1 Taylor-Maclaurin Series

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1 Taylor-Maclaurin Series Writing x = x + n x, x = (n ) x,.., we get, ( y ) 2 = y ( x) 2... and letting n, a leap of logic reduces the interpolation formula to: y = y + (x x )y + (x x ) 2 2! y +... Definition... A function f is said to be an C n function on (a, b) if f is n-times differentiable and the nth derivative f (n) is continuous on (a, b) and f is said to belong to C if every derivative of f exists (and is continuous) on (a, b). Taylor s Formula: Suppose f belongs to C on ( R, R). Then for every n N, and x ( R, R) we have: f(x) = f() + f ()x + f (2) () 2! x f (n) () n! xn + R n (x) f is said to be analytic at if the remainder R n (x) as n. There are two standard forms for the remainder.. Integral form: Proof. Integrating by parts, n! x R n (x) = n! x f (n+) (t)(x t) n dt = n! f (n) ()x n + f (n+) (t)(x t) n dt (n )! x f (n) (t)(x t) n dt Proof is completed by induction on n on observing that the result holds for n = by the Fundamental theorem of Calculus. 2. Lagrange s form: There exists c (, x) such that: R n (x) = (n + )! f (n+) (c)x n+ This form is easily derived from the integral form using intermediate value theorem of integral calculus. Lemma..2. If h( ) and g are continuous on [a, b], then there exists c (a, b) such that b a h(t)g(t)dt = g(c) b a h(t)dt Proof. If m and M denote the min and max of g on [a, b], then h implies (assuming h is not identically zero), m b a h(t)g(t)dt/ b a h(t)dt M Now the intermediate value theorem for continuous functions assures the existence of c.

2 Examples:. Let f(x) = sin(x). Then given x >, there exists ξ (, x) such that R n (x) = (n+)! f (n+) (ξ)x n+ and hence R n (x) (n+)! xn+. Thus R n (x) as n (uniformly on a finite subinterval of R). Consequently, sin(x) = x 3! x3... Note that even if the series on the r.h.s. converges for every x, to show that it converges to sin(x), it is necessary to prove that R n as n. 2. Let f(x) = log( + x), x >. Then f (n) (x) = ( ) (n+) (n )!( + x) n. Using the integral form (and assuming < x < ) we get: R n (x) x x ( (x t) ) n + t + t dt ( x n + t )dt x n log( + x) Using Lagrange s form, R n () = ( )n /( + n+ ξ) n. Thus R n (x) as n if x (, ]. log( + x) = x x3... for x (, ]. + x Let f(x) = ( + x) α where α R\N, α and x <. Then ( ) α! f () (x) = ( + x) α, =, 2... Thus, R n (x) = x n! [α(α )..(α n)] ( x t + t )n ( + t) α dt If r n = [α(α )..(α n)], then r n! n+/r n as n. Now if x <, choose ϱ such that ϱ x < and observe that r n x n c(ϱ x ) n as n. Consequently, since x t x for t (, x), as n, +t Thus, R n (x) r n x n x ( + x) α = + ( + t) α dt R n (x) = ( ) α x Example: Let α =. Then for x <, 2 ( ) 2 = ( ) (2)! 2 2.(!) 2 Hence ( + x) /2 = + ( ) (2)! 2 2.(!) 2 x = 2

