4.2. rQD fURXE DLQ PERIODI^ESKOJ S PERIODOM T = 2l FUNKCII, ZADANNOJ NA INTERWALE [;l l]
eSLI FUNKCIQ y = f(x) ZADANA NA INTERWALE [;l l] GDE l{ PROIZWOLXNOE ^ISLO, IMEET PERIOD T = 2l I UDOWLETWORQET NA RAS- SMATRIWAEMOM INTERWALE USLOWIQM TEOREMY dIRIHLE, TO UKAZANNAQ FUNKCIQ MOVET BYTX PREDSTAWLENA W WIDE SUMMY RQDA fURXE
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n x |
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cos |
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+ bn |
sin |
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n=1 |
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KO\FFICIENTY |
a0 |
an |
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bn NAHODQTSQ PO FORMULAM |
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1 l |
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an = l Z |
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f(x) cos |
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bn = l Z |
f(x) sin |
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4.3. rQDY fURXE ^ETNOJ I NE^ETNOJ FUNKCIJ |
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w SLU^AE, ESLI FUNKCIQ |
y = f(x) |
QWLQETSQ ^ETNOJ ILI NE^ET- |
NOJ, RQD fURXE DANNOJ FUNKCII BOLEE PROST. tAK, RQD fURXE ^ETNOJ FUNKCII SODERVIT TOLXKO SWOBODNYJ ^LEN I ^LENY S KOSINUSAMI. w TAKIH SLU^AQH GOWORQT O RAZLOVENII PO KOSINUSAM, A RQD NE^ETNOJ FUNKCII SODERVIT TOLXKO ^LENY S SINUSAMI I GOWORQT O RAZLOVENII W RQD fURXE PO SINUSAM.
dLQ ^ETNOJ PERIODI^ESKOJ FUNKCII WSE KO\FFICIENTY bn = 0, I RQD fURXE BUDET RQDOM PO KOSINUSAM
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f(x) = |
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an cos |
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n=1 |
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a0 = |
Z f(x) dx |
a0 = |
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f(x) dx |
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n x |
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an = |
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an = l |
Z f(x) cos |
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dx: |
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dLQ NE^ETNOJ PERIODI^ESKOJ FUNKCII KO\FFICIENTY a0 = 0 an = 0 I RQD fURXE BUDET RQDOM PO SINUSAM
f(x) = X1 bn sin nx
n=1
bn = 2 Z f(x) sin nx dx:
0
4.4. rQD fURXE NEPERIODI^ESKIH FUNKCIJ
eSLI FUNKCIQ f(x) NEPERIODI^ESKAQ I ZADANA NA PROIZWOLXNOM INTERWALE (a b) TO POD RAZLOVENIEM FUNKCII W RQD fURXE W \TOM PROMEVUTKE PONIMA@T RAZLOVENIE W RQD fURXE PERIODI^ESKOJ FUNK- CII S PERIODOM 2l = b ; a KOTORAQ NA PROMEVUTKE (a b) SOWPADAET S DANNOJ FUNKCIEJ f(x):
~ASTO PODOBNAQ ZADA^A MOVET BYTX SFORMULIROWANA TAK: rAZLOVITX W RQD fURXE FUNKCI@ f(x) NA PROMEVUTKE [0 l] (ILI,
W ^ASTNOSTI, [0 ]) W RQD PO KOSINUSAM ILI PO SINUSAM.
w \TIH SLU^AQH SMYSL ZADA^I ZAKL@^AETSQ W TOM, ^TO RASKLADYWA- ETSQ W RQD SOOTWETSTWENNO ^ETNAQ ILI NE^ETNAQ PERIODI^ESKAQ FUNK- CIQ S PERIODOM T = 2l (ILI T = 2 ), KOTORAQ NA INTERWALE [0 l] (ILI [0 ]) SOWPADAET S ZADANNOJ FUNKCIEJ.
tO ESTX FUNKCI@ DOOPREDELQ@T NA INTERWALE [;l 0] (ILI [; 0]) ^<TNYM ILI NE^<TNYM OBRAZOM I DLQ POLU^ENIQ E< RQDA fURXE IS- POLXZU@T PRIWEDENNYE WY[E FORMULY.
