4U Revising Game (1 Viewer)

qqmore · Oct 15, 2008

You can try using removing the first few terms i.e. 1 + 1/2^2 + 1/3^2 by using the exact area under the graph from integral. Then you can use the approximation method for the rest of the terms. This is should fix up and allow the answer to become closer to 1.45.

OR

you can try evaluating the first few terms and manipulating it

i.e.:
Let S = 1 + 1/2^2 + ...... + 1/99^2

By (a) S < / 2 - 1/99 < / 2- 0.01 = 1.99 (I)

But, 1 + 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2 < / S

and 1+0.25+0.1111+0.0625 +0.04 < / 1 + 1/2^2 + ... + 1/5^2

Therefore, 1 + 0.25 + 0.1111+0.0625 + 0.04 < / S

Thus, S > / 1.4646 > / 1.45 (II)

By (I) and (II), 1.45 < /S< /1.99

3.14159potato26 · Oct 15, 2008

In that case, you could just add up the first few terms, i.e. the first 5 terms to get a value > 1.45, then claim because 1 + 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2 < 1 + 1/2^2 + 1/3^2 + ... + 1/99^2, 1 + 1/2^2 + 1/3^2 + ... + 1/99^2 > 1.45.

qqmore · Oct 15, 2008

Next Question?

Five letters are chosen from the letters of the word CRICKET.

These five letters are them placed alongside one another to form a five letter arrangement.

Find the number of distinct fiver letter arrangements which are possible, considering all possible choices.

3unitz · Oct 15, 2008

u-borat said:
1.45 < 1 + 1/2^2 + 1/3^2 + ...+1/99^2 <1.99

another way you could do it is if you knew its less than zeta(2) = pi^2/6

1 + 1/2^2 + ... + 1/n^2 + ... = pi^2/6 < 1.99

1 + 1/2^2 + ... + 1/5^2 > 1.45

.'. 1.45 < 1 + 1/2^2 + 1/3^2 + ...+1/99^2 < 1.99

I-Love-Jesus · Oct 15, 2008

qqmore said:
Next Question?

Five letters are chosen from the letters of the word CRICKET.

These five letters are them placed alongside one another to form a five letter arrangement.

Find the number of distinct fiver letter arrangements which are possible, considering all possible choices.

2 letters the same - ⁵C₃x(5!/2!) = 600 (the two 'c's must be included, so there are only three spaces left, but 5 letters, hence the ⁵C₃. These letters can then be arranged (5!/2!) different ways)
All letters different - ⁶P₅ = 720 (6 different letters to go in 5 different place, hence the ⁶P₅)
Total = 600 + 720 = 1320

I apologise if this is wrong. I'm hoping it's not!
Anyway, next question:

A nine-member Fund Raising Committee consists of four students, three teachers and two parents. The Committee meets around a circular table.
(i) How many different arrangements of the nine members around the table are possible if the students sit together as a group and so do the teachers, but no teacher sits next to a student?
(ii) One student and one parent are related. Given that all arrangements in (i) are equally likely, what is the probability that these two members sit next to each other?

qqmore · Oct 15, 2008

u can also consider cases with no Cs, 1 C and 2 Cs

CRICKET

Ways for No Cs = 5C5 * 5!

Ways for 1 C = 1 * 5C4 * 5!

Ways for no Cs = 2C2 * 5C3 * 5! / 2!

Total ways = 1320 ways which equals to ur ans

Now for your question:

(i) Ways = 1 x 1 x 2 x 1 x 4! x 3! =288

(ii) no restrictions = 288 ways
with restrictions = 1 x 1 x 2 x 3! x 1 x 3! (Arrange parents first, two ways to arrange student with parent, arrange students group, one way to place teachers group, arrange teachers group) =72 ways

P = 72/288 = 1/4

I hope this is right as well...

Another QNS
a) In how many ways can 4 persons be grouped into two pairs to play a set of double tennis.

b) the eight members of a tennis club meet to play two simultaneous sets of double tennis on two separate but otherwise identical courts. In how many different ways can the members of the club be selected for these two sets of tennis?

duy.le · Oct 15, 2008

hummm just reading the "Basel problem" that 3 unitz linked; the "what u need to know section" involves everything we have learnt, i wonder if they'll ever put that in [referring to hsc paper].. would be so cool (but hard..

