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On this problem, I'm not sure what Big O definition they are referring. How would the big o definition help show this?

Use the definition of $O$ to show that if $y = y_h + O(h^p)$, then $hy = hy_h + O(h^{p+1})$.

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The first statement is saying, taking big O as $x$ approaches some $a$ for some $M>0$ $$ |y-y_h|<M|h^p|\implies \frac{|y-y_h|}{|h^p|}<M $$ for any $|x-a|<\delta$ for some $\delta>0$.

Now what happens if you multiply the LHS by $\frac{|h|}{|h|}=1$?

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  • $\begingroup$ I'm quite lost in this topic..I don't understand any of what you just wrote and can't really answer the question :( $\endgroup$ – john Jan 21 '17 at 22:33
  • $\begingroup$ @john I didn't write much more than the definition of what it means to be big o of something so maybe start with that $\endgroup$ – qbert Jan 21 '17 at 22:36

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