1: rAZLOVITX W RQD fURXE FUNKCI@, ZADANNU@ W INTERWALE (; ) |
WYRAVENIEM |
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f(x) = 8x ; 1 |
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< x + 1 |
0 < x < |
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tAK KAK DANNAQ FUNKCIQ QWLQETSQ NE^ETNOJ, TO MOVNO ZARANEE SKA-
ZATX, ^TO WSE KO\FFICIENTY an = 0 |
(n = 0 1 2 3 : : :): |
nAHODIM |
bn = 2 Z f(x) sin nx dx = 2 |
Z (x + 1) sin nx dx = |
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U = x + 1 |
dV = sin nx dx |
= 2 |
2;x +n |
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cos nx dx3= |
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dU = dx |
V = |
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cos nx |
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cos nx |
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+0# = |
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n2 |
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= n [1 ; ( + 1) cos n ] = n |
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rQD fURXE DLQ DANNOJ FUNKCII |
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f(x) = 1 bn sin nx = |
2 1 |
1 ; ( + 1) (;1)n |
sin nx: |
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zAPI[EM NESKOLXKO PERWYH KO\FFICIENTOW |
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b1 = |
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(2 + ) 3 27 |
b2 = |
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(; ) = ;1 b3 |
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(2 + ) 1 09 |
2 |
3 |
b4 = |
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(; ) = ;0 5 |
b5 = |
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(2 + ) 0 65 |
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b6 = 0 333 : : : |
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tOGDA RQD fURXE |
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f(x) = 3 27 sin x;sin 2x+1 09 sin 3x;0 5 sin 4x+0 65 sin 5x;: : :
2: rAZLOVITX W RQD fURXE FUNKCI@, ZADANNU@ W INTERWALE (; ) WYRAVENIEM
f(x) = 8;1 |
; < x < 0 |
< |
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0 < x < |
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nAJDEM KO\FFICIENTY RQDA fURXE PO SOOTWETSTWU@]IM FORMULAM, U^ITYWAQ ZADANIE FUNKCII NA KAVDOJ ^ASTI INTERWALA
a0 = 1 |
;Z |
f(x) dx = j |
INTEGRAL RAZBIWAEM NA SUMMU INTEGRALOW j = |
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Z |
( |
1) dx + |
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(2) dx |
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+2 ] = 1: |
= 1 |
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= 1 |
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an = 1 |
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f(x) cos nx dx = 1 |
2 Z |
(;1) cos nx dx + Z (2) cos nx dx |
3 = |
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= 1 2;;n1 sin nx |
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sin nx4;0 |
3 = 0 |
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T.K. sin n = sin 0 = 0 |
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bn = Z |
f(x) sin nx dx = ; |
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(;1) sin nx dx+Z (2) sin nx dx 3= |
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= 1 2 n1 cos nx |
;n2 cos nx |
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[(cos 0;cos n );2(cos n ;cos 0)] = |
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n |
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[;3 cos 0 ; |
3 cos5 n ] = |
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(1 ; cos n ) = |
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[1 ; (;1)n]: |
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zDESX MY ZAMENILI cos n = (;1)n: |
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dALEE, T.K. PRI |
n = 2k WY- |
RAVENIE [1 ; (;1)2k] = 1 ; 1 = 0 |
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TO WSE KO\FFICIENTY S ^ETNYMI |
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b2k = 0 |
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n = 2k |
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NOMERAMI OBRATQTSQ W NOLX |
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A PRI |
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WYRAVENIE |
[1;(;1) ; ] = 1+1 = 2 I WSE KO\FFICIENTY S NE^ETNYMI NOMERAMI
196
6 |
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b2k;1 = |
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b1 = |
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s U^ETOM;NAJDENNYH KO\FFICIENTOW RQD fURXE DLQ DANNOJ FUNKCII |
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a0 1 |
1 |
6 |
1 sin(2k 1)x |
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f(x) = |
2 + |
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bn sin nx = 2 |
+ |
X |
2k |
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= |
BUDET IMETX WID |
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;1 |
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n=1 |
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k=1 |
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= 0 5 + 1 91 sin x + 0 64 sin 3x + 0 38 sin 5x + ::: |
3: rAZLOVITX W RQD fURXE FUNKCI@, ZADANNU@ W INTERWALE (;2 2) WYRAVENIEM
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f(x) = 81 |
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< x |
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nAJDEM KO\FFICIENTY RQDA fURXE PO FORMULAM, SOOTWETSTWU@]IM |
SLU^A@ PROIZWOLXNOGO l |
U^ITYWAQ, ^TO W NA[EM SLU^AE l |
= 2 |
A |
TAKVE ZADANIE FUNKCII NA KAVDOJ ^ASTI INTERWALA. |
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x2 0 |
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2Z 1 dx + Z x dx3 = |
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3 = |
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a0 = 2 |
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[ 2 + 2 ] = 2 |
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6;2 |
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an = |
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f(x) cos |
