Prelim 2016 Maths Help Thread (4 Viewers)

eyeseeyou · Jun 8, 2016

leehuan said:
Will vouch for this gladly. Because without him I'd be so dead at uni.

Yeah I think everyone would be dead without him

Drongoski said:
He is the Saviour Robot, sent down from Heaven, to save all of you who are struggling in maths.

That is indeed true Drongoski, however you realise that although he helps most of us, he isn't the only person helping me. There's people like you Drongoski, David (I call him David for short instead of Davidgoes4wce), Leehuan, Kawaiipotato, Shuuya, etc. Yes he is the biggest helper out of all of us, but you all are helpers yourselves and that's something I'm grateful for. Cheers guys I don't know what I could do without you all, you're all so awesome and dedicated

davidgoes4wce · Jun 8, 2016

eyeseeyou said:
Can someone please help me with Induction (maths). I really suck at it

2. 1/(x^n(x-1)=1/(x-1)-1/x-1/x^2-...-1/x^n

For Question 2

I assume you mean

$\frac{1}{n(x-1)}=\frac{1}{x-1}-\frac{1}{x}-\frac{1}{x^2}-.........-\frac{1}{x^n}$

eyeseeyou · Jun 9, 2016

davidgoes4wce said:
For Question 2

I assume you mean

$\frac{1}{x^n(x-1)}=\frac{1}{x-1}-\frac{1}{x}-\frac{1}{x^2}-.........-\frac{1}{x^n}$

This is it

eyeseeyou · Jun 9, 2016

1.Sn=1/2!+2/3!+3/4!+...+n/(n+1)!, n=1,2,3...
a. Use the method of mathematical induction to show that Sn=1-1/(n+1)! for all positive integers n is greater than or equal to 1
b. Find the value of limit of n approaches infinity at Sn
c. Find the smallest positive integer n such that lSn-1l<10^-6

eyeseeyou · Jun 9, 2016

ANother inverse function question I am struggling with

1. Find the inverse function of the following. Also state the domain and range of the inverse function
a. y= (x+1)^2 -1, x is greater than or equal to 2
b. y=(x-2)^2 -1, x is greater than or equal to 2
c. y=8(x-1)^3

So for a and b I found out the equation and found the domain but am struggling to find the range

and for c I just can't seem to solve it for some reason

eyeseeyou · Jun 9, 2016

leehuan said:
There's that attitude again. Don't get people to help you only to then use harsh language and curse words later

The graph of y=f(x) i.e. y=tan(x/2) is simply one branch of the tan curve with period 2pi.

On this domain, since f(x) is one-to-one it is indeed invertible. To find the inverse:

Consider x=tan(y/2)
Then arctan(x)=y/2 so y=2 arctan(x)

where arctan is just inverse tan if you have not seen it before. I.e. tan^-1(x)

So we have f^-1(x) = 2arctan(x)

This is a dilated version of the regular inverse tangent curve: the asymptotes are now at pi and -pi, instead of pi/2 and -pi/2
View attachment 33257

Leehuan could you please solve this question again without arctan (since I don't know what that is)

integral95 · Jun 9, 2016

eyeseeyou said:
ANother inverse function question I am struggling with

1. Find the inverse function of the following. Also state the domain and range of the inverse function
a. y= (x+1)^2 -1, x is greater than or equal to 2
b. y=(x-2)^2 -1, x is greater than or equal to 2
c. y=8(x-1)^3

So for a and b I found out the equation and found the domain but am struggling to find the range

and for c I just can't seem to solve it for some reason

With inverse functions, the domain and range basically swap around, e.g
y = e^x would have a domain of all real x to have an inverse function, and the range is y>0
so the inverse function, y = ln(x) would have a domain x>0 and range all real y.

eyeseeyou · Jun 9, 2016

integral95 said:
With inverse functions, the domain and range basically swap around, e.g
y = e^x would have a domain of all real x to have an inverse function, and the range is y>0
so the inverse function, y = ln(x) would have a domain x>0 and range all real y.

