1/(1-x)=1+x+x²+ x³+x^4+0(x^5) and sin(x) = x + x^3/3! + x^5/5! + 0(x^7)?
Given the taylor polynomial Expression1/(1-x)=1+x+x²+ x³+x^4+0(x^5) and sin(x) = x + x^3/3! + x^5/5! + 0(x^7) determine the order of approximation for their sum and product
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There is a typo in the Taylor polynomial for sin(x): the term in x^3 must be negative.
In the sum, the order of approximation is O(x^5), and therefore doesn't make sense to keep the term x^5/5!.
In the product, the lesser terms are O(x^7)*1 = O(x^7) and O(5)*x = O(x^6). The order of approximation is therefore O(x^6).
(x/x+a million) % [ (x/x+a million)+a million ] -- equation a million Taking only the denominator first , (with the intention to simplify it), [ (x/x+a million)+a million ] = x+x+a million/x+a million so now equation a million will become (x/x+a million) % x+x+a million/x+a million - equation 2. u understand (a million/3)%(a million/3)=a million. For dividing , u would desire to alter the reciprocal to alter % to *. SO equation 2 will become (x/x+a million) * (x+a million/x+x+a million) =x/2x+a million = x+a million/x