Find the interval of convergence for the Taylor series for f(x)=sinxf\left(x\right)=\sin x centered at x=π2x=\frac{\pi}{2}.

( − ∞ , ∞ ) \left(-\infty,\infty\right)

Find the interval of convergence for the Taylor series for f ( x ) = sin x f\left(x\right)=\sin x centered at x = π 2 x=\frac{\pi}{2} .

we're asked to find the interval of convergence for the Taylor series of the function f of x equals sine of x centered at x equals pi over two. And in order to do that, we first need to determine what our Taylor series is, so let's start there. We're going to be looking here at the sum from n equals zero to infinity of the nth derivative of our function at pi over two because that's our center divided by n factorial times x minus pi over two to the power of n. Now in order to look at the first couple of terms here and see what pattern emerges, we need to look at the derivatives of our function and then also evaluate them at pi over two, that center. So starting with our original function, that's just f of x, that is sine of x. Now if we evaluate sine at pi over two, sine of pi over two, it is going to be positive one, so that's our first value. Then looking at that first derivative here, the derivative of sine of x is cosine of x, so So we can go ahead and evaluate cosine at pi over two, that's going to give us zero. Then we take the derivative again here, f double prime, that second derivative is going to be negative sine of x as we differentiate that positive cosine. So So this gives us negative sine of pi over two, which will give us here negative one. Then for that third derivative f triple prime, this is going to be the derivative of negative sine of x, which will give us negative cosine of x. Now negative cosine of pi over two is still going to be zero. And we'll look at one final derivative here, that fourth derivative, and differentiating that negative cosine is going to get us right back to positive sine. So we can event again evaluate this at pi over two. Sine of pi over two will give us positive one. So now that we've already had our derivatives evaluated at that center, we can go ahead and look at the first couple of terms of our series here or of our Taylor series. So starting for n equals zero, that's gonna be our function evaluated at its center. So that's one. So that's one over zero factorial times x minus pi over two to the power of zero. Then for that next term, we're going to take that zero, zero divided by one factorial times x minus pi over two to the power of one. Then for our next term here, we have that negative one, Negative one over two factorial times x minus pi over two squared. Then for our next term we again have zero, so that's gonna be zero over three factorial times x minus pi over two cubed. One final term to look at here, we're taking that one dividing it by four factorial times x minus pi over two to the power of four. Then, of course, this continues on infinitely, but we just need the first couple of terms to really see a pattern emerge. Emerge. Now we wanna do some simplification. So looking at that first term, one over zero factorial is going to be one, and then x minus pi over two raised to that power of zero is also still one. So our first term just becomes one. Then here we have this zero multiplying everything, so that whole term cancels out to zero. And I can look at that next term. This becomes minus one half times x minus pi over two squared. Then for our next term we again have a zero, so this entire term will cancel out to zero. And we just need to look at this final term here one over four factorial is going to give us one over 24. So one over 24 times x minus pi over two to the power of four, and of course we know this continues on. So now we wanna write this in summation notation. So this is gonna be the sum from n equals zero to infinity. Now the first thing I notice about this series is that the sign is alternating. It starts out with a positive one, then I have negative, and then I have another positive. So in order to account for the series being alternating, I need to take negative one and raise it to a particular power. Now because we're starting from n equals zero and that first term is positive, I just wanna raise this negative one to the power of n. I wouldn't wanna raise it to the power of n plus one because that would start my series with a negative. So now that I've accounted for that alternating piece of my series, I need to look at what else is going on. I can see that in my constants, I have one, one half, and then one over 24. So that's one, then one over two factorial, then one over four factorial. So in order to account for that, I need to take two n and two n factorial. So two n factorial, if we're starting from n equals zero, two times zero factorial is zero. Zero factorial is one that matches up with that first term. Then for n equals one, that gives me my next term. Two times one is just two. Two factorial is two, just like I have there. Then for n equals two, two times two is four. Four factorial is 24, which is exactly what I have there. So this accounts for everything going on with my constants and now I just need to account for that x piece. So this is going to be x minus pi over two. And what power should I be raising this to? Well because I have the same thing happening in this power that I have happening in my denominator here, I want to go ahead and raise this to the power of two n. So for n equals zero that would give me just one, which is exactly what I have here. Then for n equals one, two times one gives me two, so that's that squared term. And then for n equals two, two times two is four. That gives me that power of four. So this is my summation notation. This is my Taylor series for my function sine of x centered at pi equals two. But that's not what was asked of us here. Right? We were asked to find the interval of convergence for this Taylor series. So now that we've found our Taylor series, we can now apply convergence tests. So the first thing we wanna do here is apply that ratio test. Remember our ratio test, we take n plus one, plug it in for every instance of n, and divide it by our original series. We'll be taking the limit as n approaches infinity. So setting this up, this is going to be the limit as n approaches infinity. And since we're taking the absolute value, we don't really have to worry about the alternating piece of our series, so we're just gonna ignore that here. So I have x minus pi over two, and I'm raising that to the power of two times n plus one, plugging in n plus one where I originally had n. Then this is over two times n plus one, and all of that has that factorial. Then I'm dividing by my original series. Since I'm dividing by a fraction, that's the same thing as multiplying by its reciprocal. So that's going to be two n factorial up top, just flipping my original series, times x minus pi over two to the power of two n. Now setting up our inequality here, we're checking that this is less than one to determine convergence. Now we wanna do some simplification here, some rewriting based on exponential and factorial properties so so that we can get some stuff to cancel. So rewriting this, the limit as n approaches infinity in my numerator there, if I distribute that two, this becomes two n plus two. So I can separate that out using my exponential properties as x minus pi over two to the exponential properties as x minus pi over two to the power of two n times x minus pi over two squared. Now in my denominator here, two times n plus one, and then that is factorial, I'm going to go ahead and come off to the side here and show you what actually is happening. So if I have two times n plus one and all of that we're taking the factorial, that's the same thing as two n plus two factorial. Now because this is a little bit different than most factorial simplifications that we've seen, I wanna make sure and write this out fully. So if I have two n plus two, and then I multiply by the next term down, that's going to be two n plus one. And then that's going to multiply by two n factorial. So you can see that all of these terms just differ by a value of one. Now we want to get it down to this form specifically because we want it to cancel with that two n factorial. So make sure to put your parentheses appropriately like I now have here, and this will now allow this to cancel. So now in my denominator, I have two n plus two times two n plus one times two N factorial, then multiplying this by two N factorial over X minus PI over two to the power of two N. Now we can see a lot of stuff is going to cancel this X minus PI over two to the power of two N as well as this two n factorial. So let's see what we're left with here. The limit as n approaches infinity, the only thing left in my numerator is that x minus pi over two squared. Then in my denominator, I have two n plus two times two n plus one. Now because the only instance of n here is in my denominator, as n goes off into infinity, this denominator is going to blow up to infinity as well. As our denominator blows up to infinity, this entire value gets closer and closer to zero. So that leaves me here with the inequality that zero is less than one. Now this is a true statement, and it is always going to be a true statement. So because of that, that tells us that our series actually converges everywhere, and our radius of convergence is infinity. So with a radius of convergence of infinity, that tells us that our series converges on the interval from negative infinity to positive infinity. So this is where our Taylor series for the function sine of x centered at pi equals two converges. It converges everywhere. Let us know if you have any questions, and I'll see you in the next one.

