Extension One Revising Game (1 Viewer)

[QuLiT]

your way works too^^

 

[3.14159potato26]

lol.....don't forget to post a new question.

 

[QuLiT]

hey technically you have to you posted a solution! xD

 

[3.14159potato26]

haha....i feel like an idiot, answering my own question....
Anyways, new question:
By expressing sin x and cos x in terms of t = tan(x/2), show that the equation asinx + bcosx = c has real roots only if a^2 + b^2 >= c^2.

 

[QuLiT]

well by replacing sinx and cosx with their values in t (cbf typing this up) you end up with:

2tb + a(1-t^2) = c(1+t^2)

expand and rearange this you get:
(c+a)t^2 -2bt +c -a = 0

there for for it to have real roots discriminant >o

4b^2 -4(c-a)(c+a)>=0
4b^2 + 4a^2 >=4c^2
b^2 + a^2 >= c^2

new question:
consider the area under y=1/x between x=n and x= n+ 1

show than 1/n+1< integral between n+1 and n of 1/x < 1/n

 

[3.14159potato26]

consider the area under y=1/x between x=n and x= n+ 1
show than 1/n+1< integral between n+1 and n of 1/x < 1/n

Very classic question xD....
Proof:
I{n->n+1} 1/x represents the actual area.
By considering the graph of f(x) = 1/x,
f(n) > f(n+1)
Area of retangle with length f(n) > Area of retangle with length f(n+1), both having length of 1 unit.
Therefore, 1/n * 1 > 1/(n+1) * 1
1/n > 1/(n+1).
From the graph, 1/n * 1 > I{n->n+1} 1/x and I{n->n+1} 1/x > 1/(n+1) * 1.
Therefore,
1/(n+1) < I{n->n+1} 1/x < 1/n.

New question:
If f(x) = x * cot(x) show that f(2x) = (1-(tan(x))^2) * f(x).

 

[Azreil]

f(x)=xcotx
= x/tanx
f(2x) = 2x * (1-(tan(x))^2)/2tanx
=(1-(tan(x))^2) * 2x/2tanx
=(1-(tan(x))^2) * x/tanx
=(1-(tan(x))^2) * f(x)

New question:
f(x) is defined as f(x)= [4/(x-1)] - 2 ; x<1.
Find f^-1(x).

 

[u-borat]

Very classic question xD....
Proof:
I{n->n+1} 1/x represents the actual area.
By considering the graph of f(x) = 1/x,
f(n) > f(n+1)
Area of retangle with length f(n) > Area of retangle with length f(n+1), both having length of 1 unit.
Therefore, 1/n * 1 > 1/(n+1) * 1
1/n > 1/(n+1).
From the graph, 1/n * 1 > I{n->n+1} 1/x and I{n->n+1} 1/x > 1/(n+1) * 1.
Therefore,
1/(n+1) < I{n->n+1} 1/x < 1/n.

New question:
If f(x) = x * cot(x) show that f(2x) = (1-(tan(x))^2) * f(x).

is that even a proof?

 

[3.14159potato26]

Yes it is, draw the graph, take out the retangles and you'll find that its true.
Its done by considering the areas under the graph (which is what the question asks you to do), and since one is obviously smaller than the other, so yeah, as far as i know, its valid.

 

[3.14159potato26]

New question:
f(x) is defined as f(x)= [4/(x-1)] - 2 ; x<1.
Find f^-1(x).

f(x)= [4/(x-1)] - 2 ; x<1.
Let f(f^-1(x)) = x
x = [4/(f^-1(x)-1)] - 2
x + 2 = 4/(f^-1(x)-1)
f^-1(x) - 1 = 4/(2+x)
f^-1(x) = 1 + 4/(2+x)
f^-1(x) = (6+x)/(2+x)

New question:
Given that 1/n! > 1/e^n; n > 5, prove that 1/1! + 1/2! + 1/3! + 1/4! + .... < 103/60 + 1/((e^5)*(e-1)).

 

[u-borat]

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)).

i got the positive version of this answer =S.

wats the answer qulit?

 

[tabbaa]

f(x)= [4/(x-1)] - 2 ; x<1.
Let f(f^-1(x)) = x
x = [4/(f^-1(x)-1)] - 2
x + 2 = 4/(f^-1(x)-1)
f^-1(x) - 1 = 4/(2+x)
f^-1(x) = 1 + 4/(2+x)
f^-1(x) = (6+x)/(2+x)

New question:
Given that 1/n! > 1/e^n; n > 5, prove that 1/1! + 1/2! + 1/3! + 1/4! + .... < 103/60 + 1/((e^5)*(e-1)).[/quote]

wtfff is your question asking ??

 

[tabbaa]

is that even a proof?

yes thats a 4unit quesiton. not fair..


its simply proving from the graph that astatment is true..

its either the sum of rectangles > area under the graph > rectangles under the graph.


and thhen he goes around play with it...

 

[Tincho]

New question:
Given that 1/n! > 1/e^n; n > 5, prove that 1/1! + 1/2! + 1/3! + 1/4! + .... < 103/60 + 1/((e^5)*(e-1)).

For 1/1! + 1/2! + 1/3! + 1/4! + .... is there an end? 1/n+1 possibly?Heres another question if anyones interested:

Factorise [4b^2c^2 - (b^2 + c^2 - a^2)]as the product of four factors

 

[3.14159potato26]

For 1/1! + 1/2! + 1/3! + 1/4! + .... is there an end?

Nope, there is no end. The whole point of the ... means it goes on to infinity. However, since it is a convergent infinite series, it has a finite value, which you don't need to calcuate, but just prove that LHS < RHS, using the given formula.

 

[3.14159potato26]

New question:
Given that 1/n! > 1/e^n; n > 5, prove that 1/1! + 1/2! + 1/3! + 1/4! + .... < 103/60 + 1/((e^5)*(e-1)).
wtfff is your question asking ??

Its asking you to prove that 1/1! + 1/2! + 1/3! + 1/4! + .... < 103/60 + 1/((e^5)*(e-1)) using the given formula 1/n! > 1/e^n; n > 5. Easier to prove LHS < RHS rather than RHS > LHS.

 

[lolokay]

New question:
Given that 1/n! > 1/e^n; n > 5, prove that 1/1! + 1/2! + 1/3! + 1/4! + .... < 103/60 + 1/((e^5)*(e-1)).

I think you mean n! > eⁿ; n > 5, or 1/n! < 1/e^n; n > 5
1/1! +... + 1/5! = 103/60 [just adding]
1/6! + 1/7! +... < e^-6 + e^-7 + ... = e^-5/(e-1) [sum of geometric series]
putting together gives
1/1! +... + 1/5! + 1/6! + 1/7! +... < 103/60 + 1/e⁵(e-1)

 

[lyounamu]

Let's keep this thread alive:

Use mathematical induction to show that cos(x + nPI) = (-1)^n . cos(x)

 

[duy.le]

ive posted up my solution though i must say that maths isnt going to be important to most people until monday 12-ish, then we'll see people flooding these forum pages.

my question, rather easy.

Show that (x-1)(x-2) is a factor of

P(x)=x^n(2^m-1)+x^m(1-2^n)+(2^n-2^m)

 

[lyounamu]

ive posted up my solution though i must say that maths isnt going to be important to most people until monday 12-ish, then we'll see people flooding these forum pages.

my question, rather easy.

Show that (x-1)(x-2) is a factor of

P(x)=x^n(2^m-1)+x^m(1-2^n)+(2^n-2^m)

Yep you were spot-on.