Some examples of Mahler measures as multiple polylogarithms
|
|
- Audra Dawson
- 6 years ago
- Views:
Transcription
1 Some exmples of Mhler mesures s multiple polylogrithms Mtilde N. Llín, University of Texs t Austin. Deprtment of Mthemtics. University Sttion C. Austin, TX 787, USA Abstrct The Mhler mesures of certin polynomils of up to five vribles re given in terms of multiple polylogrithms. Ech formul is homogeneous nd its weight coincides with the number of vribles of the corresponding polynomil. Key words: Mhler mesure, L-functions, polylogrithms, hyperlogrithms, polynomils, Jensen s formul Introduction Ltely there hs been some interest in finding explicit formule for the Mhler mesure of polynomils. The (logrithmic) Mhler mesure of polynomil P C[x,...,x n ] is defined s m(p) = log P(x (πi) n,...,x n ) dx,..., dx n x x n T n where T n = {(z,...,z n ) C n z =... = z n = } is the n-torus. For the one-vrible cse, Jensen s formul π π log e iθ α dθ = log + α E-mil ddress: mllin@mth.utexs.edu Supported by Hrrington Fellowship Preprint submitted to Elsevier Science 5 October 5
2 (where log + x = log x if x nd zero otherwise), provides simple expression of the Mhler mesure s function on the roots of the polynomil: Given P(x) = j (x α j ), then m(p) = log + j log + α j The two-vrible cse is much more complicted. Severl exmples with explicit formule hve been produced. Boyd [,3] nd Smyth [9] hve computed severl exmples nd expressed some of them in terms of specil vlues of L-functions of qudrtic chrcters. Also Boyd [3] nd Rodriguez Villegs [8] hve obtined nlogous results with L-functions of certin elliptic curves. Further, Boyd nd Rodriguez Villegs [4], Millot [7], nd Vndervelde [] hve produced exmples where the Mhler mesure is expressed s combintions of dilogrithms. There re only few exmples for three vribles. Smyth [] relted the mesure of +bx +cy +(+bx+cy)z to combintions of trilogrithms nd dilogrithms. Vndervelde [] obtined the mesure of + x + z(x y) s combintions of trilogrithms. In this pper, we express the Mhler mesure of some prticulr cses of up to five vrible polynomils s combintions of multiple polylogrithms. More precisely, we determine the Mhler mesure of ( + w )...( + w n ) + ( w )...( w n )y in C[w,...,w n,y] for n =,,, 3. We will refer to these s exmples of the first kind. We lso consider ( + w )...( + w n )( + x) + ( w )...( w n )(y + z) in C[w,...,w n,x,y,z] for n =,, (exmples of the second kind). In ddition to these, we use the sme method strting from Millot s exmple [7], in order to compute the Mhler mesure of ( + w)( + y) + ( w)(x y)
3 Summry of the results for the cse = In order to be concrete nd for future reference, we summrize the results obtined for the prticulr cse of = : πm(( + x) + ( x)y) L(χ 4, ) π m(( + w)( + x) + ( w)( x)y) π 3 m(( + v)( + w)( + x) + ( v)( w)( x)y) 7ζ(3) 7πζ(3) + 4 j<k ( ) j (j+) k π m(( + x) + (y + z)) 7 ζ(3) π 3 m(( + w)( + x) + ( w)(y + z)) π 4 m(( + v)( + w)( + x) + ( v)( w)(y + z)) π L(χ 4, ) + 8 j<k 93ζ(5) ( ) j+k+ (j+) 3 k π 7 π m(( + w)( + y) + ( w)(x y)) ζ(3) + log The third nd fifth formule cn be lso written s π 3 m(( + v)( + w)( + x) + ( v)( w)( x)y) = 7πζ(3) + 6(L(χ 4,χ ;, ) L(χ 4,χ 4;, )) () π 3 m(( + w)( + x) + ( w)(y + z)) = 7 π ζ(3) 6 log L(χ 4, 3) + 6(L(χ,χ 4 ;, 3) L(χ 4,χ 4 ;, 3)) () Here χ is the principl chrcter nd χ 4 is the rel odd chrcter of conductor 4, i.e. if n mod 4 χ 4 (n) = if n mod 4 otherwise The L functions re defined by χ (k )χ (k )...χ m (k m ) L(χ,...,χ m ;n,...,n m ) := <k <k <...<k m k n k n...km nm This series is bsolutely convergent if Re(n m ) > nd Re(n i ) for i < m. 3
4 We would like to point out tht ll these formule re new, except for m(( + x) + ( x)y), proved in [9], nd m( + x + y + z). 3 Ide of the procedure nd some technicl steps Let P α C[x,...,x n ], polynomil where the coefficients depend polynomilly on prmeter α C. We replce α by α w. A polynomil +w P α C[x,...,x n,w] is obtined. The Mhler mesure of the new polynomil is certin integrl of the Mhler mesure of the former polynomil. More precisely, Proposition Let P α C[x,...,x n ] s bove, then, m( P α ) = πi ( m T P α w +w ) dw w (3) Moreover, if the Mhler mesure of P α depends only on α, then m( P α ) = π m(p x ) α dx x + α (4) PROOF. Equlity (3) is direct consequence of the definition of Mhler mesure. In order to prove equlity (4), write w = e iθ. Observe tht s long s w goes through the unit circle in the complex plne, w goes through the +w imginry xis ir, indeed, w = i tn ( ) θ +w. The integrl becomes, m( P α ) = π π m ( ) P α tn( θ ) dθ = π π ( ) m P α tn( θ ) dθ Now mke x = α tn ( ) θ, then dθ = α dx, x + α m( P α ) = π m (P x ) α dx x + α Our ide is to integrte the Mhler mesure of some polynomils in order to get the Mhler mesure of more complex polynomils. We will need the following: 4
5 Proposition Let P with > be polynomil s before, (its Mhler mesure depends only on α ) such tht F() if m(p ) = G() if > Then m( P ) = π F(x) dx x + + ( ) dx G π x x + (5) PROOF. The Proof is the sme s for eqution (4) in Proposition, with n dditionl chnge of vribles x in the integrl on the right. x Recll the definition for polylogrithms which cn be found, for instnce, in Gonchrov s ppers, [5,6]: Definition 3 Multiple polylogrithms re defined s the power series k x x k...x km m Li n,...,n m (x,...,x m ) := <k <k <...<k m k n k n...km nm which re convergent for x i <. The weight of polylogrithm function is the number w = n n m. Definition 4 Hyperlogrithms re defined s the iterted integrls I n,...,n m ( :... : m : m+ ) := m+ dt dt dt dt... dt dt dt dt t t t t }{{} t t t }{{} m t n n... dt t } {{ } n m where n i re integers, i re complex numbers, nd b k+ dt t b... dt t b k = t... t k b k+ dt t b... dt k t k b k The vlue of the integrl bove only depends on the homotopy clss of the pth connecting nd m+ on C\{,..., m }. To be concrete, when possible, we will integrte over the rel line. 5
