Very hard A Level further maths questions thread

Watch this thread
322mathsphysics
Badges: 12
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#1
Report Thread starter 2 years ago
#1
Hi guys, let’s make a thread of really hard FM problems we come up with to help prepare! I’ll upvote anyone who posts good questions
0
reply
_gcx
Badges: 21
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#2
Report 2 years ago
#2
moved this to maths for you, will try to think of some interesting questions.
1
reply
Notnek
Badges: 20
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#3
Report 2 years ago
#3
(Original post by _gcx)
moved this to maths for you, will try to think of some interesting questions.
If they're not very hard then I'll report them.
9
reply
username2998742
Badges: 22
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#4
Report 2 years ago
#4
Y12_FurtherMaths merry christmas
1
reply
Y2_UniMaths
Badges: 19
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#5
Report 2 years ago
#5
(Original post by Glaz)
Y12_FurtherMaths merry christmas
Thank you haha
1
reply
_gcx
Badges: 21
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#6
Report 2 years ago
#6
Have too many ideas, so here's a classic one. Any time I mention Taylor series it'll be about zero unless specified otherwise.

Euler was the first to prove that \displaystyle \sum_{n = 1}^\infty n^{-2} = \pi^2/6. There are many, many ways to prove this but a lot of them use quite advanced techniques, but there are a few that I can think of that can be done with only A-level techniques. A lot of them use Fourier Analysis or techniques in multivariable calculus, and one very clever way using complex numbers popped up in STEP III 2018. This first one doesn't use more than A-level integration.
  1. Use the binomial theorem to write the Taylor expansion for \displaystyle \frac 1 {\sqrt {1 - x^2}} in the form \displaystyle \sum_{n = 0}^\infty a_n x^n for some real sequence (a_n) Deduce the Taylor expansion for \arcsin x.
  2. Show that \displaystyle \int_0^1 \frac {x^{2 n + 1} } {\sqrt {1 - x^2}} \mathrm dx = \frac {2 n} {2 n + 1} \int_0^1 \frac {x^{2 n - 1} } {\sqrt{1 - x^2}} \mathrm dx for n \ge 1. Use this to find a closed for expression \displaystyle \int_0^1 \frac {x^{2 n + 1} } {\sqrt {1 - x^2}} for non-negative integer n in terms of n. (hint: when n = 0 the integral is straightforward)
  3. Show that \displaystyle \int_0^1 \frac {\arcsin x} {\sqrt {1 - x^2}} \mathrm dx = \frac{\pi^2} 8. By using the last two parts, find \displaystyle \sum_{n = 0}^\infty \frac 1 {(2 n + 1)^2}.
  4. Deduce that \displaystyle \sum_{n = 1}^\infty n^{-2} = \pi^2/6

Let me know if this needs more guidance.

* You may assume that you can differentiate and integrate power series termwise (like you can with finite sums) where they converge.**

** Strictly this is too weak and it should be within its radius of convergence, boundary cases are always tricky, (a power series with radius of convergence 1 might converge at 1 but its "termwise derivative" not, for instance \sum_{n = 1}^\infty (-x)^n/n) but you won't be familiar with this term.

Some more proofs can be found on ProofWiki but you risk spoiling the STEP question and this question for yourself! Euler's classic proof can be found here. It's not great by modern standards, pretty much because he guessed that the product should work without properly showing it. But you can rest easy now that the product has been proved to hold using modern techniques. (by Weierstrass, see Weierstrass product theorem)
Last edited by _gcx; 2 years ago
0
reply
RDKGames
Badges: 20
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#7
Report 2 years ago
#7
Oh another thread for hard questions guess I'll just copy and my paste the Q's I've posted in the other dead threads of this type. Let's hope this doesn't get polluted with STEP type / olympiad questions too quickly just like the poor dead threads before this have

Although the title's interpretation is kind of up in the air since "hard a-level problems" can either mean 'difficult problems solvable with a-level techniques' or 'difficult problems without reaching outside of the a-level specification'

Any symbols or notation you don't understand, then let me know and I will explain it.





Image
Last edited by RDKGames; 2 years ago
0
reply
RDKGames
Badges: 20
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#8
Report 2 years ago
#8
Image
0
reply
RDKGames
Badges: 20
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#9
Report 2 years ago
#9
Image
0
reply
RDKGames
Badges: 20
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#10
Report 2 years ago
#10
Image
0
reply
RDKGames
Badges: 20
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#11
Report 2 years ago
#11
Image

Image
1
reply
IrrationalRoot
Badges: 20
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#12
Report 2 years ago
#12
Find the general solution to the equation \sin x+\cos x=1+\sin 2x.

Also:

Given that z=\dfrac{x_1}{1-x_3}+i\dfrac{x_2}{1-x_3}, \ x_1^2+x_2^2+x_3^2=1, \ x_3\neq 1, express each of x_1,x_2,x_3 in terms of z.
1
reply
username2998742
Badges: 22
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#13
Report 2 years ago
#13
(Original post by _gcx)
Have too many ideas, so here's a classic one. Any time I mention Taylor series it'll be about zero unless specified otherwise.

