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    (Original post by MUN123)
    I think that they are not clever but they try hard and they have motivations from their parents etc.
    That's daft. Maths is one subject on a VERY shortlist that actually requires problem solving ability i.e. intellect.
    No essay subject requires anywhere near as much intelligence as required for doing maths to a high enough level. You're crazy to think otherwise.
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    (Original post by MUN123)
    It seems like Maths is not for me, because I just don't get the whole idea of defferentiation and Intergration. In my opinion both are useless and will not help me in the real world.
    Hardly any maths above GCSE level is applicable to your everyday life, but without it our world would be extremely different..

    http://en.wikipedia.org/wiki/Calculus#Applications
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    Ha ha I will admit it I picked As level Maths and I am strugling like f**** . I have gained a GCSE grade B and I thought oh well let me do Maths since I was the only person who got a B in my class. I am thinking about droping out since I can't even understand C1.
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    (Original post by hey_its_nay)
    can you please explain that last one please?
    Hope I get this right :p: It is true that:

     \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}  + ...

    where  n! = n(n-1)(n-2)... \times 2\times 1 eg 3! = 3x2x1 = 6

    This is an example of a Taylor Series and you derive them using differentiation.

    So: \displaystyle{\lim_{x->0} (\frac{\sin x}{x}) = \lim_{x->0} (\frac{x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}  + ...}{x}) = \lim_{x->0} (1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!}  + ...)} = 1
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    (Original post by electriic_ink)
    Hope I get this right :p: It is true that:

     \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}  + ...

    where  n! = n(n-1)(n-2)... \times 2\times 1 eg 3! = 3x2x1 = 6

    This is an example of a Taylor Series and you derive them using differentiation.

    So: \displaystyle{\lim_{x->0} (\frac{\sin x}{x}) = \lim_{x->0} (\frac{x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}  + ...}{x}) = \lim_{x->0} (1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!}  + ...)} = 1
    That's quite an interesting result, thanks for that.
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    i remember crying when i saw this @ the back of my c1 book now i just laught at it lol
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    Whats integration?
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    (Original post by mcgoohan)
    What are all the d's and what's lim?
    Apart from knowing that it finds the gradient of a curve, i haven't a clue what it is.
    Ok, when you differentiate (say) y with respect to x (dy/dx), this tells you the rate at which y changes with respect to x. Eg for the function y=2x then dy/dx = 2, ie for every unit travelled along the x axis, 2 will be travelled along the y axis (by looking at the line y=x.

    lim(x->0) read "the limit as x tends towards zero" just means what it says, the limit of the funxtion in question as x gets smaller and smaller, you can't work this out by just plugging 0 in most of the time, by the way.
 
 
 
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