrac{h(x)}{g(x)}=1-x^2f(x)=-\dfrac{1}{120}\cdot\dfrac{(x^2-1)(x^2-4)(x^2-9)(x^2-16)(x^2-25)}{(x^2+1)(x^2+2)(x^2+3)(x^2+4)(x^2+5)}$ and hence

$1-36f(6)=\dfrac{35\cdot32\cdot27\cdot20\cdot11}{120 \cdot 37\cdot38\cdot39\cdot40\cdot41}=-\dfrac{231}{374699}$,

which implies $f(6)=\dfrac{187465}{6744582}$.
 
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