)

qqmore · Oct 15, 2008

duy.le said:
hummm just reading the "Basel problem" that 3 unitz linked; the "what u need to know section" involves everything we have learnt, i wonder if they'll ever put that in [referring to hsc paper].. would be so cool (but hard.. )

They did already, look at 2002 8a)

p.s. the HSC lead up style question makes any hard questions a lot easier

khorne · Mar 5, 2009

qqmore said:
Another QNS
a) In how many ways can 4 persons be grouped into two pairs to play a set of double tennis.

There is 3 ways... I think

It's 4C2 * 2C2 (to choose the teams), but, since it doesn't matter in which way we select the teams, we divide by two, to get 3.

Example: Assume Player A is always in team 1: Therefore we get 1C1 x 3C1x2C2 = 3

This assumes of course, that they don't have to be a particular team 9eg. in a tourney), otherwise it would be 6, as the teams would be exchangeable.

Can I submit a question?

Conic Sections:

Find the area of an ellipse whose major and minor axes have lengths of 2a and 2b respectively.

azureus88 · Mar 5, 2009

[maths]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[/maths]

[maths]y=\frac{b}{a}\sqrt{a^2-x^2}[/maths] by considering top half only.

[maths]A=4\frac{b}{a}\int_{0}^{a}\sqrt{a^2-x^2}dx[/maths] due to symmetry

[maths]=\pi ab[/maths] by substitution of x=asin@ or by 1/4 area of circle radius a

New Question: Prove that [maths]\frac{x_1+x_2+...+x_n}{n}\geq \sqrt[n]{x_1.x_2...x_n}[/maths] by induction or otherwise.

Trebla · Mar 6, 2009

There are probably shorter methods out there, but this is only one that I know....
$$Consider $ f(x) = a_1+a_2+...+a_k+x-(k+1)(a_1a_2.....a_kx)^{\frac{1}{k+1}} $ where $a_1,a_2,...a_k$ are non-negative real numbers and $k$ is a positive integer$\\f'(x) = 1 - (a_1a_2.....a_k)(a_1a_2.....a_kx)^{\frac{1}{k+1}-1} \\= 1 - (a_1a_2.....a_k)(a_1a_2.....a_n)^{\frac{-k}{k+1}}x^{\frac{-k}{k+1}}\\= 1 - (a_1a_2.....a_k)^{\frac{1}{k+1}}x^{\frac{-k}{k+1}}\\$Solving $ f'(x) = 0\\x^{\frac{-k}{k+1}}=(a_1a_2.....a_k)^{\frac{-1}{k+1}}\\x=(a_1a_2.....a_k)^\frac{1}{k}\\$It can be shown by second derivative that $f((a_1a_2.....a_k)^\frac{1}{k})$ is a minimum$\\f((a_1a_2.....a_k)^\frac{1}{k})=a_1+a_2+...+a_k+(a_1a_2.....a_k)^\frac{1}{k}-(k+1)(a_1a_2.....a_k)^{\frac{1}{k}}\\=a_1+a_2+...+a_k-k(a_1a_2.....a_k)^{\frac{1}{k}}\\$Hence$\\ a_1+a_2+...+a_k+x-(k+1)(a_1a_2.....a_kx)^{\frac{1}{k+1}} \geq a_1+a_2+...+a_k-k(a_1a_2.....a_k)^{\frac{1}{k}}\\\Rightarrow -x+(k+1)(a_1a_2.....a_kx)^{\frac{1}{k+1}} \leq k(a_1a_2.....a_k)^{\frac{1}{k}}$
$$Let $x=a_{k+1}\\\Rightarrow (k+1)(a_1a_2.....a_ka_{k+1})^{\frac{1}{k+1}} \leq k(a_1a_2.....a_k)^{\frac{1}{k}}+a_{k+1}$

In the induction proof:
For n = 1, it is obvious equality holds
Assume n = k
$\dfrac{x_1+x_2+...+x_k}{k}\geq \sqrt[k]{x_1.x_2...x_k}$
Required to prove n = k + 1
$\dfrac{x_1+x_2+...+x_k+x_{k+1}}{k+1}\geq \sqrt[k+1]{x_1.x_2...x_kx_{k+1}}$

$$LHS $ = \dfrac{x_1+x_2+...+x_k+x_{k+1}}{k+1}\\\geq \dfrac{k\sqrt[k]{x_1.x_2...x_k} + x_{k+1}}{k+1} $ by assumption$\\$But from the previous part,$ (k+1)(a_1a_2.....a_ka_{k+1})^{\frac{1}{k+1}} \leq k(a_1a_2.....a_k)^{\frac{1}{k}}+a_{k+1}\\$Applying this result:\\\geq \dfrac{(k+1)(x_1x_2.....x_kx_{k+1})^{\frac{1}{k+1}}}{k+1}\\=\sqrt[k+1]{x_1.x_2...x_kx_{k+1}}\\=$RHS$\\$The statement is true for $n=k+1$ if it is true for $n=k$. Since the statement is true for n = 1, it follows by induction that it is true for all positive integers $n$

