## Which is more difficult: Precalculus or Calculus?

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### Which is more difficult: Precalculus or Calculus?

Which branch of mathematics would you say is more difficult to learn: Precalculus or Calculus?
a + bi
jonas
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### Re: Which is more difficult: Precalculus or Calculus?

Calculus will be even more difficult unless you master precalculus first.

g_edgar
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If we assume that one has learned all prerequisites of precalculus, then if he/she has mastered precalculus, then if he/she has mastered all calculus (assuming the branches of calculus I, II & III), then he/she will have concluded that calculus is definitely a harder subject than precalculus.
johnny
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### Re: Which is more difficult: Precalculus or Calculus?

I had a more difficult time with Precalculus, but I enjoyed it more than Calculus. Though if you take Physics, you will appreciate Calculus ALOT more
A theory is something nobody believes, except the person who made it. An experiment is something everybody believes, except the person who made it.
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jcmc2112
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Ah, what the heck. . .? Both precalculus and calculus are incredibly easy!
johnny
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### Re: Which is more difficult: Precalculus or Calculus?

johnny wrote:Ah, what the heck. . .? Both precalculus and calculus are incredibly easy!

*tips hat*
Numbers don't lie. They hide, but they don't lie.

greg1313
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Ha ha ha!
johnny
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Easy? In that case, find the exact value of $\sum_{n=1}^\infty\frac{1}{1 \,+\,n^2}.$
skipjack
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Perhaps I would have clarified this. Learning the concepts of precalculus and calculus for the regular educational purpose was easy, but not the special type of problems like those hard ones, like olympiad problems for example. I spent 30 minutes on this problem, but still no clue. Do you have a solution?
johnny
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I know the sum's value, but I will probably not be giving any help or stating anything as to how easy or difficult it is to obtain.
skipjack
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### Re: Which is more difficult: Precalculus or Calculus?

skipjack wrote:. . . .find the exact value of $\sum_{n=1}^\infty\frac{1}{1 \,+\,n^2}.$

$\frac{\pi}{2}$
Numbers don't lie. They hide, but they don't lie.

greg1313
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Let us assume that $\sum_{n\,=\,1}^{\infty}\,\frac{1}{n^2}\,=\,\frac{\pi^2}{6}$ is correct for the following discussion.

greg1313, if $\sum_{n\,=\,1}^{\infty}\,\frac{1}{1\,+\,n^2}\,=\,\frac{\pi}{2}$ is to be true, then $\sum_{n\,=\,1}^{\infty}\,\frac{1}{n^2(n^2\,+\,1)}\,=\,\frac{\pi^2}{6}\,-\,\frac{\pi}{2}$ must be true, since $\sum_{n\,=\,1}^{\infty}\,\frac{1}{1\,+\,n^2}\,=\,\sum_{n\,=\,1}^{\infty}\,\frac{1}{n^2}\,-\,\sum_{n\,=\,1}^{\infty}\,\frac{1}{n^2(n^2\,+\,1)}$. So, are we correct or not?
johnny
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Trivially, $\sum_{n\,=\,1}^{\infty}\,\frac{1}{1\,+\,n^2}\,<\,\frac12\,+\sum_{n\,=\,2}^{\infty}\,\frac{1}{n^2}\,=\,-\frac12\,+\,\frac{\pi^2}{6}\,= 1.14...\,<\,\frac{\pi}{2}.$
skipjack
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### Re: Which is more difficult: Precalculus or Calculus?

$\sum_{n=1}^\infty\frac1{1+n^2}=\frac12+\sum_{k=1}^\infty(-1)^{k+1}(\zeta(2k)-1)$

not that that helps much
mattpi
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### Re: Which is more difficult: Precalculus or Calculus?

$\frac{\pi}{4}\le \sum_{n=1}^\infty{\frac{1}{1 + n^2}}\le \frac{\pi}{2}$

For clarification see this.
Numbers don't lie. They hide, but they don't lie.

greg1313
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