Taylor Polynomials & Maclaurin Polynomials With Approximations

[@TheOrganicChemistryTutor]

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0 - 6:28 Taylor 6:29 Maclaurin

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[@georgesadler7830]

Professor Organic Chemistry Tutor, thank you for a detail analysis of the Taylor and Maclurin Polynomials with Approximations in Calculus Two. Taylor/Maclurin Polynomials are common/well known topics in Science and Engineering. Professor Organic Chemistry Tutor, you made a typing error at the 1:04 minute mark in the video. For the nth term in the Tayor Polynomial, you have f raised to the n (x - c) raised to the fourth divided by four factorial, instead of f raised to the n (x - c) raised to the n divided by n factorial. Professor, please correct this small error in the video.

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[@jaydenrussell7491]

2:35 I thought a negative value cannot be evaluated inside a natural log function. How did he get ln(-1) = -1 when negative number are outside the domain of ln ?

[@someone1459]

he is evaluating c = 1 at the second derivative of f(x) which is -1/x^2 , not at f(x) = ln(x). when we plug c=1 into that second derivative we get -1 as the answer. he did not take ln (-1), because as you said that would be undefined!

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[@curtain1810]

If its not stated in the problem, how so you know where is the function centered?

[@destinyobamwonyi]

I think the first formular is supposed to end with f^n(c)(x-c)^n ÷ n!

[@Uygʻun-q2c]

thank you

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[@sfcxwedsx]

does the value of the degree represent the number of accurate decimals ?

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[@kuroshitsuji2580]

Our professor gave us an assignment. And it doesn't have any polynomials on it. Just the approximate value. But we have to find its estimate value? I'm confused, is that possible.

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[@meep7895]

what does the center mean conceptually?

[@abdurrahmankursatozkan9172]

Im watching the calculus 2 playlist and im a bit confused to be honest. How did we come to this from divergent tests :d

[@janieli]

when we increase the terms, why do we need to add (x-a)^n (where n is equal to the number of derivatives)? I get that we need to substitute x and a to find approximation, but why do the exponents increase?

[@adnanmohammed4354]

As he said, the number of terms indicates more accuracy. If you have a function and take more derivatives into account, you'll end up with a better approximation. It might be difficult to put it into words. Check out 3Blue1Brown's Taylor polynomials video. His animations were really helpful.

[@internetsurfermr80]

Derivatives of 2/x^3 is not 6/x^4?

[@xkilla911]

the video put -6/x^4 because 2/x^3 is basically 2x^(-3). and so by doing simple exponential differentiation... (-3)*2x^(-3-1) = -6x^(-4) which is = -6/x^4

[@Iron_Age]

you're right, its negative

[@zacharysantana3861]

He said -6/x^4

[@jaro551]

I have an exam in 2 hours. I'm fucked...

[@katrinawalker1702]

Why didn't you subtract 1 when subbing for x=1.1 in the P(4) equation?

[@queenpenguin5685]

he did, it became .1 where the (x-1) things were

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[@krislegends]

For ln(1.1), why put 0.1 instead of 1.1?

[@babygravy7]

1.1 - 1 = 0.1. He just didn't show the work for it because it wasn't needed

[@krislegends]

@@babygravy7 Is 1 the center?

[@ShadowReaper-pu2hx]

Yes, one is the center (the problem says it’s centered at one).

[@martinhazard2025]

wrong, you must use the Landau's symbols

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[@nraishanzr3333]

if 3! = 3 x 2 x 1 why don't we just write 6? Same with 4! why isn't just written as 24

[@g.molina10]

You could. He just didn't. They are the same thing though

[@vez3834]

It's for making the general formula easier to write. So rather than listing all factorials 1, 2, 6, 24, 120... we write n! where n is the degree of our polynomial. So then when plugging something into the equation, it's nice to see where it came from.

[@qualquan]

just another cook book demonstration in a quiet no nonsense manner sadly no discussion of the beautiful underlying philosophy of the method

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can you share a link?

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thank you! T-T

[@katrinawalker1702]

I am an idiot. 1.1-1=0.1

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