Differentiating and integrating power series
Author Archives: mh225
Two applications of series to prove theorems about prime numbers
More Challenging Problems: Using power series to sum numerical series
Notes PDF
Introductory Problems
1. Sum the series ∑n=1∞((-1)nπ2n+1)/(42n+1(2n+1)!)
Answer
1. . .
Solution
1.
2. Sum the series π3/3 – π5/5⋅3! + π7/7⋅5! – π9/9⋅7! + …
Answer
1. The series sums to π.
Solution
1. The number π likely is the point at which the series is evaluated, so the series is f(x)= x3/3 – x5/5⋅3! + x7/7⋅5! – x9/9⋅7! + … So the series to sum is f(π). The x3/3, x5/5, x7/7, x9/9, … suggests an integral, so let’s differentiate the series f'(x)= x2 – x4/3! + x6/5! – x8/7! + … The denominators and alternating signs suggest sin(x), but the exponents are wrong. Try factoring out an x f'(x)= x(x – x3/3! + x5/5! – x7/7! + …) x sin(x) Then f(x)= ∫f'(x) dx= ∫x sin(x) dx= -x cos(x) + sin(x), using integration by parts. Then the series sums to f(π)= -π cos(π) + sin(π) = π.
3. Sum the series 2 + 1/3 + 2/34 + 1/35 + 2/38 + 1/39 + 2/312 + 1/313 + …
Answer
1. The series sums to 189/80.
Solution
1. This appears to be the series: 2 + x + 2x4 + x5 + 2x8 + x9 + 2x12 + x13 + … evaluated at x= 1/3. Segregate the terms with multiples of 2 from those without 2 + x + 2x4 + x5 + 2x8 + x9 + 2x12 + x13 + … = (2 + 2x4 + 2x8 + 2x12 + …) + (x + x5 + x9 + x13 + …) Factor out a 2 from the first, a x from the second, obtaining 2(1 + x4 + x8 + x12 + … ) + x(1 + x4 + x8 + x12 + … ) = (2 + x)(1 + x4 + x8 + x12 + … ) = (2 + x)/(1 – x4) At x= 1/3, the series sums to (2 + 1/3)/(1 – 1/34) = 189/80.
4. Suppose a1 = 1 and for all n > 1, an = (1/2)n + (an-1/2). Find the sum of the series a1 + a2 + a3 +…
Answer
1. The series sums to 3.
Solution
1. First, try to find an expression for an in terms of n, not an-1. a2 = (1/2)2 + a1/2 = 1/22 + 1/2 = 3/22 a3 = (1/2)3 + a2/2 = 1/23 + 3/23 = 4/23 a4 = (1/2)4 + a3/2 = 1/24 + 4/24 = 5/24 a5 = (1/2)5 + a4/2 = 1/25 + 5/25 = 6/25 and in general an = (n+1)/2n for n ≥ 1. That is, we want to sum the series : 2/21 + 3/22 + 4/23 + 5/24 + 6/25 + … Next, recall the geometric series: 1/(1 – x)= 1 + x + x2 + x3 + x4 + … . Differentiating gives: 1/(1 – x)2= 1 + 2x + 3x2 + 4x3 + 5x4 + … Except for the initial 1, when evaluated at x= 1/2 this is the series we want to sum. That is, 1/(1 – 1/2)2 – 1 = 2/21 + 3/22 + 4/23 + 5/24 + 6/25 + …= 3
More challenging problems: Differentiating and integrating power series
Notes PDF
Introductory Problems
1. Find a power series y(x) = ∑n = 0∞anxn that satisfies y’ = y and y(0) = 1.
Answer
1. 1. The series is y(x) = 1 + x + x2/2! + x3/3! + … = ex.
