Find the Maclaurin polynomials of order 0,1,20,1,2, and 33 for f(x)=sinxf\left(x\right)=\sin x

po(x)=0,p1(x)=x,p2(x)=x,p3(x)=x−16x3p_{o}(x)=0,$$p_1(x)=x$$,$$p_2(x)=x$$,$$p_3(x)=x$$-\$$frac16x^3$$

Find the Maclaurin polynomials of order 0 , 1 , 2 0,1,2 , and 3 3 for f ( x ) = sin x f\left(x\right)=\sin x

we're asked to find the Maclaurin polynomials of order zero, one, two, and three for the function f of x equals sine of x. Now remember, Maclaurin polynomials are just like Taylor polynomials, but centered at x equals zero. Now in order to find these first couple of polynomials, we need to determine the derivative of our function or the first couple of derivatives of our function, so let's start there. Derivative of our function f and then evaluate it at the center, which is x equals zero. So starting with our function itself, f equals the sine of x, we can go ahead and evaluate this at zero. The sine of zero is going to be zero. Then taking that first derivative f prime, that's going to be cosine of x. Then the cosine of zero is equal to one. Then our second derivative here, f double prime, this is going to be negative sine of x. Negative sine of zero is still zero. Then one more derivative we need to take here, that third derivative is going to be negative cosine of x. So negative cosine of zero is going to give us negative one. So So now that we have these first couple of derivatives, we can go ahead and find these Taylor polynomials or Maclaurin polynomials. So starting with the one with degree zero p sub zero of x, we can go ahead and just get that first term that's just our function evaluated at the center, which we found earlier to be zero. So that zero degree Maclaurin polynomial is just zero. Then next, we have p sub one of x. So that's our first term there, our zeroth term. So that was just zero plus this next term up f prime of a times x minus a. Now we know a is zero, so this is just going to be f prime of zero times x. So f prime of zero we calculated earlier, it's just one. So this is one times x. So this ultimately just gives us here x. Now for my next polynomial of degree two, we add together those first couple of terms we already found, zero plus x and then plus our next term up here for n equals two, that second derivative evaluated at the center, which was zero over two factorial times x squared. Now zero is still zero. Zero is multiplying everything here, so this term is also zero. So our second degree polynomial is actually still just x. Then for our third degree here, p sub three of x, we add together all of those previous terms, zero plus x, and our next term was still zero, so plus zero. And then plus that next or that third derivative evaluated at zero, which was negative one over three factorial times x cubed. Now those zeros will go away, and this leaves us with our third degree polynomial just being x minus one over three factorial, so that's going to be one six x cubed. So that is our third degree polynomial, and now we have all of them. That zero degree polynomial, p sub zero is just zero, p sub one is x, p sub two is still just x, and p sub three is that x minus one sixth x cubed. Now if you have any questions, feel free to let us know in the comments, and I'll see you in the next one.

Find the Maclaurin polynomials of order 0,1,20,1,2, and 33 for f(x)=sin⁡xf\left(x\right)=\sin x

po(x)=0,p1(x)=x,p2(x)=x,p3(x)=x−16x3p_{o}(x)=0,$$p_1(x)=x$$,$$p_2(x)=x$$,$$p_3(x)=x$$-\$$frac16x^3$$

po(x)=0,p1(x)=0,p2(x)=x,p3(x)=x−16x3p_{o}(x)=0,$$p_1(x)=0$$,$$p_2(x)=x$$,$$p_3(x)=x$$-\$$frac16x^3$$

po(x)=0,p1(x)=x,p2(x)=x,p3(x)=x−16x2p_{o}(x)=0,$$p_1(x)=x$$,$$p_2(x)=x$$,$$p_3(x)=x$$-\$$frac16x^2$$

po(x)=0,p1(x)=x,p2(x)=x,p3(x)=x−13x2p_{o}(x)=0,$$p_1(x)=x$$,$$p_2(x)=x$$,$$p_3(x)=x$$-\$$frac13x^2$$

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Multiple Choice

Find the Maclaurin polynomials of order $0 comma 1 comma 2$ , and $3$ for $f (x) = \sin x$

$p_{o} (x) = 0, p_{1} (x) = x, p_{2} (x) = x, p_{3} (x) = x - \frac{1}{6} x^{3}$

$p_{o} (x) = 0, p_{1} (x) = 0, p_{2} (x) = x, p_{3} (x) = x - \frac{1}{6} x^{3}$

$p_{o} (x) = 0, p_{1} (x) = x, p_{2} (x) = x, p_{3} (x) = x - \frac{1}{6} x^{2}$

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

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Verified step by step guidance

Step 1: Recall that the Maclaurin series is a special case of the Taylor series centered at x = 0. The general formula for the nth Maclaurin polynomial is given by: $ p_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(0)}{k!}x^k $, where $ f^{(k)}(0) $ represents the kth derivative of $ f(x) $ evaluated at x = 0.

Step 2: Start by calculating the derivatives of $ f(x) = \sin(x) $. The first few derivatives are: $ f'(x) = \cos(x) $, $ f''(x) = -\sin(x) $, $ f'''(x) = -\cos(x) $, and $ f^{(4)}(x) = \sin(x) $. Notice the derivatives repeat in a cycle every four terms.

Step 3: Evaluate each derivative at x = 0. For $ f(x) = \sin(x) $, $ f(0) = 0 $. For $ f'(x) = \cos(x) $, $ f'(0) = 1 $. For $ f''(x) = -\sin(x) $, $ f''(0) = 0 $. For $ f'''(x) = -\cos(x) $, $ f'''(0) = -1 $. For $ f^{(4)}(x) = \sin(x) $, $ f^{(4)}(0) = 0 $.

Step 4: Use the formula for the Maclaurin polynomial to construct the polynomials of various orders. For order 0, $ p_0(x) = f(0) = 0 $. For order 1, $ p_1(x) = f(0) + f'(0)x = x $. For order 2, $ p_2(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 = x $. For order 3, $ p_3(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 = x - \frac{1}{6}x^3 $.

Step 5: Verify the results for each polynomial by substituting the derivatives and simplifying. The Maclaurin polynomials for $ f(x) = \sin(x) $ are: $ p_0(x) = 0 $, $ p_1(x) = x $, $ p_2(x) = x $, $ p_3(x) = x - \frac{1}{6}x^3 $.