MATEMATICKA ANALYZA V MAPLU
Symbolicke derivovani
> restart;
Pomoci procedury diff muzeme derivovat formule (vyrazy):
> 'diff(exp(-x^2),x)';
d
d
x
( )e
( )-x
2
Apostrofy kolem predchazejiciho vyrazu zamezi vyhodnoceni.
Stejneho efektu dosahneme i procedurou Diff. Diff se pouziva pro vetsi prehlednost
a z duvodu kontroly spravnosti zadani.
> %;
-2 x e
( )-x
2
> Diff(ln(x/(x^2+1)),x):%=value(%);
=
d
d
x





ln
x
+x2
1








-
1
+x2
1
2 x2
( )+x2
1
2
( )+x2
1
x
> normal(%);
=
d
d
x





ln
x
+x2
1
-
-x2
1
x ( )+x2
1
> Diff(x^(x^x),x):%=value(%);
=
d
d
x
x
( )x
x
x
( )x
x 




+xx
( )+( )ln x 1 ( )ln x
xx
x
> collect(%,ln(x), simplify); #diva se na vyraz jako na polynom v
promenne ln(x)
=
d
d
x
x
( )x
x
+ +x
( )+x
x
x
( )ln x 2
x
( )+x
x
x
( )ln x x
( )+ -x
x
x 1
Derivace vyssich radu:
> Diff(exp(-x^2),x,x):%=value(%);
=
d
d2
x2
( )e
( )-x
2
- +2 e
( )-x
2
4 x2
e
( )-x
2
> Diff(exp(-x^2), x$5):%=value(%);
=
d
d5
x5
( )e
( )-x
2
- + -120 x e
( )-x
2
160 x3
e
( )-x
2
32 x5
e
( )-x
2
Derivace funkce dane implicitne
> restart;
> alias(y=y(x)): #y povazujeme za funkci x
> eq:=x^2+y^2=c;
:=eq =+x2
y2
c
> diff(eq,x);
=+2 x 2 y








x
y 0
> dydx:=solve(%, diff(y,x)); # 1. derivace
:=dydx -
x
y
> diff(eq,x$2);
=+ +2 2








x
y
2
2 y









2
x2
y 0
> solve(%,diff(y,x$2));
-
+1








x
y
2
y
> d2ydx2:=normal(subs(diff(y,x)=dydx,%));
:=d2ydx2 -
+x2
y2
y3
> alias(y=y):
> restart;
> implicitdiff(x^2+y^2,y,x,x);
-
+x2
y2
y3
Parcialni derivace:
> Diff(exp(a*x*y^2),x,y$2):%=value(%);
=
 
3
y2
x
( )e
( )a x y
2
+ +2 a e
( )a x y
2
10 a2
y2
x e
( )a x y
2
4 a3
y4
x2
e
( )a x y
2
> factor(%);
=
 
3
y2
x
( )e
( )a x y
2
2 a e
( )a x y
2
( )+ +1 5 a x y2
2 a2
y4
x2
> Diff(sin(x+y)/y^4, x$5, y$2):%=value(%);
=
 
7
y2
x5






( )sin +x y
y4
- + +
( )cos +x y
y4
8 ( )sin +x y
y5
20 ( )cos +x y
y6
> collect(%,cos(x+y),normal);
=
 
7
y2
x5






( )sin +x y
y4
- +
( )-y2
20 ( )cos +x y
y6
8 ( )sin +x y
y5
Pokud derivuje funkci (ve smyslu datove struktury Maplu), musime pouzit funkcniho
operatoru D.
> g:=x->x^n*exp(sin(x));
:=g x xn
e
( )sin x
> D(g);
x +
xn
n e
( )sin x
x
xn
( )cos x e
( )sin x
> D(g)(Pi/6);
+
6







