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If `f(x)`is a polynomial of degree ` |
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Answer» we have to prove that `|{:(1,,a,,(f(a))/((x-a))),( 1,,b,,(f(b))/((x-b))),(1,,c,,(f(c))/((x-c))):}| -: |{:(1,,a,,a^(2)),( 1,,b,,b^(2)),(1,,c,,c^(2)):}| =(f(x))/((x-a)(x-b)(x-c))` `L.H.S.=|{:(1,,a,,(f(a))/((x-a))),( 1,,b,,(f(b))/((x-b))),(1,,c,,(f(c))/((x-c))):}| -: [(a-b)(b-c)(c-a)]` Expanding along `C_(3)` we get `L.H.S. =(1)/((a-b)(b-c)(c-a))xx` `[(f(a)(c-b))/((x-a))+(f(b)(a-c))/((x-b))+(f(c)(b-a))/((x-c))]` Now using partial fraction method on R.H.S. we get `R.H.S. (f(x))/((x-a)(x-b)(x-c))=(A)/(x-a)+(B)/(x-b)+(C)/(x-c)` ltbr. `"(As degree of " f(x) lt 3)` `" then " A= [(f(x))/((x-b)(x-c))]_(x=a)` `=(f(a))/((b-a)(a-c))` `" Similarly "=(1)/((b-a)(b-c)) " and " C = (f(c))/((c-a)(c-b))` `:. R.H.S. =(1)/((a-b)(b-c)(c-a))xx` `[((c-b)f(a))/((x-b))+((a-c)f(b))/((x-b))+((b-a)f(c))/((x-c))]` Hence L.H.S. =R.H.S.` |
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