6 It is esy to see (for instnce, in [6]) tht, ( I n,...,n m ( :... : m : m+ ) = ( ) m Li n,...,n m, 3 m,...,, ) m+ m m Li n,...,n m (x,...,x m ) = ( ) m I n,...,n m ((x...x m ) :... : x m : ) which gives n nlytic continution to multiple polylogrithms. For instnce, with the bove convention bout integrting over rel segment, simple polylogrithms hve n nlytic continution to C \ [, ). There re modified versions of these functions which re nlytic in lrger sets, like the Bloch-Wigner dilogrithm, D(z) := Im(Li (z)) + log z rg( z) z C \ [, ) (6) which cn be extended s rel nlytic function in C \ {, } nd continuous in C. We will eventully use some properties of D(z): D( z) = D(z) ( D R ) (7) θ log sin t dt = D(e iθ ) = n= sin(nθ) n (8) More bout D(z) cn be found in []. Often we will write polylogrithms evluted in rguments of modulo greter thn, mening n nlytic continution given by the integrl. Although the vlue of these multivlued functions my not be uniquely defined, we will lwys get liner combintions of these functions which re one-vlued, since they represent Mhler mesures of certin polynomils. Now recll eqution (4). If the Mhler mesure of P α is( liner combintion α of multiple polylogrithms, nd if we write = i ) x + α x+i α x i α, then it is likely tht the Mhler mesure of P α will be lso liner combintion of multiple polylogrithms. This will be the bsis of our work. In order to express the results more clerly, we will estblish some nottion. Definition 5 Let G := σ, σ, τ ( = Z/Z Z/Z Z/Z) 6
7 n belin group generted by the following ctions in the set (R ) : σ : (,b) (,b) σ : (,b) (, b) ( τ : (,b), b) Also consider the following multiplictive chrcter: χ : G {, } χ(σ ) = χ(σ ) = χ(τ) = Definition 6 Given (,b) R,, define, log(,b) := log Definition 7 Let R, x,y C, L r(x) := Li r (x) Li r ( x) L r:(x) := log (Li r (x) Li r ( x)) L r,s(x,y) := ( χ(σ)li r,s ((x,y), ) σ ) σ G L r,s:(x,y) := σ Gχ(σ) ( log, ) σ ( ( Li r,s (x,y), ) σ ) where (x,y ) (x,y ) = (x x,y y ) is the component-wise product. Observtion 8 Let R, x,y C, then, L r,s(x,y) = L r,s(x, y) = L r,s( x,y) nd nlogously with L r,s: 7
8 Observe lso tht the weight of ny of the functions bove is equl to the sum of its subindexes. We will need some technicl Propositions: Proposition 9 Given R > L w +w n T (i) dw w = 4nL n+() + 4L n:() (9) PROOF. By definition, L w +w n T (i) dw w = T = i ( w (Li n i + w (Li n (ix) Li n ( ix)) ) ( )) w Li n i dw + w w dx x + () (which cn be proved in the sme wy s eqution (4) in Proposition ). Recll tht ( ) Li n (ix) = I n ix : = dt... dt t t + i dt t x }{{} n using this nd the fct tht t+ i t i x x = ix, the integrl in () becomes t x + 8 t t... t n x dt t x + dt t... dt n t n dx x + () But x dx (t x + )(x + ) = = ( x + t (t ) log = t t ) x dx x + t ( x + t ) x + ds s + log t () 8
9 The integrl in () becomes 8 t t... t n t ds s dt t dt t... dt n t n 8 log t t... t n dt t dt t... dt n t n Although the sum of these two integrls is well defined, ech of them is not defined for > if we choose the rel segment [, ] s the integrtion pth. We choose different integrtion pth, such s γ(θ) = eiθ + for π θ. By using t = t t, nd ds + t = t s t ds ds s t n s we get ( ) 4n (I n+ : I n+ ( )) ( ) : 4 log (I n : I n ( )) : = 4n(Li n+ () Li n+ ( )) + 4 log (Li n () Li n ( )) = 4nL n+() + 4L n:() Observe tht we will only use the bove Proposition for the cse n =. Proposition For z = e iθ, R >, L x n(z) x + dx + L x n(z) x + dx = i L n,(iz, i) (3) PROOF. By definition, the sum of the integrls is equl to ( (Li n (zx) Li n ( zx)) x + + ) dx (4) x + Using tht dt Li n (zx) = t zx dt t... dt t = t... t n x dt t z dt t... dt n t n The term in (4) with t... t n x ( is equl to x + t + z ) dt t dt t z... dt n t n x + dx 9
10 Writing ( = i ) x + x+i x i, we get i (I n, ( ) ( ) ( ) z : i : I n, z : i : + I n, z : i : I n, ( )) z : i : = i ( (Li n, iz, i ) ( Li n, iz, i ) ( + Li n, iz, i ) ( Li n, iz, i )) The other integrl cn be computed in similr wy, (or tking dvntge of the symmetry ) ( ) i iz (Li n,, i Li n, ( iz ) ( ) iz, i + Li n,, i Li n, ( iz )), i Adding both lines, we get the result. Proposition For z = e iθ, R >, T v L x +v dv n(z) x + v dx v + +v T L x n(z) v +v x +v + v dx dv v = i(l n,(z, ) + L n,:(z, )) (5) PROOF. Consider the first integrl. By definition, this is T v +v dv (Li n (zx) Li n ( zx)) x + v dx v +v We do the sme chnge of vribles s in the Proof of eqution (4) in Proposition nd we get 4i (Li n (zx) Li n ( zx)) y dx x + y dy y + (6) In the sme wy s in (), we hve: y dy (x + y )(y + ) = x log + x ds s
11 Integrl (6) becomes: 4i t... t n x ( t + z ) dt t dt... t z dt n t n log + x ds dx s x This integrl decomposes into two summnds, one with x ds nd the other s with log. But, s before, when we do this, ech summnd not longer converges if we integrte on the rel intervl [, ] nd if. So, we will chnge the pth of integrtion s we did before, to γ(θ) = eiθ + for π θ. We first compute the integrl with we get x ds s. By using tht ( = ) x x x+, ( ) i (I n, z : : I n, ( ) z : : + I n, ( ) ( )) z : : I n, z : : ( = i (Li n, z, ) ( Li n, z, ) ( + Li n, z, ) ( Li n, z, )) The term with log yields ( ) i log (I n, z : : I n, ( ) z : : + I n, ( ) ( )) z : : I n, z : : ( = i log (Li n, z, ) ( Li n, z, ) ( + Li n, z, ) ( Li n, z, )) The other integrl is bsolutely nlogous, except tht we use y dy (y x + )(y + ) = x log x ds s (we cn lso compute it using the symmetry ). 4 Exmples of the first kind The Mhler mesure of the polynomils tht we study in this section depends only on the bsolute vlue of the prmeter α, hence, we will write the formule with = α in order to simplify nottion.
