Calculus, all content (2017 edition)
Course: Calculus, all content (2017 edition) > Unit 7Lesson 13: Taylor & Maclaurin polynomials intro
- Taylor & Maclaurin polynomials intro (part 1)
- Taylor & Maclaurin polynomials intro (part 2)
- Worked example: Maclaurin polynomial
- Worked example: coefficient in Maclaurin polynomial
- Worked example: coefficient in Taylor polynomial
- Taylor & Maclaurin polynomials
- Taylor polynomial remainder (part 1)
- Taylor polynomial remainder (part 2)
- Worked example: estimating sin(0.4) using Lagrange error bound
- Worked example: estimating eˣ using Lagrange error bound
- Lagrange error bound
- Visualizing Taylor polynomial approximations
- Worked example: Taylor polynomial of derivative function
Worked example: coefficient in Maclaurin polynomial
Finding the coefficient of the x² term in a Maclaurin polynomial, given the formula for the value of any derivative at x=0.
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- in the fourth term you forget x^3(23 votes)
- You're right! In this problem, it doesn't matter, since he was just finding the coefficient to x&2, but there should have been a x^3 there. Nice catch :)(11 votes)
- While taking the root of 9, Sal said "it's positive 3".
Why does it have to be positive 3 only? If the answer was -(3/16) how will we know? -(3/16) seems like an equally logical answer to me.(3 votes)
- The formula which gives you the values of the nth derivative at x=0 really blew my mind, I don't know why but It never crossed my mind that such a thing could exist !
It is really cool, So I have two questions
The first how can I figure out such formulas, is there like some known tricks and methods to figure this out, is it easy should I try it myself or are these tricks and methods complex and I can't figure them out on my own ?
The Second is what kinda of function has this formula ? I want to know the original function that gives you the formula presented in the video, again is there a method and how hard is it , I can't think of anything except graphing the Taylor approximation and even after that the graph didn't seem that familiar, It grew very fast so I guessed factorials and exponential must be involved but that is it.
Thanks for taking the time to read this , I am sorry for making this a long question.(2 votes)
- In response to the first question: How can you find a generalized formula for the nth derivative of a function?
You can always represent the nth derivative of a function f(t) as d^n(f(t))/dt^n. Rewriting a general solution for this may not be trivial, but can sometimes be done.
Try taking the first few derivatives and seeing if you can find a pattern. For example, if f(t) = cos(t) you will find that there is a clear pattern in taking the derivatives (this is an easy example if you want to try it). If you can find a pattern, you should be able to write a formula expressing that pattern.
Often there will not be a clear pattern, and in this case it's best to just leave the nth derivative expressed as d^n(f(t))/dt^n.
In response to the second question: Given a formula for the nth derivative of a function, can the original function be found?
The answer is: sometimes yes and sometimes no.
Take the equation given in the video, for example. It's impossible to determine from this equation what g(0) is, since the equation for the nth derivative is only defined for n>=1.
The best you could do for a function like this is to plot a Taylor Series approximation. Note that you can't even plot a true Taylor Series approximation for this example since g(0) must be known, but cannot be determined.
If you continue studying math, you will eventually take a class on differential equations. This is a branch of mathematics that will offer some tools to solve problems like this.(3 votes)
- why do we use the principal root of 9(2 votes)
- At2:25Sal is evaluating the second derivative of g(x). Using the provided equation, this is sqrt(2+7)/2^3. This simplifies to sqrt(9)/2^3.
The "principal root" is another name for the square root. Since he is taking the square root of 9, he referred to the "principle root" of 9.(2 votes)
- 3/8 div 1/2 is 6/8=3/4
if it was mul then it will be 3/16(2 votes)
- I understand all of this, but usually I find all the derivatives first. But that was not done here. Instead it was just the general form g'(x), g''(x),... The derivative forms for sqrt(n+7)/n^3 was never expressed. That step was never executed.
What is my clue in the problem to know I didn't have to make a table of derived forms? Or is it that (n+7)^1/2/n^3 is already the derived function?
Because my first instinct was to start listing all the derivations of (n+7)^1/2/n^3(1 vote)
- Almost everytime you write a Taylor series, you would be required to find the derivatives. But, you don't have to here. Why? Because you can't.
