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Q14 Solutions (1 Viewer)

Fred2013

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GUYS
i manipulated the equation into
te^t - 1 = 0, and i let f(t) equals that
Then applied newton's method, i got ~ 0.570...
IS THAT WRONG?!?
I wrote 0.57... = ~ 0.56
!!!!
 

lijincheng

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no it's not fine because i got 0.57 too only because i differentiated e^x wrong, i differentiated it into xe^x for some odd reason. and that's what i got.
 

Fred2013

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no it's not fine because i got 0.57 too only because i differentiated e^x wrong, i differentiated it into xe^x for some odd reason. and that's what i got.
well i didnt differentiate wrong.. (pretty sure i didnt)
 

doublem

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You're a good man, Sy.
Thank you so much for the solutions!
 

aimpacc

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is it ok to round off the last question to 0.1??? thats wat i did....they might want 0.097 idk will i lose marks?
 

uart

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is it ok to round off the last question to 0.1??? thats wat i did....they might want 0.097 idk will i lose marks?
Ok I finally got a copy of the exam so I can now comment. :)

In my opinion the numerical approximation of 0.1 (one significant digit) should be fine. The data you were given, t1~=0.56, was itself only accurate to 2 significant digits, and in general the series of calculations required would yield a result with less accuracy than that of the input data.

There are likely two different (correct) expressions for "r" that students may have gotten in this question. The result I obtained was:



My method and results were pretty much the same as that of Sy123 in the OP. Numerically this expression yields an approximation of 0.094 for r. If however you work out a more precise value for t1, then you find that r is actually 0.09726 (to 4 s.d). So to be honest, your answer of 0.1 is probably better than leaving it as 0.094. Personally I just wrote down 0.094, but noted that this is an approximate value and not necessarily accurate to the number of significant digits written. I think this is fair enough too.

I noticed that Terry Lee started with the same two simultaneous equations but derived an alternative expression of:



Since t1 is the solution to exp(t1) = 1/t1, then clearly these two solutions are identical if an exact value of t1 is used. Quite by coincidence however, it turns out that this latter expression is less sensitive to errors in the input value t1, and thus yields an approximate value of r equal 0.097, which is correct to 2 s.d.

Hope this helps.
 
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