Answers to Selected Exercises
( \newcommand{\kernel}{\mathrm{null}\,}\)
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)! 2⋅3n/(2n+1)! 2−n(2n)!/(n!)2nn/n!\
nono
\dstAn=xnn!(lnx−∑nj=11j)
fn(x1,x2,…,xn)=2n−1max(x1,x2,…,xn), gn(x1,x2,…,xn)=2n−1min(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 61
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 rationalsS1= 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 \setxx≠0,1 \setxx≠0[1,∞)
4 12 −1 2−2
1117 −23 132
0,2 0, none −13,13none, 0
0 0 none 0 none0
0 0 none none none0
∞ −∞ ∞ ∞ ∞−∞
none ∞ ∞none
∞ ∞ ∞ −∞ none∞
32 32 ∞ −∞ ∞12
limx→∞r(x)=∞ if n>m and an/bm>0; =−∞ if n>m and an/bm<0; =an/bm if n=m; =0 if n<m. limx→−∞r(x)=(−1)n−mlimx→∞r(x)
limx→x0f(x)=limx→x0g(x)
lim supx→x0−(f−g)(x)≤lim supx→x0−f(x)−lim infx→x0−g(x); lim infx→x0−(f−g)(x)≥lim infx→x0−f(x)−lim supx→x0−g(x)
from the right continuous none continuous none continuousfrom 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 x≠0
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 x≠12 x≠1 x≠(k+12π), k= integer x≠(k+12π), k= integer x≠0x≠0
p(c)=q(c) and p′−(c)=q′+(c)
f(k)(x)=n(n−1)⋯(n−k−1)xn−k−1|x| if 1≤k≤n−1; f(n)(x)=n! if x>0; f(n)(x)=−n! if x<0; f(k)(x)=0 if k>n and x≠0; f(k)(0) does not exist if k≥n.
c′=ac−bs, s′=bc+asc(x)=eaxcosbx, s(x)=eaxsinbx
f(x)=−1 if x≤0, 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 x≠0, and f(0) =0 .
Counterexample: Let x0=0, f(x)=x/|x| if x≠0, f(0)=0.
−∞ if α≤0, 0 if α>0
∞ if α>0, −∞ if α≤0
−∞ −∞ if α≤0, 0 if α>0
0Suppose that g′ is continuous at x0 and f(x)=g(x) if x≤x0, f(x)=1+g(x) if x>x0.
1 e1 eL
f(n+1)(x0)/(n+1)!.Counterexample: Let x0=0 and f(x)=x|x|.
Let g(x)=1+|x−x0|, so f(x)=(x−x0)(1+|x−x0|).
Let g(x)=1+|x−x0|, so f(x)=(x−x0)2(1+|x−x0|).
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, 3√2, 11√2, 57√2 −1, 3, −14, 88 min neither min max min neither minmin
f(x)=e−1/x2 if x≠0, f(0)=0 (Exercise~)
None if b2−4c<0; local min at x1=(−b+√b2−4c)/2 and local max at x1=(−b−√b2−4c)/2 if b2−4c>0; if b2=4c then x=−b/2 is a critical point, but not a local extreme point.
\dst16(π20)3 \dst183 \dstπ2512√2\dst14(63)4
M3h/3, where M3=sup|x−c|≤h|f(3)(c)|\M4h2/12 where M4=sup|x−c|≤h|f(4)(c)|
k=−h/2
monotonic functionsLet [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, −1212, 1 eb−ea 1−cosb sinb
f(a)[g1−g(a)]+f(d)(g2−g1)+f(b)[g(b)−g2]
f(a)[g1−g(a)]+f(b)[g(b)−gp]+∑p−1m=1f(am)(gm+1−gm)
If g≡1 and f is arbitrary, then ∫baf(x)dg(x)=0.
