In mathematics, the Taylor series of a function is an infinite sum of terms that are expressed in terms of the function's derivatives at a single point. For most common functions, the function and the sum of its Taylor series are equal near this point. Taylor's series are named after Brook Taylor, who introduced them in 1715.
If zero is the point where the derivatives are considered, a Taylor series is also called a Maclaurin series, after Colin Maclaurin, who made extensive use of this special case of Taylor series in the 18th century.
The partial sum formed by the first n + 1 terms of a Taylor series is a polynomial of degree n that is called the nth Taylor polynomial of the function. Taylor polynomials are approximations of a function, which become generally better as n increases. Taylor's theorem gives quantitative estimates on the error introduced by the use of such approximations. If the Taylor series of a function is convergent, its sum is the limit of the infinite sequence of the Taylor polynomials. A function may differ from the sum of its Taylor series, even if its Taylor series is convergent. A function is analytic at a point x if it is equal to the sum of its Taylor series in some open interval (or open disk in the complex plane) containing x. This implies that the function is analytic at every point of the interval (or disk).
Hi ,
I have some difficulties to solve this problem. It is from my numerical methods class but the problem is about taylor series:
It is known that for 4 < x < 6, the absolute value of the m-th derivative of a certain function f(x) is bounded by m times the absolute value of the quadratic...
Hello, guys. I am studying the Taylor Theorem for functions of n variables and in one book I've found a proof based on the lemma that I am copying here. I am having some trouble in following its proof so I seek your kind assistance.
The lemma rests on two items: the definition of a function...
taylor differentition polynomials?
hi got a question here that involves this extremely difficult question anyone that can point me in the right direction on what to do will be most appreciated :)
Find Exactly the tayor polynomial of degree 4
f(x) = cos ( pi*x / 6 ) about x=-1
i know...
Hi, I have a question about Taylor series:
I know that for a function f(x), you can expand it about a point x=a, which is given by:
f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + ...
but I would like to do it for f(x+a) instead of f(x), and expand it about the very same point...
I need to find the first three terms of this series.
Am I correct in saying z^i = exp(i*Log(z)), then using the taylor series for e^z, giving me:
(i*Log(z)) - 1/2*(Log(z))^2 - i/6*(Log(z))^3 + 1/24*(Log(z))^4 + ...
I haven't worked it out, but this seems to mean that the coefficients...
How do i show that B_{x}(x+dx,y,z)-B_{x}(x,y,z)\approx \frac{\partial B_{x}(x,y,z)}{\partial x} dx
using a Taylor series to the first term. Using a Taylor series does B(x) = B(a) + B'(a)(x-a)? In that case what would B(x+dx) be and how can i obtain the desired result from this? Thanks in...
I was wondering if someone can give me some tips for finding the taylor series of functions. For example this was a test question we had:
Find the taylor series of f(x)=ln(x) about x=e
I know how to start it off but I get confused halfway through and can't seem to figure out what to do...
I understand what a linear approximation, and how it is derived using the point-slope formula:
f(x)\approx f(a)+f'(a)(x-a)
These are the first three terms of a Taylor series, so I was wondering how the rest was derived?
Thanks for your help.
Hi, I'm having trouble doing my work where I have to find the Taylor Approximation of function. My real work is the program this thing when the function, x, a, and ErrorBound is given. I don't knwo what to do with the ErrorBound to get n, where n is the number of terms. do i make any sense...
I just want to check my answer. The question asks for the Taylor polynomial of degree 6 for ln(1-x^2) for -1<x<1 with c=0.
I got tired after differentiating 6 times so I'm worried I made some mistakes along the way. The question also said: hint: evaluate the derivatives using the formula...
Show that Pn(x^2) is the 4n+2-nd Taylor polynomial of sin(x^2) by showing that
\lim_{n\rightarrow infinity} R2n+1(x^2) = 0.
note that Rn(x) represents the remainder
I'm stuck on this question, can anyone help me please?
Problem
Find the sum of the series
s(x) = \sum _{n=1} ^{\infty} \frac{1}{2^{n}} \tan \frac{x}{2^n}
Solution
If
s(x) = \sum _{n=1} ^{\infty} \frac{1}{2^{n}} \tan \frac{x}{2^n} = \frac{x}{3} + \frac{x^3}{45} + \frac{2x^5}{945} + \dotsb
\cot x = \frac{1}{x} - \frac{x}{3} -...
