Year 13 Maths Help Thread

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    I'm determining the range for which the expansion of e^{3x}+ln(1-3x) is valid and my answer is -\frac{1}{3}\leq{x}<\frac{1}{3} whereas the book has it as -\frac{1}{3}<x\leq\frac{1}{3}. Surely the book is wrong? If x is 1/3 then ln is undefined?
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    (Original post by RDKGames)
    I'm determining the range for which the expansion of e^{3x}+ln(1-3x) is valid and my answer is -\frac{1}{3}\leq{x}<\frac{1}{3} whereas the book has it as -\frac{1}{3}<x\leq\frac{1}{3}. Surely the book is wrong? If x is 1/3 then ln is undefined?
    It's surely a mistake.
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    For f(x)=\frac{e^x}{1-x} show that f(x)\rightarrow -\infty as x\rightarrow \infty

    I can see that as x goes to infinity, the numerator go to positive infinity while the denominator goes to negative infinity, so it makes sense to me that it overall goes to negative infinity. But how would I go about showing it?

    Using expansions I got e^x(1-x)^{-1}=(1+x+\frac{x^2}{2}+...)(1+x+x  ^2+x^3+...)=1+2x+\frac{5}{2}x^2+  ... which is only valid for -1<x<1 and it's in that interval where f(x) goes to positive infinity.
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    (Original post by RDKGames)
    Using expansions I got e^x(1-x)^{-1}=(1+x+\frac{x^2}{2}+...)(1+x+x  ^2+x^3+...)=1+2x+\frac{5}{2}x^2+  ... which is only valid for -1<x<1 and it's in that interval where f(x) goes to positive infinity.
    Doesn't make sense, it's not in that interval that you care about, x is going to infinity, not to a number between -1 and 1.

    Instead:

    \displaystyle 

\begin{equation*}\frac{e^x}{1-x} = \frac{1 + x + \frac{x^2}{2} + o(x^3)}{1-x} = \frac{\frac{1}{x} + 1 + \frac{x}{2} + o(x^2)}{\frac{1}{x} - 1}\end{equation*}

    So in the limit, the numerator is unbounded whilst the denominator approaches -1

    Don't use taylor expansions that are restricted to values that you're not taking the limit to, so if you're going to infinity, don't use an expansion only valid for moduli less than 1, etc... The \exp expansion is fine because that holds over all reals (and the reals are compact with \infty adjoined).
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    (Original post by Zacken)
    Doesn't make sense, it's not in that interval that you care about, x is going to infinity, not to a number between -1 and 1.

    Instead:

    \displaystyle 

\begin{equation*}\frac{e^x}{1-x} = \frac{1 + x + \frac{x^2}{2} + o(x^3)}{1-x} = \frac{\frac{1}{x} + 1 + \frac{x}{2} + o(x^2)}{\frac{1}{x} - 1}\end{equation*}

    So in the limit, the numerator is unbounded whilst the denominator approaches -1

    Don't use taylor expansions that are restricted to values that you're not taking the limit to, so if you're going to infinity, don't use an expansion only valid for moduli less than 1, etc... The \exp expansion is fine because that holds over all reals (and the reals are compact with \infty adjoined).
    Ah right, thanks that makes sense. I had a feeling I shouldn't use the expansion as I need to go to infinity and the expansion prevents that but I just couldn't think of anything else.
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    (Original post by RDKGames)
    Ah right, thanks that makes sense. I had a feeling I shouldn't use the expansion as I need to go to infinity and the expansion prevents that but I just couldn't think of anything else.
    Essentially what you should have been thinking here is that you know that \exp trumps any polynomial, so the obvious way to show that is to expand it as an unbounded polynomial and then trivially re-arrange to get the numerator unbounded (because you have an infinite degree there, you have space) whilst wearing the denominator down.

    Another option would be to L'Hopital it and handwave:

    \displaystyle 

\begin{equation*}\lim_{x \to \infty} \frac{e^x}{1-x} \stackrel{\text{DH}}{=} \lim_{x \to \infty} \frac{e^x}{-1} = -\infty\end{equation*}

    But I wouldn't recommend it.
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    How do you get from the first line to the second line? Name:  Reduction Formulae.png
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    (Original post by NotNotBatman)
    How do you get from the first line to the second line?
    Expand (1-x)\sqrt{1-x} and it should be straight forward from there.
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    (Original post by NotNotBatman)
    How do you get from the first line to the second line? Name:  Reduction Formulae.png
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    Distribution. (a-b)c = ac - bc,

    Here: \displaystyle (1-x)\sqrt{1-x} = 1 \times \sqrt{1-x} - x \times \sqrt{1-x}
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    (Original post by Zacken)
    Distribution. (a-b)c = ac - bc,

    Here: \displaystyle (1-x)\sqrt{1-x} = 1 \times \sqrt{1-x} - x \times \sqrt{1-x}
    (Original post by RDKGames)
    Expand (1-x)\sqrt{1-x} and it should be straight forward from there.
    Of course; thanks.
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    (Original post by RDKGames)
    Expand (1-x)\sqrt{1-x} and it should be straight forward from there.
    Hi, I'm doing a motion in a plane question from M2. I have V=6sint i+2j, find position vector r at time t given that r=3i+2j when t=0. I have worked this out to get
    r=(-6cost+3)i + 2(t-1)j. But apparently the answer in the book says that the answer is r=(-6cost+9)i + 2(t-1)j.