¯u=c=23 ¯u=c=0¯u=(e−2)/(e−1), c=√¯u
n! 12 divergent 1 −10
divergent convergent divergent convergent convergentdivergent
p<2 p<1 p>−1 −1<p<2 none nonep<1
p−q<1 p,q<1 −1<p<2q−1 q>−1, p+q>1 p+q>1q+1<p<3q+1
degg−degf≥2
2 1 0 1/2 1/2 1/21/2
√A 1 1 1 −∞0
If sn=1 and tn=−1/n, then (limn→∞sn)/(limn→∞tn)=1/0=∞, but limn→∞sn/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 limn→∞tn=12
limm→∞s2m=∞, limm→∞s2m+1=−∞\ limm→∞s4m=1, limm→∞s4m+2=−1, limm→∞s2m+1=0\ limm→∞s2m=0, limm→∞s4m+1=1, limm→∞s4m+3=−1\ limn→∞sn=0 limm→∞s2m=∞, limm→∞s2m+1=0\limm→∞s8m=limm→∞s8m+2=1, limm→∞s8m+1=√2,\ limm→∞s8m+3=limm→∞s8m+7=0, limm→∞s8m+5=−√2,\ limm→∞s8m+4=limm→∞s8m+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 divergentconvergent
p>1 p>1p>1
convergent convergent if 0<r<1, divergent if r≥1\ divergent convergent divergentconvergent
convergent convergent convergentconvergent
divergent convergent if and only if 0<r<1 or r=1 and p<−1 convergent convergentconvergent
divergent convergent convergentconvergent if α<β−1, divergent if α≥β−1
divergent convergent convergentconvergent
∑(−1)n ∑(−1)n/n, ∑\dst[(−1)nn+1nlogn]\ ∑(−1)n2n∑(−1)n
conditionally convergent conditionally convergent absolutely convergentabsolutely 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 s≥k+2. The series converges conditionally if s=k+1 and r=−1, and diverges in all other cases, where s≥k+1 and |r|=1.
∑(−1)n/√n 02A−a0
F(x)=0, |x|≤1 F(x)=0, |x|≤1 \ F(x)=0, −1<x≤1 F(x)=sinx, −∞<x<∞\ F(x)=1, −1<x≤1; 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 ‖FnGn‖S=∞.
3 1 12e−1
compact subsets of (−12,∞) [−12,∞) closed subsets of \dst(1−√52,1+√52) (−∞,∞) [r,∞), r>1compact 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, whilemeans that ∑|fn(x)| converges uniformly on S.
\dst∑∞n=0(−1)nx2n+1n!(2n+1)\dst∑∞n=0(−1)nx2n+1(2n+1)(2n+1)!
1/3e 1 13 1∞
1 12 14 4 1/e1
x(1+x)/(1−x)3 e−x2 \ \dst∑∞n=1(−1)n−1n2(x−1)n; R=1
Tan−1x=\dst∑∞n=0(−1)nx2n+1(2n+1); f(2n)(0)=0; f(2n+1)(0)=(−1)2(2n)!;
\dstπ6=Tan−11√3=∑∞n=0(−1)n(2n+1)3n+1/2
coshx=\dst∑∞n=0x2n(2n)!, sinhx=\dst∑∞n=0x2n+1(2n+1)!
(1−x)∑∞n=0xn=1 converges for all x
\dstx+x2+x33−3x540+⋯ \dst1−x−x22+5x36+⋯ \dst1−x22+x424−721x6720+⋯\dstx2−x32+x46−x56+⋯
\dst1+x+2x23+x33+⋯ \dst1−x−x22+3x32+⋯ \dst1+x22+5x424+61x6720+⋯ \dst1+x26+7x4360+31x615120+⋯\dst2−x2+x412−x6360+⋯
\dstF(x)=5(1−3x)(1+2x)=31−3x+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 13227
X=X0+tU (−∞<t<∞) in all cases.
…U and X1−X0 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 21/2√5
\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)z≠1 or x2+y2>1
open neitherclosed
(π,1,0)(1,0,e)
6 6 2√5 2L√n∞
\set(x,y)x2+y2=1
… if for A there is an integer R such that |Xr|>A if r≥R.
10 3 1 0 00
a/(1+a2)
∞ ∞ no −∞no
0 0 none 0none
if Df is unbounded and for each M there is an R such that f(X)>M if X∈Df and |X|>R. Replace$>M$'' by
<M’’ in
.
limX→0f(X)=0 if a1+a2+⋯+an>b; no limit if a1+a2+⋯+an≤b and a21+a22+⋯+a2n≠0; limX→0f(X)=∞ if a1=a2=⋯=an=0 and b>0.
No; for example, limx→∞g(x,√x)=0.
