Extension One Revising Game (1 Viewer)

AkaiHanabi

Thread killer
Joined
Sep 19, 2007
Messages
224
Location
Baulkham Hills
Gender
Female
HSC
2008
azureus88 said:
sum of all integers from 1 to 200 - sum of all multiples of 7

= (200/2)(1+200) - (28/2)(1+196)
=100(201) - 14(197)
= 17342

ok can someone do my question now.

prove, using induction n^2 - 11n +30 is bigger than or equal to 0 for n is bigger than or equal to 1, where n is integer.
for n=1

1 - 11 + 30
=20
20>0
true for n=1

assume for n=k
ie k^2 - 11k +30 >0

prove for n=k+1

(k+1)^2 -11(k+1) + 30
= k^2 +2k + 1 -11k -11 +30
= k^2 -9k +20
= k^2 -9k -2k +2k +30 +10 -10
= (k^2 -11k +30) +2k -10

K^2 - 11k +30>0 from assumption
and then I got stuck =/

Find the acute angle between the lines 4y=3x+8 and y=6x-9
 
Last edited:

u-borat

Banned
Joined
Nov 3, 2007
Messages
1,755
Location
Sydney
Gender
Male
HSC
2013
^^Yeah same, i have no frigging clue how to prove by induction that.

the method i used was to differentiate it and you find that the minimum point is at like 5.5, then you prove that integers 5 and 6 are >0 and hence all integers are >0 because it is a increasing function on both sides.
 

AkaiHanabi

Thread killer
Joined
Sep 19, 2007
Messages
224
Location
Baulkham Hills
Gender
Female
HSC
2008
u-borat said:
^^Yeah same, i have no frigging clue how to prove by induction that.

the method i used was to differentiate it and you find that the minimum point is at like 5.5, then you prove that integers 5 and 6 are >0 and hence all integers are >0 because it is a increasing function on both sides.
Maybe I have to show that it's true up to 6, then say since true for 1<n<6, 2k-10>0 therefore (k^2-11k+30) +2k-10 >0

gah.
 

lyounamu

Reborn
Joined
Oct 28, 2007
Messages
9,998
Gender
Male
HSC
N/A
AkaiHanabi said:
for n=1

1 - 11 + 30
=20
20>0
true for n=1

assume for n=k
ie k^2 - 11k +30 >0

prove for n=k+1

(k+1)^2 -11(k+1) + 30
= k^2 +2k + 1 -11k -11 +30
= k^2 -9k +20
= k^2 -9k -2k +2k +30 +10 -10
= (k^2 -11k +30) +2k -10

K^2 - 11k +30>0 from assumption
and then I got stuck =/

Find the acute angle between the lines 4y=3x+8 and y=6x-9
slight mistake there:

k^2 - 9k - 2k + 2k + 20 + 10 - 10
= (k^2 - 11k + 30) + 2k - 10

By the way, question is flawed.

If you look at the function, it has negative points between 5 and 6.
 

lyounamu

Reborn
Joined
Oct 28, 2007
Messages
9,998
Gender
Male
HSC
N/A
u-borat said:
na, its for integers only.
ok, in that case, I would prove by using gradient that it is increasing from 5 (to 0) and increasing from 6 to infinity.
 

QuLiT

Member
Joined
Aug 23, 2007
Messages
84
Gender
Male
HSC
2008
don't know if this is allowed but you could prove it true for n = 1, 2, 3 ,4 ,5 seperetly then use the proper induction method for n>or equal to 6
 

AkaiHanabi

Thread killer
Joined
Sep 19, 2007
Messages
224
Location
Baulkham Hills
Gender
Female
HSC
2008
Azreil said:
But that's not by induction, right?
Just out of curiousity, how many marks would you lose if you proved a question by differientiation if the question asked you to do it by induction?
 

AkaiHanabi

Thread killer
Joined
Sep 19, 2007
Messages
224
Location
Baulkham Hills
Gender
Female
HSC
2008
lyounamu said:
slight mistake there:

k^2 - 9k - 2k + 2k + 20 + 10 - 10
= (k^2 - 11k + 30) + 2k - 10

By the way, question is flawed.

