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Mathematics LibreTexts

Answers to Selected Exercises

( \newcommand{\kernel}{\mathrm{null}\,}\)

2max(a,b) 2min(a,b) 4max(a,b,c)

4min(a,b,c)

(no); 1 (yes) 3 (no); 3 (no) 7 (yes); 7 (yes)\ 2 (no); 3 (no) 1 (no); 1 (no)

7 (no); 7 (no)

2n/(2n)! 23n/(2n+1)! 2n(2n)!/(n!)2

nn/n!\

no

no

\dstAn=xnn!(lnxnj=11j)

fn(x1,x2,,xn)=2n1max(x1,x2,,xn), gn(x1,x2,,xn)=2n1min(x1,x2,,xn)

[12,1); (,12)[1,); (,0](32,); (0,32]; (,0](32,); (,12][1,) (3,2)(2,3); (,3][2,2][3,); ; (,);; (,3][2,2][3,)

; (,); ; (,); ; (,)

; (,); [1,1]; (,1)(1,); [1,1]; (,)

(0,3] [0,2] (,1)(2,)

(,0](3,)

14 16 6

1

neither; (1,2)(3,); (,1)(2,3); (,1](2,3); (,1][2,3] open; S; (1,2); [1,2] closed; (3,2)(7,8); (,3)(2,7)(8,); (3][2,7][8,)

closed; ; \set(n,n+1)n=integer; (,)

\setxx=1/n, n=1,2,; , S1= rationals, S2= irrationals any set whose supremum is an isolated point of the set , the rationals

S1= rationals, S2= irrationals

Df=[2,1)[3,), Dg=(,3][3,7)(7,), Df±g=Dfg=[3,7)(7,), Df/g=(3,4)(4,7)(7,)

, \setxx(2k+1)π/2 where k=integer \setxx0,1 \setxx0

[1,)

4 12 1 2

2

1117 23 13

2

0,2 0, none 13,13

none, 0

0 0 none 0 none

0

0 0 none none none

0

none

none

none

32 32

12

limxr(x)= if n>m and an/bm>0; = if n>m and an/bm<0; =an/bm if n=m; =0 if n<m. limxr(x)=(1)nmlimxr(x)

limxx0f(x)=limxx0g(x)

lim supxx0(fg)(x)lim supxx0f(x)lim infxx0g(x); lim infxx0(fg)(x)lim infxx0f(x)lim supxx0g(x)

from the right continuous none continuous none continuous

from the left

[0,1), (0,1), [1,2), (1,2), (1,2], [1,2] [0,1), (0,1), (1,)

tanhx is continuous for all x, cothx for all x0

No [1,1], [0,) n=(2nπ,(2n+1)π), (0,) n=(nπ,(n+1)π), (,1)(1,1)(1,)

n=[nπ,(n+12)π], [0,)

(1,1) (,) x0(2k+32π), k= integer x12 x1 x(k+12π), k= integer x(k+12π), k= integer x0

x0

p(c)=q(c) and p(c)=q+(c)

f(k)(x)=n(n1)(nk1)xnk1|x| if 1kn1; f(n)(x)=n! if x>0; f(n)(x)=n! if x<0; f(k)(x)=0 if k>n and x0; f(k)(0) does not exist if kn.

c=acbs, s=bc+as

c(x)=eaxcosbx, s(x)=eaxsinbx

f(x)=1 if x0, f(x)=1 if x>0; then f(0+)=0, but f+(0) does not exist.

continuous from the right

There is no such function (Theorem~2.3.9).

Counterexample: Let x0=0, f(x)=|x|3/2sin(1/x) if x0, and f(0) =0 .

Counterexample: Let x0=0, f(x)=x/|x| if x0, f(0)=0.

if α0, 0 if α>0

if α>0, if α0

if α0, 0 if α>0

0

Suppose that g is continuous at x0 and f(x)=g(x) if xx0, f(x)=1+g(x) if x>x0.