12 We strt with the simple polynomil + y, whose Mhler mesure is m( + y) = π π log + e iθ dθ = log + (7) The first ppliction of our procedure yields: Theorem For R >, πm(( + x) + ( x)y) = il (i) (8) PROOF. By eqution (4) in Proposition, ( πm(( + x) + ( x)y) =πm( + x) + πm + x ) + x y (we mde z = w ). = log + z dz z + = = i w ds s ( w + i w i log w ) dw dw w + (9) = i (I ( i ) ( )) i : I : = i (Li (i) Li ( i)) = il (i) Recll tht we men the nlytic continution of Li. If we wnt to void this nd work with the series, the formul should be stted in the following wy: il πm(( + x) + ( x)y) = (i) if π log il (i) if > () The cse is cler. For the > cse, m(( + x) + ( x)y) =m(( x)y + ( + x)) ( ) = log + m ( x)y + ( + x) which proves the formul ().
13 Now we pply the procedure gin: Theorem 3 For R >, π m(( + w)( + x) + ( w)( x)y) = 4L 3() L :() () PROOF. By Proposition 9, π m(( + w)( + x) + ( w)( x)y) = L w +w T (i) dw w = 4L 3() L :() As before, we cn express this with the following formul: π 4L m(( + w)( + x) + ( w)( x)y) = 3() L :() if π log + 4L3 () L : () if > Note tht we could compute the sme Mhler mesure using the formul () for the Mhler mesure of m(( + x) + ( x)y). By doing this, we obtin different formul for the Mhler mesure of the polynomil considered in Theorem 3: Theorem 4 For R >, π m(( + w)( + x) + ( w)( x)y) = iπl (i) L,(, i) PROOF. We integrte formul () nd use Proposition π m(( + w)( + x) + ( w)( x)y) = i L x (i) x + dx +π ( log x) x + dx i L x (i) x + dx () The sum of the first nd third integrls is L,(, i) = L,(, i) becuse of Proposition with n = nd z = i. The second integrl is the sme s the one tht occurs in eqution (9) nd it yields iπl (i). 3
14 Adding the three terms together we prove the sttement. If we compre the two formule tht we hve got for m(( + w)( + x) + ( w)( x)y), we hve proved the following equlity between multiple polylogrithms: Corollry 5 For R >, 4L 3() L :() = iπl (i) L,(, i) (3) PROOF. This Corollry is obtined from the two formule for the Mhler mesure of ( + w)( + x) + ( w)( x)y. See the Appendix for direct proof. Let us do the process of integrtion one more time. Theorem 6 For R >, π 3 m(( + v)( + w)( + x) + ( v)( w)( x)y) = 4πL 3() πl :() i(l,(i, ) + L,:(i, )) (4) PROOF. We will integrte eqution () of Theorem 4. The term of higher weight corresponds to T L x (i) x + v +v v +v dv dx v T L x (i) v +v x +v + v dx dv v We solve this prt by Proposition, setting z = i nd n =. The third term is: π L v +v T (i) dv v = 4πL 3() πl :() by Proposition 9. 4
15 5 Exmples of the second kind In this section we still hve tht ll the Mhler mesures only depend on = α. We strt with different polynomil, + x + y + z. Theorem 7 For R >, π L m( + x + y + z) = F() = 3() if π log + L3 () if (5) PROOF. This ws proved by Vndervelde []. It is lso possible to dpt some of the proofs of m(+x+y+z) = 7 ζ(3). For instnce, following Smyth π [], For V = m( + x + y + z) = m(( + x + y + z) V ) =m( + x z + xy + xyz) = m(x + y + (x + y)z) = π (Li 3() Li 3 ( )) = π L 3() for. Another possibility is to dpt the elementry proof given in Boyd []. For : ( ) + y π m( + x + y + z) = π m( + y + x( + w)) = π m + w + x = π π π = log + + e it + e is ds dt = (π t) log + e it dt t s π π log + e it log + e is ds dt π s log + e is ds = (here we hve used tht, nd formul (7)). t log + e it dt 5
16 Now use tht log + e it ( ) n = Re n n= n e int = n= ( ) n cos(nt) n (6) n nd pply integrtion by prts, π m( + x + y + z) = π + n= n= ( ) n sin(nt) n n dt = 4 ( ) n sin(nt) n n=(odd) n t π n n 3 = (Li 3() Li 3 ( )) When, use tht ( m( + x + y + z) = log + m + x ) + y + z If we compre with formul (), for exmple, we my wonder whether the formul in the second cse of (5) is vlue of L 3(): F(x)? = L x 3(), x > Mening, s lwys, some brnch of the nlytic continution of Li 3. We will see now tht this is flse. We should hve for x >, ( ( ) ( π log x + Li 3 Li 3? (Li3 (x) Li 3 ( x)) x x)) Differentiting, nd using tht x Li 3(x) = Li (x) π x + ( (Li ) ( ))? Li x x x x (Li (x) Li ( x)) Multiplying by x, nd differentiting gin, Li ( x ) ( Li ) = Li (x) Li ( x) x Since x >, the left term is log ( ) x + x 6
17 (using tht the principl brnch of Li is equl to log( x)). The term in the right is equl to t x t + x dt = lim α π α i e iθ dθ e iθ + x + log(x + ) (we integrted on the pth γ(θ) = eiθ + for π θ ). ( ) x = log x = log ( ) x + iπ x ( cosβ e iβ + lim log x β π x ) + log(x + ) γ(θ) = e iθ + for π θ represents the other homotopy clss nd in this cse, the integrl is = log ( ) x + + iπ x Hence both functions re not equl, which implies tht F(x) cn not be expressed s L x 3(). Theorem 8 For R >, π 3 m(( + w)( + x) + ( w)(y + z)) = iπ L (i) + il 3,(i, i) (7) PROOF. Applying Proposition to formul (5), π 3 m(( + w)( + x) + ( w)(y + z)) = 4 L x 3() x + dx +π ( log x) x + dx + 4 L x 3() x + dx (8) The sum of the first nd third integrls is il 3,(i, i) becuse of Proposition with n = 3 nd z =. The second term is the sme s the one in integrl (9) in Theorem, equl to iπ L (i). Adding the three terms together we prove the sttement. 7
18 Finlly our result with the mximum number of vribles: Theorem 9 For R >, π 4 m(( + v)( + w)( + x) + ( v)( w)(y + z)) = 4π L 3() π L :() + 4 (L 3,(, ) + L 3,:(, )) (9) PROOF. We will integrte eqution (8) of the bove Theorem s lwys. The highest weight term corresponds to i T v L x +v dv 3() x + v dx v + i +v T L x 3() v +v x +v + v dx dv v which cn be evluted using Proposition setting z =, n = 3. Finlly, π T L v +v (i) dv v = 4π L 3() π L :() by Proposition 9. 6 Integrtion of Millot s formul So fr we hve been considering the cses of + y nd ( + x) + (y + z) nd integrted them severl times. We my wonder wht hppens with n intermedite cse, nmely + x + y. This cse is not so esy to hndle, so we will consider the following vrint of the two vrible cse: +x+( )y, with C. This time the Mhler mesure will depend on the rgument of s well. According to Millot [7], α log + β log b + γ log c + D ( e iγ) b πm( + bx + cy) = π log mx{, b, c } not (3) 8