Solution
2. Find a power series y(x) = ∑n = 0∞anxn that satisfies xy” + (1 – x)y’ – y = 0 and y(0) = 2.
Answer
1. The series is y(x) = 2 + 2x + x2 + x3/3 + x4/(4⋅3) + x5/(5⋅4⋅3) +…
Solution
1. Differentiating y(x) = a0 + a1x + a2x2 + a3x3 + a4x4 + … gives: y'(x) = a1 + a22x + a33x2 + a44x3 + a55x4 + …y”(x) = a22 + a33⋅2x + a44⋅3x2 + a55⋅4x3 + a66⋅5x4 + Then: xy'(x) = a1x + a22x2 + a33x3 + a44x4 + a55x5 + …xy”(x) = a22x + a33⋅2x2 + a44⋅3x3 + a55⋅4x4 + a66⋅5x5 + … Equating coefficients of like powers of x gives: xy” y’ – xy’ -y = 0 gives x0 0 a1 – 0 -a0 = 0 a1 = a0 x1 2a2 2a2 – a1 -a1 = 0 4a2 – 2a1 = 0, so a2 = a1/2 x2 3⋅2a3 3a3 – 2a2 -a2 = 0 9a3 – 3a2 = 0, so a3 = a2/3 x3 4⋅3a4 4a4 – 3a3 -a3 = 0 16a4 – 4a3 = 0, so a4 = a3/4
All the coefficients can be expressed in terms of a0, giving an = a0/n! and consequently:
y(x) = a0(1 + x + x2/2 + x3/3! + x4/4! + …) = a0ex.The condition y(0) = 2 implies a0 = 2.
3. Prove that on its domain, (1 – sin(x))-2 = 1 + 2sin(x) + 3sin2(x) + 4sin3(x) + …
Solution
4. By dividing the series, find the coefficients, through the cubic term, of (1 – x)/ex.
Answer
1. The series for (1 – x)/ex begins 1 – 2x + (3/2)x2 – (2/3)x3.
Solution
1. Mimic long division of polynomials.
More Challenging Problems: Taylor’s theorem
Notes PDF
Introductory Problems
1. Find the Taylor series for ln(1 + x) expanded about x = 0. Find the radius of convergence of this series.
Answer
1. The series is x – x2/2 + x3/3 – x4/4 + … . The radius of convergence is 1.
Solution
1. To find the taylor series, compute the first few derivatives of f(x) = ln(1 + x) and look for patterns. f'(x)= 1/(1 + x) f”(x)= -1/(1 + x)2 f”'(x)= 2/(1 + x)3 f(4)(x)= -3!/(1 + x)4 and in general f(n)(x)= (-1)n-1(n-1)!/(1 + x)n . Then f(n)(0)= (-1)n-1(n-1)! and so f(n)(0)/n! = (-1)n-1/n. Recalling f(0)= 0, the Taylor series is ln(1 + x)= x – x2/2 + x3/3 – x4/4 + … + (-1)n-1xn/n + … Apply the Ratio Test to find the radius of convergence limn → ∞|((-1)nxn+1/(n+1)) / ((-1)n-1xn/n)| = limn → ∞|(n/(n+1)) x| = |x| The series converges for |x| < 1, so the radius of convergence is 1.
2. Without computing any derivatives, find the Taylor series of cos(√(x2 + 1)) expanded about x = 0.
Answer
1. The series is 1 – (x2 + 1)/2! + (x2 + 1)2/4! – (x2 + 1)3/6! + … + (-1)n(x2 + 1)n/(2n)! + …
Solution
1. Recall the Taylor series for cos(x): cos(x) = 1 – x2/2! + x4/4! – x6/6! + … + (-1)nx2n/(2n)! + … Substitute √(x2 + 1) for x, obtaining 1 – (x2 + 1)/2! + (x2 + 1)2/4! – (x2 + 1)3/6! + … + (-1)n(x2 + 1)n/(2n)! + …
3. Evaluate the limit limx → 0(tan(x) – sin(x))/x3
Answer
1. The limit is 1/2.
Solution
1. Because the denominator is x3, we need the series for the numerator through the cubic term. The series for sin(x) is familiar: sin(x) = x – x3/3! + … . To find the terms of the Taylor series for tan(x) up through the cubic term, first compute the derivatives (tan(x))’= sec2(x) (tan(x))”= 2 sec(x) sec(x) tan(x) = 2 sec2(x) tan(x) (tan(x))”’= 2(2 sec(x) sec(x) tan(x))tan(x) + 2 sec2(x) sec2(x) = 4sec2(x)tan2(x) + 2sec4(x) Evaluating these derivatives at x = 0 gives tan'(0)= 1 tan”(0)= 0 tan”'(0)= 2 Then the first few terms of the Taylor series for tan(x) are x + x3/3 Up through cubic terms, tan(x) – sin(x) = (1/3 – (-1/3!))x3 = x3/2 Then the limit is 1/2.