1 e

1 eL

f(n+1)(x0)/(n+1)!.

Counterexample: Let x0=0 and f(x)=x|x|.

Let g(x)=1+|xx0|, so f(x)=(xx0)(1+|xx0|).

Let g(x)=1+|xx0|, so f(x)=(xx0)2(1+|xx0|).

1, 2, 2, 0 0, π, 3π/2, 4π+π3/2\ π2/4, 2π, 6+π2/4, 4π

2, 5, 16, 65

0, 1, 0, 5

0, 1, 0, 5 1, 0, 6, 24 2, 32, 112, 572 1, 3, 14, 88 min neither min max min neither min

min

f(x)=e1/x2 if x0, f(0)=0 (Exercise~)

None if b24c<0; local min at x1=(b+b24c)/2 and local max at x1=(bb24c)/2 if b24c>0; if b2=4c then x=b/2 is a critical point, but not a local extreme point.

\dst16(π20)3 \dst183 \dstπ25122

\dst14(63)4

M3h/3, where M3=sup|xc|h|f(3)(c)|\

M4h2/12 where M4=sup|xc|h|f(4)(c)|

k=h/2

monotonic functions

Let [a,b]=[0,1] and P={0,1}. Let f(0)=f(1)=12 and f(x)=x if 0<x<1. Then s(P)=0 and S(P)=1, but neither is a Riemann sum of f over P.

12, 12

12, 1 ebea 1cosb sinb

f(a)[g1g(a)]+f(d)(g2g1)+f(b)[g(b)g2]

f(a)[g1g(a)]+f(b)[g(b)gp]+p1m=1f(am)(gm+1gm)

If g1 and f is arbitrary, then baf(x)dg(x)=0.

¯u=c=23 ¯u=c=0

¯u=(e2)/(e1), c=¯u

n! 12 divergent 1 1

0

divergent convergent divergent convergent convergent

divergent

p<2 p<1 p>1 1<p<2 none none

p<1

pq<1 p,q<1 1<p<2q1 q>1, p+q>1 p+q>1

q+1<p<3q+1

deggdegf2

2 1 0 1/2 1/2 1/2

1/2

A 1 1 1

0

If sn=1 and tn=1/n, then (limnsn)/(limntn)=1/0=, but limnsn/tn=.

, 0 , if |r|>1; 2, 2 if r=1; 0, 0 if r=1; 1, 1 if |r|<1 , if r<1; 0, 0 if |r|<1; 12, 12 if r=1; , if r>1 \ ,

|t|, |t|

1, 1 2, 2 3, 1

3/2, 3/2

If {sn}={1,0,1,0,}, then limntn=12

limms2m=, limms2m+1=\ limms4m=1, limms4m+2=1, limms2m+1=0\ limms2m=0, limms4m+1=1, limms4m+3=1\ limnsn=0 limms2m=, limms2m+1=0\

limms8m=limms8m+2=1, limms8m+1=2,\ limms8m+3=limms8m+7=0, limms8m+5=2,\ limms8m+4=limms8m+6=1

{1,2,1,2,3,1,2,3,4,1,2,3,4,5,}

Let {tn} be any convergent sequence and {sn}={t1,1,t2,2,,tn,n,}.

No; consider 1/n

convergent convergent divergent divergent \ convergent convergent divergent

convergent

p>1 p>1

p>1

convergent convergent if 0<r<1, divergent if r1\ divergent convergent divergent

convergent

convergent convergent convergent

convergent

divergent convergent if and only if 0<r<1 or r=1 and p<1 convergent convergent

convergent

divergent convergent convergent

convergent if α<β1, divergent if αβ1

divergent convergent convergent

convergent

(1)n (1)n/n, \dst[(1)nn+1nlogn]\ (1)n2n

(1)n

conditionally convergent conditionally convergent absolutely convergent

absolutely convergent

%\begin{exercisepatch1}

Let k and s be the degrees of the numerator and denominator, respectively. If |r|=1, the series converges absolutely if and only if sk+2. The series converges conditionally if s=k+1 and r=1, and diverges in all other cases, where sk+1 and |r|=1.

