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First I tried to get the C2 by this: n=1

1/3n^5- 10n^3+7n-25 <= 1/3n^5+7n^5 = 22/3n^5 , C2=22/3 (I removed all the negative terms and multiplied the positive so I can get the max value, is this correct?)

And then I tried to get the min value by this: 1/3n^5- 10n^3+7n-25 >= - 10n^3-25n^3= -35n^3 (Removed all the positive and multiplied negative, but then the C1=-35 is negative too, which is clearly wrong. How shoud I solve this?)

DaDA
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    Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Nov 08 '22 at 16:13
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    To show it's big theta, just need positive constants A,B so that for sufficiently large n, $An^5 \le P(n) \le Bn^5.$ Since only need for sufficiently large n, there's really no need to be as careful as you seem to be doing. – coffeemath Nov 08 '22 at 16:14
  • But how do I find the A,B,n... ? – DaDA Nov 08 '22 at 16:25
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    Use that $\lim P(n)/n^5 = 1/3$ – lhf Nov 08 '22 at 16:33
  • Trying to do it straight from the definition is fine too. Again... we don't need tight bounds! It is a waste of time trying to get the tightest bounds here. It should be clear that $\frac{1}{3}n^5-10n^3+7n-25\leq \frac{1}{3}n^5+7n^5$ since subtracting fewer things of course makes it bigger, and since adding bigger things instead of smaller things makes it bigger. Now, for the other side... $\frac{1}{6}n^5+\frac{1}{6}n^5-10n^3-25n^3\leq \frac{1}{3}n^5-10n^3+7n-25$ by similar logic. Now, notice that we were able to split the $n^5$ term in two. – JMoravitz Nov 08 '22 at 16:41
  • Now... can we eventually reach a point where $\frac{1}{6}n^5-35n^3\geq 0$? With large enough $n$ we should be able to. Can you find it and finally finish the problem? Again, we don't need to be tight with our bounds. – JMoravitz Nov 08 '22 at 16:42

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