3 2 Power Series 2. Prerequisites Definition 2... (a)if {f n } is a sequence of functions defined on a common domain A, we say that {f n } converges to a function f defined on A if for every x A, f n (x) f(x) as n. (b) We say that f n converges uniformly to f on A if sup { f n (x) f(x) : x A} asn. What are the advantages of having uniform convergence? Recall that g is said to be continuous at x if g(x + x) g(x) as x. Theorem Suppose f n f uniformly on A R. The following hold.. If each f n is continuous on A, then so is f. 2. If A = [a, b], then A f n A f. Proof.. Fix x A. Given ε >, choose N such that f(x) f N (x) < ε/3 for every x A. In particular, f(x ) f N (x ) < ε/3 and f(x + x ) f N (x + x ) < ε/3. By hypothesis, f N is continuous, so there exists δ such that if x < δ, then Now, if x < δ f N (x + x ) f N (x ) < ε/3 f(x + x ) f(x ) f(x ) f N (x ) + f N (x + x ) f N (x ) + f N (x + x ) f N (x ) i.e. f is continuous at x. 2. Simply observe that b a < ε/3 + ε/3 + ε/3 f n fdx (b a)max{ f n (x) f(x) : a x b} Examples:. Let A = [, ],and f n (x) = x n. If f() = and f(x) = on [, ), then f n f, but not uniformly. 2. Let A = [, ) f n be the (isosceles) triangular graph with base [, n] and height 2/n at x = n/2. Then f n (x) uniformly on A. 3. Let A = [, ) f n be the (isosceles) triangular graph with base [, /n] and height at x = /n. Then f n (x) on A, but not uniformly. 3

4 2.2 Power Series A power seriescentered at a is an infinite series of the form c n(x a) n where {c n } R. For our purpose, it suffices to set a = though the results are true for any a R. There exists a number R, so that the series converges whenever x < R and diverges whenever x > R. For clear reasons, R is called the radius of convergence. If c n+ c n /R or equivalently c n /n /R, then c i x i ( x R )i If x < R, then by comparison to the geometric series, c x is convergent. Let f(x) = c + c x + c 2 x What maes power series very useful is that they are (almost) as easy to manipulate as polynomials. The principal reason is the following. Theorem If [a, b] ( R, R), and x [a, b], then s n (x) = n c ix i converges uniformly to f(x). Proof: As before, c i x i b/r i and b/r <. The following corollary is now immediate. Corollary If [a, b] ( R, R), then, b a f(x)dx = c + b c i x i dx a Also, nc n /n c n /n /R and hence for x < R, i=n+ ic ix i as n. Hence s n g s n g But s n = s n f and thus f = g or g = f. Thus, s n f. It follows by induction that f is in C on ( R, R). Discussing the behavior of the power series at x = ±R taes wor. Abel s Theorem: If f(x) = c nx n is convergent on (, ) and c n is a convergent series of numbers, then lim f(x) = c n x Example: Recall that, = x 2 + x 4... Integrating term-by-term we have: +x 2 tan (x) = x x3 + x5... if x <. 3 5 By alternating series theorem, we now that +... is convergent. 3 5 So by Abel s theorem, using the fact that tan () = π we get 4 π 4 = Question: Let f(x) = e /x if x > and otherwise. Show that f C on R. Is f analytic at? 4

5 3 Applications 3. Differential Equations Consider the second order linear differential equation of second order. y + p(x)y + q(x)y = Definition 3... A point x in R is said to be a regular point if p(x) and q(x) have power-series expansion in an interval around x. Otherwise, it is a singular point. Example : y + 2xy + 2y = We loo for a solution of the form y = = c x for x ( R, R). Differentiate and substitute, Thus, Equivalently, In general,for, = Thus the general solution is: Example 2 y = c x, y = = ( )c x 2 =2 c +2 ( + )( + 2)x + 2 c x + 2 c x = ( + )( + 2)c ( + )c = c +2 = c, = c 2 = ( ) c andc 2+ =! y = c = ( ) x 2 + c! = xy + y + xy = = ( ) (2 + ) c ( ) (2 + ) x2+ Here x = is a singular point. Nonetheless, the power-series solution is possible. In this case we get, c = and for, c +2 = ( + 2) 2 c Thus, the solution to Bessel s equation is: J (x) = x x x

