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Is this only a desirable condition to make the job easier or a strict constraint of the method itself?

("Proper fraction" = a fraction where numerator is smaller then denominator)

ribitskiyb
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1 Answers1

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Try to find constants $A$ and $B$ here: $$ \frac{x^{10}}{x(1-x)} = \frac{A}{x}+\frac{B}{1-x} $$ If you cannot do it, you have your reason...

(I assume "poper fraction" means; degree of numerator is less than degree of denominator.)

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All the terms for a partial fraction decomposition go to $0$ as $x \to \infty$. So if you add them all together, still you have something at goes to $0$ as $x \to \infty$. The only rational functions with that property are the "proper" rational functions.

GEdgar
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  • I see that in this example trying to find constant values for A and B leads to a nonvalid solution, but why does it happen so? – ribitskiyb Oct 31 '16 at 17:25