V = 2\pi \sigma(\sqrt{R^2+a^2}-R)
Show that for large R,
V \approx \frac{\pi a^2 \sigma}{R}
I figured if I could develop the MacLaurin serie with respect to an expression in R such that when R is very large, this expression is near zero, then the first 1 or 2 terms should be a fairly...
I am having difficulty understanding Taylor and MacLaurin series. I need someone to go through step by step and explain a problem from beginning to end. You could use the function f(x) = cos x. Also, could someone find the MacLaurin series of 1/(x^2 + 4) ? I just don't understand the basics of...
1. Let f(x) = (1+x)^{2/3}
(a) find the taylor polynomial T_2(x) of f expanded about a = 0.
i got 1 + (1/3)x - (1/9)x^{2}
For the rest, i have no idea how to do...any help would be greatly appreciated.
(b) For the givven f write the lagrange remainder formula for the error term...
In order to obtain with the aid of Mathematica, say, an 8-th degree Taylor polynomial of \sqrt{x} centered at 4 , I use the following command:
Normal[Series[Sqrt[x], {x, 4, 8}]]
and I get
\sqrt{x} \approx 2 + \frac{1}{2^2} \left( x - 4 \right) - \frac{1}{2^6} \left( x - 4 \right) ^2...
Consider the following:
f(x) = 1 + x + x^2 = 7 + 5 (x-2) + (x-2)^2
which is a Taylor series centered at 2. My question is: what is the radius of convergence? The answer in my book is R=\infty , but take a look at this:
f(x) = 7 + 5 (x-2) + (x-2)^2 = \sum _{n=0} ^{\infty} b_n (x-2)^n...
I am a bit confused about taylor approximation. Taylor around x_0 yields
f(x) = f(x_0) + f'(x_0)(x-x_0) + O(x^2)
which is the tangent of f in x_0, where
f'(x) = f'(x_0) + f''(x_0)(x-x_0) + O(x^2)
which adds up to
f(x) &=& f(x_0) + (f'(x_0) + f''(x_0)(x-x_0) +...
Help me out with this Taylor series problem:
The Taylor series for sin x about x = 0 is x-x^3/3!+x^5/5!-... If f is a function such that f '(x)=sin(x^2), then the coefficient of x^7 in the Taylor series for f(x) about x=0 is?
thanks
Taylor Series in x-a
Hi,
I've got a question about the use of dummy variables in Taylor Series.
We are asked to expand:
g(x) = xlnx
In terms of (x-2). So originally, I used a dummy variable approach to try and find an answer.
Let t = x-2, so x = t+2.
g(x) = (t+2)ln(2+t)...
This problem has been bugging me and I can't seem to figure it out:
y'' = e^y where y(0)= 0 and y'(0)= -1
I'm supposed to get the first 6 nonzero terms
I know the form is:
y(x) = y(0) + y'(0)x/1! + y''(0)x^2/2! + y'''(0)x^3/3! +...
and I know the first two terms are
y(x) = 0 -...
Let f be a function that has derivatives of all orders for all real numbers. Assume f(1)=3, f'(1)=-2, f"(1)=2, and f'''(1)=4
a. Write the second-degree Taylor polynomial for f about x=1 and use it to approximate f(0.7).
b. Write the third-degree Taylor polynomial for f about x=1 and use it...
Find the 4th term of the Taylor series centerd at x=1 for f(x)=ln(x+1)
f(x)=ln(1=x)
f'(x)=(1+x)^-1
f"(x)=(-1)[(1+x)^-2]
f"'(x)=(2)[(1+x)^-3]
f""(x)=-6[(1+x)^-4)]
Plug in 1:
.6931
.5
-.25
.25
-.375
What do I do next? (Also, is the 4th term the 4th term starting with f(x)? or the...
We were gievn a question in tutorial last week asking us to calculate the Taylor series of the function f(x,y) = e^(x^(2) + y^(2)) to second order in h and k about the point x=0, y=0
I've got the forumla here with all the h's and k's in it and have it written down, but it's actually how to...
basic taylor expansion...
Hi, could some one explain how i could use the taylor series to expand out:
f(x)= 1/sqrt(1-x^2)
Any help would be appreciated, thanks.
I'm having some problems expanding i^i, could anyone help? I know it becomes a real number somehow, and I'm familiar with the e^(i * pi) expansion, but is the i^i done in the same way?