6
n
n e
( )/1 2

1
2







6
n
3 e
( )/1 2
diff derivuje vzorec a na vystupu vraci vzorec, D derivuje funkci a na vystupu vraci funkci.
Priklady:
> Eval(Diff(g(x), x), x=Pi/6);








x
( )xn
e
( )sin x
=x

6
> value(%);
+
6







6
n
n e
( )/1 2

1
2







6
n
3 e
( )/1 2
> diff(cos(t),t); #derivace vzorce
- ( )sin t
> D(cos); #derivace funkce
-sin
> (D@@2)(cos); #pro druhou derivaci funkce musime pouzit operatoru
skladani funkci
-cos
> D(cos)(t); # derivace funkce v danem bode
- ( )sin t
Vsimnete si rozdilu mezi nasledujicimi dvema prikazy:
> D(cos(t));
( )D ( )cos t
Maple povazuje cos(t) za slozeni funkci cos a t, spravny zapis je tedy:
> D(cos @ t);
( )@( )-sin t ( )D t
Derivace implicitni funkce pomoci operatoru D:
> eq:=x^2+y^2=c:
> D(eq);
=+2 ( )D x x 2 ( )D y y ( )D c
> solve(%,D(y));
-
1
2
-2 ( )D x x ( )D c
y
> dydx:=subs(D(x)=1, D(c)=0, %);
:=dydx -
x
y
> (D@@2)(eq);
=+ + +2 ( )( )D
( )2
x x 2 ( )D x 2
2 ( )( )D
( )2
y y 2 ( )D y 2
( )( )D
( )2
c
> solve(%,(D@@2)(y));
-
1
2
+ + -2 ( )( )D
( )2
x x 2 ( )D x 2
2 ( )D y 2
( )( )D
( )2
c
y
> d2ydx2:=normal(subs(D(x)=1, (D@@2)(x)=0, (D@@2)(c)=0, D(y)=dydx,
%));
:=d2ydx2 -
+x2
y2
y3
Operatoru D je mozno pouzit i pro vypocet parcialnich derivaci:
> h:=(x,y,z)->1/(x^2+y^2+z^2)^(1/2);
:=h ( ), ,x y z
1
+ +x2
y2
z2
> 'D[1](h)'=D[1](h);
=( )D1
h








( ), ,x y z -
x
( )+ +x2
y2
z2
( )/3 2
Zde D[1](h) je parcialni derivace vzhledem k x.
> 'D[1,2](h)'=D[1,2](h);
=( )D ,1 2
h








( ), ,x y z
3 y x
( )+ +x2
y2
z2
( )/5 2
D[1,2](h) je vlastne D[1](D[2](h)) - smisena parcialni derivace vzhledem k x a y.
> 'D[1,1](h)'=D[1,1](h);
=( )D ,1 1
h








( ), ,x y z -
3 x2
( )+ +x2
y2
z2
( )/5 2
1
( )+ +x2
y2
z2
( )/3 2
Druha parcialni derivace vzhledem k x.
> L[h]:=(D[1,1]+D[2,2]+D[3,3])(h);
Lh








( ), ,x y z -
3 x2
( )+ +x2
y2
z2
( )/5 2
1
( )+ +x2
y2
z2
( )/3 2
:=








( ), ,x y z -
3 y2
( )+ +x2
y2
z2
( )/5 2
1
( )+ +x2
y2
z2
( )/3 2
+








( ), ,x y z -
3 z2
( )+ +x2
y2
z2
( )/5 2
1
( )+ +x2
y2
z2
( )/3 2
+
> normal(L[h](x,y,y));
0
Maple muze derivovat i po castech definovane funkce:
> F:=x->piecewise(x>0, sin(x), arctan(x));
:=F x ( )piecewise , ,<0 x ( )sin x ( )arctan x
> Fp:=D(F);
:=Fp x





piecewise , , ,x 0
1
+1 x2
<0 x ( )cos x
> plot({F, Fp}, -3*Pi..3*Pi);
­1.5
­1
­0.5
0.5
1
­8 ­6 ­4 ­2 2 4 6 8
> F:=x->if x>0 then sin(x) else arctan(x) fi;
:=F proc( ) end procx option ;,operator arrow if then else end if<0 x ( )sin x ( )arctan x
> Fp:=D(F);
:=Fp proc( ) end procx option ;,operator arrow if then else end if<0 x ( )cos x / ( )1 +1 ^x 2
> plot({F, Fp}, -3*Pi..3*Pi);
­1.5
­1
­0.5
0.5
1
­8 ­6 ­4 ­2 2 4 6 8
> restart;
Integrace a sumace
Maple pouziva k integraci specialni algoritmy (napr. tzv. Rischuv algoritmus).
Procedura Int integral nevyhodnocuje, pouze prepisuje.
> Int(x/(x^3+1), x):%=value(%);
=d






x
+x3
1
x - + +
1
3
( )ln +x 1
1
6
( )ln - +x2
x 1
1
3
3







arctan
( )-2 x 1 3
3
> diff(rhs(%), x);
- + +
1
3 ( )+x 1
-2 x 1
6 ( )- +x2
x 1
2
3





