Numerical Analysis Questions & Answers | |
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This is fine, I'll give you 2 points bonus. I would have given 3 if you would have typeset correctly your equations with proper equation numbers as specified in the LATEX mini HOWTO in the Skylounge. But, this proof is still not what I was looking for.. There's a much easier way to prove it using Taylor series (without power series). Three bonus points are still up for grabs.
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This is what I was looking for. I'll give you 2.5 points bonus. The only problem here is that you don't define $f(x)$ prior to doing the expansion with the Taylor series.
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Well, in this case, it's clear what the notation means. It doesn't matter if it's a superscript or subscript..
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I don't understand this part: $$ k^{\sum_{m=0}^{n-1} p^{m-1}}=k^{1+\sum_{m=0}^{n-1} p^{m-1}} $$ This doesn't make sense to me. Is this a typo or a misunderstanding? Please fix this and retype your proof below.
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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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$\pi$ |