19 c α b π / θ / β γ = - i tn ( θ / ) A) B) C) Fig.. A) Reltion mong the prmeters in Millot s formul. B) Tringle for the generl cse of + + ( )y. C) Tringle for the cse = i tn ( θ ). Where stnds for the sttement tht, b, nd c re the lengths of the sides of tringle; nd α, β, nd γ re the ngles opposite to the sides of lengths, b nd c respectively (Figure.A). In our prticulr cse, πm( + x + ( )y) = rg log + rg( ) log D() if Im() + D(ā) if Im() < (3) since, nd re the lengths of the sides of the tringle whose vertices re, nd in the complex plne (Figure.B). For the rgument in the dilogrithm, γ = rg(), then we hve to tke or ā so γ is lwys positive. We will integrte s lwys. We replce by w +w. Theorem We hve the following: π m(( + w)( + y) + ( w)(x y)) = L 3() + π log (3) PROOF. We will pply Proposition to eqution (3) nd then the chnge of vribles w = e iθ, which implies w = i tn ( ) θ +w. With tht chnge, will be lwys pure imginry, so rg = π (Figure.C). π m(( + w)( + y) + ( w)(x y)) = = π ( π π log cos ( ) θ + ( ) ( ( ))) θ log θ θ tn + D i tn dθ 9
20 Using the definition (6) of Bloch Wigner dilogrithm, = π ( π ( ) ( ( )))) θ θ log cos + Im Li (i tn π dθ For the first term, mke τ = π θ, this prt becomes, π π by eqution (8). log sin τ dτ = π π (log sin τ log ) dτ = π 4 D(eiπ ) + π log = π log For the second term, mke x = tn ( ) θ, then dθ = dx, the integrl becomes, x + dx Im(Li (i x )) x + = i dx (Li (ix) Li ( ix)) x + = L 3() The lst equlity is prticulr cse of the vlue computed for expression () in Proposition 9 Adding both terms we obtin the result. 7 Concluding remrks To conclude, let us observe tht ll the presented formule shre common feture. Let us ssign weight to ny Mhler mesure, to π nd to ny logrithmic function. Then ll the formule re homogeneous, mening ll the monomils hve the sme weight, nd this weight is equl to the number of vribles of the corresponding polynomil. Appendix : Some dditionl remrks bout the cse = In this ppendix, we would like to give some dditionl detils bout the computtion of the specific formule tht occur in the tble of results for =. All but two of these formule cn be directly deduced from the Theorems we
21 hve proved. These exceptions re: the formul with the term ζ(5) nd formul (). For the formul with the term ζ(5): Theorem We hve the following, π 4 m(( + v)( + w)( + x) + ( v)( w)(y + z)) = 93ζ(5) (33) PROOF. By Theorem 9 we hve to prove tht 4π L 3() + 4 L 3,(, ) = 93ζ(5) (34) i.e., 7π ζ(3) + 8(Li 3, (, ) Li 3, (, ) + Li 3, (, ) Li 3, (, )) = 93ζ(5) Now we use formul (75) of [], which in this prticulr cse, sttes tht Li 3, (x,y) = Li 5(xy) + Li 3 (x) Li (y) + 3Li 5 (x) + Li 5 (y) Li (xy)(li 3 (x) + Li 3 (y)) for x,y = ±. Tking into ccount tht Li k () = ζ(k) nd Li k ( ) = ( k ) ζ(k) (35) we get Li 3, (, ) Li 3, (, ) + Li 3, (, ) Li 3, (, ) = 93 ζ()ζ(3) ζ(5) We obtin the result by using tht ζ() = π 6 For formul () we hve the following Proposition We hve j<k ( ) j+k+ (j + ) 3 k = 7 4 ζ(3)l(χ 4, ) 3 ζ()l(χ 4, ) log L(χ 4, 3) + <m<n (m even) χ 4 (n) m n 3
22 PROOF. Writing l = j + nd l + n = k, the left side is equl to j<k ( ) j+k+ (j + ) 3 k = <l,n (l,n odd) ( ) (l )+(l+n) + l 3 (l + n) = <l,n (l odd) χ 4 (n) l 3 (l + n) Now write: l 3 (l + n) = l 3 n l n + l n 3 n 3 (l + n) Using formule (35), we get the first two terms of the sttement. We need to look t the lst two terms together in order to ensure convergence, χ 4 (n) ( n 3 l ) = χ 4 (n) n log + l + n <n n 3 m <n <l (l odd) <m (m even) nd the sttement follows. Formul () cn be obtined from the bove Proposition nd the fct tht L(χ 4, ) = π 4. Appendix : A direct proof for Corollry 5 Recll the sttement: Corollry 5 For R >, 4L 3() L :() = iπl (i) L,(, i) PROOF. First observe tht fter the chnge the equlity (3) remins the sme with the cnceltion of term of the form π log in ech side. Then, it is enough to prove eqution (3) for <. Eqution (3) is equivlent to 8 n=(odd) n n 3 4 log dt + 4 t + dt t n=(odd) n n? = iπ n=(odd) ( ) dt t + + dt t + i n n n
23 = π +4 n=(odd) rctn s s ( ) n n n ( π rctn ( s ) ) rctn(s) ds (36) Our strtegy will be s follows: we will prove the equlity for the derivtives nd for the prticulr cse =. In order to prove the equlity for the derivtives, we will do the sme, i.e., we will exmine the cse = nd differentite gin nd compre the second derivtives. Let us strt with =. The term in (36) becomes = π n=(odd) ( ) n n + π rctn s s ds 8 rctn s s ds Mke s = tnx: rctn s s ds = π 4 x sin x cos x dx = x log(tnx) π 4 π 4 x(log( sinx) log( cosx)) dx The first term is zero. For the second term mke y = π x = = π 4 π x log( sinx) dx + s log( sins) ds + π π π 4 π π 4 (π y) log( siny) dy log( sins) ds Using properties (7) nd (8), of the Bloch Wigner dilogrithm, = s D(e is ) π π n= sin(ns) n ds π D(eπis ) π π 4 3
24 = n=(odd) n 3 + π n=(odd) ( ) n n Using the power series of +s nd integrting, it is esy to see tht rctn s s ds = n=(odd) ( ) n n Putting ll of this together in eqution (36), we conclude: iπl (i) L,(, i) = 8 n=(odd) n 3 = 4L 3() s we expected. We differentite the originl eqution (36) nd multiply it by : 4 n=(odd) n n 4 log n=(odd) n n? = π n=(odd) i n n n + 4 ( rctn s s + s + ) ds (37) Set =, we get 4 3 ζ() lim 4 log (log( + ) log( )) =? π (log( + i) log( i)) i This is n equlity, becuse the first term in the left nd the term in the right re equl to π nd the other term is zero. Apply integrtion by prts on the lst term of (37): 4 ( rctn s s + s + ( ( ) = π rctn rctn() ) ) ds 4 ( ( ) s ds rctn rctn(s) ) s + 4
