Extension One Revising Game (2 Viewers)

[AkaiHanabi]

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

 

[u-borat]

^^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]

^^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]

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.

 

[u-borat]

na, its for integers only.

 

[lyounamu]

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.

 

[Azreil]

But that's not by induction, right?

 

[QuLiT]

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]

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?

 

[QuLiT]

all but one i think

 

[AkaiHanabi]

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]

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]

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.

 

[tabbaa]

Conics likes men. I love you Conics

 

[azureus88]

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

 

[pLuvia]

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.]

 

[QuLiT]

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)

 

[3.14159potato26]

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.

 

[QuLiT]

isn't taht just simply because a^x is the inverse function of log_a(x) ?

 

[3.14159potato26]

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.