Find the Taylor polynomials of order 0,1,20,1,2, and 33 for f(x)=ln(x)f\left(x\right)=\ln\left(x\right) centered at x=1x=1.

p0(x)=0,p1(x)=x−1,p2(x)=(x−1)−12(x−1)2,p3(x)=(x−1)−12(x−1)2+13(x−1)3p_0\left(x\right)=0,$$p_1$$\left(x\right)=x-1,$$p_2$$\left(x\right)=\left(x-1\right)-\frac12\left(x-1\$$right)^2$$,$$p_3$$\left(x\right)=\left(x-1\right)-\frac12\left(x-1\$$right)^2$$+\frac13\left(x-1\$$right)^3$$

Find the Taylor polynomials of order 0 , 1 , 2 0,1,2 , and 3 3 for f ( x ) = ln ( x ) f\left(x\right)=\ln\left(x\right) centered at x = 1 x=1 .

we're asked to find the Taylor polynomials of order zero, one, two, and three for the function f of x equals the natural log of x centered at x equals one. Now in order to find these Taylor polynomials, we're gonna need the first couple of derivatives of our function. So that's the first, second, and third derivatives. So let's start there. We want to find the derivatives of our function f and then specifically at x equals one, that center that's given to us in the problem. So starting with just our function itself, that's the natural log of x. We can evaluate that quickly for x equals one. The natural log of one is zero. Then taking our first derivative here f prime is going to give us a one over x, then evaluating that at one. One over one gives us one. Then as we take the derivative again here, f double prime, that's going to end up giving us negative one over x squared. So evaluating that at one, negative one over one squared is negative one. Then finally here f triple prime, that third derivative taking the derivative one more time, is going to give us positive two over x cubed. Evaluating that at one here will be two over one cubed, which is simply two. So now that we have evaluated our derivatives for those first couple of values, we can go ahead and start finding those Taylor polynomials. So for that zeroth degree polynomial, so that's p zero of x, this is just going to be equal to that first term. So f of a, that's just our function evaluated at the center, which we found over here to be zero. So that first polynomial is just zero. And for my second polynomial, p sub one of x, this is going to be this first term, or rather my zeroth term, and my first term added together. So this is going to be zero, that previous term, plus the first derivative of my function evaluated at a, which we found over here as one. So that's one times x minus one. So this simplifies here to just x minus one, and that gives me my first degree Taylor polynomial. Then we want to find p sub two of x, so that's gonna be f of a plus this next term up plus this next term up as well. So I can go ahead and add those past two terms together. That's zero plus x minus one. Then for our next term, this is going to be the second derivative of our function evaluated at a. So that's negative one. Negative one over two factorial times x minus one squared. Doing a bit of simplification here, this ends up giving us x minus one minus one half times x minus one squared. So that gives us our second degree polynomial. Finally, the last one we need to find here, that third degree Taylor polynomial, that's everything we've added together so far plus one additional term. So I already know that those past terms all added together are going to be x minus one minus one half times x minus one squared. So we just need to add one more term here. So we're adding that together with our third derivative of our function evaluated at a. So we found that to be two over three factorial and then times x minus one cubed. So simplifying here, x minus one minus one half times x minus one squared, plus two over three factorial simplifies to one third x minus one cubed. So these are first couple of Taylor polynomials. So p sub zero of x is zero, then our next one up is x minus one. Then that second degree Taylor polynomial is right here, x minus one minus one half times x minus one squared, and that third degree Taylor polynomial is right here, x minus one minus one half times x minus one squared plus one third times x minus one cubed. And that's our final answer. Let us know if you have any questions and I'll see you in the next one.

Find the Taylor polynomials of order 0,1,20,1,2, and 33 for f(x)=ln⁡(x)f\left(x\right)=\ln\left(x\right) centered at x=1x=1.

p0(x)=0,p1(x)=x−1,p2(x)=(x−1)−12(x−1)2,p3(x)=(x−1)−12(x−1)2+13(x−1)3p_0\left(x\right)=0,$$p_1$$\left(x\right)=x-1,$$p_2$$\left(x\right)=\left(x-1\right)-\frac12\left(x-1\$$right)^2$$,$$p_3$$\left(x\right)=\left(x-1\right)-\frac12\left(x-1\$$right)^2$$+\frac13\left(x-1\$$right)^3$$