+1
( )-2 x 1 2
3
Pomoci rhs() se odkazujeme na pravou stranu rovnice, lhs() na levou.
> normal(%,'expanded');
x
+x3
1
> infolevel[int]:=2:
> Int(x/(x^5+1),x): %=value(%);
int/indef1: first-stage indefinite integration
int/ratpoly: rational function integration
int/ratpoly: rational function integration
d






x
+x5
1
x
1
5
( )ln +x 1
1
20
( )ln - + + -2 x2
x 5 x 2
1
20
( )ln - + + -2 x2
x 5 x 2 5- + -=
2
5







arctan
- + +4 x 1 5
-10 2 5
5
-10 2 5
1
20
( )ln - + +2 x2
x 5 x 2 5
1
20
( )ln - + +2 x2
x 5 x 2- + +
2
5







arctan
- +4 x 1 5
+10 2 5
5
+10 2 5
>
normal(diff(rhs(%),x), 'expanded');
x
+x5
1
> infolevel[int]:=0:
Maple pri vypoctech automaticky voli nulovou integracni konstantu.
> Int(2*x*(x^2+1)^24, x);
d



2 x ( )+x2
1
24
x
> value(%);
208012 x24
81719 x32
7084 x38
19228 x14
19228 x36
653752
5
x30
7084 x12
43263 x34
+ + + + + + +
x2
92 x44
506 x42
653752
5
x20
81719 x18
43263 x16
10626
5
x40
178296 x22
+ + + + + + + +
208012 x26
1
25
x50
x48
10626
5
x10
178296 x28
12 x46
12 x4
92 x6
506 x8
+ + + + + + + + +
> factor(%);
x2
( )+ + + +x8
5 x6
10 x4
10 x2
5 x40
20 x38
190 x36
1140 x34
4845 x32
15505 x30
+ + + + +(
38775 x28
77625 x26
126425 x24
169325 x22
187760 x20
172975 x18
132450 x16
+ + + + + + +
84075 x14
43975 x12
18760 x10
6425 x8
1725 x6
350 x4
50 x2
5+ + + + + + + + ) 25/
> factor(%+1/25);
( )+x2
1
25
25
> restart;
Urcity integral:
> Integrate(x/(x^3+1), x=1..2): %=value(%);
=d






1
2
x
+x3
1
x + -
1
3
( )ln 2
3 
18
1
6
( )ln 3
> Int(1/((1+x^2)*(1+2*x^2)), x=0..1):
> %=value(%);
=d






0
1
1
( )+1 x2
( )+1 2 x2
x - +

4
2 ( )arctan 2
> Int(1/x^2, x=-1..1): %=value(%);
=d






-1
1
1
x2
x 
> Int(1/x^2, x): %=value(%);
=d






1
x2
x -
1
x
> subs(x=1, rhs(%)) - subs(x=-1, rhs(%));
-2
> plot(1/x^2, x=-1..1, y=-10..10);
­10
­8
­6
­4
­2
0
2
4
6
8
10
y
­1 ­0.8 ­0.6 ­0.4 ­0.2 0.2 0.4 0.6 0.8 1
x
Maple kontroluje nespojitosti integrandu na zadanem intervelu.
Nevlastni integraly:
> Int(t^4*ln(t)^2/(1+3*t^2)^3, t=0..infinity): %=value(%);
=d






0

t4
( )ln t 2
( )+1 3 t2
3
t + - +
3 
216

3
3
576
1
108
 3 ( )ln 3
1
576
 3 ( )ln 3 2
V pripade, ze Maple neni schopen najit reseni symbolicky, muzeme pouzit numerickeho
integrovani:
> int(exp(arcsin(x)), x=0..1);
d



0
1
e
( )arcsin x
x
> evalf(%);
1.905238690
Nekdy je vyhodne Maplu pri reseni asistovat (pokud chceme i postup reseni):
> with(student):
> Int(sqrt(9-x^2), x);
d


