HSC Mathematics Marathon (2 Viewers)

Drongoski · Sep 22, 2011

apollo1 said:
Im going to post a question on Drongoski's behalf:

By using geometric evidence, explain why:

$\int_{0}^{1}x^{n}dx +\int_{0}^{1}x^{\frac{1}{n}}dx = 1$

(n E Z)

Interesting question. Also didn't realise I'm reqd to post follow-up question. Thanks for posting on my behalf apollo1.

RANK 1 · Sep 22, 2011

apollo1 said:
Im going to post a question on Drongoski's behalf:

By using geometric evidence, explain why:

$\int_{0}^{1}x^{n}dx +\int_{0}^{1}x^{\frac{1}{n}}dx = 1$

(n E Z)

when i integrate that i get n+1, not 1........

Drongoski · Sep 22, 2011

RANK 1 said:
when i integrate that i get n+1, not 1........

No - you should get 1. Check again.

RANK 1 · Sep 22, 2011

i get (1/n+1) + (n/n+1) which equals n+1.....can u show your working?

AAEldar · Sep 22, 2011

RANK 1 said:
i get (1/n+1) + (n/n+1) which equals n+1.....can u show your working?

$\frac{1}{n+1}+\frac{n}{n+1} \\\\ =\frac{1+n}{n+1} \\\\ =\frac{n+1}{n+1} \\\\ =1$

Drongoski · Sep 22, 2011

RANK 1 said:
i get (1/n+1) + (n/n+1) which equals n+1.....can u show your working?

$\int _{0} ^{1} (x^n + x^{\frac {1}{n}}) dx = \left [ \frac {x^{n+1}}{n+1} + \frac {nx^{\frac {n+1}{n}}}{n+1}\right ]^1 _0 = (\frac {1}{n+1} + \frac {n}{n+1}) - (0 + 0) = 1$

Geometrically: $x^{\frac {1}{n}} and x^n$ are mutual inverses.

The 2 functions intersect at x=1. If you sketch the usual graphs of the 2 mutually-inverse functions which are symmetric about the line y = x you will see the area below the graphs between x=0 and x=1 can be put together as a 1-by-1 square whose area obviously is 1.

apollo1 · Sep 22, 2011

AAEldar said:
$\frac{1}{n+1}+\frac{n}{n+1} \\\\ =\frac{1+n}{n+1} \\\\ =\frac{n+1}{n+1} \\\\ =1$

this isnt using geometric evidence. drongoski's answer is correct.

AAEldar · Sep 23, 2011

apollo1 said:
this isnt using geometric evidence. drongoski's answer is correct.

I know, I wasn't doing the integration I was just showing him that the addition equaled 1.

hup · Sep 23, 2011

$if\ L =\frac{1}{2}^{\frac{1}{2}^{\frac{1}{2}^{\frac{1}{2}}^{...}}}\\ \\ and\ it\ is\ known\ that\ L \ lies\ between\ 0.3\ and\ 0.7, \\ show\ that\ L\ is\ approximately\ equal\ to\ 0.64$
thats 0.5 to the power of 0.5 repeatedly

apollo1 · Sep 23, 2011

hup said:
$if\ L =\frac{1}{2}^{\frac{1}{2}^{\frac{1}{2}^{\frac{1}{2}}^{...}}}\\ \\ and\ it\ is\ known\ that\ L \ lies\ between\ 0.3\ and\ 0.7, \\ show\ that\ L\ is\ approximately\ equal\ to\ 0.64$
thats 0.5 to the power of 0.5 repeatedly

is this a hsc style question? i mean does it use hsc concepts?

hup · Sep 23, 2011

apollo1 said:
is this a hsc style question? i mean does it use hsc concepts?

idk about hsc style
but yes it does use stuff youve learnt in the hsc course

apollo1 · Sep 23, 2011

can u give us the solution plz

hup · Sep 24, 2011

apollo1 said:
can u give us the solution plz

try this one it might help with the other
$L =\frac{1}{1+\frac{1}{1+\frac{1}{1+...}}}\\ Find\ L$

apollo1 · Sep 24, 2011

hup said:
try this one it might help with the other
$L =\frac{1}{1+\frac{1}{1+\frac{1}{1+...}}}\\ Find\ L$

