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### Course: Integral Calculus>Unit 5

Lesson 11: Taylor and Maclaurin polynomials intro

# Worked example: Maclaurin polynomial

Finding the second degree Maclaurin polynomial of 1/√(x+1).

## Want to join the conversation?

• I still confuse about the function of X in Taylor formula.
This formula uses for estimate at point A, so what is the X use for?
• a is the point where you base the approximation, but you can vary x in order to get an approximation of the function itself using the polynomial.
For example, f(a) = P(a) (because you know the value of a), but f(x) ~ P(x) (because P(x) gets you an APPROXIMATION of f(x))
• What kinds of functions are exactly the same as their taylor series and what kinds of functions are not? In other words, what are the conditions for a function to equal its taylor infinite sum?
• Greetings!

If we could take the Taylor Series to an infinite number of terms, it would be exactly equal to the original function.

Since we aren't able to get an infinite number of terms, it is only an increasingly better approximation.

I hope this helps!
• Why would one want to approximate the function with a polynomial at a specific x if you could just fill in x in the original function?
• If you don't have the original function, and are, instead, using a graph, then a polynomial would be necessary. (Am I understanding your question right?)
(1 vote)
• When writing out the Taylor/Maclaurin series, what is the purpose of dividing each term by a factorial? How does that make the polynomial a better approximation as opposed to not dividing by the factorial? He probably answered this in the intro videos but I may have missed it or simply didn't understand.
• It comes from the fact that you differentiate several times to get eg the second or third derivative. If you know the value you want eg. the third derivative to have, you need to work out what original term would give you that value when differentiated 3 times.
Say you know at the point you are centering you the third derivative is a, then the original coefficient for the term in the polynomial to give that would be a/(3*2*1). Try for a Maclaurin series: a/(3*2*1) * x^3.
differentiate once: a/(2 * 1) * x^2
differentiate second time: ax
differentiate third time: a
• we take the principle root for f(0), because the function is defined for positive root , isnt it ?
• How can you find `limit` to 0 and to positive infinity using Taylor's polynomial formula?
like: `lim(x->0)` of [ `(a^x-6)/(x^3)`] ?
• Limit to 0 can be found by replacing functions with their Taylor polynomials to high enough degree to evaluate directly plugging in x=0. So (cos(x)-1) / x^2 would be approximated by (-x^2/2 +x^4/4! -...) / x^2 = - 1/2 + x^2-stuff = -1/2 as x->0.
(1 vote)
• what is [1-(-1)^n] as n goes to infinity
(1 vote)
• It doesn't converge, it oscillates between 2 and 0.
• Does it make sense to have a Taylor polynomial for a function at it's vertical asymptote line?
For Example: A Maclaurin polynomial for f(x)=1/x
(1 vote)
• No, the Taylor series is a series of successive polynomial approximations at a point. You can't approximate a function at point where it doesn't exist.