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Черновицкий национальный университет им. Ю. Федьковича

Предмет:

[НЕСОРТИРОВАННОЕ]

Файл:

Lab_4

.pdf

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1 / 21 2 > Следующая >>>

’ – ++ © . .

: . . – . .

. : , while, do – while, for,goto, break continue. ! – "# . # " ! "$.

1. . ! " #$

- , ! .

! while

" while ( while ( ) ;

) – , + ! true ( + (). . + / ( ( .

! do – while

" do – while (

do ; while ( );

) – , / ! true ( + (). . + / .

! for

" for (

for ( _1 ; _2 ; _3 ) ;

+ (+:

_1;

while (_2)

{

_3;

}

. ! - _1, +; + - _2,/ +, + +, 7 8+- false ( + ); – – , 8+- + ( ; (+ - _3 , ( ( , / . ; / , 8 ( /+ 8+ +7 .

) + _2 + ( true , 8 “ (+” +. > 8-( + ( .

2. ! goto, break continue. ! break.

" break 8 / while, do-while, for, 8 switch (+ ; + + (+ +, ( (. ) +! + 8 switch, 7 8 .

! continue

" continue !+ + + while, do-while 8 for, 7 , (+ + + + +, 8 +. " continue + .

! goto #

’ – ++ © . .

D / 8 +( + + + ( goto ( ( . D ( 8 ( ( () F , ( :

! 7_ :

G F ( 8+ + ( + F+ . D +8' goto. " goto ( :

goto ! 7_;

I ( goto ( / + 8+- ( F+ , 7 ! 7 , 8 / .

> ( ( / 8 ( . > ( + 8 8 , + ( , ( / ( ! ( ( ( : break continue .

3. 0 1 . 2 34 $ – 56 1 #$.

J ( - , 8 ’ . L , + ( 8 (+ ’ , ( ,

+( ( ’ + / , 7 + (. . 8 / (+ + 8-( + ( ( ( F + ( .

M-( (+ 8 / (+ ( ( ( F + . > 8 ! / ( F + ( 8 / (+ :

8 8 , (;

(-)

(+)

( ( ( 8 ) ( , 7 / , +

8	+	(’
( ( ;

8 +( , ( . W (8 + +( . X 7 +( + , ( 8+ + “ ”, + + (+ + “ ”;

8 ,	8
F ( ;

8 + 8 ( , + 8 + (;

( 8-( ( ( ( ( +

M ( 8 + + + (+ (+) / +, (+ + ( +, ! - .

. 1. O ( 8-(+ (+, (+ ( zTRUE( ), 7 M(x,y) / F + ( 8 7 ),

’ – ++

8( / / ( ( (F+ ( ) FALSE( 8 ) – + + (+ + :

x = y2 ,

y = x2 .

. F + ( 8 7 ) . 1.

. 1.

M+ + ( 8-(+ (+ ( . 2).

.

) x,y
		+
y ≥ x2	i x ≥ y2	+
y ≥ x2	i x ≥ y2		z:=true

z:=false

_+ z

;

! 2. O (+ + + 8 ( + ( ( + ( ` (.

( . ( ( (+ O ).

’ – ++		© . .
.
l o w e r = 0 u p p e r = 3 0 0 s t e p = 1 0
s u m :)= 1
+ 8
i : = 2
f a h r = l o w e r
c e l s i u s = 5 . / 9 . * ( f a h r - 3 2 . 0 )
)
f a h r = f a h r + s t e p
f a h r < = u p p e r
;

#include <stdio.h>
void main()
{
int lower=0,upper=300,step=10;
float fahr,celsius;
printf("	Таблиця відповідності температур по Фарангейту \n ");
printf("		температурам по Цельсію. \n ");
printf("\|	Фарангейт	\|	Цельсію \| \n ");
fahr=lower;
while(fahr<=upper)		{ celsius=5./9.*( fahr-32.0);
printf("\|	%4.0	\|	%6.1 \|\n ",fahr,celsius);
	fahr+=step;

}

! 3. I + n. O (+ 8 8+ ++ 1 n, 8 12 22 32 42 ... n2.

’ – ++

) n

m u l:= 1

i:= 2



	m u l:= m u l* i2
		+
		+
	i< n		i:= i+ 1
	i< n		i:= i+ 1
_
_

)m u l

#include<iostream.h> void main()

{

int n,i; double mul;

cout<<" Введіть натуральне число n"; cin<<n;

mul:=1;

for( i=2;i<= n; i++ ) mul*=i*i; cout<<" Добуток квадратів перших "<<n;

cout<<" натуральних чисел дорівнює ="<<mul;

}

. 4. I +			n x. "8 ε = 0.001
+(+ :
	100	1		x	2s+n
	100	1		x
	∑				.
	∑				.
	s=1 s!(n + s)! 2s +1
#include <stdio.h>
#define	EPSILON 0.001
#define	MAXSUM 100
void main()
{
int n,s,	i,j,fs,fsn;
double x , px,Summa;
printf(" Введіть натуральне число n та дійсне число x \n ");
scanf("%d%lf" ,&n,&x);
fs=1;fns=1; Summa=0;
for(i=2; i<=n; i++) fns*=i;
for(s=1; s<=MAXSUM; s++) {
	fsn*=(s+n);
	fs*=s;		px=1;
	for(i=1; i<=2s+n; i++) px=(x/(2*s+n));
	px/=(fs*fsn);

Summa+=px; if(fabs(px)<EPSILON) break;

}

printf(" Результат %f з точністю EPSILON. \n ",Summa);

}

5. 2 1 3$; 3 6 ! 5 #4$5 <.

