I want to find the roots of x^3+2x^2-3x-1=0 using bisection method. But Im not sure how to select the initial approximation. Can you tell me how select a suitable initial approximation to this?
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OpenStudy (anonymous):
i wl tell u d trick
OpenStudy (neo92):
ok. tl me.. my buk menyons something called taylor method. But i hav no ida abt it.
OpenStudy (neo92):
mentions
OpenStudy (anonymous):
frst select the interval where sign of the equation changes
OpenStudy (anonymous):
exactly you have to choose smallest interval of two points say a and b such that f(a)<0 and f(b)>0
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OpenStudy (anonymous):
absolutely right
OpenStudy (anonymous):
so that interval is (a,b)
OpenStudy (neo92):
can u tell me how to select the intervals?
OpenStudy (anonymous):
first method is hit n trial..................
OpenStudy (neo92):
can i differentiate and find it?
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OpenStudy (anonymous):
gd there u cn find the points where slope is zero
OpenStudy (anonymous):
by differntiating u gt x= 0.58 n - 2.58
OpenStudy (neo92):
yep. so what should i do after that?
OpenStudy (anonymous):
then take points nearer to one point u wl see sign of equation changes similarly repeat for other point that u hav find after differntiation....
OpenStudy (neo92):
|dw:1358689959945:dw|
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OpenStudy (neo92):
but does the sign of the eqn changes on a occassion like this?
OpenStudy (anonymous):
i didn't get wat r u saying????
OpenStudy (neo92):
You said I should select tha interval where the sign of the equation changes.
OpenStudy (anonymous):
w8 i m clearing
OpenStudy (anonymous):
first put x = 0 in the original cubic equation tl me wat u get ?
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OpenStudy (anonymous):
asap
OpenStudy (neo92):
1
OpenStudy (anonymous):
dear it's negative keep all the expression on the left while putting so u gt -1
OpenStudy (anonymous):
then put x = 1 u wl get -1 again .........till now sign of the function is negative kk
OpenStudy (neo92):
yep.. i got it
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OpenStudy (anonymous):
nw put x = 2 u wl gt 9 which is positive so ur initial approximation is (1,2)
OpenStudy (anonymous):
hope u gt it dear NEO
OpenStudy (anonymous):
THEN APPLY BISECTION
OpenStudy (neo92):
Okey.. i gt it now. so is that the only way.. ?
OpenStudy (anonymous):
yup dear n thanks fr awarding me the medal
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OpenStudy (neo92):
:)
OpenStudy (neo92):
so I hav to apply the same method for an eqn like \[3x-e ^{x}=0\]
OpenStudy (anonymous):
u can dear........
OpenStudy (anonymous):
here interval is wat tl me asap?
OpenStudy (neo92):
wait a min
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OpenStudy (anonymous):
i wl
OpenStudy (neo92):
is it (1,2) ?
OpenStudy (anonymous):
w8
OpenStudy (anonymous):
by putting 1 u get 1.74 n fr x = 0 u gt -1 so changes here first interval is (0,1)
OpenStudy (neo92):
oh yes.. i hav missed the -sign when calculating
and i think 1 gives 0.28..... (3-e)
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