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Why bother with a post grad? Are they even worth it? Have your say! 26-10-2016

    (Original post by thefatone)
    seems pretty dangerous... do i learn this in c3 and c4?
    The implicit differentiation he uses is learned in c4, modulus function is learnt in fp1(for OCR at least) or c3

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    (Original post by Zacken)
    The other users have provided you with perfectly valid proofs so I'm not going to go on with that, I'll give a few counter examples (remember, testing out special cases is always a good way of making sure what you're writing down is correct) and then go on and talk/waffle on a bit about something called linearity. I'm going to write in a very informal and unrigorous way that is meant to supplement your intuition, so nobody hate on me.

    Okay - first off, try and drill the practice of checking special cases in your head as you write something down. So if you're writing \sqrt{a^2 + b^2 } = c \Rightarrow a + b = c try and practice to get your head to test the case a=b=1 for example.

    It's fairly obvious that right away this gives you \sqrt{a^2 + b^2} = \sqrt{2} but it is not the case that a+b = 2 =^? \sqrt{2} so that should set off alarm bells right away.

    Okay, now onto me waffling about linearity: so things are linear when they satisfy (amongst other things) the fact that f(x+y) = f(x) + f(y), it's quite a privilege to work with linear things, it makes life very easy.

    A few examples of linear things is, well... a straight line, f(x) = ax is linear because f(x+y) = a(x+y) = ax + ay = f(x) + f(y), makes life very easy, you don't need to worry about stuff so much.

    Another example of linear things is integration or differentiation, although we're moving away from the realm of functions and into the realm of operators here (this needn't concern us, though) we can call differentiation linear because \frac{\mathrm{d}}{\mathrm{d}x}(f  (x) + g(x)) = \frac{\mathrm{d}}{\mathrm{d}x} f(x) + \frac{\mathrm{d}}{\mathrm{d}x} g(x) - see, life is easy! We don't need to worry about whether we add then differentiate or differentiate and add, meh, it all works out to be the same thing.

    Okay, linear is easy... that kinda makes nearly everything not being linear kinda make sense, because when is maths easy, eh? Functions, in general aren't linear. f(x) = \sqrt{x} isn't a linear function, unfortunately. I've just said that square rooting isn't a linear function! So it does not hold that \sqrt{a +b} = \sqrt{a} + \sqrt{b}.

    In general, square rooting is a messy business, it works from \mathbb{R}^{+} \to \mathbb{R}^{+} (we're constraining ourself to the reals here, it we want to go complex, bleurgh), we need to take care of the signs, restrict our domains, makes sure to use moduli bars, etc... so it hardly stands to reason that square rooting is going to be linear and make life easy for us, oh no, that's not what little Mr. Square Root is going to do. He's going to stand there and make life hard.

    Square rooting and logarithiming (made that word up, hi) kind of provoke the same reaction in me, whenever I see \sqrt{ab} or \ln xy I know right away I can split that up into \sqrt{a}\sqrt{b} or \ln x + \ln y and make life easy, but as soon as I see \sqrt{a+b} or \ln (x+y), I close my eyes and go cry a little bit because there's nothing I can really do now!

    So, hopefully, that mildly entertaining drivel has let you intuit that \sqrt{a^2 + b^2} \neq \sqrt{a^2} + \sqrt{b^2} = |a| + |b|.
    Thank you so much that's brilliant I really appreciate all your effort!

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    Thank you all so much!!

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    (Original post by Zacken)
    We define the absolute value function:


\begin{equation*} |x| = \begin{cases} x \quad \text{for } \, x \geq 0 \\ -x \quad \text{for } \, x < 0 \end{cases}\end{equation*}

    So plug in x=3 into the function |x| we get that since 3 > 0 so we take the top bit of the definition: |3| = 3

    Plug in x=-3 into the function |x| we get that since -3 < 0 so we take the bottom bit of the definition |-3| = -(-3) = 3.

    Essentially, the absolute value functions strips away the sign of the number and makes it positive.

    Because the absolute value function is defined piecewise, it stands to reason why it is not differentiable at its cusp, more specifically: y = |x| \Rightarrow y^2 = x^2 \Rightarrow 2yy' = 2x \Rightarrow y' = \frac{x}{|x|} so the derivative is undefined for |x| =0 \iff x=0 and that is because it is undefined there.

    Most of this post is useless extra knowledge, but... meh.
    Ooooh love your explanation really interesting!

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    (Original post by maths_4_life)
    Ooooh love your explanation really interesting!

    Posted from TSR Mobile
    Thank you, I appreciate that!
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