(1)n/n 0

2Aa0

F(x)=0, |x|1 F(x)=0, |x|1 \ F(x)=0, 1<x1 F(x)=sinx, <x<\ F(x)=1, 1<x1; F(x)=0, |x|>1 F(x)=x, <x<\ F(x)=x2/2, <x< F(x)=0, <x<\

F(x)=1, <x<

F(x)=0 F(x)=1, |x|<1; F(x)=0, |x|>1 \

F(x)=sinx/x

Fn(x)=xn; Sk=[k/(k+1),k/(k+1)]

[1,1] [r,r]{1}{1}, 0<r<1 [r,r]{1}, 0<r<1 \ [r,r], r>0 (,1/r][r,r][1/r,){1}, 0<r<1\ [r,r], r>0 [r,r], r>0 (,r][r,){0}, r>0 \

[r,r], r>0

Let S=(0,1], Fn(x)=sin(x/n), Gn(x)=1/x2; then F=0, G=1/x2, and the convergence is uniform, but FnGnS=.

3 1 12

e1

compact subsets of (12,) [12,) closed subsets of \dst(152,1+52) (,) [r,), r>1

compact subsets of (,0)(0,)

Let S=(,), fn=an (constant), where an converges conditionally, and gn=|an|.

``absolutely"

means that |fn(x)| converges pointwise and fn(x) converges uniformly on S, while

means that |fn(x)| converges uniformly on S.

\dstn=0(1)nx2n+1n!(2n+1)

\dstn=0(1)nx2n+1(2n+1)(2n+1)!

1/3e 1 13 1

1 12 14 4 1/e

1

x(1+x)/(1x)3 ex2 \ \dstn=1(1)n1n2(x1)n; R=1

Tan1x=\dstn=0(1)nx2n+1(2n+1); f(2n)(0)=0; f(2n+1)(0)=(1)2(2n)!;

\dstπ6=Tan113=n=0(1)n(2n+1)3n+1/2

coshx=\dstn=0x2n(2n)!, sinhx=\dstn=0x2n+1(2n+1)!

(1x)n=0xn=1 converges for all x

\dstx+x2+x333x540+ \dst1xx22+5x36+ \dst1x22+x424721x6720+

\dstx2x32+x46x56+

\dst1+x+2x23+x33+ \dst1xx22+3x32+ \dst1+x22+5x424+61x6720+ \dst1+x26+7x4360+31x615120+

\dst2x2+x412x6360+

\dstF(x)=5(13x)(1+2x)=313x+21+2x=n=0[3n+1(2)n+1]xn 1

(3,0,3,3) (1,1,4)

(16,1112,2324,536)

15 65/12 31

3

89 166/12 3

31

12 132

27

X=X0+tU (<t<) in all cases.

U and X1X0 are scalar multiples of V.

X=(1,3,4,2)+t(1,3,5,3)

X=(3,1,2,1,4,)+t(1,1,1,3,7)

X=(1,2,1)+t(1,3,0)

5 2

1/25

\set(x1,x2,x3,x4)|xi|3 (i=1,2,3) with at least one equality \set(x1,x2,x3,x4)|xi|3 (i=1,2,3)

S

\set(x1,x2,x3,x4)|xi|>3 for at least one of i=1,2,3

S S

\set(x,y,z)z1 or x2+y2>1

open neither

closed

(π,1,0)

(1,0,e)

6 6 25 2Ln

\set(x,y)x2+y2=1

if for A there is an integer R such that |Xr|>A if rR.

