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 2 years ago
On clip 4 of this link: http://ocw.mit.edu/courses/mathematics/1801scsinglevariablecalculusfall2010/partadefinitionandbasicrules/session2examplesofderivatives/
the teacher explains how he uses the binomial theorem to substitute for (x+dx)^n. What I don't get is: how did he simplify the binomial theorem down to x^n + (nx^(n1))dx?
In other words, I don't understand how he simplified the binomial theorem in the video at around 46:30.
 2 years ago
On clip 4 of this link: http://ocw.mit.edu/courses/mathematics/1801scsinglevariablecalculusfall2010/partadefinitionandbasicrules/session2examplesofderivatives/ the teacher explains how he uses the binomial theorem to substitute for (x+dx)^n. What I don't get is: how did he simplify the binomial theorem down to x^n + (nx^(n1))dx? In other words, I don't understand how he simplified the binomial theorem in the video at around 46:30.

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Nefarim
 2 years ago
Best ResponseYou've already chosen the best response.0\[x^n + nx^{n1}\Delta x \] is the problem I wrote in my post.

MattBenjamins
 2 years ago
Best ResponseYou've already chosen the best response.0He actually doesn't. He writes down the entire binomial expansion in the form: \[x^{n}+n ^{n1}\Delta x+ O((\Delta x)^{2})\] The O((delta x)^2) (read Big O) simply means that there are an undetermined but large amount of terms that all include delta x to some (rising) power, and that the first one of those terms includes delta x squared (say). This means that even if we divide all terms by delta x, these "junk" terms will all still include delta x as a factor. The reason why he does this is because when you take the limit of resulting expression as delta x goes to zero, every term with delta x still in them is going to tend towards zero and can therefore be safely ignored as adding nothing to the entire expression.

bob7185
 2 years ago
Best ResponseYou've already chosen the best response.0you can easily understand it from the note :One way to begin to understand this is to think about multiplying all the x’s together from (x + Δx) n = (x + Δx)(x + Δx)...(x + Δx) n times. There are n of these x’s, so multiplying them together gives you one term of x n . What if you only multiply together n − 1 of the x’s? Then you have one (x+Δx) left that you haven’t taken an x from, and you can multiply your x n−1 by Δx. (If you multiplied by x, you’d just have the x n that you already got.) There were n diﬀerent Δx’s that you could have chosen to use, so you can get this result n diﬀerent ways. That’s where the n(Δx)x n−1 comes from. We could keep going, and ﬁgure out how many diﬀerent ways there are to multiply n − 2 x’s by two Δx’s, and so on, but it turns out we don’t need to. Every other way of multiplying together one thing from each (x + Δx) gives you at least two Δx’s, and Δx · Δx is going to be too small to matter to us as Δx → 0.

kb9agt
 2 years ago
Best ResponseYou've already chosen the best response.0I just watched this about 40 minutes ago. He did not simplify the Binomial Theorem, he used it to simplify the derivative (x+dx)^n into a binomial expression. With that said you should have enough information to use what is posted here to help you understand it better I hope. Perhaps you need to do some research on the Binomial Theorem is all. Just a thought.
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