25 Now we differentite (37), 4 log +4 ( n=(odd) n? = π n=(odd) ) s + + s ds s + s + (i) n π + And this is n equlity indeed, which cn be seen from the fct: 4 log = 4 ( ) s + s ds s + Acknowledgements I m deeply grteful to Fernndo Rodriguez Villegs for his constnt guidnce nd support nd for shring severl ides tht hve enriched this work. I would lso like to thnk Sm Vndervelde for severl helpful discussions, nd the Referee, whose suggestions led to severl improvements. References [] J. M. Borwein, D. M. Brdley, D. J. Brodhurst, Evlutions of k-fold Euler/Zgier sums: compendium of results for rbitrry k, Electronic J. Combin. 4 (997), no., #R5. [] D. W. Boyd, Specultions concerning the rnge of Mhler s mesure, Cnd. Mth. Bull. 4 (98), [3] D. W. Boyd, Mhler s mesure nd specil vlues of L-functions, Experiment. Mth. 7 (998), [4] D. W. Boyd, F. Rodriguez Villegs, Mhler s mesure nd the dilogrithm (I), Cnd. J. Mth. 54 (), [5] A. B. Gonchrov, Polylogrithms in rithmetic nd geometry, Proc. ICM-94 Zurich (995), [6] A. B. Gonchrov, Multiple polylogrithms nd mixed Tte motives, (Preprint). 5
26 [7] V. Millot, Géométrie d Arkelov des vriétés toriques et fibrés en droites intégrbles. Mém. Soc. Mth. Fr. (N.S.) 8 () 9pp. [8] F. Rodriguez Villegs, Modulr Mhler mesures I, in Topics in number theory (University Prk, PA 997), Mth. Appl., 467, Kluwer Acd. Publ. Dordrecht, 999, pp [9] C. J. Smyth, On mesures of polynomils in severl vribles, Bull. Austrl. Mth. Soc. Ser. A 3 (98), Corrigendum (with G. Myerson): Bull. Austrl. Mth. Soc. 6 (98), [] C. J. Smyth, An explicit formul for the Mhler mesure of fmily of 3-vrible polynomils, J. Nombres Bordeux 4 (), [] S. Vndervelde, A formul for the Mhler mesure of xy + bx + cy + d. J. Number Theory (3), no., 84. [] D. Zgier, The Dilogrithm function in Geometry nd Number Theory, Number Theory nd relted topics, Tt Inst. Fund. Res. Stud. Mth. Bomby (988),
Physics 116C Solution of inhomogeneous ordinary differential equations using Green s functions
Physics 6C Solution of inhomogeneous ordinry differentil equtions using Green s functions Peter Young November 5, 29 Homogeneous Equtions We hve studied, especilly in long HW problem, second order liner
More informationAM1 Mathematical Analysis 1 Oct Feb Exercises Lecture 3. sin(x + h) sin x h cos(x + h) cos x h
AM Mthemticl Anlysis Oct. Feb. Dte: October Exercises Lecture Exercise.. If h, prove the following identities hold for ll x: sin(x + h) sin x h cos(x + h) cos x h = sin γ γ = sin γ γ cos(x + γ) (.) sin(x
More informationA REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007
A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus
More informationThe Regulated and Riemann Integrals
Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue
More informationExam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH-1105 Instructor: Attila Máté 1
Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH-115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixed-point itertion to converge when solving the eqution
More informationTheoretical foundations of Gaussian quadrature
Theoreticl foundtions of Gussin qudrture 1 Inner product vector spce Definition 1. A vector spce (or liner spce) is set V = {u, v, w,...} in which the following two opertions re defined: (A) Addition of
More informationP 3 (x) = f(0) + f (0)x + f (0) 2. x 2 + f (0) . In the problem set, you are asked to show, in general, the n th order term is a n = f (n) (0)
1 Tylor polynomils In Section 3.5, we discussed how to pproximte function f(x) round point in terms of its first derivtive f (x) evluted t, tht is using the liner pproximtion f() + f ()(x ). We clled this
More informationConvergence of Fourier Series and Fejer s Theorem. Lee Ricketson
Convergence of Fourier Series nd Fejer s Theorem Lee Ricketson My, 006 Abstrct This pper will ddress the Fourier Series of functions with rbitrry period. We will derive forms of the Dirichlet nd Fejer
More informationMath 1B, lecture 4: Error bounds for numerical methods
Mth B, lecture 4: Error bounds for numericl methods Nthn Pflueger 4 September 0 Introduction The five numericl methods descried in the previous lecture ll operte by the sme principle: they pproximte the
More informationExamples of Mahler Measures as Multiple Polylogarithms
Examples of Mahler Measures as Multiple Polylogarithms The many aspects of Mahler s measure Banff International Research Station for Mathematical Innovation and Discovery BIRS, Banff, Alberta, Canada April
More informationReversing the Chain Rule. As we have seen from the Second Fundamental Theorem ( 4.3), the easiest way to evaluate an integral b
Mth 32 Substitution Method Stewrt 4.5 Reversing the Chin Rule. As we hve seen from the Second Fundmentl Theorem ( 4.3), the esiest wy to evlute n integrl b f(x) dx is to find n ntiderivtive, the indefinite
More informationProperties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives
Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums - 1 Riemnn
More informationUnit #9 : Definite Integral Properties; Fundamental Theorem of Calculus
Unit #9 : Definite Integrl Properties; Fundmentl Theorem of Clculus Gols: Identify properties of definite integrls Define odd nd even functions, nd reltionship to integrl vlues Introduce the Fundmentl
More informationConvex Sets and Functions
B Convex Sets nd Functions Definition B1 Let L, +, ) be rel liner spce nd let C be subset of L The set C is convex if, for ll x,y C nd ll [, 1], we hve 1 )x+y C In other words, every point on the line
More informationMath& 152 Section Integration by Parts
Mth& 5 Section 7. - Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible
More informationp(t) dt + i 1 re it ireit dt =
Note: This mteril is contined in Kreyszig, Chpter 13. Complex integrtion We will define integrls of complex functions long curves in C. (This is bit similr to [relvlued] line integrls P dx + Q dy in R2.)