6 3.2 Cauchy s derivation of Newton s binomial series Given α R and x (, ), let r n (α) = α.(α )..(α n+) n!. Lemma Then for a fixed x (, ) and any α R r n+ (α) / r n (α) = α n + 2 / n + and further the convergence is uniform over α in any closed subinterval of R. The following fact is well-nown. If {t n } is a sequence of positive numbers, then Consequently, t n+ /t n l (t n ) /n l r n (α) /n [ r n (α) x n ] /n x and the convergence is uniform over α in any closed subinterval of R. Corollary If R n (α, x) = =n+ r (α)x, then for a fixed x (, ), R n (α, x) and the convergence is uniform over α in any closed subinterval of R. Proof. Choose ϱ ( x, ) such that r n (α) x n ϱ n, n N and α [ M, M]. R n (α, x) =n+ ϱ ϱn+ ϱ Now define a family of functions f(α, x) by the following convergent series and observe its properties. α.(α ) f(α, x) = + αx + x !. f(α + α 2, x) = f(α, x).f(α 2, x) for all x (, ). 2. f(n, x) = [f(, x)] n = ( + x) n, if n N. 3. f(α, x) = [f(, x)] α, if α Q. 4. For a fixed x (, ), f(α, x) is a continuous function of α. 6

7 Proof. () follows from a simple computation. (2) It follows by induction on n that, f(n, x) = f n (, x) By the same argument, f(βn, x) = f n (β, x) for any β R. (3) If α = m n, then f(m, x) = [f(αn, x)] = [f(α, x)]n. However, f(m, x) = [f(, x)] m by (2). Thus, f(α, x) = [f(, x)] m/n. (4) If S n (α) = n = r (α)x, then S n (α, x) f(α, x). Thus, for every fixed x (, ), f(α, x) is a continuous function of α since α belongs to the closed interval [α, α + ]. Proposition For α R and x (, ) we have: f(α, x) = ( + x) α Proof. If β Q, then using f(, x) = + x and Property (3) we get, f(β, x) = ( + x) β Now, given α R\Q, choose {β n } Q which converges to α. By continuity, f(α, x) = lim n f(β n, x) = lim n [f(, x)] βn = lim n ( + x) βn = ( + x) α Thus, for x (, ) and α R, ( + x) α = + αx + α.(α ) x ! Note: Observe that ( α ) [M/ +α ] implies that for α >, n r n(α) <. Hence R n (α, x) uniformly on [, ]. Thus f(α, x) converges on [, ]. Historical note: This is basically Cauchy s proof of Newton s expression of the binomial series. However, Cauchy had simply assumed the continuity of the limit function f(α, x) without realizing that a nonuniform limit of continuous functions can fail to be continuous. The introduction of the notion of uniform convergence and its application here are attributed to a young Norwegian mathematician named Abel. Abel called Cauchy a bigoted Catholic but praised his mathematical contributions. 7

8 3.3 Evaluation of ζ(2) Now consider the special case of the binomial series with x = t 2 and α =. Then 2 ( t 2 ) /2 = + a t 2, t < = (2)! 2 2.(!) 2. where a = By Wallis s product formula, a π, Integrating the power-series term-by-term we have: sin (x) = x + = a 2 + x2+, x < Now = a /(2 + ) = /3/2 <. Hence by Weierstass s M-test, the r.h.s. is continuous on [, ]. If θ = sin (x), then θ = sin θ + = a 2 + (sin θ)2+, θ π 2 and the series converges uniformly on [/2, π/2]. Integrate term-by-term, Now recall that Hence /2 θdθ = Term-by-term integration yields /2 I 2+ = sin(θ)dθ + /2 π 2 8 = + = a 2 + /2 (sin θ) 2+ dθ = 22 (!) 2 (2 + )! a 2 + I 2+ = = (2 + ) 2 (2 + ) = 2 = (2 + ) 2 (sin θ) 2+ dθ Now observe that Thus, n= n 2 = 4 n= n + 2 (2 + ) 2 ζ(2) = π2 6 = 8