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dx= 22 |
Z |
1 cos |
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dx+ |
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x cos |
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dx3= |
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4; |
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U = x |
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dU = dx 5 |
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wTOROJ INTEGRAL BEREM PO ^ASTQM: |
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dV = cos n x dx V = |
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sin |
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sin |
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0 ; |
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cos |
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3 = |
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[1+(;1) |
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5: |
rAZLOVITX FUNKCI@ |
y = e; |
x |
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ZADANNU@ W INTERWALE |
(0 1) |
W RQD fURXE PO SINUSAM. |
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fUNKCIQ ZADANA NA POLUINTERWALE. |
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pRODOLVIM EE NA INTERWAL (;1 0) NE- |
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^ETNYM OBRAZOM. tOGDA WSE |
KO\FFICI- |
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ENTY |
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an = 0 |
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A KO\FFICIENTY bn |
NAHODQTSQ PO FORMULAM, SOOTWETSTWU@]IM SLU- |
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^A@ NE^ETNOJ FUNKCII W INTERWALE [;l |
l ] PRI^EM l = 1: |
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l |
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bn = |
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e;x sin n x dx = |
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l |
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0 |
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dANNYJ INTEGRAL QWLQETSQ CIKLI^ESKIM. wOSPOLXZUEMSQ TABLICEJ
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Z eax sin bx dx = |
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eax |
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(a sin bx ; b cos bx) : |
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a2 + b2 |
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1+n2 2 |
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he;1(;1)n+1 +1i : |
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1+n2 2 |
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rQD fURXE DLQ FUNKCII |
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f(x) = 1 bn sin n x |
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1 + n2 2 i |
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z A M E ^ A N I E. oTMETIM, ^TO MOVNO RASKLADYWATX W RQD fURXE FUNKCII, ZADANNYE W PROIZWOLXNOM INTERWALE [a a + 2l] DLINOJ 2l: w \TIH SLU^AQH W FORMULAH NAHOVDENIQ KO\FFICIENTOW RQDA fURXE
(2) NUVNO PREDELY INTEGRIROWANIQ ZAMENITX NA a I a+ 2l: nAIBOLEE ^ASTO \TA SITUACIQ WOZNIKAET, KOGDA FUNKCIQ ZADANA NA INTERWALE [0 2 ] TOGDA PREDELY ; I ZAMENQ@TSQ SOOTWETSTWENNO NA 0 I 2 :
6: w INTERWALE (0 2 ) RAZLOVITX W RQD fURXE FUNKCI@ y = x2:
tAK KAK DANNAQ FUNKCIQ NE QWLQETSQ ^ETNOJ ILI NE^ETNOJ, TO NAHODIM WSE KO- \FFICIENTY RQDA.
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a0 = |
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f(x) dx = |
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3 j0 |
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an = 1 |
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U = x2 |
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U = x |
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( ; nx cos nxj02 + n1 |
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1 cos nx = |
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U = x2 |
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dV = sin nx dx |
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1 cos nx |
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4 2 |
+ 0] = |
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= [ |
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0 |
n |
x cos nx dx] = |
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j |
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iTAK, RQD fURXE |
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4 2 |
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f(x) = |
a0 |
+ |
1 |
an cos nx+ bn sin nx = |
+ 4 |
1 cos nx |
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1 sin nx |
: |
2 |
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3 |
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n2 |
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n=1 |
n |
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n=1 |
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nAJDEM AMPLITUTU An I FAZU 'n GARMONIK, IME@]IH ODINAKOWU@ ^ASTOTU. iSPOLXZUEM IZWESTNOE IZ FIZIKI PRAWILO SLOVENIQ DWUH GARMONIK
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nX=1 an cos nx+bn sin nx= nX=1 An sin(nx+'n) = nX=1 qan2 +bn2 sin(nx+'n): |
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An = v |
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p |
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16 |
+ 16 2 |
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4 |
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aMPLITUDA n |
; |
OJ GARMONIKI |
= |
1 + 2n2 |
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n2 |
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un4 |
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an |
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4=n2 |
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1 |
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fAZA |
'n = arctg bn |
= arctg |
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;4 =n |
n |
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