+27 Hi guys i have the above book and the sum at the bottom of page 116 is getting me down! You see he starts the sum but then gets a load of figures with no explanation of where they're coming from. Then he admits he didn't show all his work! Well i'm a beginner at Calculus and so am a bit of a dummy and i need it explained to me. I understand the first line and where those numbers come from but then from the second line i'm lost. He's using the summation sign and it eventually is boiling down to the area under a curve. If no one knows then i'll try to write out the full sum it's just a bit difficult with not knowing how to work out all the options on here. I have an account on here but have forgotten my password again. I got a lot of help from Alan before who answered all my crazy trig questions!
Gary
+9479 Is it called the Simpson's Rule, Trapezoidal Rule, or Midpoint Rule? I'm just guessing...
+27 Hi mate, thanks. It's a Riemann sum or at least in that section of the book and it's for calculating the area under a curve so it's integration. I'll try to write it out here, the first line of it anyway. He starts off with using the sigma summation sign and then on the next page he uses the integration sign and takes it to the limit of the function. I think basically he then takes the limit as infinity in order to get the exact area under the curve.
27/n3 . n(n+1)(2n+1)/6 + 3.n/n
That's the basic sum. The last bit is 3 times ndivided by n. The dot for multiplication is hard to see but i didn't want to use an x because it can cause confusion. Now the second line in the book has a breakdown of the middle bit there, the n+1 bit, where he gets n3/3 + n2/2 + n/6 but that's the bit where he writes at the side that he didn't include all his work. So it's not broken down how he gets there and for a beginner it's just impossible trying to figure it out. I've looked on YT videos for the sum of the squares that the middle part of the sum concerns but most go into proving the theory or induction or something which is way above my head and too advanced for me. I'm just looking for a kind of simple breakdown if it's possible. I think he omitted the explanation because he didn't want to have to write all the extra steps out!!
Gary
+9479 I managed to get a screen shot of the page you're talking about:
I'm not claiming to understand what all this is... but I think I can show how he got from step 1 to step 2.
\(\frac{27}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}+\frac{3}{n}\cdot n\) Multiply out the numerator.
\( \frac{27}{n^3}\cdot\frac{n(2n^2+3n+1)}{6}+\frac{3}{n}\cdot n \\~\\ \frac{27}{n^3}\cdot\frac{2n^3 + 3n^2 +n}{6}+\frac{3}{n}\cdot n\)
\(\frac{27}{n^3}\cdot(\frac{2n^3}{6} + \frac{3n^2}{6} +\frac{n}{6})+\frac{3}{n}\cdot n \\~\\ \frac{27}{n^3}\cdot(\frac{n^3}{3} + \frac{n^2}{2} +\frac{n}{6})+\frac{3}{n}\cdot n\) Distribute the 1/6 and reduce.
\(\frac{27}{n^3}\cdot(\frac{n^3}{3} + \frac{n^2}{2} +\frac{n}{6})+3\) In the last term the n's cancel :)
+27 Thanks so much hectictar i appreciate it. That's the exact bit i'm talking about. I looked at your post and saw it and said "That's it!" Your post helps me out a lot. I wonder why he didn't explain it all out in the book? I thought that was the idea for the dummies books to completely break things down in easily understood language as Calculus can get complicated. Thanks again.
Gary
+9479 Haha, I guess the author was just lazy. I feel you though, there's been multiple times where I want to have a word with the crazy people who wrote the calculus book I have to use!
Btw, for the next step, looks like he just distributed the 27/n3 to each term in the parenthesees. I wasn't sure if you were wondering about that too.
+27 Thanks mate. Yeah i'm wondering about the whole sum apart from the first line! That bit where he wrote that about not showing all the work really freaked me out and i'm saying: "But it's a dummies book! You're supposed to be explaining everything, every step." I have two Algebra books I and II by Mary Jane Sterling in the dummies series and she explains everything a lot better and doesn't leave anything out. I was going to ask how he got from the three 6's in the denominators to the 3 2 in the next line but looking at it i take it he's just divided but in like the opposite direction using the 2 and 3 in the numerators into 6? I forgot you could do that and am so brainwashed into just dividing the standard way. Yeah i think the author was being a bit lazy and trying to save himself some work in explaining how he derived each line in the sum!
Gary
+9479 Yeah, all it really is is just reducing a fraction like we did back in elementary school. Just with letters mixed in!
\(\frac{5}{10}=\frac{1}{2}\)
because
\(\frac{5}{10}=\frac{5\cdot1}{5\cdot2}\)
And the fives cancel.
So
\(\frac{2n^3}{6}=\frac{n^3}{3}\)
because
\(\frac{2n^3}{6}=\frac{2\cdot n^3}{2\cdot 3}\)
And the twos cancel.
It's pretty crazy how math is just simple stuff piled on top of each other until it gets so confusing and hard to understand!
+27 Thanks for all the help mate it's appreciated. I wanted to give your posts the thumbs up but i'll have to wait until i can reset my password to get into my account. I'm glad i came back on and made a new account on the forum as i see me having a lot of questions from this calculus book!
Gary