10 3 1 0 0

0

a/(1+a2)

no

no

0 0 none 0

none

if Df is unbounded and for each M there is an R such that f(X)>M if XDf and |X|>R. Replace $>M$'' by<M’’ in

.

limX0f(X)=0 if a1+a2++an>b; no limit if a1+a2++anb and a21+a22++a2n0; limX0f(X)= if a1=a2==an=0 and b>0.

No; for example, limxg(x,x)=0.

R3 R2 R3 R2 \set(x,y)xy

Rn

R3{(0,0,0)} R2 R2 R2

R2

f(x,y)=xy/(x2+y2) if (x,y)(0,0) and f(0,0)=0

\dst23(x+ycosxxysinx)223(xcosx) \dst12y3ex+y2+2z \dst2n(x1+x2++xn)

1/(1+x+y+z)

ϕ21ϕ2 5π/6 2e 0

0

fx=fy=1/(x+y+2z), fz=2/(x+y+2z)

fx=2x+3yz+2y, fy=3xz+2x, fz=3xy fx=eyz, fy=xzeyz, fz=xyeyz

fx=2xycosx2y, fy=x2cosx2y, fz=1

fxx=fyy=fxy=fyx=1/(x+y+2z)2, fxz=fzx=fyz=fzy= 2/(x+y+2z)2, fzz=4/(x+y+2z)2

fxx=2, fyy=fzz=0, fxy=fyx=3z+2, fxz=fzx=3y, fyz=fzy=3x

fxx=0, fyy=xz2eyz, fzz=xy2eyz, fxy=fyx=zeyz, fxz=fzx=yeyz, fyz=fzy=xeyz

fxx=2ycosx2y4x2y2sinx2y, fyy=x4sinx2y, fzz=0, fxy=fyx=2xcosx2y2x3ysinx2y, fxz=fzx=fyz=fzy=0

fxx(0,0)=fyy(0,0)=0, fxy(0,0)=1, fyx(0,0)=1\

fxx(0,0)=fyy(0,0)=0, fxy(0,0)=1, fyx(0,0)=1

f(x,y)=g(x,y)+h(y), where gxy exists everywhere and h is nowhere differentiable.

df=(3x2+4y2+2ysinx+2xycosx)dx+(8xy+2xsinx)dy,\ dX0f=16dx, (dX0f)(XX0)=16x \ df=exyz(dx+dy+dz), dX0f=dxdydz, \ (dX0f)(XX0)=xyz \ df=(1+x1+2x2++nxn)1nj=1jdxj, dX0f=nj=1jdxj, \ (dX0f)(XX0)=nj=1jxj,\

df=2r|X|2r2nj=1xjdxj, dX0f=2rnr1nj=1dxj, \ (dX0f)(XX0)=2rnr1nj=1(xj1),

The unit vector in the direction of (fx1(X0),fx2(X0),,fxn(X0)) provided that this is not 0; if it is 0, then f(X0)/Φ=0 for every Φ.

z=2x+4y6 z=2x+3y+1 z=(πx)/2+yπ/2

z=x+10y+4

5du+34dv 0 6du18dv

8du

hr=fxcosθ+fysinθ, hθ=r(fxsinθ+fycosθ), hz=fz

hr=fxsinϕcosθ+fysinϕsinθ+fzcosϕ, hθ=rsinϕ(fxsinθ+fycosθ), hϕ=r(fxcosϕcosθ+fycosϕsinθfzsinϕ)

hy=gxxy+gy+gwwy, hz=gxxz+gz+gwwz

hrr=fxxsin2ϕcos2θ+fyysin2ϕsin2θ+fzzcos2ϕ+fxysin2ϕsin2θ+fyzsin2ϕsinθ+fxzsin2ϕcosθ,

\dsthrθ=(fxsinθ+fycosθ)sinϕ+r2(fyyfxx)sin2ϕsin2θ+rfxysin2ϕcos2θ+r2(fzycosθfzxsinθ)sin2ϕ

\dst1+x+x22y22+x36xy22

\dst1xy+x22+xy+y22x36x2y2xy22y36\ 0

xyz

(d2(0,0)p)(x,y)=(d2(0,0)q)(x,y)=2(xy)2

[346242723]

[24327461]

[8816240041212162844]

[260024226]

[226673026]

[1735514]

[132516311625]

[2950]

A and B are square of the same order.