 -9 x2
x
> changevar(x=3*sin(t), Int(sqrt(9-x^2), x),t);
d


3 -9 9 ( )sin t 2
( )cos t t
> value(%);
+
9
2
-1 ( )sin t 2
( )sin t
9 t
2
> simplify(subs(t=arcsin(x/3), %));
+
-9 x2
x
2
9
2





arcsin
x
3
Pro metodu per - partes:
> i:=Int((x^2+1)*ln(x), x);
:=i d


( )+x2
1 ( )ln x x
> i=intparts(i, ln(x));
=d


( )+x2
1 ( )ln x x -( )ln x





+x
1
3
x3
d








+x
1
3
x3
x
x
> i=value(lhs(%));
=d


( )+x2
1 ( )ln x x - + -x ( )ln x x
1
3
x3
( )ln x
x3
9
Zobrazeni postupu vypoctu:
> read "VypocetIntegralu.txt":
> VypocetIntegralu(exp(x)*sin(x), czech);
=d


ex
( )sin x x
[ ], ,uijeme metodu per partes =u ex
=v - ( )cos x
= - -ex
( )cos x d


-ex
( )cos x x
[ ], ,uijeme metodu per partes =u ex
=v ( )sin x
= - + -ex
( )cos x ex
( )sin x d


ex
( )sin x x
[ ]integrl spotme algebraicky
= - +
1
2
ex
( )cos x
1
2
ex
( )sin x
[ ]upravme v raz
=
1
2
ex
( )- +( )cos x ( )sin x
> with(Student[Calculus1]):
> IntTutor();
Initializing Java runtime environment.
=d


ex
( )sin x x - +
1
2
ex
( )cos x
1
2
ex
( )sin x
> Int(exp(x)*sin(x), x);
d


ex
( )sin x x
> Hint(%);
[ ], ,parts ex
- ( )cos x
> restart;
Konecne a nekonecne soucty:
> Sum(k^7, k=1..20): %=value(%);
=
=k 1
20
k7
3877286700
> Sum(k^7, k=1..n): %=value(%);
=
=k 1
n
k7
- + - +
( )+n 1 8
8
( )+n 1 7
2
7 ( )+n 1 6
12
7 ( )+n 1 4
24
( )+n 1 2
12
> factor(%);
=
=k 1
n
k7
n2
( )+ - - +3 n4
6 n3
n2
4 n 2 ( )+n 1 2
24
> Sum(1/(k^2-4), k=3..infinity): %=value(%);
=
=k 3

1
-k2
4
25
48
Tayloruv rozvoj
> taylor(sin(tan(x))-tan(sin(x)), x=0, 25);
1
30
x7
29
756
x9
1913
75600
x11
95
7392
x13
311148869
54486432000
x15
10193207
4358914560
x17
1664108363
1905468364800
- - - - - - -
x19
2097555460001
7602818775552000
x21
374694625074883
6690480522485760000
x23
( )O x25
- - +
> whattype(%);
series
> order(%%);
25
> 25;
25
> Order;
6
> Order:=3: taylor(f(x), x=a);
+ + +( )f a ( )( )D f a ( )-x a
1
2
( )( )( )D
( )2
f a ( )-x a 2
( )O ( )-x a 3
> sin_series:=taylor(sin(x), x=0, 6);
:=sin_series - + +x
1
6
x3
1
120
x5
( )O x7
> sin_series := series(1*x-1/6*x^3+1/120*x^5+O(x^7),x,7);
:=sin_series - + +x
1
6
x3
1
120
x5
( )O x7
I kdyz struktura rozvoje nam pripomina polynom, interni datova reprezentace je jina:
> subs(x=2, sin_series);
Error, invalid substitution in series
> op(sin_series);
, , , , , , ,1 1
-1
6
3
1
120
5 ( )O 1 7
> sin_series*sin_series;





- + +x
1
6
x3
1
120
x5
( )O x7
2
> expand(%);





