Year 13 Maths Help Thread

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Managed to finish STEP I 2007 Q5 :smile:

!

One area of maths which I'm particularly weak at is recurrence relations, particularly the types of questions which define

u_{n+1}

in terms of

u_n

and then ask you to define

u_n

in terms of

n

.

What are some good questions that I can do to practice?

What part of it are you weak at? Making sense of breaking down the recurrence system? If so 15, 17 and 35 of advanced problems in mathematics are good

http://www.mathshelper.co.uk/110501_Advanced_Problems_in_Mathematics.pdf

Reply 561

Palette

Original post by ValerieKR

It's questions like 10b) here
http://madasmaths.com/archive/iygb_practice_papers/c1_practice_papers/c1_t.pdf

which I find difficult. Out of interest, do you prefer the old Siklos booklet or the new one?

(edited 7 years ago)

Reply 562

ValerieKR

Original post by Palette

That question's similar to the loans and savings STEP ones (in the late 90s there's a few of those)

I didn't use them too much when I was studying for STEP because I got tempted to scroll down - I'm working through them both now and haven't really gone through the other one properly yet

Reply 563

ValerieKR

Original post by Palette

If I give you the answer to that one the method becomes clear, so i've put it in the spoiler

Spoiler

if you're stuck you can spot most common ones by taking the difference between terms (here the difference was the sequence 1,2,4,8 etc)

(edited 7 years ago)

Examsolutions

Let L be the value of

\sum^{\infty}_{r=1} \frac{1}{r^2}

.

Consider the series

S=\frac{1}{1^2}+\frac{1}{2^2}-\frac{8}{3^2}+\frac{1}{4^2}+

\frac{1}{5^2}-\frac{8}{6^2}+\frac{1}{7^2}

+\frac{1}{8^2}-\frac{8}{9^2}+...

.

I wrote that

S

is equal to

\sum^{\infty}_{r=1} \frac{1}{r^2}-\sum^{\infty}_{r=1} \frac{9}{(3r)^2}=0

; am I right?

Reply 566

ValerieKR

Original post by Palette

Let L be the value of

\sum^{\infty}_{r=1} \frac{1}{r^2}

.

Consider the series

S=\frac{1}{1^2}+\frac{1}{2^2}-\frac{8}{3^2}+\frac{1}{4^2}+

\frac{1}{5^2}-\frac{8}{6^2}+\frac{1}{7^2}

+\frac{1}{8^2}-\frac{8}{9^2}+...

.

I wrote that

S

is equal to

\sum^{\infty}_{r=1} \frac{1}{r^2}-\sum^{\infty}_{r=1} \frac{9}{(3r)^2}=0

; am I right?

Yeah, looks it

Reply 567

Zacken

Original post by Palette

Let L be the value of

\sum^{\infty}_{r=1} \frac{1}{r^2}

.

Consider the series

S=\frac{1}{1^2}+\frac{1}{2^2}-\frac{8}{3^2}+\frac{1}{4^2}+

\frac{1}{5^2}-\frac{8}{6^2}+\frac{1}{7^2}

+\frac{1}{8^2}-\frac{8}{9^2}+...

.

I wrote that

S

is equal to

\sum^{\infty}_{r=1} \frac{1}{r^2}-\sum^{\infty}_{r=1} \frac{9}{(3r)^2}=0

; am I right?

Your justification is very fishy, you'd need to prove that

S

converges absolutely.

Reply 568

Palette

Original post by Zacken

Your justification is very fishy, you'd need to prove that

S

converges absolutely.

http://madasmaths.com/archive/iygb_practice_papers/sp_practice_papers/sp_t.pdf

I think the question (Q16) assumes convergence of

S]

otherwise the 'limit of the following infinite series' would not be asked. If I am wrong on this, I may try and exploit the direct comparison test when I'm not in a half-asleep state.

Reply 569

Zacken

Original post by Palette

http://madasmaths.com/archive/iygb_practice_papers/sp_practice_papers/sp_t.pdf

I think the question (Q16) assumes convergence of

S]

otherwise the 'limit of the following infinite series' would not be asked. If I am wrong on this, I may try and exploit the direct comparison test when I'm not in a half-asleep state.

It's not too hard to prove it anyway, just note that

\displaystyle S < \sum_{\geq 1} \frac{1}{k^2}

and

\displaystyle s_m = \sum_{1 \leq r \leq m} \frac{1}{r^2}

is a bounded monotone sequence or apply Cauchy's Condensation Test to get convergence of

\displaystyle \sum_{\geq 1}\frac{1}{k^2}

Reply 570

Palette

Original post by Zacken

It's not too hard to prove it anyway, just note that

\displaystyle S < \sum_{\geq 1} \frac{1}{k^2}

and

\displaystyle s_m = \sum_{1 \leq r \leq m} \frac{1}{r^2}

is a bounded monotone sequence or apply Cauchy's Condensation Test to get convergence of

\displaystyle \sum_{\geq 1}\frac{1}{k^2}

Would the direct comparison test work too?

Reply 571

Zacken

Original post by Palette

Would the direct comparison test work too?

That is the comparison test?

Reply 572

Palette

Original post by Zacken

That is the comparison test?

Oops, I failed to spot that you just wrote it out in a different form to what I learnt it by; my mistake!

Reply 573

Palette

Can you be deducted a mark by writing

\int x^{-1} dx

\ln x +C

instead of

\ln |x|+ C

Reply 574

B_9710

Original post by Palette

Can you be deducted a mark by writing

\int x^{-1} dx

\ln x +C

instead of

\ln |x|+ C

You could but I don't think it matters too much. It could be important for a definite integral though where the argument inside the log is negative because of one of the limits. It's best to put it when it's needed.

Reply 575

Palette

Consider

I=\int^b_a f(x) dx, b>a

. How can one intuitively explain why

\int^a_b f(x) dx=-I

Reply 576

RDKGames

Study Forum Helper

Original post by Palette

Consider

I=\int^b_a f(x) dx, b>a

. How can one intuitively explain why

\int^a_b f(x) dx=-I

I just imagine measuring the area from b to a, which would make the difference between the two negative as you're moving backwards (so to speak), and the area would be multiplied by a negative, hence the negative I.

Feels weird to explain it like this...

I think you'd have to refer back to what integration actually does by considering infinitesimal changes on the rectangles which approximate the area, and the width each rectangle is a negative infinitesimal while height is positive. Hope that makes sense.

(edited 7 years ago)

Reply 577

Zacken

Original post by Palette

Consider

I=\int^b_a f(x) dx, b>a

. How can one intuitively explain why

\int^a_b f(x) dx=-I

We have, by FToC:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}I = \int_a^b f(x) \, \mathrm{d}x= F(x)\bigg|_{a}^{b} = F(b) - F(a) \end{equation*}

whilst

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\int_b^a f(x) \, \mathrm{d}x = F(x)\bigg|_b^a = F(a) - F(b) = -(F(b) - F(a)) = -I \end{equation*}

See, definite integration is only defined for

b \geq a

so at some, we need to define that flipping the limits gives us the negative of the integral. One justification for this is that we'd like the integral identity:

Unparseable latex formula:

\displaystyle [br]\begin{equation*} \int_a^b f(x) \mathrm{d}x + \int_b^c f(x) \mathrm{d}x = \int_a^c f(x) \mathrmm{d}x \end{equation*}

to hold for arbitrary