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Nefarim

On clip 4 of this link: http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-a-definition-and-basic-rules/session-2-examples-of-derivatives/ 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^(n-1))dx? In other words, I don't understand how he simplified the binomial theorem in the video at around 46:30.

  • one year ago
  • one year ago

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  1. Nefarim
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    \[x^n + nx^{n-1}\Delta x \] is the problem I wrote in my post.

    • one year ago
  2. MattBenjamins
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    He actually doesn't. He writes down the entire binomial expansion in the form: \[x^{n}+n ^{n-1}\Delta x+ O((\Delta x)^{2})\] 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.

    • one year ago
  3. bob7185
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    you 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.

    • one year ago
  4. kb9agt
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    I 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.

    • one year ago
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