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# OCR (non mei) FP2 Monday 27th June 2016 Watch

1. (Original post by duncanjgraham)
Off the top of my head - deduce the Newton raphson formula :P
think I messed up the notation a bit but I think that's the gist of it

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2. (Original post by drandy76)
think I messed up the notation a bit but I think that's the gist of it

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seems a good method, but i obviously can't legitimately say how good it is

i like this method :

3. (Original post by duncanjgraham)
seems a good method, but i obviously can't legitimately say how good it is

i like this method :

yeah i think that way makes much more intuitive sense, but no fancy expansions so its not for me
4. (Original post by drandy76)
yeah i think that way makes much more intuitive sense, but no fancy expansions so its not for me
haha and this is why maths is brilliant
5. Ahhh let's just get this one out of the way, shall we?
6. Can someone help me with the very last part of the last question. I calculated g'(alpha) but do not know where to go from here. Neither do I really understand the situation/mathematics so could someone explain..
7. (Original post by Mathematicus65)
Can someone help me with the very last part of the last question. I calculated g'(alpha) but do not know where to go from here. Neither do I really understand the situation/mathematics so could someone explain..
hope this helps. 😃😃
8. OCR FP2 Jan 2013, Question 7. You have to draw graph of y=(x2+1)/((x+1)(x+7)).
I found the equation of asymptotes as y=1, x=7 and x=-1 and stationary points (1/3,-1/8) and (-3,1/2)
Curve meets asymptote at (-4/3,1).
Now I need to sketch the curve but I can't move any further. Can anyone explain this to me please?

Thanks in advance.
9. (Original post by Mathematicus65)
Can someone help me with the very last part of the last question. I calculated g'(alpha) but do not know where to go from here. Neither do I really understand the situation/mathematics so could someone explain..
We know that delta 2 can be calculated by g'(alpha) x delta 1 and that each successive term can be generalised by g'(alpha) x the previous term.

It follows that any given term delta n can be found be g'(alpha)^n-1 x delta1 you then use this knowledge to find n

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10. (Original post by tangotangopapa2)
OCR FP2 Jan 2013, Question 7. You have to draw graph of y=(x2+1)/((x+1)(x+7)).
I found the equation of asymptotes as y=1, x=7 and x=-1 and stationary points (1/3,-1/8) and (-3,1/2)
Curve meets asymptote at (-4/3,1).
Now I need to sketch the curve but I can't move any further. Can anyone explain this to me please?

Thanks in advance.
Investigate how the values change either side of the asymptotes to get a rough idea of the general shape of the graph

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11. (Original post by tangotangopapa2)
hope this helps. 😃😃
Thank you!! I'm still unsure where g'(alpha) ^(n-1) comes from?
12. (Original post by drandy76)
We know that delta 2 can be calculated by g'(alpha) x delta 1 and that each successive term can be generalised by g'(alpha) x the previous term.

It follows that any given term delta n can be found be g'(alpha)^n-1 x delta1 you then use this knowledge to find n

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Where is the g'(alpha)^n-1 coming from. I understand where to go from that point but I don't understand where this comes from and why you can multiply by delta 1. Thank you for your response though!!
13. (Original post by Mathematicus65)
Where is the g'(alpha)^n-1 coming from. I understand where to go from that point but I don't understand where this comes from and why you can multiply by delta 1. Thank you for your response though!!
It comes from the fact the delta 2=g'(alpha) x delta 1

And delta 3= g'(alpha) x delta 2
Which we can substitute to get
Delta 3 = g'(alpha) ^2 x delta 1
From here it is just generalising this result

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14. (Original post by drandy76)
It comes from the fact the delta 2=g'(alpha) x delta 1

And delta 3= g'(alpha) x delta 2
Which we can substitute to get
Delta 3 = g'(alpha) ^2 x delta 1
From here it is just generalising this result

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Oh I see! Thank you! That question has been puzzling me for weeks
15. (Original post by Mathematicus65)
Thank you!! I'm still unsure where g'(alpha) ^(n-1) comes from?
Since, delta(n+1)=g'(alpha)Xdelta(n)

delta(2)= g'(alpha) X delta(1)
delta(3) = g'(alpha) X delta(2) = g'(alpha) X g'(alpha) X delta(1) and so on.
You get g'(alpha) ^(n-1) when you generalize these sequence of equations.

Always welcome.
16. (Original post by Mathematicus65)
Oh I see! Thank you! That question has been puzzling me for weeks
No problemo

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17. (Original post by drandy76)
Investigate how the values change either side of the asymptotes to get a rough idea of the general shape of the graph

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Thanks you so much. These graphs seem to be killing me. Is there resource in the internet which helps in sketching similar curves?
18. (Original post by tangotangopapa2)
Thanks you so much. These graphs seem to be killing me. Is there resource in the internet which helps in sketching similar curves?
You can put the functions into Desmos to get an idea of how curves you behave around asymptotes I guess

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19. June 2013, Q 8) A line (x + 2y = 2) cuts curve with polar equation (r=1+cos theta ) is two parts. Find the ratio of areas above and below the line enclosed by the curve.
Mark Scheme does not seem to be helpful either. It simply states points of intersection as (0,1) and (2,0) out of nowhere and then proceeds in alien fashion. Could someone explain to me? I don't even know where to begin.
The Cartesian equation of curve is x2 + y2 - x = sqrt( x2 + y2).
20. (Original post by tangotangopapa2)
Thanks you so much. These graphs seem to be killing me. Is there resource in the internet which helps in sketching similar curves?
Sub in really big postive X values eg X=100 - see where the curve approaches

Sub in really negative X values eg X= -100- see where the curve approaches

Sub in an X value extremely close to one side of a known asymptote , eg if an asymptote is X=1 sub in X=1.001 and see where the curve is going

Sub in an X value extremely close to the OTHER side of the known asymptote , eg for the above you'd sub in X=0.009 and this gives you a very very very good idea of the nature of the curve either side of the asymptotes

You then find points of intersections and any stationary points, and this is enough information to then intuitively join up the curve correctly

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