More informationOrthogonal Polynomials and Least-Squares Approximations to Functions
Chpter Orthogonl Polynomils nd Lest-Squres Approximtions to Functions **4/5/3 ET. Discrete Lest-Squres Approximtions Given set of dt points (x,y ), (x,y ),..., (x m,y m ), norml nd useful prctice in mny
More informationUNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3
UNIFORM CONVERGENCE Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 Suppose f n : Ω R or f n : Ω C is sequence of rel or complex functions, nd f n f s n in some sense. Furthermore,
More informationAdvanced Calculus: MATH 410 Uniform Convergence of Functions Professor David Levermore 11 December 2015
Advnced Clculus: MATH 410 Uniform Convergence of Functions Professor Dvid Levermore 11 December 2015 12. Sequences of Functions We now explore two notions of wht it mens for sequence of functions {f n
More informationAnti-derivatives/Indefinite Integrals of Basic Functions
Anti-derivtives/Indefinite Integrls of Bsic Functions Power Rule: In prticulr, this mens tht x n+ x n n + + C, dx = ln x + C, if n if n = x 0 dx = dx = dx = x + C nd x (lthough you won t use the second
More informationImproper Integrals, and Differential Equations
Improper Integrls, nd Differentil Equtions October 22, 204 5.3 Improper Integrls Previously, we discussed how integrls correspond to res. More specificlly, we sid tht for function f(x), the region creted
More informationEuler-Maclaurin Summation Formula 1
Jnury 9, Euler-Mclurin Summtion Formul Suppose tht f nd its derivtive re continuous functions on the closed intervl [, b]. Let ψ(x) {x}, where {x} x [x] is the frctionl prt of x. Lemm : If < b nd, b Z,
More informationSection 17.2 Line Integrals
Section 7. Line Integrls Integrting Vector Fields nd Functions long urve In this section we consider the problem of integrting functions, both sclr nd vector (vector fields) long curve in the plne. We
More information20 MATHEMATICS POLYNOMIALS
0 MATHEMATICS POLYNOMIALS.1 Introduction In Clss IX, you hve studied polynomils in one vrible nd their degrees. Recll tht if p(x) is polynomil in x, the highest power of x in p(x) is clled the degree of
More informationMORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.)
MORE FUNCTION GRAPHING; OPTIMIZATION FRI, OCT 25, 203 (Lst edited October 28, 203 t :09pm.) Exercise. Let n be n rbitrry positive integer. Give n exmple of function with exctly n verticl symptotes. Give
More information63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1
3 9. SEQUENCES AND SERIES 63. Representtion of functions s power series Consider power series x 2 + x 4 x 6 + x 8 + = ( ) n x 2n It is geometric series with q = x 2 nd therefore it converges for ll q =
More informationF (x) dx = F (x)+c = u + C = du,
35. The Substitution Rule An indefinite integrl of the derivtive F (x) is the function F (x) itself. Let u = F (x), where u is new vrible defined s differentible function of x. Consider the differentil
More information1 The Riemann Integral
The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re
More informationTHE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS.
THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrd-lindelof-theorem/ This document is proof of the existence-uniqueness theorem
More informationMath Calculus with Analytic Geometry II
orem of definite Mth 5.0 with Anlytic Geometry II Jnury 4, 0 orem of definite If < b then b f (x) dx = ( under f bove x-xis) ( bove f under x-xis) Exmple 8 0 3 9 x dx = π 3 4 = 9π 4 orem of definite Problem
More informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 203 Outline Riemnn Sums Riemnn Integrls Properties Abstrct
More informationARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac
REVIEW OF ALGEBRA Here we review the bsic rules nd procedures of lgebr tht you need to know in order to be successful in clculus. ARITHMETIC OPERATIONS The rel numbers hve the following properties: b b
More informationWe partition C into n small arcs by forming a partition of [a, b] by picking s i as follows: a = s 0 < s 1 < < s n = b.
Mth 255 - Vector lculus II Notes 4.2 Pth nd Line Integrls We begin with discussion of pth integrls (the book clls them sclr line integrls). We will do this for function of two vribles, but these ides cn
More informationMath 100 Review Sheet
Mth 100 Review Sheet Joseph H. Silvermn December 2010 This outline of Mth 100 is summry of the mteril covered in the course. It is designed to be study id, but it is only n outline nd should be used s
More information7.2 The Definite Integral
7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where
More informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 2013 Outline 1 Riemnn Sums 2 Riemnn Integrls 3 Properties
More informationA basic logarithmic inequality, and the logarithmic mean
Notes on Number Theory nd Discrete Mthemtics ISSN 30 532 Vol. 2, 205, No., 3 35 A bsic logrithmic inequlity, nd the logrithmic men József Sándor Deprtment of Mthemtics, Bbeş-Bolyi University Str. Koglnicenu
More informationImproper Integrals. Type I Improper Integrals How do we evaluate an integral such as
Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph
More informationTaylor Polynomial Inequalities
Tylor Polynomil Inequlities Ben Glin September 17, 24 Abstrct There re instnces where we my wish to pproximte the vlue of complicted function round given point by constructing simpler function such s polynomil
More informationAbsolute values of real numbers. Rational Numbers vs Real Numbers. 1. Definition. Absolute value α of a real
Rtionl Numbers vs Rel Numbers 1. Wht is? Answer. is rel number such tht ( ) =. R [ ( ) = ].. Prove tht (i) 1; (ii). Proof. (i) For ny rel numbers x, y, we hve x = y. This is necessry condition, but not
More informationf(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral
Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one
More informationPolynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230
Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given
More information1 1D heat and wave equations on a finite interval
1 1D het nd wve equtions on finite intervl In this section we consider generl method of seprtion of vribles nd its pplictions to solving het eqution nd wve eqution on finite intervl ( 1, 2. Since by trnsltion
More informationWe divide the interval [a, b] into subintervals of equal length x = b a n
Arc Length Given curve C defined by function f(x), we wnt to find the length of this curve between nd b. We do this by using process similr to wht we did in defining the Riemnn Sum of definite integrl:
More informationMATH 185: COMPLEX ANALYSIS FALL 2009/10 PROBLEM SET 5 SOLUTIONS. cos t cos at dt + i
MATH 85: COMPLEX ANALYSIS FALL 9/ PROBLEM SET 5 SOLUTIONS. Let R nd z C. () Evlute the following integrls Solution. Since e it cos t nd For the first integrl, we hve e it cos t cos t cos t + i t + i. sin
More informationChapter 4 Contravariance, Covariance, and Spacetime Diagrams
Chpter 4 Contrvrince, Covrince, nd Spcetime Digrms 4. The Components of Vector in Skewed Coordintes We hve seen in Chpter 3; figure 3.9, tht in order to show inertil motion tht is consistent with the Lorentz
More informationChapter 6 Techniques of Integration
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Chpter 6 Techniques of Integrtion Recll: Some importnt integrls tht we hve lernt so fr. Tle of Integrls n+ n d = + C n + e d = e + C ( n ) d = ln
More informationReview of basic calculus
Review of bsic clculus This brief review reclls some of the most importnt concepts, definitions, nd theorems from bsic clculus. It is not intended to tech bsic clculus from scrtch. If ny of the items below
More informationLecture 1. Functional series. Pointwise and uniform convergence.