Observe the question tackled in the video. You don't have a function to take a derivative for. What's given is the general formula for the derivative itself. So, if you wanted the first derivative, plug in n=1. If you wanted the 59th derivative, plug in n=59. And so on. You can't take the derivatives of the given function as n there is a constant. So, taking the first derivative itself would end up giving 0.
So, your clue is essentially this: See if you have a function to even take the derivative of(1 vote)
- Is it very common to get question like this, say on exams or tests? I cannot really tell what is being asked in the example.(1 vote)
- let's say I want to find the complete maclaurin expansion of g(x) for the given question. Is the given info enough, or do we need a little more to solve it?
I have a feeling we need g(0) provided to us, but I'm not sure whether there's some clever way of finding that from the given question.(1 vote)
- You could add any value of g(0) that you please, and still have a function g(x) that meets the criteria given in the question - nothing more can be established about g(x). However the Maclaurin expansion is complete after adding arbitrary g(0); it converges for x in [-1,+1].(1 vote)
- Why do you not have to take the second derivative to find the solution to this problem?(1 vote)
- The function in the question
g^n(0)=sqrt(n + 7)/n^3actually gives you the derivatives for the function g at x = 0; whatever you plug in for n will result in that order of derivative for the function g at x = 0. If you plugged in 1 for n, you would get g(0), if you plugged in 3, you would get g^3(0), etc. Normally, you would have to manually take the derivative twice, but in this case, you would just have to plug in 2, like in the video.(0 votes)
- Why isn't there an x in the function
g^n(0)=sqrt(n + 7)/n^3
How is the expression a function of x in the first place?(1 vote)
- In the proposition of the exercise, it says that g at x = 0 is given by
g^n(0)=sqrt(n + 7)/n^3. This is indeed a function of x, just evaluated at x = 0. Maybe you should watch the previous videos on the Taylor & Maclaurin polynomials to understand whats going on.(1 vote)
- [Instructor] Nth derivative of g at x equals zero is given by, so the nth derivative of g evaluated x equals zero is equal to square root of n plus seven over n to the third for n is greater than or equal to one. What is the coefficient for the term containing x squared in the Maclaurin series of g? Let's just think about the Maclaurin series for g. If I were to have my function g of x. The Maclaurin series, I could say approximately equal to especially if I'm not gonna list out all of the terms, is going to be equal to, well it's going to be equal to g of zero plus g prime of zero times x plus g prime prime of zero divided by, I could say two factorial but that's just two, times x squared, and that's about as far as we go. Because we just have to think about what is the coefficient for the term containing x squared. If they said what's the coefficient for the term containing x to the third, I would keep going. I'd go g prime. I would take the third derivative evaluated at zero over three factorial. I could do this as a factorial too, but that just evaluates to two. I could do this as one factorial. I could do this as zero factorial just so you see it's a consistent idea here. I could, of course, keep on going, but we just care about, they're just asking us what is the coefficient for the term containing x squared? They just want us to figure out this. What is this thing right over here? To know that, we need to figure out what is the second derivative of g evaluated x equals zero? Well they tell us that over here. It's a little bit unconventional where they give us a formula, a general formula for any derivative evaluated x equals zero, but that's what they're telling us here. So in this case, the n isn't zero. The n is the derivative we're taking, and that's going to be our second derivative, so this is, so if I wanted to figure out g. If I am figuring out the second derivative, and I could write it like that evaluated zero, or I could write it like this just so the notation is consistent. I could write it like that. The second derivative evaluated x equals zero is going to be equal to, well our n is two, so this is going to be the square root of two plus seven over two to the third power. Two plus seven is nine, take the principle root of that. It's gonna give us positive three over two to the third which is eight, so this part right over here is 3/8, so the whole coefficient is going to be 3/8, that's this numerator, divided by, divided by two. Which of course is equal to 3/16, and we're done. They didn't want us to figure out a couple of terms of this which we could call the Maclaurin polynomial and nth degree Maclaurin polynomial. They didn't want us to find the entire, you know, keep going with this series. They just wanted to find one coefficient right here. The coefficient on the second degree term which we just figured out is 3/16.