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 August 25th, 2009, 06:39 AM #1 Joined: Jun 2009 Posts: 22 Thanks: 0 Which is more difficult: Precalculus or Calculus? Which branch of mathematics would you say is more difficult to learn: Precalculus or Calculus?
 August 25th, 2009, 04:35 PM #2 Joined: Jun 2009 Posts: 150 Thanks: 0 Re: Which is more difficult: Precalculus or Calculus? Calculus will be even more difficult unless you master precalculus first.
 August 25th, 2009, 10:25 PM #3 Joined: Apr 2007 Posts: 2,141 Thanks: 0 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.
 August 27th, 2009, 01:27 PM #4 Joined: Jul 2009 From: New Jersey Posts: 65 Thanks: 0 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
 August 27th, 2009, 03:40 PM #5 Joined: Apr 2007 Posts: 2,141 Thanks: 0 Ah, what the heck. . .? Both precalculus and calculus are incredibly easy!
August 27th, 2009, 03:46 PM   #6
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Re: Which is more difficult: Precalculus or Calculus?

Quote:
 Originally Posted by johnny Ah, what the heck. . .? Both precalculus and calculus are incredibly easy!
*tips hat*

 August 27th, 2009, 03:48 PM #7 Joined: Apr 2007 Posts: 2,141 Thanks: 0 Ha ha ha!
 August 28th, 2009, 01:39 AM #8 Global Moderator   Joined: Dec 2006 Posts: 10,487 Thanks: 29 Easy? In that case, find the exact value of $\sum_{n=1}^\infty\frac{1}{1 \,+\,n^2}.$
 August 28th, 2009, 02:53 AM #9 Joined: Apr 2007 Posts: 2,141 Thanks: 0 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?
 August 28th, 2009, 11:04 AM #10 Global Moderator   Joined: Dec 2006 Posts: 10,487 Thanks: 29 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.
August 28th, 2009, 11:55 AM   #11
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Re: Which is more difficult: Precalculus or Calculus?

Quote:
 Originally Posted by skipjack . . . .find the exact value of $\sum_{n=1}^\infty\frac{1}{1 \,+\,n^2}.$
$\frac{\pi}{2}$

 August 28th, 2009, 04:10 PM #12 Joined: Apr 2007 Posts: 2,141 Thanks: 0 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?
 August 29th, 2009, 01:19 AM #13 Global Moderator   Joined: Dec 2006 Posts: 10,487 Thanks: 29 Trivially, $\sum_{n\,=\,1}^{\infty}\,\frac{1}{1\,+\,n^2}\,<\,\ frac12\,+\sum_{n\,=\,2}^{\infty}\,\frac{1}{n^2}\,= \,-\frac12\,+\,\frac{\pi^2}{6}\,= 1.14...\,<\,\frac{\pi}{2}.=$
 August 29th, 2009, 04:08 AM #14 Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 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
 August 30th, 2009, 02:38 PM #15 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 5,537 Thanks: 19 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.

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