Find the interval of convergence for the Taylor series for f ( x ) = sin ⁡ x f\left(x\right)=\sin x centered at x = π 2 x=\frac{\pi}{2} .

( − ∞ , ∞ ) \left(-\infty,\infty\right)

( − π 2 , 3 π 2 ) \left(-\frac{\pi}{2},\frac{3\pi}{2}\right)

[ − π 2 , 3 π 2 ] \left\lbrack-\frac{\pi}{2},\frac{3\pi}{2}\right\rbrack

( − π 2 , π 2 ) \left(-\frac{\pi}{2},\frac{\pi}{2}\right)

15. Power Series

Taylor Series & Taylor Polynomials

Calculus

15. Power Series

Taylor Series & Taylor Polynomials

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Multiple Choice

Find the interval of convergence for the Taylor series for $f (x) = \sin x$ centered at $x = \frac{π}{2}$ .

$(- \frac{π}{2}, \frac{3 π}{2})$

$[- \frac{π}{2}, \frac{3 π}{2}]$

$(- \frac{π}{2}, \frac{π}{2})$

$(- \infty, \infty)$

Verified step by step guidance

Step 1: Recall the formula for the Taylor series expansion of a function f(x) centered at x = c. The general form is: f(x) = Σ (f^(n)(c) / n!) * (x - c)^n, where n ranges from 0 to infinity, and f^(n)(c) represents the nth derivative of f evaluated at c.

Step 2: For f(x) = sin(x), calculate the derivatives of sin(x). The derivatives follow a cyclic pattern: f'(x) = cos(x), f''(x) = -sin(x), f'''(x) = -cos(x), and f''''(x) = sin(x). This pattern repeats every four derivatives.

Step 3: Substitute x = π/2 into the derivatives to find the coefficients of the Taylor series. For example, sin(π/2) = 1, cos(π/2) = 0, -sin(π/2) = -1, and -cos(π/2) = 0. These values will determine the terms of the series.

Step 4: Write the Taylor series for sin(x) centered at x = π/2 using the formula from Step 1 and the coefficients calculated in Step 3. The series will be Σ ((-1)^n / (2n+1)!) * (x - π/2)^(2n+1), where n ranges from 0 to infinity.

Step 5: Determine the interval of convergence for the Taylor series. Since the Taylor series for sin(x) converges for all x (as sin(x) is an entire function), the interval of convergence is (-∞, ∞).

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