4. For what value of k is this limit limx → 0(1 – kx2 – cos(3x))/x4 finite? Find the limit for that k.
Answer
1. The value of k is 9/4. For that value of k, the limit is 27/8.
Solution
1. The Taylor series for cos(3x) is 1 – (3x)2/2! + (3x)4/4! – (3x)6/6! + … . Because the denominator is x4, in order for the limit to exist the kx2 term must subtract out of the numerator, so k = 32/2! = 9/2. For k = 9/2, the value of the limit is the ratio of the coefficients of the quartic terms, so the limit is 34/4! = 81/24 = 27/8.
More Challenging Problems: The radius and interval of convergence
Notes PDF
Introductory Problems
Find the radius and interval of convergence of these power series.
1. ∑n=1∞ n! xn/(2n)!
Answer
1. The radius of convergence is ∞.
Solution
1. Factorials suggest the Ratio test. limn → ∞|((n+1)!xn+1/(2n+2)!) / (n!xn/(2n)!)| = limn → ∞|((n+1)!/n!)⋅((2n)!/(2n+2)!)x| = limn → ∞(n+1)/((2n+2)(2n+1))|x| = 0 for all x. This power series converges for all x, so the radius of convergence is ∞.
2. ∑n=1∞ nnxn/2n
Answer
1. The radius of convergence is 0. The interval of convergence is [0].
Solution
1. Exponents suggest the Root Test. limn → ∞(nn|x|n/2n)1/n= limn → ∞(n/2)|x| = ∞ for all x ≠ 0. Then the radius of convergence is 0, and the inverval of convergence is [0].
3. ∑n=1∞ n2xn/2n
Answer
1. The radius of convergence is 2. The interval of convergence is (-2, 2).
Solution
1. The exponent of n2 suggests the Root Test will not be effective. Try the Ratio Test. limn → ∞|((n+1)2xn+1/2n+1) / (n2xn/2n)| = limn → ∞((n+1)2/n2)⋅(2n/2n+1)|x| = |x|/2. The interval of convergence includes |x|/2 < 1, that is, (-2,2), and so the radius of convergence is 2. To find the interval of convergence, test the endpoints of (-2,2). For x = 2 the series becomes ∑n=1∞ n22n/2n = ∑n=1∞ n2. This diverges by the nth term test. For x = -2 the series becomes ∑n=1∞ n2(-2)n/2n = ∑n=1∞ (-1)nn2. This diverges by the nth term test. Then the interval of convergence is (-2,2).
4. ∑n=1∞ (2x – 1)2n+1/n1/2
Answer
1. The radius of convergence is 1/2. The interval of convergence is (0, 1).
Solution
1. The exponent of n1/2 suggests the Root Test will not be effective. Try the Ratio Test. limn → ∞|((2x – 1)2n+3/(n+1)1/2) / ((2x – 1)2n+1/n1/2)| = limn → ∞(n/(n+1))1/2|(2x – 1)2| = |(2x – 1)2| The interval of convergence includes |(2x – 1)2| < 1, that is, |2x – 1| < 1, which gives 0 < x < 1. Then the radius of convergence is 1/2. To find the interval of convergence, test the endpoints of (0,1). For x = 1 the series becomes ∑n=1∞ 12n+1/n1/2 = ∑n=1∞ 1/n1/2, a diverging p-series. For x = 0 the series becomes ∑n=1∞ (-1)2n+1/n1/2 = -∑n=1∞ 1/n1/2, a diverging p-series. Then the interval of convergence is (0,1).
More Challenging Problems: Geometric series
Notes PDF
Introductory Problems
1. Write 1 + 2x + 4x2 + 8x3 + 16x4 + … as a ratio of polynomials.
Answer
1. The series sums to 1/(1 – 2x).
Solution
1. Rewrite the terms of the series 1 + 2x + 4×2 + 8×3 + 16×4 + … = 1 + 2x + (2x)2 + (2x)3 + (2x)4 + …So we see this series converges to 1/(1 – 2x).
2. Write x + x2 + x3 + x4 + … as a ratio of polynomials.
Answer
1. The series sums to x/(1 – x).
Solution
1. The series is the geometric series for 1/(1 – x), missing the 1, so x + x2 + x3 + x4 + … = (1 + x + x2 + x3 + x4 + … ) – 1= 1/(1 – x) – 1 = x/(1 – x).