Yeah I did that but can't find the range for some reason and therefore need help

davidgoes4wce · Jun 10, 2016

davidgoes4wce said:
For Question 2

I assume you mean

$\frac{1}{n(x-1)}=\frac{1}{x-1}-\frac{1}{x}-\frac{1}{x^2}-.........-\frac{1}{x^n}$

I'm actually not sure how to do this induction type. Maybe someone else can give us a kick start?

davidgoes4wce · Jun 10, 2016

eyeseeyou · Jun 10, 2016

What's the difference between ln and log? I get confused at times

leehuan · Jun 10, 2016

eyeseeyou said:
What's the difference between ln and log? I get confused at times

Nothing.

$$The natural logarithm represents the logarithm to the base $e$, i.e. $\log_e{x}$

$But in mathematics, when you type $\log{x}$ without a base you naturally assume it's to the base $e$ anyway$

authenticity · Jun 10, 2016

leehuan said:
Nothing.

$$The natural logarithm represents the logarithm to the base $e$, i.e. $\log_e{x}$

$But in mathematics, when you type $\log{x}$ without a base you naturally assume it's to the base $e$ anyway$

I got taught that log x is base 10 LOL

InteGrand · Jun 10, 2016

authenticity said:
I got taught that log x is base 10 LOL

The Casio calculators have "log" as the button for log-base-10, but actually in mathematics log refers to base e.

Trebla · Jun 10, 2016

eyeseeyou said:
ANother inverse function question I am struggling with

1. Find the inverse function of the following. Also state the domain and range of the inverse function
a. y= (x+1)^2 -1, x is greater than or equal to 2
b. y=(x-2)^2 -1, x is greater than or equal to 2
c. y=8(x-1)^3

So for a and b I found out the equation and found the domain but am struggling to find the range

and for c I just can't seem to solve it for some reason

Is this from tutor? What topics have you already done at school (or tutor for that matter)? This isn't exactly prelim stuff.

davidgoes4wce · Jun 11, 2016

davidgoes4wce said:
$$ My thinking is for S(1) , we could let the LHS $ = \frac{1}{x-1} $ and RHS= $ \frac{1}{x-1}$

$$ For S(k) $ : \frac{1}{k(x-1)}=\frac{1}{x-1}-\frac{1}{x}-\frac{1}{x^2}-........-\frac{1}{x^k}$

BUMP does anyone understand this induction process?

Trebla · Jun 11, 2016

You got the question wrong.

Assume that for some n=k

$\frac{1}{x^k(x-1)}=\frac{1}{x-1}-\frac{1}{x}-\frac{1}{x^2}-.........-\frac{1}{x^k}$

Now we can show that

$\frac{1}{x-1}-\frac{1}{x}-\frac{1}{x^2}-.........-\frac{1}{x^{k+1}}\\\\=\frac{1}{x^k(x-1)}-\frac{1}{x^{k+1}}\\\\=\dfrac{1}{x^k}\left(\dfrac{1}{x-1}-\dfrac{1}{x}\right)$

and the result follows

eyeseeyou · Jun 11, 2016

Trebla said:
Is this from tutor? What topics have you already done at school (or tutor for that matter)? This isn't exactly prelim stuff.

Yeah I know it isn't prelim stuff. Am trying to self learn it (using other ppl tutor resources)

I posted this here b/c I didn't know where else to post it and I'm in yr 11 atm

eyeseeyou · Jun 11, 2016

Trebla said:
You got the question wrong.

Assume that for some n=k

$\frac{1}{x^k(x-1)}=\frac{1}{x-1}-\frac{1}{x}-\frac{1}{x^2}-.........-\frac{1}{x^k}$

Now we can show that

$\frac{1}{x-1}-\frac{1}{x}-\frac{1}{x^2}-.........-\frac{1}{x^{k+1}}\\\\=\frac{1}{x^k(x-1)}-\frac{1}{x^{k+1}}\\\\=\dfrac{1}{x^k}\left(\dfrac{1}{x-1}-\dfrac{1}{x}\right)$

and the result follows

Me or David?

davidgoes4wce · Jun 11, 2016

eyeseeyou said:
Me or David?

I couldn't read your writing very cleary and had to check

Prelim 2016 Maths Help Thread (4 Viewers)

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