9 4 Infinite Products This is a tae-off on the concept of infinite sums and uses many of the same ideas. We do not explore it in full generality, but deal with cases that provide sufficient structure for examples that interest us. Definition 4... Given {a n } R, and a n, we say that the infinite product Π( + a ) is convergent if the sequence of partial products p n = Π n = ( + a ) converges to a non-zero limit p in R. Otherwise, the infinite product is said to diverge. When the infinite product Π = ( + a ) converges, then a. Theorem (a) If a n for all n (or a n for all n), then Π = ( + a ) is convergent if and only if the series a n is convergent. (b) If a >, then the infinite product Π = ( + a ) is convergent if and only if log( + a ) is convergent. Proof. (a) If a n, then p n p n+ and s n s n+. Also it follows by induction that p n + s n. Moreover, + x e x p n e sn. It now follows from monotone convergence theorem that {p n } is convergent if and only if {s n } is. Corollary = ( x2 / 2 ) converges for all x R. 4. Application to ζ(2) Recall that if {a, a 2,..., a n } are roots of a polynomial p(x) and p() =, then we may write: Consider the series for sin(x) x p(x) = ( x a )...( x a n ) = x( a a n ) +.. and recall that lim x sin(x) x =. sin(x) = x2 x 3! +... is a representation of a polynomial of infinite degree with roots x = ±nπ with n N. In analogy with a polynomial,(and ignoring many questions of rigor, which were resolved much later), Euler claimed: sin(x) = ( x2 x2 )( x π2 4π )... 2 Now comparing the coefficients of x 2 in the factored form and the infinite series, we have: π 2 n= n 2 = 3! Note: If we substitute x = π/2 in Euler s factorisation formula and then reciprocate both sides, we get: 2 π = 2n ( 2n ).( 2n 2n + ) which is the formula for Wallis s product. n= 9

10 5 Weierstrass s Approximation Theorem Theorem 5... If f is a continuous function on [a, b], there exists a sequence of polynomials {p n } which converges to f uniformly on [a, b]. Bernstein s Proof: Let [a, b] = [, ]. The tools involve the following combinatorial formulae.. n = ( n ) x ( x) n = (x + x) n =. 2. n = (x )2( n n ) x ( x) n = x( x). n Proof. Write M = max { f(x) : x } and define the Bernstein polynomials: B n (x) = f( ( ) n n ) x ( x) n = Given ε >, choose δ > such that whenever x y < δ, then f(x) f(y) < ε, which is possible since f is continuous on a closed sub-interval of R. Now B n (x) f(x) = [f( ( ) n n ) f(x)] x ( x) n = Let S = { : x < δ} and S n 2 = { : x δ}. n Now f( ( n n ) f(x) S ) x ( x) n ε S ( n ε ε = Moreover, if S 2, then (x n )2 δ 2 and if n δ 4, then f( ( ) n n ) f(x) x ( x) n 2M ( n S 2 S 2 ) x ( x) n ( ) n x ( x) n ) x ( x) n 2Mδ 2 S 2 (x n )2 ( n 2Mδ 2 = 2Mδ 2 x( x) n 2M n x( x) n 2M n ) x ( x) n (x ( ) n n )2 x ( x) n Thus B n (x) f(x) ε + 2M n for every x [, ] and max{ B n (x) f(x) : x [, ]} as n.

