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MAT-10432 Insinöörimatematiikka B3u - 28.03.2011

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Original exam
SYN -
EI TAMPEREEN TEKNILLINEN YLIOPISTO KULLO onen

MAT-10432 Insinöörimatematiikka B 3u
Tentti 28.3.2011

 

 

 

Ei laskinta eikä taulukkokirjoja. Kaavaliite ohessa.

+ Integroi
/ 2
(7-1

1831 dx.

2. a) Muodosta funktion f(x) = Inx kolmannen asteen Taylorin polynomi
pisteen 1 suhteen.
Suppeneeko vai hajaantuuko
+ sin(2n) ,
2 1+2? '
n=1

 

x Ratkaise alkuarvotehtävä
x Ratkaise alkuarvotehtävä

y" +y"—6y'=0. ylo)=-1. y'(0) = 0, y"(0) =3.

 

Tehtäväkohtaiset tulokset julkaistaan POPissa toteutuskerran sivulla.

 
(117214) W] *(y1y2 A) U IMSTYNIVÄ 119 FO = IV '6
[194 nie ja9TA jyTa] = (MX OH) X = (NX :xp = x "s
9 ad [yt **(2G)S09 2,9 ((TJ)809 11,9

(29 )UIS 199] yt ** LYS 779L (LG) US 779
JUSTONJVA 11G/ FO = Y LVÄIMIITIELUTSETTT UVUTLITDN-Y (F)

 

 

 

—y? av? avata?
JUSTEHJILI :Y LINNÄTELDI UOUTVIION-Y (€)
(TG)809 ,,,9 el (nzg)us

JUSTOJIVLI 126) FO = [ 11PdIINNITLLUTJTLUN UVUTLIVYUISYÄ (Z)

 

)

 

ay? OSTEYJLI :Y LINNÄÖTELDI UOUTEJIYUISIA (T)

0=00+KI10+---+ ju VIT UD + XD
:0 = 00 + /ilD + --- + (1-4) 6140 + (upi"D L

(2 + "Lay? | ) o = (0), x)f = fi(z)0+ fi 9

= 2 s09

; = L UIS

I
v

 

 

a+ tal)

 

ep;

D=

 

=P ()f / 4=4
ay |

(OS + IMG [ sa=v apiloyp+an | =» s
, OLTV

2p (e)6X)] | — (a)öl2)7 = ap (a)0le) [=

1-8
u[-=Z T YUB) 18
a hmm ARS

 

+ 2| u] = 9 + x y4s00 1e
+ 2) U] = E + e yusae
2 + T uvGare

I+ L UsVae

 

 

 

 

2+ 100 —
J + 2 uv)
TUI
2+|vuis|uj s = 1109
2 + |z soo | u] — TUI
sp [ (2)

LloALeNTNUTo1d)U] *T

 

(TT0Z-010z/€ ipod) a11jenee) unuon
nE G OHAHIELIDIEUMOOUISU] ZEPOT-LVW

 

 

OLSIJOITA NINITIINNII NIIHIAWVL =


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