Euler was the first to prove that \displaystyle \sum_{n = 1}^\infty n^{-2} = \pi^2/6. There are many, many ways to prove this but a lot of them use quite advanced techniques, but there are a few that I can think of that can be done with only A-level techniques. A lot of them use Fourier Analysis or techniques in multivariable calculus, and one very clever way using complex numbers popped up in STEP III 2018. This first one doesn't use more than A-level integration.
  1. Use the binomial theorem to write the Taylor expansion for \displaystyle \frac 1 {\sqrt {1 - x^2}} in the form \displaystyle \sum_{n = 0}^\infty a_n x^n for some real sequence (a_n) Deduce the Taylor expansion for \arcsin x.
  2. Show that \displaystyle \int_0^1 \frac {x^{2 n + 1} } {\sqrt {1 - x^2}} \mathrm dx = \frac {2 n} {2 n + 1} \int_0^1 \frac {x^{2 n - 1} } {\sqrt{1 - x^2}} \mathrm dx for n \ge 1. Use this to find a closed for expression \displaystyle \int_0^1 \frac {x^{2 n + 1} } {\sqrt {1 - x^2}} for non-negative integer n in terms of n. (hint: when n = 0 the integral is straightforward)
  3. Show that \displaystyle \int_0^1 \frac {\arcsin x} {\sqrt {1 - x^2}} \mathrm dx. By using the last two parts, find \displaystyle \sum_{n = 0}^\infty \frac 1 {(2 n + 1)^2}.
  4. Deduce that \displaystyle \sum_{n = 1}^\infty n^{-2} = \pi^2/6

Let me know if this needs more guidance.

* You may assume that you can differentiate and integrate power series termwise (like you can with finite sums) where they converge.**

** Strictly this is too weak and it should be within its radius of convergence, boundary cases are always tricky, (a power series with radius of convergence 1 might converge at 1 but its "termwise derivative" not, for instance \sum_{n = 1}^\infty (-x)^n/n) but you won't be familiar with this term.

Some more proofs can be found on ProofWiki but you risk spoiling the STEP question and this question for yourself! Euler's classic proof can be found here. It's not great by modern standards, pretty much because he guessed that the product should work without properly showing it. But you can rest easy now that the product has been proved to hold using modern techniques. (by Weierstrass, see Weierstrass product theorem)
god this looks so good :drool:
wish i took fm :bawling:
0
reply
Cat?
Badges: 18
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#14
Report 2 years ago
#14
Calculate

\displaystyle \int_{0}^{\frac{\pi}{4}} \dfrac{\mathrm{d}x}{(\cos x + \sin x)\cos x}
1
reply
_gcx
Badges: 21
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#15
Report 2 years ago
#15
(Original post by Your Local Cat)
Calculate

\displaystyle \int_{0}^{\frac{\pi}{4}} \dfrac{\mathrm{d}x}{(\cos x + \sin x)\cos x}
Bit of a softball

Spoiler:
Show
just multiplying through \sec^2 x
0
reply
_gcx
Badges: 21
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#16
Report 2 years ago
#16
A silly question.

Say we know that:

\displaystyle \int_0^1 x^n \mathrm dx = \frac 1 {n + 1}

for real n > -1. We could derive this using Riemann sums or similar. By knowing this one definite integral, we can work out some indefinite integrals. See if you can prove that:

\displaystyle \int x^n \mathrm dx = \frac {x^{n + 1}} {n + 1} + C

and

\displaystyle \int \ln x \mathrm dx = x \ln x - x + C

using only this fact and standard integration theorems.
Last edited by _gcx; 2 years ago
1
reply
322mathsphysics
Badges: 12
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#17
Report Thread starter 2 years ago
#17
(Original post by _gcx)
A silly question.

Say we know that:

\displaystyle \int_0^1 x^n \mathrm dx = \frac 1 {n + 1}

for real n > -1. We could derive this using Riemann sums or similar. By knowing this one definite integral, we can work out some indefinite integrals. See if you can prove that:

\displaystyle \int x^n \mathrm dx = \frac {x^{n + 1}} {n + 1} + C

and

\displaystyle \int \ln x \mathrm dx = x \ln x - x + C

using only this fact and standard integration theorems.
Thanks for these contributions!! I’ll have a go now
0
reply
RDKGames
Badges: 20
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#18
Report 2 years ago
#18
Image
0
reply
RDKGames
Badges: 20
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#19
Report 2 years ago
#19
Image
0
reply
RDKGames
Badges: 20
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#20
Report 2 years ago
#20
Image
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest

Y13s: How will you be receiving your A-level results?

In person (69)
69%
In the post (4)
4%
Text (13)
13%
Something else (tell us in the thread) (14)
14%

Watched Threads

View All