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    im quite confused about c3 logarithms as i missed all my lessons on it due to a course, and now im having to teach myself.. but the book isnt very explanatory.. im having trouble finding the gradient.. im pretty sure its not just differentiating the expression y = e^x?

    Thanks in advance..
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    \displaystyle\lim_{h\to0}[\frac{e^{x+h}-e^x}{h}]=\frac{e^x(e^h-1)}{h}=e^x\lim_{h\to0}[\frac{e^h-1}{h}] And by L'Hoptial once the limit will be \lim_{h\to0}e^h=1 and thus the derivative of e^x is e^x...

    Sorry don't really know what your question was
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    Nope, that was his question

    The differential of e^x is e^x. Odd
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    (Original post by nota bene)
    \displaystyle\lim_{h\to0}[\frac{e^{x+h}-e^x}{h}]=\frac{e^x(e^h-1)}{h}=e^x\lim_{h\to0}[\frac{e^h-1}{h}] And by L'Hoptial once the limit will be \lim_{h\to0}e^h=1 and thus the derivative of e^x is e^x...
    How can you use l'Hopital's rule when it relies on knowing what the derivative of e^x is in the first place??
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    exactly.. :rolleyes:

    edit: for the post above the one above..


    well in the book they just seem to have a gradient already.. they dont explain how they got it..


    for example.. the gradient of y = 2^x is 0.7 x 2^x...

    why?!?
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    firstly, for A level, you're not going to need to "prove" anything about the derivative of e^x - you just take it as given that its derivative is e^x.

    now, if y = a^x, then taking logs gives you ln y = x ln a so you can rewrite the original equation as y = e^(x ln a).

    since ln a is just a constant, you can differentiate this to get:

    dy/dx = (ln a) e^(x ln a) = (ln a) a^x.

    In your case, a= 2 so the constant that appears when you differentiate is just ln 2 which is approx 0.7.
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    (Original post by nota bene)
    \displaystyle\lim_{h\to0}[\frac{e^{x+h}-e^x}{h}]=\frac{e^x(e^h-1)}{h}=e^x\lim_{h\to0}[\frac{e^h-1}{h}] And by L'Hoptial once the limit will be \lim_{h\to0}e^h=1 and thus the derivative of e^x is e^x...

    Sorry don't really know what your question was
    No. I think the proof of e^x consists of it's maclaurin expansion being differentiated.
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    (Original post by Pencil King)
    No. I think the proof of e^x consists of it's maclaurin expansion being differentiated.
    But the Maclaurin series still relies on knowing f'(0), f''(0) etc and plugging these into a formula - i.e you have to differentiate your function f!!

    A proper proof would start from defining e^x as a power series, proving that it converges, and showing that it can be differentiated / integrated using theorems of analysis. That's why you won't find it in an A level course...
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    umm.. thanks for your help.. but im guessing it doesn't yet have anything to do with maclaurin or l'hoptial.. as i have no idea what they are and the book doesnt seem to mention them anyway... ?
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    (Original post by davros)
    But the Maclaurin series still relies on knowing f'(0), f''(0) etc and plugging these into a formula - i.e you have to differentiate your function f!!

    A proper proof would start from defining e^x as a power series, proving that it converges, and showing that it can be differentiated / integrated using theorems of analysis. That's why you won't find it in an A level course...
    Agreed.
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    (Original post by n0b0dy)
    umm.. thanks for your help.. but im guessing it doesn't yet have anything to do with maclaurin or l'hoptial.. as i have no idea what they are and the book doesnt seem to mention them anyway... ?
    That's right - if you look at my post at 11:44, that explains what you need to know / do...

    Post back if you're still confused
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    hmm.. uve explained it really well.. but i think i need to look over logarithms again.. summer holidays r not good.. :rolleyes:

    Thank u very much all..
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    (Original post by davros)
    A proper proof would start from defining e^x as a power series, proving that it converges, and showing that it can be differentiated / integrated using theorems of analysis. That's why you won't find it in an A level course...
    Yeah, I was talking crap up there

    e^x=\displaystyle\sum_{n=0}^{\in  fty}\frac{x^n}{n!} and by the ratio test \frac{1}{n+1}, which converges by comparison with \sum \frac{1}{n} (converges by integral test).

    Now, is this along the right lines?
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    (Original post by nota bene)
    \sum \frac{1}{n} (converges by integral test).

    Now, is this along the right lines?
    Er, are you sure that sum converges? I was under the impression that the harmonic series DIVERGED :confused:
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    (Original post by davros)
    Er, are you sure that sum converges? I was under the impression that the harmonic series DIVERGED :confused:
    it does......


    I'm not having a good day
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    (Original post by nota bene)
    it does......


    I'm not having a good day
    no problem...

    the ratio test will give you what you want - you just need to re-read how to use it. Basically, it's the LIMIT of the ratio that tells you whether the original series converges (not the SUM of the ratios)...
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    (Original post by nota bene)
    Yeah, I was talking crap up there

    and by the ratio test , which converges by comparison with [IMG]http://thestudentroom.co.uk/latexrender/pictures/1651c3444f1b741558fb6****bec9e80 .png[/IMG] (converges by integral test).

    Now, is this along the right lines?
    the first rule of math club: the harmonic series diverges! ;yes;

    i don't think you need the integral test though, because there is a general result about sum (1/n)^p, converging if p is more 1, diverging if p is less than or equal to one, and i don't really remember about zero! (which you can use once proved).
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    (Original post by h.b)
    the first rule of math club: the harmonic series diverges! ;yes;
    You don't have to rub it in my face! :p:
    i don't think you need the integral test though, because there is a general result about sum (1/n)^p, converging if p is more 1, diverging if p is less than or equal to one, and i don't really remember about zero! (which you can use once proved).
    You use the integral test to prove for which values of p the 1/n^p series converges, so yes I think I need the integral test unless I can assume the result to be true. About 0, what can we say about n^0 ? (it's 1), hence the convergence/divergence would seemingly only depend on the numerator...?
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    (Original post by nota bene)
    You don't have to rub it in my face! :p:


    You use the integral test to prove for which values of p the 1/n^p series converges, so yes I think I need the integral test unless I can assume the result to be true. About 0, what can we say about n^0 ? (it's 1), hence the convergence/divergence would seemingly only depend on the numerator...?
    the first rule of... ach fine i won't.

    (i don't really know much about the integral test so i'll take your word for it. but if we can prove that the harmonic series diverges and that (1/n)^2 converges, then the comparison test can always be used i think).

    about the zero thing - i'm not sure. i was told something about this but can't remember. :rolleyes:
 
 
 
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