The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

This discussion is now closed.

Check out other Related discussions

Mega A Level Maths Thread MK II

Scroll to see replies

Original post by DJMayes
STEP, although the prerequisite knowledge is FP3.

I see. Thanks for the reply :biggrin: I always feel honored :angelblush::angelblush: as your undoubtedly constantly busy!
Original post by justinawe
Yes :colone:

I keep forgetting to try it :tongue:. Something maths related: what's the easiest STEP Q ever? :cool:
Original post by reubenkinara
I keep forgetting to try it :tongue:. Something maths related: what's the easiest STEP Q ever? :cool:


Assuming you have the prerequisite knowledge, Q10 of 1994 STEP II can be done in a line.
Original post by DJMayes
Assuming you have the prerequisite knowledge, Q10 of 1994 STEP II can be done in a line.

Thanks DJ!
Original post by MAyman12
Hint

Spoiler


I was focusing far too much on your hint to have any sense like Justin and take a step (:colone:) back but hey ho :redface:

Spoiler

(edited 10 years ago)
Original post by Felix Felicis
I was focusing far too much on your hint to have any sense like Justin and take a step (:colone:) back but hey ho :redface:

Spoiler



:clap2: :bumps:

Oh god thats one hell of an answer:eek: Is it weird that I actually understood what you wrote?:tongue:
(edited 10 years ago)
Original post by MAyman12
:clap2: :bumps:

Oh god thats one hell of an answer:eek: Is it weird that I actually understood what you wrote?:tongue:

If the hint you gave me is part of Ramanujan's original proof then I applaud him :redface: There's no way in hell I would've gotten that otherwise. :redface:
Original post by Felix Felicis
If the hint you gave me is part of Ramanujan's original proof then I applaud him :redface: There's no way in hell I would've gotten that otherwise. :redface:


Yes, it was part of Ramanujan's original proof. :tongue:
Original post by MAyman12
Yes, it was part of Ramanujan's original proof. :tongue:


Are you okay with this now, or do you want an explanation for my solution?

I did have another part explaining the iteration fully when I wrote it down, but I didn't want to bother typing it out :tongue:
Original post by Felix Felicis
I was focusing far too much on your hint to have any sense like Justin and take a step (:colone:) back but hey ho :redface:

Spoiler



Considering the fact that you're the one sitting for STEP this time around, you'd better learn to take a step (:wink:) back :colone:
Original post by justinawe
Considering the fact that you're the one sitting for STEP this time around, you'd better learn to take a step (:wink:) back :colone:


Thinking questions through is overrated. :tongue:
Original post by Felix Felicis

Spoiler



Of course - you weren't supposed to do it though, it is too easy for you. :wink:



Here is an alternate way of tackling the 1+21+3⋯\sqrt{1+2\sqrt{1+3\sqrt{\cdots}}} someone posted:

f(x)=1+x1+(1+x)⋯⇒f2(x)=1+xf(x+1)    (⋆)f(x)=\sqrt{1+x\sqrt{1+(1+x)\sqrt{\cdots}}}\Rightarrow f^2(x)=1+xf(x+1)\;\;(\star )

Clearly f(x)≥xxx⋯=x≥12(x+1)f(x)\geq \sqrt{x\sqrt{x\sqrt{x\sqrt{ \cdots}}}}=x\geq\frac{1}{2}(x+1)

and
Unparseable latex formula:

\begin{aligned} f(x)\leq\sqrt{(x+1)\sqrt{2(x+1) \sqrt{3(x+1) \sqrt{ \cdots}}}}=(x+1) \sqrt{1 \sqrt{2 \sqrt{3 \cdots}}} \leq (x+1) \sqrt{1 \sqrt{2 \sqrt{4 \cdots}}}=2(x+1)



By the preceding inequalities and using (⋆)(\star ) we get:

2−12(x+1)<f(x)<212(x+1)2^{-\frac{1}{2}}(x+1)<f(x)<2^{ \frac{1}{2}}(x+1), which inductively yields 2−12n(x+1)<f(x)<212n(x+1)2^{-\frac{1}{2^n}}(x+1)<f(x)<2^{ \frac{1}{2^n}}(x+1)

Letting n→∞n\to \infty gives f(x)=x+1⇒f(2)=3f(x)=x+1\Rightarrow f(2)=3
Original post by Lord of the Flies
Of course - you weren't supposed to do it though, it is too easy for you. :wink:



Here is an alternate way of tackling the 1+21+3⋯\sqrt{1+2\sqrt{1+3\sqrt{\cdots}}} someone posted:

f(x)=1+x1+(1+x)⋯⇒f2(x)=1+xf(x+1)    (⋆)f(x)=\sqrt{1+x\sqrt{1+(1+x)\sqrt{\cdots}}}\Rightarrow f^2(x)=1+xf(x+1)\;\;(\star )

Clearly f(x)≥xxx⋯=x≥12(x+1)f(x)\geq \sqrt{x\sqrt{x\sqrt{x\sqrt{ \cdots}}}}=x\geq\frac{1}{2}(x+1)

and
Unparseable latex formula:

\begin{aligned} f(x)\leq\sqrt{(x+1)\sqrt{2(x+1) \sqrt{3(x+1) \sqrt{ \cdots}}}}=(x+1) \sqrt{1 \sqrt{2 \sqrt{3 \cdots}}} \leq (x+1) \sqrt{1 \sqrt{2 \sqrt{4 \cdots}}}=2(x+1)



By the preceding inequalities and using (⋆)(\star ) we get:

2−12(x+1)<f(x)<212(x+1)2^{-\frac{1}{2}}(x+1)<f(x)<2^{ \frac{1}{2}}(x+1), which inductively yields 2−12n(x+1)<f(x)<212n(x+1)2^{-\frac{1}{2^n}}(x+1)<f(x)<2^{ \frac{1}{2^n}}(x+1)

Letting n→∞n\to \infty gives f(x)=x+1⇒f(2)=3f(x)=x+1\Rightarrow f(2)=3

:eek: I shall save this for the future :biggrin:
Original post by justinawe
Considering the fact that you're the one sitting for STEP this time around, you'd better learn to take a step (:wink:) back :colone:


Spoiler

:zomg: STEP :cry2:
Original post by justinawe
Are you okay with this now, or do you want an explanation for my solution?

I did have another part explaining the iteration fully when I wrote it down, but I didn't want to bother typing it out :tongue:


Sorry for replying very late but I actually understand it now. It just took me a while.:tongue: Thank you.
Funny how maths students procrastinate by doing more maths...

:lol:
Reply 3377
Avatar for Faye_m
Faye_m
14
Original post by ThatRandomGuy
Funny how maths students procrastinate by doing more maths...

:lol:


OMG THAT IS SO ME!!! I had to study my American APs and I'm taking A level math next year! I studied math to make self feel less guilty procrastinating! OMG I don't feel alone in the world anymore haha
Original post by Lord of the Flies
Of course - you weren't supposed to do it though, it is too easy for you. :wink:



Here is an alternate way of tackling the 1+21+3⋯\sqrt{1+2\sqrt{1+3\sqrt{\cdots}}} someone posted:

f(x)=1+x1+(1+x)⋯⇒f2(x)=1+xf(x+1)    (⋆)f(x)=\sqrt{1+x\sqrt{1+(1+x)\sqrt{\cdots}}}\Rightarrow f^2(x)=1+xf(x+1)\;\;(\star )

Clearly f(x)≥xxx⋯=x≥12(x+1)f(x)\geq \sqrt{x\sqrt{x\sqrt{x\sqrt{ \cdots}}}}=x\geq\frac{1}{2}(x+1)

and
Unparseable latex formula:

\begin{aligned} f(x)\leq\sqrt{(x+1)\sqrt{2(x+1) \sqrt{3(x+1) \sqrt{ \cdots}}}}=(x+1) \sqrt{1 \sqrt{2 \sqrt{3 \cdots}}} \leq (x+1) \sqrt{1 \sqrt{2 \sqrt{4 \cdots}}}=2(x+1)



By the preceding inequalities and using (⋆)(\star ) we get:

2−12(x+1)<f(x)<212(x+1)2^{-\frac{1}{2}}(x+1)<f(x)<2^{ \frac{1}{2}}(x+1), which inductively yields 2−12n(x+1)<f(x)<212n(x+1)2^{-\frac{1}{2^n}}(x+1)<f(x)<2^{ \frac{1}{2^n}}(x+1)

Letting n→∞n\to \infty gives f(x)=x+1⇒f(2)=3f(x)=x+1\Rightarrow f(2)=3


You're probably going to have enough time for an hours nap at the end of the STEP exam...
Original post by Lord of the Flies
Of course - you weren't supposed to do it though, it is too easy for you. :wink:



Here is an alternate way of tackling the 1+21+3⋯\sqrt{1+2\sqrt{1+3\sqrt{\cdots}}} someone posted:

Spoiler


Damn, you never cease to amaze me :L PRSOM :adore:

Related discussions

Latest

Trending

Trending

Articles for you