Trebla · Mar 6, 2009

Question:

$$Suppose$ \displaystyle\int_{-\infty}^{\infty} e^{-\frac{x^2}{2}}\, dx = \sqrt{2\pi}\\\\$Find the value of $\displaystyle\int_{0}^{\infty} e^{-x^2}\, dx$

cyl123 · Mar 6, 2009

Trebla said:
Question:

$$Suppose$ \displaystyle\int_{-\infty}^{\infty} e^{-\frac{x^2}{2}}\, dx = \sqrt{2\pi}\\\\$Find the value of $\displaystyle\int_{0}^{\infty} e^{-x^2}\, dx$

$$Sub into the first integral $ u = \frac{x}{\sqrt{2}}\\\\ $So $ \sqrt{2}du=dx $then$ \displaystyle\int_{-\infty}^{\infty} e^{-\frac{x^2}{2}}\, dx=\sqrt{2} \int_{-\infty}^{\infty} e^{-u^{2}}du\\\\ $Using change of variable$\displaystyle\sqrt{2}\int_{-\infty}^{\infty} e^{-u^{2}}du = \sqrt{2}\int_{-\infty}^{\infty} e^{-x^{2}}dx\\\\ $Thus $ \displaystyle\int_{-\infty}^{\infty} e^{-\frac{x^2}{2}}\, dx= \sqrt{2}\int_{-\infty}^{\infty} e^{-x^{2}}dx\\\\ \sqrt{2\pi}=\sqrt{2}\int_{-\infty}^{\infty} e^{-x^{2}}dx\\\\ \sqrt{\pi} = \int_{-\infty}^{\infty} e^{-x^{2}}dx\\\\ $Using the fact $ e^{-x^2} = e^{-\left (-x \right )^2}\\\\ $ie it is an even function. So $ \int_{-\infty}^{\infty} e^{-x^{2}}dx = 2\int_{0}^{\infty} e^{-x^{2}}dx\\\\ $Thus $ \int_{0}^{\infty} e^{-x^{2}}dx=\frac{\sqrt{\pi}}{2}\\\\$

Need to learn how to use latex properly

edit: NEW PROBLEM: (i) Suppose a1,a2,...,an are n positive integers. Prove by induction or otherwise:
(a1+a2+...+an)(1/a1+1/a2+...+1/an) >= n^2
(ii) Hence prove that ((secx)^2+(cotx)^2+(cosecx)^2)>=9(cosx)^2

sorry it this question has already been asked... but i cbf to go thru 20 pages of threads to check

gurmies · May 6, 2009

Thought it's about time to start this thread up again:

$A \,\, sequence \,\, {a_{n}} \,\, is \,\, defined \,\, by \,\, a_{1}=1, \,\, a_{2}=3 \,\, and \,\, a_{n+2}=2a_{n+1}+a_{n} \,\, for \,\, n=1,2,3... \,\, . Prove \,\, by \,\, mathematical \,\, induction \,\, that \,\,a_{n}=\frac{(1+\sqrt{2})^{n}+(1-\sqrt{2})^{n}}{2} \,\, for \,\, n=1,2,3... \,\, .$

harism · May 7, 2009

$a_{a}=1;\,\,a_{2}=3\,\,and\,\,a_{n+2}=2a_{n+2}+a_{n+1}\,\, for\,\, n=1,2,3...\,\,\,\,\,\,\,\,this\,\,can\,\,be\,\,turned\,\,into\,\,a_{n}=2a_{n-1}+a_{n-2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, now,\,\,proving\,\,a_{n}=\frac{(1+\sqrt{2})+(1-\sqrt{2})}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,for\,\,n=1:\,\,L.H.S.:\,\,a_{1}=1\,\,and\,\,R.H.S.:a_{1}=1\,\,(using\,\, given)\,\,\,\,\therefore L.H.S.=R.H.S.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$
$now,\,\,assume\,\,for\,\,n=k:\,\,\frac{(1+\sqrt{2})^{k}+(1-\sqrt{2})^{k}}{2}\,\,\,\,k\,\,being\,\,an\,\,integer.\,\,\,\,\,\, thus\,\,proving\,\,for\,\,n=k+1:\,\,\,\,a_{k+1}=\frac{(1+\sqrt{2})^{k+1}+(1-\sqrt{2})^{k+1}}{2}\,\,\,\,\,\,$