- + +x
1
6
x3
1
120
x5
( )O x7
2
> taylor(%,x);
+x2
( )O x4
> mtaylor(sin(x), x=0,6);
- +x
1
6
x3
1
120
x5
mtaylor pocita Taylorovy rozvoje i pro funkce vice promennych a vysledkem je datova
struktura typu polynom.
> subs(x=2, %);
14
15
> whattype(%%);
+
Muzeme urcovat i koeficienty u danych mocnin x bez nutnosti pocitat cely rozvoj:
> coeftayl(sin(x), x=0, 19);
-1
121645100408832000
> sin_series;
- + +x
1
6
x3
1
120
x5
( )O x7
> diff(sin_series, x);
- + +1
1
2
x2
1
24
x4
( )O x6
> integrate(sin_series, x);
- + +
1
2
x2
1
24
x4
1
720
x6
( )O x8
> convert(sin_series, 'polynom');
- +x
1
6
x3
1
120
x5
> restart;
> with(powseries); #procedury, slouzici k praci s mocninnymi
radami
compose evalpow inverse multconst multiply negative powadd powcos powcreate powdiff, , , , , , , , , ,[
powexp powint powlog powpoly powsin powsolve powsqrt quotient reversion subtract, , , , , , , , , ,
template tpsform, ]
> powcreate(f(n)=a^n/n!);
> powcreate(g(n)=(-1)^(n+1)/n,g(0)=0);
> f(2);
a2
2
> f_series:=tpsform(f,x,5);
:=f_series + + + + +1 a x
a2
2
x2
a3
6
x3
a4
24
x4
( )O x5
> g_series:=tpsform(g,x,5);
:=g_series - + - +x
1
2
x2
1
3
x3
1
4
x4
( )O x5
> s:=powadd(f,g): tpsform(s,x,3);
+ + +1 ( )+a 1 x





-
a2
2
1
2
x2
( )O x3
> series(1+(a+1)*x+(1/2*a^2-1/2)*x^2+O(x^3),x,3);
+ + +1 ( )+a 1 x





-
a2
2
1
2
x2
( )O x3
> p:=multiply(f,g): tpsform(p,x,4);
+ + +x





- +
1
2
a x2





- +
1
3
1
2
a
1
2
a2
x3
( )O x4
> d:=powdiff(f): tpsform(d,x,4);
+ + + +a a2
x
a3
2
x2
a4
6
x3
( )O x4
> i:=powint(f): tpsform(i,x,4);
+ + +x
a
2
x2
a2
6
x3
( )O x4
Vypocty limit
> Limit( cos(x)^(1/x^3), x=0): %=value(%);
=lim
x 0
( )cos x








1
x
3
undefined
Jednostranne limity:
> Limit(cos(x)^(1/x^3), x=0, 'right'):
> %=value(%);
=lim
 +x 0
( )cos x








1
x
3
0
> Limit(cos(x)^(1/x^3), x=0, 'left'):
> %=value(%);
=lim
 -x 0
( )cos x








1
x
3

> y:=exp(a*x)*cos(b*x);
:=y e
( )a x
( )cos b x
> limit(y, x=-infinity);
lim
x ( )-
e
( )a x
( )cos b x
> assume(a>0):
> limit(y, x=-infinity);
0
> limit(y, x=-infinity) assuming a>0;
0
Student Calculus Package
> restart;
> with(student):
> f:=x->-2/3*x^2+x;
:=f x - +
2
3
x2
x
> (f(x+h)-f(x))/h;
- + +
2 ( )+x h 2
3
h
2 x2
3
h
> Limit(%,h=0);
lim
h 0
- + +
2 ( )+x h 2
3
h
2 x2
3
h
> value(%);
- +
4 x
3
1
> eval(%, x=0);
1
> showtangent(f(x), x=0);
­60
­40
­20
0
­10 ­8 ­6 ­4 ­2 2 4 6 8 10
x
> intercept(y=f(x), y=0);
,{ },=y 0 =x 0 { },=y 0 =x
3
2
> middlebox(f(x), x=0..3/2);
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.2 0.4 0.6 0.8 1 1.2 1.4
x
> middlebox(f(x), x=0..3/2, 10);
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.2 0.4 0.6 0.8 1 1.2 1.4
x
> middlesum(f(x), x=0..3/2, 10);
3
20















=i 0
9











- + +
2





+
3 i
20
3
40
2
3
3 i
20
3
40
> value(%);
603
1600
> middlesum(f(x), x=0..3/2, n);
3
2

=i 0
-n 1













- +
3





+i
1
2
2
2 n2
3





+i
1
2
2 n
n
> Limit(%, n=infinity);
lim
n 
3
2

=i 0
-n 1













- +
3





+i
1
2
2
2 n2
3





+i
1
2
2 n
n
> value(%);
3
8
> Int(f(x), x=0..3/2);
d





0
/3 2
- +
2
3
x2
x x
> value(%);
3
8
> with(Student[Calculus1]):
> DiffTutor();
=
d
d
x
( )x ( )sin x +( )sin x x ( )cos x
> IntTutor();
Initializing Java runtime environment.
=d


x2
( )sin x x - + +x2
( )cos x 2 x ( )sin x 2 ( )cos x
>