    I'm not sure where they got 9 from in the i component bracket. I'd appreciate any insight. Thank you.
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    (Original post by student1856)
    Hi, I'm doing a motion in a plane question from M2. I have V=6sint i+2j, find position vector r at time t given that r=3i+2j when t=0. I have worked this out to get
    r=(-6cost+3)i + 2(t-1)j. But apparently the answer in the book says that the answer is r=(-6cost+9)i + 2(t-1)j.

    I'm not sure where they got 9 from in the i component bracket. I'd appreciate any insight. Thank you.
    When you integrate you should get r=(-6cost)i+(2t)j+c.

    r=3i+2j at t=0 so if you sub in t=0, replace r by its given vector, and solve for c, you should get c=9i+2j. Put that back into the position vector and you should have r=(-6cost+9)i+(2t+2)j

    Edit: you either gave me the wrong position vector (should it be 3i-2j?) at t=0 or the final answer should have +2j rather than -2j.
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    (Original post by RDKGames)
    When you integrate you should get r=(-6cost)i+(2t)j+c.

    r=3i+2j at t=0 so if you sub in t=0, replace r by its given vector, and solve for c, you should get c=9i+2j. Put that back into the position vector and you should have r=(-6cost+9)i+(2t+2)j
    Ahh, right. I replaced the entire i bracket with 0, I forgot that cos(0)=1. Stupid mistake. Thanks for the help.
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    (Original post by student1856)
    Ahh, right. I replaced the entire i bracket with 0, I forgot that cos(0)=1. Stupid mistake. Thanks for the help.
    No problem, happens to anyone
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    (Original post by RDKGames)
    When you integrate you should get r=(-6cost)i+(2t)j+c.

    r=3i+2j at t=0 so if you sub in t=0, replace r by its given vector, and solve for c, you should get c=9i+2j. Put that back into the position vector and you should have r=(-6cost+9)i+(2t+2)j

    Edit: you either gave me the wrong position vector (should it be 3i-2j?) at t=0 or the final answer should have +2j rather than -2j.
    Hello again, I just read your edit. Yeah, the final answer should be +2j, The position vector was 3i+2j which would make the final answer also +2j. Sorry about that, my brain is running on fumes. I should really eat something.Thanks again
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    This may seem really stupid but i have to answer questions about algebraic fractions to help me transition to A2. i did them and got them all wrong (i havent had to solve these kind of questions before) please can someone tell me where im going wrong?

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    you have to solve the equation. i cross multiplied then worked from there to get 1(x+4) - 2(x-2) = 1/3
    i got x = 7.6666

    but that is wrong - there should be 2 solutions?

    IGNORE BELOW ATTACHMENT - i cant delete it
    Attached Images
     
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    (Original post by kiiten)
    This may seem really stupid but i have to answer questions about algebraic fractions to help me transition to A2. i did them and got them all wrong (i havent had to solve these kind of questions before) please can someone tell me where im going wrong?
    you have to solve the equation. i cross multiplied then worked from there to get 1(x+4) - 2(x-2) = 1/3
    i got x = 7.6666

    but that is wrong - there should be 2 solutions?

    IGNORE BELOW ATTACHMENT - i cant delete it
    You forgot to multiply the right hand side by (x+4) and (x-2). Cross multiplication would mean you get those two fractions under the same denominator, thus turning it into a single one with denominator (x+4)(x-2) which shows that it's a quadratic; thus 2 solutions.

    If you struggle with questions of type \frac{a}{b}+\frac{c}{d}=A just simply do what I do and multiply everything by b first and then d (or other way round). This will get rid off the fractions. Giving ad+cb=Abd
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    (Original post by RDKGames)
    You forgot to multiply the right hand side by (x+4) and (x-2). Cross multiplication would mean you get those two fractions under the same denominator, thus turning it into a single one with denominator (x+4)(x-2) which shows that it's a quadratic; thus 2 solutions.
    Ughhhh that makes sense - thanks

    What about simplifying. (see attachedName:  1.png
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    (Original post by kiiten)
    Ughhhh that makes sense - thanks

    What about simplifying. (see attached)
    Get it under the same denominator. I would multiply first fraction's numerator by the other fraction's denominator, and exactly the same for the other fraction. So \frac{3}{x+1} \mapsto \frac{3(x^2+3x+2)}{(x+1)(x^2+3x+  2)} and of course that quadratic factorises as you've shown so you can cancel some factors once you combine the two fractions. I'll post a pic to show what I mean.
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    (Original post by kiiten)
    ...
    1. Get everything under same denominator
    2. Tidy up the numerator
    3. Cancel any common factors

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