11 5. Landau s proof This is similar in character to the techniques used in Fourier series. Step : Given f C([, ]), let g(x) = f(x) f() [f() f()]x and write g(x) M for all x. Now g C([, ]) and g() = g() =. Extend g by setting g = outside [, ]. Step 2: Define the ernel function by K n (x) = c n ( x 2 ) n, K n (x)dx = By Bernoulli s inequality, ( x 2 ) n nx 2 whenever x (, ). Now, c n = / n / n 4/ n It follows that if < δ <, then for x [δ, ], ( x 2 ) n dx ( x 2 ) n dx ( nx 2 )dx K n (x) n( δ 2 ) n Thus K n (x) uniformly on δ x and hence as n, [ δ + Step 3: Now for x [, ], define the polynomial P n by P n (x) = δ ]K n (t)dt K n (x t)g(t)dt If ε >, choose δ such that if t [ δ, δ], and x [, ], g(x t) g(x) < ε. This is possible since g is continuous (and hence uniformly continuous) on [, ]. P n (x) g(x ) = K n (t)[g(x + t) g(x)] dt δ 2M[ 2M[ δ + + δ δ δ ]K n (t)dt + ε K n (t)dt δ ]K n (t)dt + ε Thus P n (x) + f() + [f() f()]x f(x) uniformly on [, ]

12 6 Bernoulli Numbers Observe that ζ(2) = It follows that t dt, indicating the integrand is a function of interest. Now write: e t x e x = n! B nx n n= n= x = [ n! B nx n ].[ By comparing coefficients, we conclude that: Exercise: Show that However, = m= m! xm ] B =, B = n 2, ( ) n B = n 2 x + x is an even function and hence B e x 2 2n+ = for n. ζ(2n) = ( )n (2π) 2n B 2n 2(2n)! Note: Mathematica displays the Bernoulli numbers under the code BernoulliB[n]. As a point of historical interest, Lady Lovelace who was Byron s daughter and a contemporary of Queen Victoria is credited with the computation of B Extension to Bernoulli polynomials Given t R, write xe tx e x = n= b n (t) x n n! Clearly, b n () = B n and b (t) = and for n, (by comparing coefficients), b n (t) n! = = B n (n )!! t It is now easy to chec that the Bernoulli polynomials {b n (t)} satisfy the following properties for n : b n(t) = nb n (t), b n (t)dt = These polynomials are of great interest in Fourier Analysis and we will revisit them later. 2

13 6.2 Sums of positive powers The following result is due to Jacob Bernoulli. For p =, 2,... Proof: For x, = p = p+ p + j= ( p + j ) B p j+ (n + ) j e x = = = = p= ( p= = p! (x)p p ) xp p! However, the l.h.s. =reduces to: Now, (e (n+)x )/(e x ) = [x/(e x )].[(e (n+)x )/x] ( p ) xp+ p! p= = = [x/(e x )].[(e (n+)x )] = [ = i= B i i! xi ].[ j= a p x p+ p= (n + ) j x j ] j! Collecting coefficients we see that: a p = p+ j= Equating the coefficients of x p+, we have: p = p!a p = = B p j+ (n + )j (p j + )!j! = p + [(p + )!]a p = p+ ( ) p + B p j+ (n + ) j p + j j= 3

14 7 Fourier Series As the theorems on Taylor series suggest, the study of functions was more or less restricted to C functions. Polynomials, and more generally analytic functions (including transcendental functions lie the exponential and logarithmic functions) have this property. Fourier s notion that a discontinuous function could be represented as a sum of sines and cosines was inconceivable and received with septicism. Even the arguments presented in the Analytic Theory of Heat (822) were not rigorous. But the paradoxes contained in it led Cauchy and others to mae precise many concepts contained therein. We mae our tas easier by restricting ourselves to functions which are piecewise-c on [, π] and assume that f() = f(π). The Fourier series of f is defined by: F (θ) = a 2 + a cos(θ) + = b sin(θ) = where Write a = f(θ) cos(θ)dθ; b = π π f(θ) sin(θ)dθ s n (f, θ) = a 2 + a cos(θ) + = b sin(θ) and σ N (f, θ) = N = N = s n (f, θ) This begs the question: When does F (θ) converge? When it does, how do f(θ) and F (θ) relate? We collect and state basic results. Theorem 7... : Suppose f is piecewise-c on [, π] and assume that f() = f(π).. If f is continuous at θ, then s n (f, θ) f(θ). 2. If f is continuous at every point of [, π], then convergence is uniform. 3. The sequence a 2 θ + n on [, π]. = a sin(θ) n b = cos(θ) converges to θ f(t)dt uniformly Caution: Mere continuity of f s n (f, θ) f(θ) for every θ [, π]. There exists a function f C([, 2π]) whose Fourier series diverges at the point. This example is due to Dubois-Raymond. Corollary If p is a polynomial, then for every t (, π), p(t) = a cos(t) + b sin(t) = 4