$L.H.S.=a_{k+1}$
$=2a_{k}+a_{k-1}\,\,(using given)$
$=(1+\sqrt{2})^{k}+(1-\sqrt{2})^{k}+\frac{(1+\sqrt{2})^{k-1}+(1-\sqrt{2})^{k-1}}{2}$
$=(1+\sqrt{2})^{k}(1+\frac{1}{2(1+\sqrt{2})})+(1-\sqrt{2})^{k}(1+\frac{1}{2(1-\sqrt{2})})$
$=(1+\sqrt{2})^{k}(\frac{(1+\sqrt{2})}{2})+(1-\sqrt{2})^{k}(\frac{(1-\sqrt{2})}{2})$
$=\frac{(1+\sqrt{2})^{k+1}+(1-\sqrt{2})^{k+1}}{2}$
$=R.H.S.$
$Hens\,\,it\,\,is\,\,proven\,\,for\,\,n=k+1.$

i hope this is the right way about it. Im still taking tentative steps at this kinda induction.

hermand · May 7, 2009

harism said:
$a_{a}=1;\,\,a_{2}=3\,\,and\,\,a_{n+2}=2a_{n+2}+a_{n+1}\,\, for\,\, n=1,2,3...\,\,\,\,\,\,\,\,this\,\,can\,\,be\,\,turned\,\,into\,\,a_{n}=2a_{n-1}+a_{n-2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, now,\,\,proving\,\,a_{n}=\frac{(1+\sqrt{2})+(1-\sqrt{2})}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,for\,\,n=1:\,\,L.H.S.:\,\,a_{1}=1\,\,and\,\,R.H.S.:a_{1}=1\,\,(using\,\, given)\,\,\,\,\therefore L.H.S.=R.H.S.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$
$now,\,\,assume\,\,for\,\,n=k:\,\,\frac{(1+\sqrt{2})^{k}+(1-\sqrt{2})^{k}}{2}\,\,\,\,k\,\,being\,\,an\,\,integer.\,\,\,\,\,\, thus\,\,proving\,\,for\,\,n=k+1:\,\,\,\,a_{k+1}=\frac{(1+\sqrt{2})^{k+1}+(1-\sqrt{2})^{k+1}}{2}\,\,\,\,\,\,$

$L.H.S.=a_{k+1}$
$=2a_{k}+a_{k-1}\,\,(using given)$
$=(1+\sqrt{2})^{k}+(1-\sqrt{2})^{k}+\frac{(1+\sqrt{2})^{k-1}+(1-\sqrt{2})^{k-1}}{2}$
$=(1+\sqrt{2})^{k}(1+\frac{1}{2(1+\sqrt{2})})+(1-\sqrt{2})^{k}(1+\frac{1}{2(1-\sqrt{2})})$
$=(1+\sqrt{2})^{k}(\frac{(1+\sqrt{2})}{2})+(1-\sqrt{2})^{k}(\frac{(1-\sqrt{2})}{2})$
$=\frac{(1+\sqrt{2})^{k+1}+(1-\sqrt{2})^{k+1}}{2}$
$=R.H.S.$
$Hens\,\,it\,\,is\,\,proven\,\,for\,\,n=k+1.$

i hope this is the right way about it. Im still taking tentative steps at this kinda induction.

post a question??

and it's henCE.

sorry to sound like a bitch, but wrong spelling is frustrating.

gurmies · May 7, 2009

Well done Musht

Btw, I think it should have been KHEEEEENCE. Post a question n00b0t.

harism · May 7, 2009

hermand said:
post a question??

and it's henCE.

sorry to sound like a bitch, but wrong spelling is frustrating.

sorry to sound like a bitch but it was on purpose.
doing 4 unit maths, you think people dont joke around?

harism · May 7, 2009

looking for question gimme a sec gurms.

gurmies · May 7, 2009

Sorry hermand - it's an in-joke we have xD

$Let \,\, f(x)=ax^{2}+bx+c \,\, where \,\, a, b, c \,\, are \,\, real \,\, numbers \,\, and \,\, a\neq 0. \,\, Show \,\, that \,\, if \,\, f[f(x)]=[f(x)]^{2} \,\, for \,\, all \,\, x, \,\, then \,\, f(x)=x^{2}$

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