1 Introduction. Lecture 1. Functionl series. Pointwise nd uniform convergence. In this course we study mongst other things Fourier series. The Fourier series for periodic function f(x) with period 2π is
More informationNUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by.
NUMERICAL INTEGRATION 1 Introduction The inverse process to differentition in clculus is integrtion. Mthemticlly, integrtion is represented by f(x) dx which stnds for the integrl of the function f(x) with
More informationThe First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).
The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples
More informationAdvanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004
Advnced Clculus: MATH 410 Notes on Integrls nd Integrbility Professor Dvid Levermore 17 October 2004 1. Definite Integrls In this section we revisit the definite integrl tht you were introduced to when
More informationS. S. Dragomir. 2, we have the inequality. b a
Bull Koren Mth Soc 005 No pp 3 30 SOME COMPANIONS OF OSTROWSKI S INEQUALITY FOR ABSOLUTELY CONTINUOUS FUNCTIONS AND APPLICATIONS S S Drgomir Abstrct Compnions of Ostrowski s integrl ineulity for bsolutely
More informationODE: Existence and Uniqueness of a Solution
Mth 22 Fll 213 Jerry Kzdn ODE: Existence nd Uniqueness of Solution The Fundmentl Theorem of Clculus tells us how to solve the ordinry differentil eqution (ODE) du = f(t) dt with initil condition u() =
More informationAP Calculus Multiple Choice: BC Edition Solutions
AP Clculus Multiple Choice: BC Edition Solutions J. Slon Mrch 8, 04 ) 0 dx ( x) is A) B) C) D) E) Divergent This function inside the integrl hs verticl symptotes t x =, nd the integrl bounds contin this
More informationAbstract inner product spaces
WEEK 4 Abstrct inner product spces Definition An inner product spce is vector spce V over the rel field R equipped with rule for multiplying vectors, such tht the product of two vectors is sclr, nd the
More informationChapter 3 Polynomials
Dr M DRAIEF As described in the introduction of Chpter 1, pplictions of solving liner equtions rise in number of different settings In prticulr, we will in this chpter focus on the problem of modelling
More informationOn the Generalized Weighted Quasi-Arithmetic Integral Mean 1
Int. Journl of Mth. Anlysis, Vol. 7, 2013, no. 41, 2039-2048 HIKARI Ltd, www.m-hikri.com http://dx.doi.org/10.12988/ijm.2013.3499 On the Generlized Weighted Qusi-Arithmetic Integrl Men 1 Hui Sun School
More informationPhil Wertheimer UMD Math Qualifying Exam Solutions Analysis - January, 2015
Problem 1 Let m denote the Lebesgue mesure restricted to the compct intervl [, b]. () Prove tht function f defined on the compct intervl [, b] is Lipschitz if nd only if there is constct c nd function
More informationHow can we approximate the area of a region in the plane? What is an interpretation of the area under the graph of a velocity function?
Mth 125 Summry Here re some thoughts I ws hving while considering wht to put on the first midterm. The core of your studying should be the ssigned homework problems: mke sure you relly understnd those
More informationDiscrete Least-squares Approximations
Discrete Lest-squres Approximtions Given set of dt points (x, y ), (x, y ),, (x m, y m ), norml nd useful prctice in mny pplictions in sttistics, engineering nd other pplied sciences is to construct curve
More informationPartial Derivatives. Limits. For a single variable function f (x), the limit lim
Limits Prtil Derivtives For single vrible function f (x), the limit lim x f (x) exists only if the right-hnd side limit equls to the left-hnd side limit, i.e., lim f (x) = lim f (x). x x + For two vribles
More informationIntegration Techniques
Integrtion Techniques. Integrtion of Trigonometric Functions Exmple. Evlute cos x. Recll tht cos x = cos x. Hence, cos x Exmple. Evlute = ( + cos x) = (x + sin x) + C = x + 4 sin x + C. cos 3 x. Let u
More information1 Techniques of Integration
November 8, 8 MAT86 Week Justin Ko Techniques of Integrtion. Integrtion By Substitution (Chnge of Vribles) We cn think of integrtion by substitution s the counterprt of the chin rule for differentition.
More informationNew Expansion and Infinite Series
Interntionl Mthemticl Forum, Vol. 9, 204, no. 22, 06-073 HIKARI Ltd, www.m-hikri.com http://dx.doi.org/0.2988/imf.204.4502 New Expnsion nd Infinite Series Diyun Zhng College of Computer Nnjing University
More informationIndefinite Integral. Chapter Integration - reverse of differentiation
Chpter Indefinite Integrl Most of the mthemticl opertions hve inverse opertions. The inverse opertion of differentition is clled integrtion. For exmple, describing process t the given moment knowing the
More informationMath Solutions to homework 1
Mth 75 - Solutions to homework Cédric De Groote October 5, 07 Problem, prt : This problem explores the reltionship between norms nd inner products Let X be rel vector spce ) Suppose tht is norm on X tht
More informationJournal of Inequalities in Pure and Applied Mathematics
Journl of Inequlities in Pure nd Applied Mthemtics GENERALIZATIONS OF THE TRAPEZOID INEQUALITIES BASED ON A NEW MEAN VALUE THEOREM FOR THE REMAINDER IN TAYLOR S FORMULA volume 7, issue 3, rticle 90, 006.
More informationMath 113 Exam 1-Review
Mth 113 Exm 1-Review September 26, 2016 Exm 1 covers 6.1-7.3 in the textbook. It is dvisble to lso review the mteril from 5.3 nd 5.5 s this will be helpful in solving some of the problems. 6.1 Are Between
More information13.4. Integration by Parts. Introduction. Prerequisites. Learning Outcomes
Integrtion by Prts 13.4 Introduction Integrtion by Prts is technique for integrting products of functions. In this Section you will lern to recognise when it is pproprite to use the technique nd hve the
More informationAn approximation to the arithmetic-geometric mean. G.J.O. Jameson, Math. Gazette 98 (2014), 85 95
An pproximtion to the rithmetic-geometric men G.J.O. Jmeson, Mth. Gzette 98 (4), 85 95 Given positive numbers > b, consider the itertion given by =, b = b nd n+ = ( n + b n ), b n+ = ( n b n ) /. At ech
More informationBernoulli Numbers Jeff Morton
Bernoulli Numbers Jeff Morton. We re interested in the opertor e t k d k t k, which is to sy k tk. Applying this to some function f E to get e t f d k k tk d k f f + d k k tk dk f, we note tht since f
More information4.4 Areas, Integrals and Antiderivatives
. res, integrls nd ntiderivtives 333. Ares, Integrls nd Antiderivtives This section explores properties of functions defined s res nd exmines some connections mong res, integrls nd ntiderivtives. In order
More informationSection 6.1 INTRO to LAPLACE TRANSFORMS
Section 6. INTRO to LAPLACE TRANSFORMS Key terms: Improper Integrl; diverge, converge A A f(t)dt lim f(t)dt Piecewise Continuous Function; jump discontinuity Function of Exponentil Order Lplce Trnsform
More informationDefinition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim
Mth 9 Course Summry/Study Guide Fll, 2005 [1] Limits Definition of Limit: We sy tht L is the limit of f(x) s x pproches if f(x) gets closer nd closer to L s x gets closer nd closer to. We write lim f(x)
More informationSOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 2014
SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 014 Mrk Scheme: Ech prt of Question 1 is worth four mrks which re wrded solely for the correct nswer.