3. Write 2 + x2 + x3 + x4 + 2x6 + x8 + x9 + x10 + 2x12 + x14 + x15 + x16 + 2x18 + … as a ratio of polynomials.
Answer
1. The series sums to (2 – x2 – x3)/(1 – x2 – x3 + x5).
Solution
1. The terms with coefficient 2 are x0, x6, x12, x18, … , in general, x6n. This suggests rearranging the series into terms of the form x2n and terms of the form x3n, obtaining: 1 + x2 + x4 + x6 + x8 + x10 + x12 + x14 + x16 + … +1 + x3 + x6 + x9 + x12 + x15 + x18 + …= 1/(1 – x2) + 1/(1 – x3)(2 – x2 – x3)/(1 – x2 – x3 + x5)
4. Write 1 + 2x + 3x2 + 4x3 + 5x4 + … as a ratio of polynomials.
Answer
1. The series converges to 1/(1 – 2x + x2).
Solution
1. First observe the sum 1 + 2x + 3x2 + 4x3 + 5x4 +… can be reorganized:1 + x + x2 + x3 + x4 +…= 1/(1 – x) x + x2 + x3 + x4 +… = x/(1 – x) x2 + x3 + x4 +… = x2/(1 – x) x3 + x4 +… = x3/(1 – x) x4 +… = x4/(1 – x) Consequently, 1 + 2x + 3x2 + 4x3 + 5x4 + … = (1 + x + x2 + x3 + …)/(1 – x) = 1/(1 – x)2 = 1/(1 – 2x + x2)
Power Series: Using power series to sum numerical series
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More Challenging Problems
Sum these series.
1. 1 + 1/22 + 1/24 + 1/26 + 1/28 + …
Answer
1. The series sums to 4/3.
Solution
1. The powers of 2 in the denominators suggest trying x = 1/2. Then we see the numerical series is x = 1/2 substituted into 1 + x2 + x4 + x6 + x8 + … We recognize this as the series expansion for 1/(1 – x2). Then, 1 + 1/22 + 1/24 + 1/26 + 1/28 + … = 1/(1 – (1/2)2) = 4/3.
2. 2/3 + 4/33 + 6/35 + 8/37 + 10/39 + …
Answer
1. The series sums to 27/32.
Solution
1. The powers of 3 in the denominators suggest trying x = 1/3. Then we see the numerical series is x = 1/3 substituted into 2x + 4x3 + 6x5 + 8x7 + 10x9 + …The relation between the coefficient and the exponent of each term suggests viewing this as a derivative: 2x + 4x3 + 6x5 + 8x7 + 10x9 + … = (x2 + x4 + x6 + x8 + x10 + …)’ Of course, we can add any constant to the series being differentiated, without changing its derivative. Adding 1 gives a familiar series. 2x + 4x3 + 6x5 + 8x7 + 10x9 + … = (1 + x2 + x4 + x6 + x8 + x10 + …)’ = (1/(1 – x2))’ = 2x/(1 – x2)2 Substituting in x = 1/3, we find 2/3 + 4/33 + 6/35 + 8/37 + 10/39 + … = (2/3)/(1 – 1/32)2 = 27/32
3. 1 – 1/42 + 1/44 – 1/46 + 1/48 – …
Answer
1. The series sums to 16/17.
Solution
1. The powers of 4 in the denominators suggest trying x = 1/4. Then we see the numerical series is x = 1/4 substituted into 1 – x2 + x4 – x6 + x8 + …We recognize this as the series expansion for 1/(1 + x2). Then 1 – 1/42 + 1/44 – 1/46 + 1/48 – … = 1/(1 + 1/42) = 16/17.
4. 1 – 1/3 + 1/5 – 1/7 + 1/9 – 1/11 +
Answer
1. The series sums to π/4.
Solution
1. Here we have fewer clues; no obvious choice for x. But in order to view this as a power series evaluated at some x = a, then the numerical series should contain some number raised to different powers. With this assumption, the only choice for x is x = 1. But what powers should we use? Trying 1, x, x2, x3, … gives 1 – x/3 + x2/5 – x3/7 + x4/9 – x5/11 + …This doesn’t look like a familiar series. If we match the exponent of x to the denominator, then we might recognize this series as the integral of a familiar series.x – x3/3 + x5/5 – x7/7 + x9/9 – x11/11 + 2 + x4 – x6 + x8 – x10 + …)dx = ∫1/(1 + x2) dx= arctan(x). Substituting x = 1 we find: 1 – 1/3 + 1/5 – 1/7 + 1/9 – 1/11 + …= arctan(1) = π/4.