p0(x)=0,p1(x)=x−1,p2(x)=(x−1)−(x−1)2,p3(x)=(x−1)−12(x−1)2+13(x−1)3p_0\left(x\right)=0,$$p_1$$\left(x\right)=x-1,$$p_2$$\left(x\right)=\left(x-1\right)-\left(x-1\$$right)^2$$,$$p_3$$\left(x\right)=\left(x-1\right)-\frac12\left(x-1\$$right)^2$$+\frac13\left(x-1\$$right)^3$$

p0(x)=1,p1(x)=x−1,p2(x)=(x−1)−12(x−1)2,p3(x)=(x−1)−12(x−1)2+13(x−1)3p_0\left(x\right)=1,$$p_1$$\left(x\right)=x-1,$$p_2$$\left(x\right)=\left(x-1\right)-\frac12\left(x-1\$$right)^2$$,$$p_3$$\left(x\right)=\left(x-1\right)-\frac12\left(x-1\$$right)^2$$+\frac13\left(x-1\$$right)^3$$

p0(x)=0,p1(x)=1,p2(x)=(x−1),p3(x)=(x−1)2+13(x−1)3p_0\left(x\right)=0,$$p_1$$\left(x\right)=1,$$p_2$$\left(x\right)=\left(x-1\right),$$p_3$$\left(x\right)=\left(x-1\$$right)^2$$+\frac13\left(x-1\$$right)^3$$

15. Power Series

Taylor Series & Taylor Polynomials

Calculus

15. Power Series

Taylor Series & Taylor Polynomials

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Multiple Choice

Find the Taylor polynomials of order $0 comma 1 comma 2$ , and $3$ for $f (x) = \ln (x)$ centered at $x equals 1$ .

$p_{0} (x) = 0, p_{1} (x) = x - 1, p_{2} (x) = (x - 1) - {(x - 1)}^{2}, p_{3} (x) = (x - 1) - \frac{1}{2} {(x - 1)}^{2} + \frac{1}{3} {(x - 1)}^{3}$

$p_{0} (x) = 1, p_{1} (x) = x - 1, p_{2} (x) = (x - 1) - \frac{1}{2} {(x - 1)}^{2}, p_{3} (x) = (x - 1) - \frac{1}{2} {(x - 1)}^{2} + \frac{1}{3} {(x - 1)}^{3}$

$p_{0} (x) = 0, p_{1} (x) = x - 1, p_{2} (x) = (x - 1) - \frac{1}{2} {(x - 1)}^{2}, p_{3} (x) = (x - 1) - \frac{1}{2} {(x - 1)}^{2} + \frac{1}{3} {(x - 1)}^{3}$

$p_{0} (x) = 0, p_{1} (x) = 1, p_{2} (x) = (x - 1), p_{3} (x) = {(x - 1)}^{2} + \frac{1}{3} {(x - 1)}^{3}$

검증된 단계별 안내

Step 1: Recall the formula for the Taylor polynomial of a function f(x) centered at x=a. The nth-order Taylor polynomial is given by: $ p_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f^{(3)}(a)}{3!}(x-a)^3 + \dots + \frac{f^{(n)}(a)}{n!}(x-a)^n $.

Step 2: For the function $ f(x) = \ln(x) $, calculate the derivatives of $ f(x) $ at $ x=1 $. The first derivative is $ f'(x) = \frac{1}{x} $, the second derivative is $ f''(x) = -\frac{1}{x^2} $, the third derivative is $ f^{(3)}(x) = \frac{2}{x^3} $, and so on.

Step 3: Evaluate the derivatives at $ x=1 $. For $ f(x) = \ln(x) $: $ f(1) = \ln(1) = 0 $, $ f'(1) = \frac{1}{1} = 1 $, $ f''(1) = -\frac{1}{1^2} = -1 $, $ f^{(3)}(1) = \frac{2}{1^3} = 2 $, and so on.

Step 4: Substitute these values into the Taylor polynomial formula for different orders. For $ p_0(x) $, use $ f(1) $. For $ p_1(x) $, use $ f(1) $ and $ f'(1) $. For $ p_2(x) $, include $ f(1) $, $ f'(1) $, and $ f''(1) $. For $ p_3(x) $, include up to $ f^{(3)}(1) $.

Step 5: Write the Taylor polynomials explicitly. For example: $ p_0(x) = 0 $, $ p_1(x) = (x-1) $, $ p_2(x) = (x-1) - (x-1)^2 $, $ p_3(x) = (x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3 $. Continue this process for higher-order polynomials as needed.

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Find the Taylor polynomials of order 0,1,20,1,2, and 33 for f(x)=ln(x)f\(\left\)(x\(\right\))=\(\ln\]\left\)(x\(\right\)) centered at x=1x=1.

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