15 7. Examples. Let f(x) = on (, π) and f(x) = on (, ) (the square-wave function). Then, f(x) = 4 π = sin(2 + )x 2 + For what values of x does the derived series converge? Graph f(x), s (x) and s 2 (x). 2. Let f(x) = x, x π. Graph f(x) and s 5 (x). f(x) = π 2 4 π = cos(2 + )x (2 + ) 2 3. Let f(x) = x, < x < π. Then Derive Leibniz s series for π/4. x = 2 n+ sin(nx) ( ) n 4. Integrate the previous series term-by-term to show that for x [, π], (a) Derive the formula for ζ(2). (b) Graph f and s 5. n= x 2 = π ( ) n cos(nx) n 2 n= 5

16 7.2 Euler s Theorem We start with finding Fourier series for the Bernoulli polynomials. Let b (x) = x /2 for x [, ] and observe that b (t)dt =. Then b (x) = α cos(2πx) + β sin(2πx), < x < = since b (t)dt = implies α =. A simple computation shows that for, α = 2 = (t 2 ) cos(2πt)dt = Thus, β = 2 = 2 b (t) sin(2πt)dt (t 2 ) sin(2πt)dt = π [t 2 cos(2πt)] + π = π b (x) = π = cos(2πt)dt sin(2πnx), < x < n Integrating term-by-term results in a series which converges uniformly on [, ] (by the M-test). Conclude that B 2n+ = if n. Of greater importance is the next case. b 2n+ (x) (2n + )! = 2 ( )n+ (2π) 2n+ m= sin(2πmx) m 2n+ b 2 (x) (2)! = ( ) 2 (2π) 2 n= cos(2πnx) n 2 To derive Euler s theorem by plugging in x =. Thus, B 2 (2)! = 2 ( ) (2π) 2 n= n 2 Consequently, (2π)2 B 2 ζ(2) = ( ) 2 (2)! 6

17 7.3 Orthogonality At heart is the theory of orthogonal basis in a vector space with a dot-product and the fundamental identities governing it. Recall that: Definition Two vectors w and w 2 are said to be orthogonal if their dot-product is zero and norm of a vector w is the square-root of dot-product of w with itself. It is nown that if { w,...w n } are pairwise orthogonal, i.e. wi w j = whenever i j, then their linear combination satisfies, a w 2 = a 2 w 2 = An example of an infinite-dimensional vector space is the space of sequences {α = {a n } : a 2 n < } and the inner-product (dot-product) of α and β is defined as < α, β >= a n b n = It is nown that if { w,...w n } are pairwise orthogonal, i.e. wi w j = whenever i j, then their linear combination satisfies, a w 2 = a 2 w 2 = Now consider the vector-space V of piecewise-c functions on [, π] which are periodic of period 2π. Next, if f, g V, we define the inner-product)as follows: < f, g >= π fgdx and f 2 = π f 2 dx Using basic trigonometric identities we prove the following:{cos(nx), sin(mx : n, m } is an orthogonal family of vectors in V and { So, } 2π, 2π, sin(x) π cos 2 (nx)dx = sin 2 (nx)dx = π, n is an orthonormal basis of V and s n (f, t) is the finite-dimensional projection of f onto the span of {cos(x), sin(x) : n}. An extension of the finite-dimensional identity is nown as Plancherel s Theorem: π f 2 dx = a2 π 2 + ( a ) 2 + b 2 ) = 7

