Numerical Analysis Questions & Answers | |
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Hm, but if that is true, then I don't understand how you obtain the following on the denominator on the LHS: $$ \frac{k^{2^n}*\epsilon_{0}^{2^n}}{k*k^{p^n-1}*\epsilon_{0}^{p^n}}=\frac{k^{2^n}*\epsilon_{0}^{2^n}}{k^{p^n}*\epsilon_{0}^{p^n}} $$ Please fix this and rewrite your proof better below. And please never use the newline command in your posts (two blackslashes). This is not necessary and is a bad habit.
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Obtaining convergence to 4 significant numbers means that iterating more won't change the first 4 significant numbers. Only one question per post please. 1 point bonus.
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There is still the same problem with your proof. You claim that $$ \frac{\epsilon_{(n)(Newton)}}{\epsilon_{(n)(secant)}}=\frac{k^{2^{n}-1} \epsilon_{0}^{2n}}{k^{\sum_{m=0}^{n-1} p^{m-1}} \epsilon_{0}^{p^{n}}}=\frac{\frac{k^{2^n} \epsilon_{0}^{2n}}{k}}{k^{\sum_{m=0}^{n-1} p^{m-1}} \epsilon_{0}^{p^n}}=\frac{k^{2^n} \epsilon_{0}^{2n}}{k^1 k^{\sum_{m=0}^{n-1} p^{m-1}} \epsilon_{0}^{p^n}} =\frac{k^{2^n} \epsilon_{0}^{2n}}{k^{1+\sum_{m=0}^{n-1} p^{m-1}} \epsilon_{0}^{p^n}} $$ and that $$ k^{\sum_{m=0}^{n-1} p^{m-1}} \epsilon_{0}^{p^n}=k^{1+p^{-1}+p^0+p^1+...+p^{n-2}} \epsilon_{0}^{p^n}=k^{1+p^n-1} \epsilon_{0}^{p^n}=k^{p^n} \epsilon_{0}^{p^n} $$ But if you substitute the latter in the former, you will not get $$ \frac{\epsilon_{(n)(Newton)}}{\epsilon_{(n)(secant)}}=\frac{k^{2n} \epsilon_{0}^{2n}}{k^{p^n} \epsilon_{0}^{p^n}}= \frac{{(k \epsilon_{0})}^{2^n}}{{(k \epsilon_{0})}^{p^n}}={(k \epsilon_{0})}^{2^n-p^n} $$ Thus your proof is wrong. I gave you 3 chances already, and you didn't succeed. I can not give you one more chance. 1 point bonus for the effort. Let others now try.
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I don't understand your question. Please formulate this better.
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I'm not sure what you mean. The number of iterations depends on the initial guess. If the initial guess is the root, then the number of iterations is 1. I'll give you 0.5 point for the effort.
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$\pi$ |