5. 3 – 32/3 + 33/5 – 34/7 + 35/9 – …
Answer
1. The series sums to π/√3.
Solution
1. The previous problem has the same set of denominators as this problem, which saw the exponent of x increasing by 2 between successive terms. The powers of 3 in the numerators suggest taking x = 3, but since the powers of 3 increase by 1 between successive terms of the series, maybe x = √3 is a better choice. Then the series has the form x2 – x4/3 + x6/5 – x8/7 + x10/9 – … This isn’t exactly a series we recognize, but it’s close. Try factoring out an x (NOT an x2, because the series from the previous problem, which looks pretty similar to this, starts with an x). x⋅(x – x3/3 + x5/5 – x7/7 + x9/9 – … ) = x⋅arctan(x) where we’ve used the arctan(x) series from the previous problem. Substituting x = √3 we see 3 – 32/3 + 33/5 – 34/7 + 35/9 – … = (√3)⋅arctan(√3) = (√3)⋅(π/3) = π/√3.
Power Series: Differentiating and integrating power series
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More Challenging Problems
1. Find the Taylor series, expanded about x = 0, for f(x) = x⋅cos(x2).
Answer
1. The series is x – x5/2 + x9/4! – x13/6! + … .
Solution
1. First, recognize that x⋅cos(x2) is 1/2 the derivative of sin(x2). Substituting u= x2 in the series for sin(u), we find sin(x2)= x2 – x6/3! + x10/5! – x14/7! + … Then 2x⋅cos(x2) = (sin(x2))’= (x2 – x6/3! + x10/5! – x14/7! + …)’ = 2x – 6x5/3! + 10x9/5! – 14x13/7! + … and so x⋅cos(x2)= x – x5/2 + x9/4! – x13/6! + … Alternately, we could substitute x2 for x in the series expansion of cos(x), them multiply that series by x.
2. Find the Taylor series, expanded about x = 0, for f(x) = ex + x2.
Answer
1. The series is 1 + x + 3x2/2 + 7x3/6 + 25x4/24 + 27x5/40 + 331x6/720 +…
Solution
1. Substitute u = x + x2 in the series eu = 1 + u + u2/2! + u3/3! + u4/4! + … obtaining ex + x2 = 1 + (x + x2) + (x + x2)2/2! + (x + x2)3/3! + (x + x2)4/4! + …= 1 + (x + x2) + (x2 + 2x3 + x4)/2! + (x3 + 3x4 + 3x5 + x6)/3! + (x4 + 4x5 + 6x6 + 4x7 + x8)/4! + …Now group together like powers of x. ex + x2 = 1 + x + (1 + 1/2!)x2 + (1 + 1/3!)x3 + (1/2! + 3/3! + 1/4!)x4 + (3/3! + 4/4! + 1/5!)x5 + … = 1 + x + 3x2/2 + 7x3/6 + 25x4/24 + 27x5/40 + 331x6/720 + …
3. Find the Taylor series, expanded about x = 0, for f(x) = 1/(1 – x)3.
Answer
1. The series is 1 + 3x + 6×2 + 10×3 + 15×4 + 21×5 + …
Solution
1. First, recognize that 1/(1 – x)3 is 1/2 the second derivative of 1/(1 – x). Recall the series for 1/(1 – x) is 1/(1 – x) = 1 + x + x2 + x3 + x4 + x5 + … Then (1/(1 – x))’ = 1 + 2x + 3x2 + 4x3 + 5x4 + 6x5 + … (1/(1 – x))” = 2 + 3⋅2x + 4⋅3x2 + 5⋅4x3 + 6⋅5x4 + 7⋅6x5 + …Then the series for 1/(1 – x)3 is: 1 + (3⋅2/2)x + (4⋅3/2)x2 + (5⋅4/2)x3 + (6⋅5/2)x4 + (7⋅6/2)x5 + …= 1 + 3x + 6x2 + 10x3 + 15x4 + 21x5 + …
4. Find the Taylor series, expanded about x = 0, for cos(x)⋅sin(x).
Answer
1. The series is x – 2x3/3 + 2x5/15 – 4x7/315 + 2x9/2835 – 4x11/155925 + …
Solution
1. Multiply the series for cos(x) by the series for sin(x) and group together terms having like powers of x. cos(x)⋅sin(x) = (1 – x2/2! + x4/4! – x6/6! + x8/8! – …)⋅(x – x3/3! + x5/5! – x7/7! + …) = x – (1/3! + 1/2!)x3 + (1/5! + (1/2!)⋅(1/3!) + 1/4!)x5 – (1/7! + (1/2!)⋅(1/5!) + (1/4!)⋅(1/3!) + 1/6!)x7 – (1/9! + (1/2!)⋅(1/7!) + (1/4!)⋅(1/5!) + (1/6!)⋅(1/3!) + 1/8!)x9 + ..