[733477691]

[141062142]

[76497135014], [5604123403]

[6xyz3xz23x2y];
[633]

cos(x+y)[11]; [00]

[(1xz)yexzxexzx2yexz]; [212]\

sec2(x+2y+z)[121]; [242]

|X|1[x1x2xn]; \dst1n[111]

(2,3,2) (2,3,0) (2,0,1)

(3,1,3,2)

\dst110[4231]

\dst12[112314112]

\dst125[4356853410]

\dst12[111111111]

\dst17[3200210000230012]

\dst110[1205141810202122102517241025]

\boldsymbol{\mathbf{F}'(\mathbf{X})=\left[\begin{array}{ccc} 2x&1&2\\-\sin(x+y+z)&-\sin(x+y+z)&-\sin(x+y+z)\\[2\jot] yze^{xyz}&xze^{xyz}&xye^{xyz}\end{array}\right]};\[2] JF(X)=exyzsin(x+y+z)[x(12x)(yz)z(xy)];\[2] G(X)=[011]+[212000001][x1y+1z]

F(X)=[excosyexsinyexsinyexcosy]; JF(X)=e2x;\[2] G(X)=[01]+[0110][xyπ/2]\[2]

F(X)=[2x2y002y2z2x02z]; JF=0;\[2] G(X)=[220022202][x1y1z1]

F(X)=[(x+y+z+1)exexex(2xx2y2)ex2yex0]

F(X)=[g1(x)g2(x)gn(x)] \

F(r,θ)=[exsinyzzexcosyzyexcosyzzeycosxzeysinxzxeycosxzyezcosxyxezcosxyezsinxy]

F(r,θ)=[cosθrsinθsinθrcosθ]; JF(r,θ)=r\[2]

F(r,θ,ϕ)=[cosθcosϕrsinθcosϕrcosθsinϕsinθcosϕrcosθcosϕrsinθsinϕsinϕ0rcosϕ];\ JF(r,θ,ϕ)=r2cosϕ\[2]

F(r,θ,z)=[cosθrsinθ0sinθrcosθ0001]; JF(r,θ,z)=r

[0040120] [18020]

[933810]

[431011] [2020]

[51091848]

[1,π/2] [1,2π] [1,π] [22,9π/4]

[2,3π/4]

[1,3π/2] [1,2π] [1,π] [22,7π/4]

[2,5π/4]

Let f(x)=x (0x12), f(x)=x12 (12<x1); then f is locally invertible but not invertible on [0,1].

F(S)=\set(u,v)π+2ϕ<arg(u,v)<π+2ϕ, where ϕ is an argument of (a,b);

\boldsymbol{\mathbf{F}_S^{-1}(u,v)=(u^2+v^2)^{1/4} \left[\begin{array}{r}\cos(\arg(u,v)/2) \\[2\jot]\sin(\arg(u,v)/2)\end{array}\right],\
2\phi-\pi<\arg(u,v)<2\phi+\pi}

[xy]=\dst110[3u2v3u+4v]; (F1)=\dst110[1234]

[xyz]=\dst12[u+2v+3wuwu+v+2w]; (F1)=\dst12[123101112]

G1(u,v)=\dst12[u+vuv], G1(u,v)=\dst122[1/u+v1/u+v1/uv1/uv]

G2(u,v)=\dst12[u+vuv], G2(u,v)=\dst122[1/u+v1/u+v1/uv1/uv]

G3(u,v)=\dst12[u+vuv], G3(u,v)=\dst122[1/u+v1/u+v1/uv1/uv]

G4(u,v)=\dst12[u+vuv], G4(u,v)=\dst122[1/u+v1/u+v1/uv1/uv]

From solving x=rcosθ, y=rsinθ for θ=arg(x,y). Each equation is satisfied by angles that are not arguments of (x,y), since none of the formulas identifies the quadrant of (x,y) uniquely. Moreover,

does not hold if x=0.