More informationTHE QUADRATIC RECIPROCITY LAW OF DUKE-HOPKINS. Circa 1870, G. Zolotarev observed that the Legendre symbol ( a p
THE QUADRATIC RECIPROCITY LAW OF DUKE-HOPKINS PETE L CLARK Circ 1870, Zolotrev observed tht the Legendre symbol ( p ) cn be interpreted s the sign of multipliction by viewed s permuttion of the set Z/pZ
More informationChapters 4 & 5 Integrals & Applications
Contents Chpters 4 & 5 Integrls & Applictions Motivtion to Chpters 4 & 5 2 Chpter 4 3 Ares nd Distnces 3. VIDEO - Ares Under Functions............................................ 3.2 VIDEO - Applictions
More information1 The fundamental theorems of calculus.
The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl- new nme for nti-derivtive. Differentiting integrls. Tody we provide the connection
More informationMATH 144: Business Calculus Final Review
MATH 144: Business Clculus Finl Review 1 Skills 1. Clculte severl limits. 2. Find verticl nd horizontl symptotes for given rtionl function. 3. Clculte derivtive by definition. 4. Clculte severl derivtives
More informationComplex integration. L3: Cauchy s Theory.
MM Vercelli. L3: Cuchy s Theory. Contents: Complex integrtion. The Cuchy s integrls theorems. Singulrities. The residue theorem. Evlution of definite integrls. Appendix: Fundmentl theorem of lgebr. Discussions
More informationGoals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite
Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite
More informationChapter 8: Methods of Integration
Chpter 8: Methods of Integrtion Bsic Integrls 8. Note: We hve the following list of Bsic Integrls p p+ + c, for p sec tn + c p + ln + c sec tn sec + c e e + c tn ln sec + c ln + c sec ln sec + tn + c ln
More informationMath 360: A primitive integral and elementary functions
Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth 360 001 2017C: Integrl/functions 1 / 32 Setup for the integrl prtitions Definition:
More informationACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019
ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS MATH00030 SEMESTER 208/209 DR. ANTHONY BROWN 7.. Introduction to Integrtion. 7. Integrl Clculus As ws the cse with the chpter on differentil
More informationDefinite integral. Mathematics FRDIS MENDELU
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová Brno 1 Motivtion - re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function defined on [, b]. Wht is the re of the
More informationChapter 0. What is the Lebesgue integral about?
Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous
More informationReview of Riemann Integral
1 Review of Riemnn Integrl In this chpter we review the definition of Riemnn integrl of bounded function f : [, b] R, nd point out its limittions so s to be convinced of the necessity of more generl integrl.
More informationMath Advanced Calculus II
Mth 452 - Advnced Clculus II Line Integrls nd Green s Theorem The min gol of this chpter is to prove Stoke s theorem, which is the multivrible version of the fundmentl theorem of clculus. We will be focused
More information1.9 C 2 inner variations
46 CHAPTER 1. INDIRECT METHODS 1.9 C 2 inner vritions So fr, we hve restricted ttention to liner vritions. These re vritions of the form vx; ǫ = ux + ǫφx where φ is in some liner perturbtion clss P, for
More informationBest Approximation in the 2-norm
Jim Lmbers MAT 77 Fll Semester 1-11 Lecture 1 Notes These notes correspond to Sections 9. nd 9.3 in the text. Best Approximtion in the -norm Suppose tht we wish to obtin function f n (x) tht is liner combintion
More informationOverview of Calculus I
Overview of Clculus I Prof. Jim Swift Northern Arizon University There re three key concepts in clculus: The limit, the derivtive, nd the integrl. You need to understnd the definitions of these three things,
More informationSome estimates on the Hermite-Hadamard inequality through quasi-convex functions
Annls of University of Criov, Mth. Comp. Sci. Ser. Volume 3, 7, Pges 8 87 ISSN: 13-693 Some estimtes on the Hermite-Hdmrd inequlity through qusi-convex functions Dniel Alexndru Ion Abstrct. In this pper
More informationReview of Calculus, cont d
Jim Lmbers MAT 460 Fll Semester 2009-10 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some
More information1. Gauss-Jacobi quadrature and Legendre polynomials. p(t)w(t)dt, p {p(x 0 ),...p(x n )} p(t)w(t)dt = w k p(x k ),
1. Guss-Jcobi qudrture nd Legendre polynomils Simpson s rule for evluting n integrl f(t)dt gives the correct nswer with error of bout O(n 4 ) (with constnt tht depends on f, in prticulr, it depends on
More informationDefinite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová (Mendel University) Definite integrl MENDELU / Motivtion - re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function
More informationMAA 4212 Improper Integrals
Notes by Dvid Groisser, Copyright c 1995; revised 2002, 2009, 2014 MAA 4212 Improper Integrls The Riemnn integrl, while perfectly well-defined, is too restrictive for mny purposes; there re functions which
More informationFurther integration. x n nx n 1 sinh x cosh x log x 1/x cosh x sinh x e x e x tan x sec 2 x sin x cos x tan 1 x 1/(1 + x 2 ) cos x sin x
Further integrtion Stndrd derivtives nd integrls The following cn be thought of s list of derivtives or eqully (red bckwrds) s list of integrls. Mke sure you know them! There ren t very mny. f(x) f (x)
More information4. Calculus of Variations
4. Clculus of Vritions Introduction - Typicl Problems The clculus of vritions generlises the theory of mxim nd minim. Exmple (): Shortest distnce between two points. On given surfce (e.g. plne), nd the
More information6.2 CONCEPTS FOR ADVANCED MATHEMATICS, C2 (4752) AS
6. CONCEPTS FOR ADVANCED MATHEMATICS, C (475) AS Objectives To introduce students to number of topics which re fundmentl to the dvnced study of mthemtics. Assessment Emintion (7 mrks) 1 hour 30 minutes.
More information