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MIT 18.01 David Jerison talks about a symmetry shortcut in the first lecture. he says its basically switching what we call the axis, tring to find the x and y intercept, wouldnt this just get the same point just calling it something else? is there a name for this trick or something i could search to understand whats happening better
 one year ago
 one year ago
MIT 18.01 David Jerison talks about a symmetry shortcut in the first lecture. he says its basically switching what we call the axis, tring to find the x and y intercept, wouldnt this just get the same point just calling it something else? is there a name for this trick or something i could search to understand whats happening better
 one year ago
 one year ago

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hynsteinBest ResponseYou've already chosen the best response.0
he starts talking about it at 31 min in.
 one year ago

StaceyBest ResponseYou've already chosen the best response.0
Can you give a link to the lecture?
 one year ago

rosinaBest ResponseYou've already chosen the best response.1
At 35min he explains the reason. I think the function y=1/x is very special as y=1/x and x=1/y. The symmetry here means the computing process of yintercept is totally the same as for xintercept, so if you get the xintercept is 2x0, then yintercept must be 2y0. So he used the shortcut. But it doesn't mean the point is same as you can see y0=1/x0, the xintercept is 2x0, the yintercept is 2y0=2/x0. They're different points. Hope this answer your question. If it's very hard to understand, you can use the common method to compute the yintercept to understand the trick.
 one year ago

Data_GeneBest ResponseYou've already chosen the best response.0
The function is y = 1/x, so multiply both sides by x and we get xy = 1. Now divide both sides by y and we now have x = 1/y. So the x and they y are interchangeable. It is a special property of this function. That's also why he can say in the end that 2x0y0 = 2. From algebraic manipulation above we saw that xy = 1 and x and y can refer to any point on the function graph, including x0y0 then 2x0y0 must equal 2xy which is 2(1) hitch is 2. To see that we are not just renaming things, suppose the function was y = 2x. Dividing both sides by x gives y/x = 2. Now divide by y and you get 1/x = 2/y. Multiply by x^2 and you get x =( 2x^2)/y. Clearly x and y are not symmetrical here.
 one year ago
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