\boldsymbol{\left[\begin{array}{c}x\\y\end{array}\right]= \mathbf{G}(u,v)=(u^2+v^2)^{1/4} \left[\begin{array}{r}\cos[\frac{1}{2}\arg(u,v)] \\[2\jot] \sin(\arg(u,v)/2)\end{array}\right]},

where βπ/2<arg(u,v)<β+π/2 and β is an argument of (a,b);

G(u,v)=\dst12(x2+y2)[xyyx]

If F(x1,x2,,xn)=(x31,x32,,x3n), then F is invertible, but\ JF(0) =0.

A(U)=[11]\dst125[5538][u+5v4]

A(U)=[11]+\dst16[4233][u2v3]

A(U)=[011]+[011110100][u1v1w2]

A(U)=[1π/2π]+[010100001][uv+1w]

\boldsymbol{\mathbf{G}'(x,y,z)= \left[\begin{array}{ccc} \cos\theta\cos\phi&\sin\theta\cos\phi&\sin\phi\\[2\jot] -\dst\frac{\sin\theta}{ r\cos\phi}&\dst\frac{\cos\theta}{ r\cos\phi}&0\\[2\jot] -\dst\frac{1}{ r}\cos\theta\sin\phi&-\dst\frac{1}{ r}\sin\theta\sin\phi&\dst\frac{1}{ r}\cos\phi\end{array}\right]}

\boldsymbol{\mathbf{G}'(x,y,z)= \left[\begin{array}{ccc}\cos\theta&\sin\theta&0\\[2\jot] -\dst\frac{1}{ r}\sin\theta&\dst\frac{1}{ r}\cos\theta&0\\[2\jot] 0&0&1\end{array}\right]}

[uv]=\dst12[3412][xy]

[uvw]=\dst12[331223][xy]

[uv]=\dst15[2113][y+sinxx+siny]

u=x, v=y, z=w

fi(X,U)=\dst(nj=1aij(xjxj0))r(uiui0)s, 1im, where r and s are positive integers and not all aij=0. r=s=3; r=1, s=3;

r=s=2

ux(1,1)=58, uy(1,1)=12

ux(1,1,1)=58, uy(1,1,1)=98, uz(1,1,1)=12

u(1,2)=0, ux(1,2)=uy(1,2)=4

u(1,2)=2, ux(1,2)=1, uy(1,2)=12

u(π/2,π/2)=ux(π/2,π/2)=uy(π/2,π/2)=0

u(1,1)=1, ux(1,1)=uy(1,1)=1

u1(1,1)=1,\dstu1(1,1)x=5, \dstu1(1,1)y=2

u2(1,1)=2, \dstu2(1,1)x=14; \dstu2(1,1)y=2

uk(0,π)=(2k+1)π/2, \dstuk(0,π)x=0, \dstuk(0,π)y=1,k= integer

\dst15[121121] u(0)=3, v(0)=1

\dst16[555566]

U1(1,1)=[31], U1(1,1)=[1312];

U2(1,1)=[31], U2(1,1)=[1312]

ux(0,0,0)=2, vx(0,0,0)=wx(0,0,0)=2

yx=\dst\dst(f,g,h)(x,z,u)\dst(f,g,h)(y,z,u), yv=\dst\dst(f,g,h)(v,z,u)\dst(f,g,h)(y,z,u), zx=\dst\dst(f,g,h)(y,x,u)\dst(f,g,h)(y,z,u),

zv=\dst\dst(f,g,h)(y,v,u)\dst(f,g,h)(y,z,u), ux=\dst\dst(f,g,h)(y,z,x)\dst(f,g,h)(y,z,u), uv=\dst\dst(f,g,h)(y,z,v)\dst(f,g,h)(y,z,u)

