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    (Original post by ntrik)
    Damn I must be using the wrong approach then.

    I have to prove that x = 1.66 to 3sf.

    I've also got an interation formula which is xn+1 = (1/3)e^(1+(1/x)) where x0 = 2 but I dunno if that helps or anything.

    Edit - Oh wait, does that mean I need to subsitute 1.66 into the equation to get a value that is close to 0?
    No you sub. x = 2 in until you get something like 1.66.
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    In the real exam the spec states that for iteration questions "leads will be given".

    ie. the iterative formula.
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    f(x) = x4 – x – 1
    f(x) = 0 has a solution such that n < x < n + 1 where n is a positive integer.
    a)i)Find a positive value of n such that the inequality is true. [3]
    ii)Construct a simple logical argument to prove that such a solution exists.[3]
    b)Using an iteration based on the equation x = , find a solution to f(x) = 0 to 3 decimal places.[4]
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    (Original post by samdavyson)
    In the real exam the spec states that for iteration questions "leads will be given".

    ie. the iterative formula.
    not necessarily, they may just give you x0 but you'l have to devise the formula yourself, it shouldnt be too hard tho
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    (Original post by Featherflare)
    not necessarily, they may just give you x0 but you'l have to devise the formula yourself, it shouldnt be too hard tho
    from practice edexcel papers it usually says show that the following iteration formula can be derived and then says use this iteration formula to calculate the root etc.
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    (Original post by mackin boi)
    f(x) = x4 – x – 1
    f(x) = 0 has a solution such that n < x < n + 1 where n is a positive integer.
    a)i)Find a positive value of n such that the inequality is true. [3]
    ii)Construct a simple logical argument to prove that such a solution exists.[3]
    b)Using an iteration based on the equation x = , find a solution to f(x) = 0 to 3 decimal places.[4]
    a) i) n= 1 because f(1) = -1 and f(2) = 13.. sign change so root between 1 & 2
    ii) sign change, f(x) continuous so root change.. (dnt really knw wat els to say for dis)
    b) x = 4 √(x +1)
    x0 = 1.5
    x1 = 1.257
    x2 = 1.225
    x3 = 1.221
    x4 = 1.2208
    x5 = 1.2207
    x6 = 1.2207

    so to 3 d.p, its 1.221
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    (Original post by mackin boi)
    f(x) = x4 – x – 1
    f(x) = 0 has a solution such that n < x < n + 1 where n is a positive integer.
    a)i)Find a positive value of n such that the inequality is true. [3]
    ii)Construct a simple logical argument to prove that such a solution exists.[3]
    b)Using an iteration based on the equation x = , find a solution to f(x) = 0 to 3 decimal places.[4]
    Is that from a ZigZag Paper?

    (a) (i) You just have to find where there is a sign change

    i.e. f(1) = -1 and f(2) = 13

    So n = 1

    (ii) f(x) is continuous, there is a sign change in the interval so there must be s root of f(x) = 0.

    (b) I can't see the equation it is supposed to be based on, but I think it is 4th√(1 + x)

    x0 = 1

    x1 = 1.189207...

    x2 = 1.216386...

    x3 = 1.220144...

    x4 = 1.220661...

    x5 = 1.220732...

    x6 = 1.220742...

    So to 3dp: 1.221.
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    (Original post by Widowmaker)
    from practice edexcel papers it usually says show that the following iteration formula can be derived and then says use this iteration formula to calculate the root etc.
    Well it should be fine either way
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    I am going now to do a few papers.

    See everyone back here tomorrow for full C3 reaction.

    GOOD LUCK !

    Sam.
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    Thanks. Good luck! =)
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    Yah, good luck dude!
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    has anyone done the edexcel mock paper for C3, its really easy comapred to edexcel's other papers.
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    I found the specimen paper easier
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    (Original post by sb1986)
    has anyone done the edexcel mock paper for C3, its really easy comapred to edexcel's other papers.
    i agree.
    c3 mock is easy, specimen is slightly harder, but then i did do the specimen about 1 month ago so it might seem easier now.
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    I havent done the mock but i found the specimen easy
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    the specimin paper was so annoying, that question 3 on trig was a killer. lol
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    I did the mock paper like 2 months ago though, so that's probably why I found the specimen easier. And yup, Q3 was a toughie but it wasn't too bad on the whole. C4 though, that's a whole new story..
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    Peeps - you learning back c1 c2 stuff? Which stuff do you think we need? Personally, I dont think therell be stuff like arithmetic progression..etc, it'd have to be something related to the c3 syllabus. =o
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    3b) was tough, in the end i did it halfway from th LHS n halfway from the RHS and got a common equation lol. As for C4, it hasnt been touched yet
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    (Original post by TheWolf)
    Peeps - you learning back c1 c2 stuff? Which stuff do you think we need? Personally, I dont think therell be stuff like arithmetic progression..etc, it'd have to be something related to the c3 syllabus. =o
    nope i ain goin over it, despite wat they say, thers no way they'l waste a whole question testing us on c1 & c2 knowledge on this the first ever C3 exam. Even if it comes up it will be something ridiculously easy
 
 
 
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