’ – ++

. + , 7 +

/ .

. 5. I + + xn = 1.25xn−1 + 0.125, x0 = 1. O

8-(+ (+ / n -

)

x n 1 = 1 , 0

i : =

x n = 1 . 2 5 x n 1 + 0 , 1 2 5

x n 1 = x n

i < n

i : = i + 1

_ +

x n

;

#include <iostream>

using namespace std;

void main()

{

int n,i;

double xn,xn1;

do {

cout<<"Input n \n";

cin>>n;

} while(n<=1);

xn1=1.0;

for(i=1;i<n;i++)

{

xn=1.25*xn1+0.125;

xn1=xn;

}

cout<<" x ("<<n<<")="<<xn<<endl;

}

. 6. I +

xn = 1.25xn−1 + 0.125yn−1,

x0 = 1,

yn = 1.25yn−1 + 0.125xn,

y0 = 0,

_ n (+ / yn - + .

#include <iostream> using namespace std; void main()

{

int n,i;

double xn,xn1; double yn,yn1;

’ – ++

do { cout<<"Input n \n";

cin>>n; } while(n<=0);

xn1=1.0;

yn1=1.0; for(i=0;i<n;i++)

{

xn=1.25*xn1+0.125*yn1;

yn=1.25*yn1+0.125*xn;

xn1=xn;

yn1=yn;

}

cout<<" y ("<<n<<")="<<yn<<endl;

}

. 7. I + + xn = 1.25xn−1 + 0.125, x0 = 1.

_ [5,8] [10,14], +( ( + , / (.

#include <iostream> using namespace std; void main()

{

double xn, Sum1,Sum2; Sum1=0; Sum2=0;

xn=1.0; while(1)

{

xn=1.25*xn+0.125; if( xn>14 ) break;

if( xn>=10) Sum2+=xn; if( xn>8) continue; if( xn<=5) Sum1+=xn;

}

cout<<" Summa [5,8] = "<<Sum1<<" Summa [10,14] = "<<Sum2<<endl;

}

			. / 0 /
= 4 > 1.
1.1.	I		+		n. > (+ 8	8+ +
+		1 n, 8 n!=1 2 3 4 ... n.
1.2.	I +				n. > (+ 8 + :
1 2 +	2 2 + 3 2 +	4	2 + ... + n 2
				.
	1 2 3 4			.
	1 2 3 4		... n

1.3.I + n x. > (+ 8

+ : cos x + cos3x + cos5x + ... + cos(2 n + 1) x.

1.4.I + n x. > (+ 8+ : ln 2x + ln 4x + ln 6x + ... + ln 2nx

1.5.I + n x. > (+ 8

+ : tgx + tg9x + tg17x + ... + tg(8n +1)x .

1.6.I + n x. > (+ 8

+ : xtgx + 2xtg3x + 3xtg5x +...+ nxtg(2n −1)x.

1.7. I + n. > (+ 8 + :

e0 + e1 + e2 + e3 + ...+ en . 1 2 3 4 ... n

1.8. I + n. > (+ 8 + :

1+	1	+	1	+	1	+ ... +	1	.

2		22		23			2n


’ – ++		© . .
1.9.	I + n. > (+ 8 + :

e0 + e1 + e2 + e3 + ... + en . 1 2 3 4 ... n

1.10. I + n. > (+ 8 + :

1+	1	+	1	+	1	+ ... +	1	.

2		22		23			2n

1.11.I + n x. > (+ 8

+ : x + x 2 + x3 + ... + x 2n .


1.12.					I +									n. > (+ 8 + :
	1		+	1		+	1	+ ... +		1		.
	2			2			2			2
1		2			3					n
1.13.					I +									n. > (+ 8 + :
		1		+	1			+	1		+ ... +			1	.
													nln(n +1)
1ln 2						2ln3			3ln 4

1.14.I + n. > (+ 8 + :

cos1	+	cos2	+	cos3	+ ... +	cos n	.

1 2 2 3 3 4						n (n +1)

1.15.I + n. > (+ 8 + :

sin 2	+	sin	3	+	sin	4	+ ... +	sin n	.
								(n −1) n (n +1)
1 2 3 2 3 4 3 4 5

= 4 > 2.

2.1.> + + 8 F+ y = x3 [2,5] ( 0.5.