R3 R2 R3 R2 \set(x,y)x≥yRn
R3−{(0,0,0)} R2 R2 R2R2
f(x,y)=xy/(x2+y2) if (x,y)≠(0,0) and f(0,0)=0
\dst2√3(x+ycosx−xysinx)−2√23(xcosx) \dst1−2y√3e−x+y2+2z \dst2√n(x1+x2+⋯+xn)1/(1+x+y+z)
ϕ21ϕ2 −5π/√6 −2e 00
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=xyeyzfx=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=3xfxx=0, fyy=xz2eyz, fzz=xy2eyz, fxy=fyx=zeyz, fxz=fzx=yeyz, fyz=fzy=xeyz
fxx=2ycosx2y−4x2y2sinx2y, fyy=−x4sinx2y, fzz=0, fxy=fyx=2xcosx2y−2x3ysinx2y, 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)(X−X0)=16x \ df=−e−x−y−z(dx+dy+dz), dX0f=−dx−dy−dz, \ (dX0f)(X−X0)=−x−y−z \ df=(1+x1+2x2+⋯+nxn)−1∑nj=1jdxj, dX0f=∑nj=1jdxj, \ (dX0f)(X−X0)=∑nj=1jxj,\df=2r|X|2r−2∑nj=1xjdxj, dX0f=2rnr−1∑nj=1dxj, \ (dX0f)(X−X0)=2rnr−1∑nj=1(xj−1),
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+4y−6 z=2x+3y+1 z=(πx)/2+y−π/2z=x+10y+4
5du+34dv 0 6du−18dv8du
hr=fxcosθ+fysinθ, hθ=r(−fxsinθ+fycosθ), hz=fz
hr=fxsinϕcosθ+fysinϕsinθ+fzcosϕ, hθ=rsinϕ(−fxsinθ+fycosθ), hϕ=r(fxcosϕcosθ+fycosϕsinθ−fzsinϕ)
hy=gxxy+gy+gwwy, hz=gxxz+gz+gwwz
hrr=fxxsin2ϕcos2θ+fyysin2ϕsin2θ+fzzcos2ϕ+fxysin2ϕsin2θ+fyzsin2ϕsinθ+fxzsin2ϕcosθ,
\dsthrθ=(−fxsinθ+fycosθ)sinϕ+r2(fyy−fxx)sin2ϕsin2θ+rfxysin2ϕcos2θ+r2(fzycosθ−fzxsinθ)sin2ϕ
\dst1+x+x22−y22+x36−xy22
\dst1−x−y+x22+xy+y22−x36−x2y2−xy22−y36\ 0xyz
(d2(0,0)p)(x,y)=(d2(0,0)q)(x,y)=2(x−y)2
[3462−42723][243−27−461]
[8816240041212162844][−2−600−2−4−22−6]
[−22667−30−26][−1735514]
[132516311625][2950]
A and B are square of the same order.
[7334776−91][14106−2142]
[−764−971350−14], [−5604−123403]
[6xyz3xz23x2y];
[−63−3]
cos(x+y)[11]; [00]
[(1−xz)ye−xzxe−xz−x2ye−xz]; [21−2]\
sec2(x+2y+z)[121]; [242]
|X|−1[x1x2⋯xn]; \dst1√n[11⋯1]
(2,3,−2) (2,3,0) (−2,0,−1)(3,1,3,2)
\dst110[42−31]\dst12[−11231−4−1−12]
\dst125[43−56−85−3410]\dst12[1−11−11111−1]
\dst17[3−2002100002−30012]\dst110[−1−205−14−1810202122−10−251724−10−25]
\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(1−2x)(y−z)−z(x−y)];\[2] G(X)=[011]+[21200000−1][x−1y+1z]
F′(X)=[excosy−exsinyexsinyexcosy]; JF(X)=e2x;\[2] G(X)=[01]+[0−110][xy−π/2]\[2]
F′(X)=[2x−2y002y−2z−2x02z]; JF=0;\[2] G(X)=[2−2002−2−202][x−1y−1z−1]
F′(X)=[(x+y+z+1)exexex(2x−x2−y2)e−x2ye−x0]
F′(X)=[g′1(x)g′2(x)⋮g′n(x)] \F′(r,θ)=[exsinyzzexcosyzyexcosyzzeycosxzeysinxzxeycosxzyezcosxyxezcosxyezsinxy]
F′(r,θ)=[cosθ−rsinθsinθrcosθ]; JF(r,θ)=r\[2]
F′(r,θ,ϕ)=[cosθcosϕ−rsinθcosϕ−rcosθsinϕsinθcosϕ−rcosθcosϕ−rsinθsinϕsinϕ0rcosϕ];\ JF(r,θ,ϕ)=r2cosϕ\[2]
F′(r,θ,z)=[cosθ−rsinθ0sinθ−rcosθ0001]; JF(r,θ,z)=r
[0040−120] [−18020][9−33−810]
[4−31011] [2020][510918−4−8]
[1,π/2] [1,2π] [1,π] [2√2,9π/4][√2,3π/4]
[1,−3π/2] [1,−2π] [1,−π] [2√2,−7π/4][√2,−5π/4]
Let f(x)=x (0≤x≤12), f(x)=x−12 (12<x≤1); 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[3u−2v3u+4v]; (F−1)′=\dst110[1−234]
[xyz]=\dst12[u+2v+3wu−wu+v+2w]; (F−1)′=\dst12[12310−1112]
G1(u,v)=\dst1√2[√u+v√u−v], G′1(u,v)=\dst12√2[1/√u+v1/√u+v1/√u−v−1/√u−v]
G2(u,v)=\dst1√2[−√u+v√u−v], G′2(u,v)=\dst12√2[−1/√u+v−1/√u+v1/√u−v−1/√u−v]
G3(u,v)=\dst1√2[√u+v−√u−v], G′3(u,v)=\dst12√2[1/√u+v1/√u+v−1/√u−v1/√u−v]
G4(u,v)=\dst1√2[−√u+v−√u−v], G′4(u,v)=\dst12√2[−1/√u+v−1/√u+v−1/√u−v1/√u−v]
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 β−π/2<arg(u,v)<β+π/2 and β is an argument of (a,b);
G′(u,v)=\dst12(x2+y2)[xy−yx]
If F(x1,x2,…,xn)=(x31,x32,…,x3n), then F is invertible, but\ JF(0) =0.