5. Find the Taylor series, expanded about x = 0, for ∫1/(1 – x5) dx.
Answer
1. The series is x + x6/6 + x11/11 + x16/16 + x21/21 + x26/26 + …
Solution
1. The series for 1/(1 – x5) can be found by substituting u = x5 into the series for 1/(1 – u). 1/(1 – x5) = 1 + x5 + (x5)2 + (x5)3 + (x5)4 + (x5)5 + …= 1 + x5 + x10 + x15 + x20 + x25 + …Then the integral ∫1/(1 – x5) dx has series obtained by integrating this, so ∫1/(1 – x5) dx = x + x6/6 + x11/11 + x16/16 + x21/21 + x26/26 + …
Power Series: Taylor’s theorem
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More Challenging Problems
1. Find the Taylor series, expanded about x = 0, for f(x) = 1/(1 + x)3
Answer
1. The series is ∑n=0∞anxn,where an = (-1)n(n+1)(n+2)/2
Solution
1. For f(x) = 1/(1 + x)3 we compute some derivatives and seek a pattern. f'(x)= -3/(1 + x)4 f”(x)= 3⋅4/(1 + x)5 f”'(x)= -3⋅4⋅5/(1 + x)6 and so on. Evaluating the function and its derivatives at x = 0 we find f(0)= 1 f'(0)= -3 f”(0)= 3⋅4 f”'(0)= -3⋅4⋅5 and so on. Then the coefficients of the Taylor series, f(n)(0)/n!, are f(0) = 1 f'(0)/1! = -3 f”(0)/2! = 3⋅4/2! = 3⋅4/2 f”'(0)/3! = -3⋅4⋅5/3! = – 4⋅5/2 We need a few more terms to find the pattern. f(4)(0)/4! = 3⋅4⋅5⋅6/4! = 5⋅6/2 f(5)(0)/5! = -3⋅4⋅5⋅6⋅7/5! = -6⋅7/2 Now we see it: f(n)(0)/n! = (-1)n(n+1)(n+2)/2
2. Find the Taylor series for f(x) = 1 + x – x3, expanded about x = 1.
Answer
1. The expansion is f(x)= 1 -2(x-1) -3(x-1)2 – (x-1)3.
Solution
1. We compute the derivatives. Only a few are nonzero, so we need to expend much energy looking for the pattern. f(1)= 1 f'(1)= -2 f”(1)= -6 f”'(x)= -6 f(n)(x) = 0 for all n > 3. Then the Taylor series is f(x) = 1 – (2/1!)(x-1) – (6/2!)(x-1)2 – (6/3!)(x – 1)3 = 1 -2(x-1) -3(x-1)2 – (x-1)3
3. Without computing any derivatives, find the Taylor series, expanded about x = 0, for x3ex
Answer
1. The series is x3 + x4 + x5/2! + x6/3! + x7/4! + … + xn+3/n! + …
Solution
1. Recall the Taylor series for ex= 1 + x + x2/2! + x3/3! + x4/4! + … Multiplying by x3 raises all the exponents by 3: x3ex= x3 + x4 + x5/2! + x6/3! + x7/4! + … + xn+3/n! + …
4. Without computing any derivatives, find the Taylor series, expanded about x = 0, for sin2(x). Hint: trigonometric identities.
Answer
1. sin2(x)= -x2 + 23x4/4! – 25x6/6! + …
Solution
1. Recall sin2(x) = (1 – cos(2x))/2. The Taylor series for cos(2x) is cos(2x)= 1 – (2x)2/2! + (2x)4/4! – (2x)6/6! + ….So 1 – cos(2x) has series expansion 1 – cos(2x)= -22x2/2! + 24x4/4! – 26x6/6! + …and we find sin2(x)= -x2 + 23x4/4! – 25x6/6! + …
5. Evaluate the limit limx → 0(sin(x) – x)/x3. Use Taylor series, not l’Hôpital’s rule.
Answer
1. The limit is -1/6.
Solution
1. Substitute in the Taylor series for sin(x), obtaining sin(x) – x = (x -x3/3! + x5/5! – …) – x= -x3/3! + x5/5! – … and so (sin(x) – x)/x3 = -1/3! + x2/5! – … where all the omitted terms have powers of x higher than 2, specifically, x4, x6, x8 and so on. As x → 0, only the -1/3! remains.