2.2.	I n>0. > + + 8 F+ y =			x2	− 7
2.2.	I n>0. > + + 8 F+ y =
					2
[0, n ] ( 0.5.
2.3.	I n>0. > + + 8 F+	y = xsin(x)
[0, n ] ( 1.
2.4.	I h>0. > + + 8 F+ y = cos(x)
[0,π ] ( h.
2.5.	I b>0 h>0. > + + 8 F+ y = ex
[0, b] ( h.
2.6.	I a, b h (b>a,h>0). > + + 8 F+
y = tg(x) [a, b] ( h.
2.7.	I a, b h(b>a,h>0). > + + 8 F+
y = xe−x [a, b] ( h.
		y =		ex	+ e	− x
2.8.	I n>0. > + + 8 F+
2.8.	I n>0. > + + 8 F+				2
					2
[0, n ] ( 1.
		y =		ex	− e	− x
2.9.	I n>0. > + + 8 F+
2.9.	I n>0. > + + 8 F+				2
					2
[0, n ] ( 0.5.
		y =		ex	− e− x
2.10.	I n>0. > + + 8 F+
2.10.	I n>0. > + + 8 F+		ex
			ex		+ e− x
[-n, n ] ( 0.1.
2.11.	I h>0. > + + 8 F+ y = arctg(x)
[0, 3 ] ( h.
2.12.	I a b(b>a). > + + 8 F+
y = 5 x3 − 2 x [a, b] ( 0.1.


’ – ++		© . .
2.13.	I n>0. > + + 8 F+ y = x5 − x3 + x

[-n, n ] ( 2.

2.14. I h>0. > + + 8 F+ y = (π − x)tg(x) [0, π ] ( h.

2.15. I h>0. > + + 8 F+

y = (π − 2x)tg(2x) [0, π ] ( h.

2 8

= 4 > 3.

3.1. _ + n x. > (+ 8 +( ε = 0.001:

	20		1	x		2s			20	1		x		2s
	20								20
a)	s=1				,			)	s=1				,
a)					,			)					,
				2s								2s
	∑	(n + s)!		2s					∑	(\| s − n \|)!		2s
	20		xs+1						10	1 x		s
8)	∑						,	)	∑			, .
8)			(n + s +10)!−s!				,	)				, .
			(n + s +10)!−s!							(n + s)!	2
	s=1								s=1

3.2. _ + x 8 ε . > (+ 8 +(+ :

∞

1 x 2s

∞

1 x s

∑

)

(s −1)!

s=1

∞

xs+1

∞

1 x s

)

∑

)

∑

(s +1)!

(5 + s)!

s=1

∞	x s
∑			,
∑			,
s=1	s!
∞		x − s		s
∑
				.
s=1	(s +1)!

3.3. _ + n x 8 ε . > (+ 8 +(+ :

∑

) ∑

(2 + s)!

(| s − 5 |)!

s=1

xs+1

x s

∑

) ∑

, .

(s +10)!−s!

(|10 − s |)!

s=1

’ – ++		© . .
= 4 ># 4-6 .
J K L ! ! -"# .
1 1
1. I + + xn = xn−1 + 0.5,		x0 = −5. > (+
/ n - + .
2. I +
xn = 1.5xn−1 − 0.25yn−1,	x0 = 1,
yn = 1.05yn−1 +1.025xn,	y0 = 0.5,
_ n (+ / yn		- + .
3. I + + xn = xn−1 + 0.5,		x0 = −5.

_ [5,12] [18,22], +( ( + , / (.

1 2
1.	I + + xn = xn−1 + 0.5xn−2, x0 = 1,			x1 = 2.	> (+
/ n - + .
2.	I +
	xn = 1.75xn−1 − 0.275yn−1,	x0 = 1.25,
	yn = 1.05yn−1 +1.025xn −1.05,	y0	= 0.5 .
_ n (+ /		yn	- + .
3.	I + + xn = xn−1 + 0.5xn−2 + 0.25,			x0 = 0.5,	x1 = 0.75.

_ [5,12] [30,38], +( ( + , / (.

1 3.
1.	I	+ + xn = axn−1 + 0.5xn−2, x0 = 1, x1 = a.
> (+ / n - + .
2.	I	,b,c +
	xn	= axn−1 − 0.275yn ,	x0 = 1.5	*b,
	yn = byn−1 +1.025xn−1 − c,		y0 = 0.5 .
_ n (+ / yn				- + .
3.	I	(a>0) ,b,(c>0),d(d>2*c) + +

xn = xn−1 + axn−2 + 0.25, x0 = 0.5*b, x1 = 0.75.

_ [c,2*c] [d,3*d], +( ( + , / (.

1 4.
1.	I	+ +
xn = axn−1 + 0.25xn−2 − 0.25a,			x0 = a, x1 = 1. > (+ / n -
	+ .
2.	I	,b,c,d	+
	xn	= axn−1 − 0.75byn−1 + 2.5d +1,		x0 = 1.5*b,
	yn	= byn−1 − 0.5axn − 2.5c + 2.76,		y0	= 0.5*d .
_ n (+ / xn				-	+ .
3.	I	(a>3) , b (b > a) + +
xn = 0.5xn−1 + 0.1axn−2 + 0.1bxn−3 + 0.25a, x0 = 0.25,				x1 = 0.75, x2 = 0.51.

I +(+ ( , / ( [0, 1+ ], [3+2 ,3+5*b] [100+5*b, +∞ [.

1 / 21 2 > Следующая >>>

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