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Year 13 Maths Help Thread

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I'm determining the range for which the expansion of e3x+ln(13x)e^{3x}+ln(1-3x) is valid and my answer is 13x<13-\frac{1}{3}\leq{x}<\frac{1}{3} whereas the book has it as 13<x13-\frac{1}{3}<x\leq\frac{1}{3}. Surely the book is wrong? If x is 1/3 then ln is undefined?
Original post by RDKGames
I'm determining the range for which the expansion of e3x+ln(13x)e^{3x}+ln(1-3x) is valid and my answer is 13x<13-\frac{1}{3}\leq{x}<\frac{1}{3} whereas the book has it as 13<x13-\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.
For f(x)=ex1xf(x)=\frac{e^x}{1-x} show that f(x)f(x)\rightarrow -\infty as xx\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 ex(1x)1=(1+x+x22+...)(1+x+x2+x3+...)=1+2x+52x2+...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<1-1<x<1 and it's in that interval where f(x) goes to positive infinity.
(edited 7 years ago)
Original post by RDKGames
Using expansions I got ex(1x)1=(1+x+x22+...)(1+x+x2+x3+...)=1+2x+52x2+...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<1-1<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, xx is going to infinity, not to a number between -1 and 1.

Instead:

Unparseable latex formula:

\displaystyle [br]\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\exp expansion is fine because that holds over all reals (and the reals are compact with \infty adjoined).
Original post by Zacken
Doesn't make sense, it's not in that interval that you care about, xx is going to infinity, not to a number between -1 and 1.

Instead:

Unparseable latex formula:

\displaystyle [br]\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\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.
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\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:

Unparseable latex formula:

\displaystyle [br]\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.
How do you get from the first line to the second line? Reduction Formulae.png
Original post by NotNotBatman
How do you get from the first line to the second line?


Expand (1x)1x(1-x)\sqrt{1-x} and it should be straight forward from there.
Original post by NotNotBatman
How do you get from the first line to the second line? Reduction Formulae.png


Distribution. (ab)c=acbc(a-b)c = ac - bc,

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

Here: (1x)1x=1×1xx×1x\displaystyle (1-x)\sqrt{1-x} = 1 \times \sqrt{1-x} - x \times \sqrt{1-x}


Original post by RDKGames
Expand (1x)1x(1-x)\sqrt{1-x} and it should be straight forward from there.


Of course; thanks.
Original post by RDKGames
Expand (1x)1x(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.
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+cr=(-6cost)i+(2t)j+c.

r=3i+2jr=3i+2j at t=0t=0 so if you sub in t=0, replace r by its given vector, and solve for c, you should get c=9i+2jc=9i+2j. Put that back into the position vector and you should have r=(6cost+9)i+(2t+2)jr=(-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.
(edited 7 years ago)
Original post by RDKGames
When you integrate you should get r=(6cost)i+(2t)j+cr=(-6cost)i+(2t)j+c.

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

r=3i+2jr=3i+2j at t=0t=0 so if you sub in t=0, replace r by its given vector, and solve for c, you should get c=9i+2jc=9i+2j. Put that back into the position vector and you should have r=(6cost+9)i+(2t+2)jr=(-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
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) :frown: please can someone tell me where im going wrong?

1.png
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
(edited 7 years ago)
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) :frown: 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 ab+cd=A\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=Abdad+cb=Abd
(edited 7 years ago)
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 :smile:

What about simplifying. (see attached1.png)
Original post by kiiten
Ughhhh that makes sense - thanks :smile:

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 3x+13(x2+3x+2)(x+1)(x2+3x+2)\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.
Original post by kiiten
...


1. Get everything under same denominator
2. Tidy up the numerator
3. Cancel any common factors

ImageUploadedByStudent Room1469984130.963724.jpg


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