L = 1/1+L
L^2+L-1 = 0
therefore L = (-1+sqrt(5))/2

hup · Sep 24, 2011

apollo1 said:
L = 1/1+L
L^2+L-1 = 0
therefore L = (-1+sqrt(5))/2

now try the one before

ohexploitable · Sep 24, 2011

hup said:
$if\ L =\frac{1}{2}^{\frac{1}{2}^{\frac{1}{2}^{\frac{1}{2}}^{...}}}\\ \\ and\ it\ is\ known\ that\ L \ lies\ between\ 0.3\ and\ 0.7, \\ show\ that\ L\ is\ approximately\ equal\ to\ 0.64$
thats 0.5 to the power of 0.5 repeatedly

$\\L=\frac{1}{2}^L$

now use newton's method a couple of times

hup · Sep 24, 2011

ohexploitable said:
$\\L=\frac{1}{2}^L$

now use newton's method a couple of times

nais
now give us a question

apollo1 · Sep 24, 2011

let w be any non real solutions to $z^{5}=1$

fully simplify:

$w*w^{2}*w^{3}*....*w^{5k-1}-w-w^{2}-w^{3}-.....-w^{5k-1}$

k is an integer (k greater than or equal to 1)

largarithmic · Sep 24, 2011

apollo1 said:
let w be any non real solutions to $z^{5}=1$

fully simplify:

$w*w^{2}*w^{3}*....*w^{5k-1}-w-w^{2}-w^{3}-.....-w^{5k-1}$

k is an integer (k greater than or equal to 1)

$w*w^2*w^3*...*w^{5k-1} = w^{1+2+...+5k-1} = w^{\frac{5k(5k-1)}{2}} = (w^5)^{\frac{k(5k-1)}{2}} = 1$
and
$w+w^2+w^3+...+w^{5k-1} = \frac{w(w^{5k-1}-1)}{w-1} = \frac{w(\frac{w^{5k}}{w}-1)}{w-1} = \frac{w(\frac{1}{w}-1)}{w-1} = -1$

So combining the answer is 2.

Now try this question, it's from my trial. It's pretty neat too.

$P(x)$ is a polynomial of degree $n$ with real coefficients, where $n$ is an odd positive integer, satisfying:

$P(k) = \frac{k}{k+1}$ for each of $k = 0, 1, 2, ..., n$ .
(In other words, $P(0) = 0, P(1) = \frac{1}{2}, P(2) = \frac{2}{3}, P(3) = \frac{3}{4}, ..., P(n) = \frac{n}{n+1}$ ).

By considering the polynomial $Q(x) = (x+1)P(x) - x$ , find the leading coefficient of $P(x)$ and hence find $P(n+1)$ .

apollo1 · Sep 24, 2011

largarithmic said:
$w*w^2*w^3*...*w^{5k-1} = w^{1+2+...+5k-1} = w^{\frac{5k(5k-1)}{2}} = (w^5)^{\frac{k(5k-1)}{2}} = 1$
and
$w+w^2+w^3+...+w^{5k-1} = \frac{w(w^{5k-1}-1)}{w-1} = \frac{w(\frac{w^{5k}}{w}-1)}{w-1} = \frac{w(\frac{1}{w}-1)}{w-1} = -1$

So combining the answer is 2.

Now try this question, it's from my trial. It's pretty neat too.

$P(x)$ is a polynomial of degree $n$ with real coefficients, where $n$ is an odd positive integer, satisfying:

$P(k) = \frac{k}{k+1}$ for each of $k = 0, 1, 2, ..., n$ .
(In other words, $P(0) = 0, P(1) = \frac{1}{2}, P(2) = \frac{2}{3}, P(3) = \frac{3}{4}, ..., P(n) = \frac{n}{n+1}$ ).

By considering the polynomial $Q(x) = (x+1)P(x) - x$ , find the leading coefficient of $P(x)$ and hence find $P(n+1)$ .

which school do u go to?

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