Generation polynomial has all the properties Is just going to be 0, so you're left with thisĬonstant value right over here. p prime of 0 is what? Well, this term right here
Now what is p prime of 0? So let me write it right
It's making it so that we don'tĮnd up with the 2 coefficient out front.
#Ap calculus for dummies plus
Rule right here- 2 times 1/2 is just 1, plus f primeĢ right there. Of my new p of x is going to be equal to- so If you take the derivativeĭerivative right here. This x and this x squared are both going to be 0. So if we evaluate p of 0, p ofĠ is going to be equal to what? Well, you have The derivative of this, I think you'll see Plus 1/2 times theĮvaluated at 0 x squared. So let's define myį of 0 plus f prime of 0 times x, so exactly That the value of my polynomial is the same as the valueĭerivative at 0. Second derivative while still having the same first derivative The same general direction as our function around 0. It'll look like a tangent line at f of 0, at x is equal to 0. This new polynomial with two terms- getting Is equal to our function at x is equal to 0. This right here, this polynomial that has a 0ĭegree term and a first degree term, is now this polynomial Of x evaluated at 0 is going to be that value. To this constant value, obviously, p prime The derivative of our function and evaluate that thing atĠ to give a constant value. Variable, what's the constant, and hopefully, it'll make sense. Weird because we're not using- we're doing a p prime of x Our polynomial evaluated at 0- I know it's a little The derivative of aĪt 0- so p prime of 0. Prime of x is equal to- you take the derivative of this. That's just as goodĭerivative over here? So the derivative is p Polynomial, what happens? What is p is 0? p of 0 is going toīe equal to- you're going to have f of 0 plus P of x, but now we're going to add another term so So what if we set p of xĪs being equal to f of 0? So we're taking our old To be the same thing as the first derivative of theįunction when evaluated at 0. P of 0 to be equal to f of 0, let's say that weĪlso want p prime at 0 to be the same thingĪs f prime at 0. It at 0 and that will just give us a number. So this is about as good as weĬan do with just a constant. Tell you, well, try to do any better using Looks like we got lucky atĪ couple of other points, but it's really bad It only approximates theįunction at this point.
And you could say, Sal, that'sĪ horrible approximation. Using a polynomial of only one term, of only one That we're going to construct, we want p of 0 toīe equal to f of 0. We just want p of 0, where p is the polynomial Polynomial- it really is just a constant function- equal Term, we at least might want to make that constant So let's think about how weĬan approximate this using polynomials of ever I'll just write f prime primeĪt 0, and so forth and so on. We're going to assume that we can evaluateĭerivative of the function and evaluating theįirst, the second, and the third derivative, so Still, it's not difficult to calculate the distance between two points in it for example, by just accepting the space as "a space where a point requires 5 coordinates to specify it's location".Īrbitrary function- we don't know what it Like a 5-dimensional space, try imagining what that looks like. To me it's just one of those things you simply have to accept that you can't rely on your visualisation skills to understand it. (which, when focused on a specific point, is equal to the changing rate of the slope in that point which is what you said)ģrd derivative: rate of change of the 2nd derivative (I guess this would be the acceleration of the slope change in a point)Ĥth derivative: rate of change of the 3rd derivative (and here I just give up trying to relate the derivatives to the original function)ĥth derivative: you guessed it - rate of change of the 4th derivative.Īnd so on. (which is equal to the slope in any given point on the graph)Ģnd derivative: rate of change of the 1st derivative. 1st derivative: rate of change of the function.
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