18 If x [, π] and f is periodic of period 2π, and f(x) = a /2 + then orthogonality implies that: a cos(x) + b sin(x), a.e. = f(x) cos(x)dx = πa, f(x) sin(x)dx = πb If f(x) = = then orthogonality of { e ix, x π } implies that Plancherel s theorem now reads, c e ix f(x)e ix dx = c (2π) π f(x) 2 dx = 2π = Thining of e it as the basic independent variable on the unit circle T, by De-Moiver s theorem, T n (t) = = n c e it c 2 is called a trigonometric polynomial. In real form, assuming b =, T n (t) = a cos(t) + b sin(t) = Every Fourier series of a piecewise-continuous function is a trigonometric series. But the converse is false, even if class of piecewise-continuous functions is expanded considerably. Example n=2 sin(nt)/ log n cannot be a Fourier series of a (Lebesgue)-integrable function. But n=2 cos(nt)/ log n is. Exercise: Use Plancherel s theorem to show that if f is piecewise-c, then as n, f(x) s n (x) 2 dx 8

19 7.4 Dirichlet and Fejer ernels The function (nown as the Dirichlet ernel)is defined as: Then, D n (x) = sin(n + 2 )x 2 sin( x 2 ) D n (x) = 2 + n = cos(x) D n(t)dt = π. s n (x) = /π D n(t)f(x + t)dt Recall that σ n (x) = n + The function (nown as the Fejer ernel) is defined as: Then, F n (x) = n+ s n (x) = F n (x) = sin 2 (n + )x 2 n + 2 sin 2 ( x) 2 n = D n(x) F n(t)dt = π. σ n (x) = /π F n(t)f(x + t)dt, if f is periodic of period 2π. If δ (, π), then δ F n(t)dt + δ F n(t)dt as n. We can now prove Fejer s theorem, which is a trigonometric version of Weierstrass s theorem. Theorem If f is continuous on [, π] and periodic of period 2π, then σ n (x) f(x) uniformly for x [, π]. Proof. By hypothesis, f is uniformly continuous and bounded. Let M = max{ f(x) : x π}. Also, given ε >, we can choose δ (, π) so that for all t ( δ, δ) and x [, π] f(x + t) f(x) < ε Now choose N (depending on δ) such that for n N, Now for x [, π] and n N, σ n (x) f(x) π δ π [ + δ F n (t)dt + F n (t) f(x + t) f(x) dt ε + (2Mε/π) 2ε δ δ F n (t)dt < ε ]F n (t) f(x + t) f(x) dt + π F n (t)dt δ δ F n (t) f(x + t) f(x) )dt 9

20 7.5 Applications. If g is piecewise-continuous on [, π], then Plancherel s theorem implies that its Fourier coefficients are square-summable. In particular, we have the Riemann-Lebesgue lemma; As as n, g(t) cos(nt)dt and g(t) sin(nt)dt. Consequently, as n, g(t) sin((n + )t)dt Use the Dirichlet ernel to show that: Solution: Write D n (x)dx = sin x x dx = π 2 sin((n + 2 )x)( 2 sin(x/2) x )dx + sin((n + 2 )x)/xdx Now the first integral goes to by the Riemann-Lebesgue lemma. Hence π 2 = lim n [sin((n + 2 )x)/x]dx Mae a change of variable t = (n + )x. It follows that 2 sin x x dx = lim n (n+ 2 )π sin(t) dt = π t 2 3. However, as n, D n (t) dt Observe that for t [, π/2], sin t t (π/2) sin t. /2 [ sin(2n + )t / sint ]dt c /2 = 2 Hence [ sin(2n + )t / t ]dt 2 (+)π 2(2n+) = 2 = π 2(2n+) (2n + )π c + + [ sin(2n + )t / t ]dt (+)π 2(2n+) π 2(2n+) sin(2n + )t dt 2