untouchablecuz
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Partial Fractions (1 Viewer)
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bwahahahha(x-x_{2})...(x-x_{n})}=\frac{c_{1}}{x-x_{1}}+\frac{c_{2}}{x-x_{2}}+...+\frac{c_{n}}{x-x_{n}}\text {where} \ x\neq [x_{1},x_{2},..., x_{n}]\\\text {The next sp in splitting the fraction into partial fractions is to multiply through by} \ (x-x{1})(x-x_{2})...(x-x_{n})\\\text{So, } 1=c_{1}(x-x_{2})...(x-x_{n})+c_{2}(x-x_{1})(x-x_{3})...(x-x_{n})+...+c_{n}(x-x_{2})...(x-x_{n-1})\\\text {How is substituting x=[x_{1},x_{2},...,x_{n}] \ \text {to find } [c_{1},c_{2},...,c_{n}] \ \text {justified, since initially } x\neq [x_{1},x_{2},..., x_{n}] \ \text {?})
untouchablecuz
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that was quick :|
study-freak
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It's because when you changed it, you effectively isolated the numerator, which does not have any restrictions.(x-x_{2})...(x-x_{n})}=\frac{c_{1}}{x-x_{1}}+\frac{c_{2}}{x-x_{2}}+...+\frac{c_{n}}{x-x_{n}}\text {where} \ x\neq [x_{1},x_{2},..., x_{n}]\\\text {The next step in splitting the fraction into constituent partial fractions is to multiply through by} \ (x-x_{1})(x-x_{2})...(x-x_{n})\\\text{So, } 1=c_{1}(x-x_{2})...(x-x_{n})+c_{2}(x-x_{1})(x-x_{3})...(x-x_{n})+...+c_{n}(x-x_{2})...(x-x_{n-1})\\\text {How is substituting } x=[x_{1},x_{2},...,x_{n}] \ \text {to find } [c_{1},c_{2},...,c_{n}] \ \text {justified, since initially } x\neq [x_{1},x_{2},..., x_{n}] \ \text {?})
What I mean is, take this as an example (instead of your complicated one lol):
1/(x-1)(x-2)=A/(x-1)+B/(x-2)=[A(x-2)+B(x-1)]/[(x-1)(x-2)]
Now, eliminate the denominator (i.e. equate the numerator)
1=A(x-2)+B(x-1)
This is only possible since they shared the same denominator and hence the same restrictions. The numerator is now made to be independent of the denominator. (Hence x=1, 2 can be substituted into the identity.)
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