If you look at the function, it has negative points between 5 and 6.
hehe woops ^^"

yeah, the question does seem a bit flawed >_>
 

lyounamu

Reborn
Joined
Oct 28, 2007
Messages
9,998
Gender
Male
HSC
N/A
QuLiT said:
don't know if this is allowed but you could prove it true for n = 1, 2, 3 ,4 ,5 seperetly then use the proper induction method for n>or equal to 6
That's a brilliant idea! I will definitely use that for this question.
 

AkaiHanabi

Thread killer
Joined
Sep 19, 2007
Messages
224
Location
Baulkham Hills
Gender
Female
HSC
2008
lyounamu said:
That's a brilliant idea! I will definitely use that for this question.
=( I suggested that earlier, but I wasn't sure. anyway, good luck guys. I'm going to tackle some english. rawr.
 

azureus88

Member
Joined
Jul 9, 2007
Messages
278
Gender
Male
HSC
2009
lyounamu said:
That's a brilliant idea! I will definitely use that for this question.
Theres another method which u could consider using (it doesnt make use of assumption tho),

for n=k+1
LHS=(k+1)^2 - 11(k+1) + 30
=k^2 -9k+20
=(k-4.5)^2 - 0.25
>or= 0 since (k-4.5)^2 >or= 0.25 for k is an integer
 
P

pLuvia

Guest
Seeing as though this is similar to something that has already been done before by past/present maths veterans

Check out these threads for questions to practice with
Marathon 1
Marathon 2
Marathon 3

Question from one of the threads

Question 37:

Given that when applying the iterative method of Newton's to find a zero of a function f(x), the output values of successive iterations follow a linear pattern, irrespective of the starting input value, given by:
y = mx + b ; where 'y' is the output value, and 'x' is the input value. and 'm' & 'b' are any reals.

Find:

(a) ONE such function f(x) where m = 1 and b is non-zero.

(b) ONE such function f(x) where b = 0 and m is non-zero.

[HINT: f(x) need not be a linear function.]
 
Last edited by a moderator:

QuLiT

Member
Joined
Aug 23, 2007
Messages
84
Gender
Male
HSC
2008
a) would f(x) = e^x be one?
b) f(x) = 3x? duno seems too easy:S

new question, out of a textbook:

integral between 1 -> 0 2^(log x)
 
Last edited:

3.14159potato26

New Member
Joined
Aug 11, 2008
Messages
28
Location
Malaysia
Gender
Male
HSC
2008
QuLiT said:
integral between 1 -> 0 2^(log x)
I{1->0} 2^(log_e(x))
Notation: log_a(...) means log base a of .... .
Note: Writing it on paper makes things a lot less confusing.
Consider log_e(x).
log_e(x) = log_2(x) / log_2(e) --(change base)
log_e(x) = (1/log_2(e)) * log_2(x)
log_e(x) = log_2(x^(1/log_2(e)))
Therefore,
I{1->0} 2^(log_e(x))
= I{1->0} 2^(log_2(x^(1/log_2(e))))
= I{1->0} x^(1/log_2(e)) ---(Note:Refer to comments after solution)
= [(x^(1+1/log_2(e))/(1+1/log_2(e))] {1->0}
= 0 - 1/(1+1/log_2(e))
= -1/(1+log_e(2)/log_e(e)) --(change base)
= -1/(1+log_e(2)).

New question:
In this proof, the assumption a^log_a(b) = b was made (has been proven).
Prove that a^log_a(b) = b.
 
Last edited:

QuLiT

Member
Joined
Aug 23, 2007
Messages
84
Gender
Male
HSC
2008
isn't taht just simply because a^x is the inverse function of log_a(x) ?
 

3.14159potato26

New Member
Joined
Aug 11, 2008
Messages
28
Location
Malaysia
Gender
Male
HSC
2008
xD....yeah, I guess that you could look at it that way.
My proof was like this:
Let a^log_a(b) = y.
log_a(a^log_a(b)) = log_a(y)
log_a(b) * log_a(a) = log_a(y)
log_a(b) = log_a(y)
b = y.
Therefore, a^log_a(b) = b.
Yours is a better alternative to solving it though.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top