x=2yu, z=2v; x=2yu, v=\dstz2; y=\dstx2\dstu2, z=2v; y=\dstx2\dstu2, v=\dstz2; z=2v, u=x2y; u=x2y, v=\dstz2

yx(1,1,2)=12, vu(1,1,2)=1

uw(0,1)=56, uy(0,1)=0, vw(0,1)=56, vy(0,1)=0,\ xw(0,1)=1, xy(0,1)=1

ux(1,1)=0, uy(1,1)=0, vx(1,1)=1, vy(1,1)=1, uxx(1,1)=2,\ uxy(1,1)=1, uyy(1,1)=2, vxx(1,1)=2, vxy(1,1)=1, vyy(1,1)=2

ux(1,1)=0, uy(1,1)=\dst12, vx(1,1)=\dst12, vy(1,1)=0,\ uxx(1,1)=\dst18, uxy(1,1)=\dst18, uyy(1,1)=\dst18, vxx(1,1)=\dst18,\ vxy(1,1)=\dst18, vyy(1,1)=\dst18

28

14 3(ba)(dc), 0 \set(m,n)m,n=integers

12 7920 1

(1log2)/2

74 17 23(21)

1/4π

38, 58 38, 58 34, 54 34(z+12), 54(z+12)

z+12, 1

285 0 0

14(e52)

324 16

1 5215

36 1 643

(e6+17)/2

227 12(e52) 124

136

16π 16 12821

π2

12(b1a1)(bnan)nj=1(aj+bj)

13(b1a1)(bnan)nj=1(a2j+ajbj+b2j)

2n(b21a21)(b2na2n)

3/23/2dx1x21/2f(x,y)dy 12

Let S1 and S2 be dense subsets of R such that S1S2=R.

1; c (constant); 1 (u2u1)(v2v1)/|adbc|

56 49

log52 3 12 54e(e1)

43πabc 2π(e25e9) 16π/3 21/64

(π/8)log5 (π/4)(e41)

2π/15

π2a4/2

(β1α1)(βnαn)/|det(A)| |a1a2an|Vn

\end{document}

{}

Suppose fC[a,b]. Acording to the , if ϵ>0, there is a polynonmial p such that |f(x)p(x)|<ϵ, axb. Now suppose that ax1n<x2n<<xnnb,ay1n<y2n<<ynnb,n1.

and limn1nni=1|xinyin|=0.
Show that if fC[a,b], then limn1nni=1|f(xin)f(yin)|=0.

By the triangle inequality, |f(xin)f(yin)||f(xin)p(xin)|+|p(xin)p(yin)|+|p(yin)f(yin)|,

so |f(xin)f(yin)|<|p(xin)p(yin)|+2ϵ.
Let M=maxaxb|p(x)|. By the mean value theorem, |p(xin)p(yin)|M|xinyin|.
This and (A) imply that 1nni=1|f(xin)f(yin)|<2ϵ+Mnni=1|xinyin|.
From this and (A), lim supn1nni=1|f(xin)f(yin)|2ϵ.
Since ϵ is arbitrary, this implies the conclusion.

{} In the setting of Exercise~1, let yin=a+inni=1(ba),1in,n1.

Show that limn1nni=1f(xin)=1babaf(x)dx.

{Solution 2.} Since fC[a,b], baf(x)dx exists (Theorem~3.2.8, p.~133).

Therefore, from Definition~3.1.1 (p.114), limnni=1f(yin)=1babaf(x)dx.

Since 1n|ni=1f(xin)1babaf(x)dx|1nni=1|f(xin)f(yin)|+|1nni=1f(yin)1babaf(x)dx|,
Exercise1 implies the conclusion.

Suppose g is continuous and nondecreasing on [c,d]. For

\end{document}

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