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# Edexcel C4 Revision Thread (28/06 14:00) watch

1. (Original post by Gaz031)

Clearly ie for the series to converge.

So basically if it was larger then it wudnt be converging, it would be diverging? ie each term would be getting larger and larger?
2. (Original post by melbourne)
So basically if it was larger then it wudnt be converging, it would be diverging? ie each term would be getting larger and larger?
That's it. If the y value in is such that |y|>1, then etc. progressively get larger and larger. Only in the case |y|<1 is the approximation sensible.
3. (Original post by Gaz031)
That's it. If the y value in is such that |y|>1, then etc. progressively get larger and larger. Only in the case |y|<1 is the approximation sensible.
ah that makes sense, it explains why in c2 we get those questions "now use your answer to make an approximation............." so the more terms you get the closer to the answer, whereas if it was diverging the answer would go beyond.
4. Integration By Parts

I'm just doing the Integration by parts - and i am a lil stuck.

This is what i have done so far:

int X(x-1)^½

V=x dv/dx=1

du/dx= (x-1)^½ u=2/3(x-1)^3/2

I whack them into the eqn. and get

2/3x(x-1)^3/2 - 4/15(x-1)^5/2

I think this is correct, im doing it from the P3 Heinemann book pg.119 q 8....Any help would be appreciated!

5. Looks right!
6. in the book they simplify it and i am unsure of how to do this?
7. (Original post by bavman)
in the book they simplify it and i am unsure of how to do this?
What do they simplify it to?

[[ don't have copy of book ]]
8. lol sorry - they make it

2/15(3x+2)(x-1)^3/2

my algebra is awful
9. Right so you have got:

2/3x(x-1)3/2 - 4/15(x-1)5/2 (1)

and you want:

2/15(3x+2)(x-1)3/2 (2)

I'll just to a check to see if they are the same (I am aware this will not prove anything):

Put x = 2: (1) gives: 16/15

(2) gives: 16/15.

Alright they could be equivalent.

EDIT: poor maths removed.

=> Anyone else help me out?
10. 2/3x(x-1)3/2 - 4/15(x-1)5/2
2/3[x(x-1)3/2 - 2/5(x-1)5/2]
2/3(x-1)3/2[x - 2/5(x-1)]
2/3(x-1)3/2[3/5x+2/5]
2/3(x-1)3/2 x 1/5 (3x+2)]
= 2/15(3x+2)(x-1)3/2

EDIT: would be much easier if you just take out 2/15(x-1)3/2 first.
11. thanks mate, your a legend!

Nah, i understood better because you left it in. I just didnt realise that if youi take the indices out the 5/2 becomes to the power of 1 thus eliminatin gthe indicies on the inside of the brackets
12. (Original post by bavman)
np. just play around with it, and you'll soon find what you need!
13. (Original post by samdavyson)
=> (x-1)1/2[2/3x(x-1)3 - 4/15(x-1)5]

=> (x-1)3/2[2/3x(x-1) - 4/15(x-1)5/3]

=> (x-1)3/2[2/3 - 2/3x - 4/15(x-1)5/3]
when you took out (x-1)3, you have to MINUS the powers of the rest inside the brackets (while you divided).
14. (Original post by Zuber)
when you took out (x-1)3, you have to MINUS the powers of the rest inside the brackets (while you divided).

[[ think sam think ]]
15. Crapp i thought this'd be piss

im basically implicitly differentiating:

x³-2xy-4x+y³-51 =0

when I differentiate -2xy using chain rule i get -2x dy/dx - 2y, but it should be -2x dy/dx + 2y. Whys that? Cheers
16. (Original post by TheWolf)
Crapp i thought this'd be piss

im basically implicitly differentiating:

x³-2xy-4x+y³-51 =0

when I differentiate -2xy using chain rule i get -2x dy/dx - 2y, but it should be -2x dy/dx + 2y. Whys that? Cheers
It is a brackets problem I think.

x³-2xy-4x+y³-51 =0

=> 3x² - (2x(dy/dx) + 2y) - 4 - 3y²(dy/dx) - 51 = 0

=> 3x² - 2x(dy/dx) - 2y - 4 - 3y²(dy/dx) - 51 = 0
17. (Original post by TheWolf)
Crapp i thought this'd be piss

im basically implicitly differentiating:

x³-2xy-4x+y³-51 =0

when I differentiate -2xy using chain rule i get -2x dy/dx - 2y, but it should be -2x dy/dx + 2y. Whys that? Cheers
you should differentiate it via the product rule.
18. (Original post by samdavyson)
It is a brackets problem I think.

x³-2xy-4x+y³-51 =0

=> 3x² - (2x(dy/dx) + 2y) - 4 - 3y²(dy/dx) - 51 = 0

=> 3x² - 2x(dy/dx) - 2y - 4 - 3y²(dy/dx) - 51 = 0

naz i do use the product rule
19. (Original post by TheWolf)

naz i do use the product rule
They have just missed out the brackets.

If you look at the working for the rest it follows what I (& you) got.
20. Ah coolio.. thanx sam =)

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