A(U)=[1−1]−\dst125[5538][u+5v−4]
A(U)=[11]+\dst16[4−2−33][u−2v−3]
A(U)=[011]+[0−11−110100][u−1v−1w−2]
A(U)=[1π/2π]+[0−1010000−1][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[−341−2][xy]
[uvw]=\dst−12[33−1223][xy][uv]=\dst15[2−1−13][−y+sinx−x+siny]
u=−x, v=−y, z=−w
fi(X,U)=\dst(∑nj=1aij(xj−xj0))r−(ui−ui0)s, 1≤i≤m, 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,\dst∂u1(1,1)∂x=5, \dst∂u1(1,1)∂y=2
u2(1,1)=2, \dst∂u2(1,1)∂x=−14; \dst∂u2(1,1)∂y=−2
uk(0,π)=(2k+1)π/2, \dst∂uk(0,π)∂x=0, \dst∂uk(0,π)∂y=−1,k= integer
\dst15[−1−21−1−21] u′(0)=3, v′(0)=−1
\dst16[55−5−566]
U1(1,1)=[31], U′1(1,1)=[13−12];
U2(1,1)=−[31], U′2(1,1)=−[13−12]
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=−2y−u, z=−2v; x=−2y−u, v=−\dstz2; y=−\dstx2−\dstu2, z=−2v; y=−\dstx2−\dstu2, v=−\dstz2; z=−2v, u=−x−2y; u=−x−2y, 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
2814 3(b−a)(d−c), 0 \set(m,n)m,n=integers
12 7920 −1(1−log2)/2
74 17 23(√2−1)1/4π
38, 58 38, 58 34, 54 34(z+12), 54(z+12)z+12, 1
−285 0 014(e−52)
324 161 5215
36 1 643(e6+17)/2
227 12(e−52) 124136
16π 16 12821π2
12(b1−a1)⋯(bn−an)∑nj=1(aj+bj)
13(b1−a1)⋯(bn−an)∑nj=1(a2j+ajbj+b2j)
2−n(b21−a21)⋯(b2n−a2n)
∫√3/2−√3/2dx∫√1−x21/2f(x,y)dy 12
Let S1 and S2 be dense subsets of R such that S1∪S2=R.
−1; c (constant); 1 (u2−u1)(v2−v1)/|ad−bc|
56 49log52 3 12 54e(e−1)
43πabc 2π(e25−e9) 16π/3 21/64
(π/8)log5 (π/4)(e4−1)2π/15
π2a4/2
(β1−α1)⋯(βn−αn)/|det(A)| |a1a2⋯an|Vn
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Suppose f∈C[a,b]. Acording to the , if ϵ>0, there is a polynonmial p such that |f(x)−p(x)|<ϵ, a≤x≤b. Now suppose that a≤x1n<x2n<⋯<xnn≤b,a≤y1n<y2n<⋯<ynn≤b,n≥1.
By the triangle inequality, |f(xin)−f(yin)|≤|f(xin)−p(xin)|+|p(xin)−p(yin)|+|p(yin)−f(yin)|,
{} In the setting of Exercise~1, let yin=a+inn∑i=1(b−a),1≤i≤n,n≥1.
{Solution 2.} Since f∈C[a,b], ∫baf(x)dx exists (Theorem~3.2.8, p.~133).
Therefore, from Definition~3.1.1 (p.114), limn→∞n∑i=1f(yin)=1b−a∫baf(x)dx.
Suppose g is continuous and nondecreasing on [c,d]. For
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