, which is impossible since this states f(1) is not in the codomain. A similar argument at x=0 shows we cannot have f(x)<sin(x*pi/2) for all x in [0,1] either.

Thus there must exist

a\in [0,1]

for which

f(x)\geq \sin(\frac{x\pi}{2})

, and

b\in [0,1]

such that

f(x)\leq \sin(\frac{x\pi}{2})

. Suppose without loss of generality we have

a\leq b

. Let

g(x)=f(x)-sin(\frac{x\pi}{2})

. Then at a,

g(a)\geq 0

and at b

g(b)\leq 0

. Thus by the intermediate value theorem, there exists a

c

such that

a\leq c \leq b

for which

g(c)=0

, which implies by construction of g

f(c)=\sin(\frac{c\pi}{2})

Edit: A nice intuitive picture for how this theorem works: Draw a square, and from the bottom left corner to the top right corner, draw a curve resembling the graph of sin from 0 to pi/2. Now try and join the left side to the right side with a continuous curve that doesn't intersect the sine curve :wink:

.

(edited 8 years ago)

Reply 3389

DFranklin

18

Original post by FireGarden

Solution 539.

Suppose that

f(x)>sin(\frac{x\pi}{2})

for all

x\in [0,1]

. Then we must have

f(1)>sin(\frac{\pi}{2})=1

, which is impossible since this states f(1) is not in the codomain. A similar argument at x=0 shows we cannot have f(x)<sin(x*pi/2) for all x in [0,1] either.

Thus there must exist

a\in [0,1]

for which

f(x)\geq \sin(\frac{x\pi}{2})

, and

b\in [0,1]

such that

f(x)\leq \sin(\frac{x\pi}{2})

. Suppose without loss of generality we have

a\leq b

. Let

g(x)=f(x)-sin(\frac{x\pi}{2})

. Then at a,

g(a)\geq 0

and at b

g(b)\leq 0

. Thus by the intermediate value theorem, there exists a

c

such that

a\leq c \leq b

for which

g(c)=0

, which implies by construction of g

f(c)=\sin(\frac{c\pi}{2})

All fine, but a bit verbose:

Define g(x) = sin(pi x / 2) - f(x). Then g(0) = -f(0), g(1) = 1- f(1) and so since we know 0<=f(x)<=1, g(0)<=0, g(1)>=0. If we have equality in either case we're done, otherwise by IVT we can find k in (0,1) with g(k) = 0 and we're done.

[Questions like this come up often enough that you should make sure you know how to answer them concisely. The defn of IVT I use requires f(a) < 0, f(b) > 0 (i.e. it doesn't allow equality) or it could go a little shorter still].

Reply 3390

atsruser

11

Problem 540 (**/***)

a) Show that:

\displaystyle \int_0^{2\pi} \frac{r^2\sin^2 2\theta}{1-2r^2\cos 2\theta+r^4} \ d\theta = \pi r^2

stating any conditions that must hold for this to be valid.

b) Find the value of

\displaystyle \int_0^{2\pi} \frac{4\sin^2 2\theta}{17-8\cos 2\theta} \ d\theta

(Hints available if necessary)

Reply 3391

atsruser

11

Original post by atsruser

Problem 531: */**

Give a geometrical argument to show that, if

x^2+y^2=1

, then

\displaystyle \int_{\frac{\sqrt{2}}{2}}^{ \frac{\sqrt{2+\sqrt{2}}}{2} } \frac{1}{\sqrt{1-x^2}} \ dx = \frac{\pi}{8}

Bumping with hints:

Spoiler

Reply 3392

Zacken

22

Original post by atsruser

x

That's quite an interesting problem, unfortunately, it's three in the morning here and I'm heading off to bed, but I'll give these a try first thing in the morning, thanks for your contribution!

I can't quite wrap my head around the whole

\frac{\mathrm{d}x}{y}

thing.

Reply 3393

username1162972

18

Can't believe I've never used this thread before.

Posted from TSR Mobile

Reply 3394

username1162972

18

UProblem 541
I am on a IPad so not really sure about this bold stuff etc.
Prove the for natural n the fraction (21n+4)/(14n+3) is irreducible.
I will post a solution later on today.

This is a problem from an old IMO problem!
Posted from TSR Mobile

(edited 8 years ago)

Reply 3395

joostan

13

Solution 541:
Observe that

3(14n+3)=2(21n+4)+1

.
Thus if

p|(14n+3)

and

p|(21n+4)

for

p \in \mathbb{N}

then

p|(3(14n+3)-2(21n+4))

so

p|1 \Leftrightarrow p=1

.
Hence we have

14n+3

and

21n+4

coprime, so:

\dfrac{21n+4}{14n+3}

is irreducible.

Reply 3396

username1162972

18

Original post by joostan

Solution 541:
Observe that

3(14n+3)=2(21n+4)+1

.
Thus if

p|(14n+3)

and

p|(21n+4)

for

p \in \mathbb{N}

then

p|(3(14n+3)-2(21n+4))

so

p|1 \Leftrightarrow p=1

.
Hence we have

14n+3

and

21n+4

coprime, so:

\dfrac{21n+4}{14n+3}

is irreducible.

Mine essentially the same but slight different at the start I think.
Gcd(21n+4,14n+3)=gcd(7n+1,14n+3)
=gcd(7n+1,14n+3-2(7n+1))
=gcd(7n+1,1)=1 Q.E.D

Posted from TSR Mobile

Reply 3397

Zacken

22

Original post by physicsmaths

Mine essentially the same but slight different at the start I think.
Gcd(21n+4,14n+3)=gcd(7n+1,14n+3)
=gcd(7n+1,14n+3-2(7n+1))
=gcd(7n+1,1)=1 Q.E.D

Posted from TSR Mobile

Sorry to be a buzzkill, but it's specified in the OP to Latex all solutions/problems (I understand that you're on an iPad, but it's not hard to use Latex) - just to maintain austerity. :smile:

Also - it's fine to post a solution to your own problem once somebody else has already answered it, but not before.

(edited 8 years ago)

Reply 3398

16Characters....

13

Original post by 16Characters....

Problem 508

M

is an N x N matrix with N distinct eigenvalues all in the range

- 1 < \lambda < 1

. Prove that:

I + \displaystyle \sum_{r = 1}^{\infty} M^r

Converges. {NB: I denotes the N x N identity matrix}

Bumping with a hint:

Spoiler

Reply 3399

joostan

13

Original post by 16Characters....

Problem 508

M

is an N x N matrix with N distinct eigenvalues all in the range

- 1 < \lambda < 1

. Prove that:

I + \displaystyle \sum_{r = 1}^{\infty} M^r

Converges. {NB: I denotes the N x N identity matrix}

Let

S=I+\displaystyle\sum_{r=1}^ { \infty} M^r

Since

M

has

N

distinct eigenvalues, we can write

M=P^{-1}DP

for some fixed, invertible

N \times N

matrix

P

, and where

D=diag(\lambda_1, . . .,\lambda_N)

.
Notice that:

M^r=P^{-1}D^rP[br]\Rightarrow S=I+\displaystyle\sum_{r=1}^ { \infty} P^{-1}D^rP[br]\Rightarrow S=\displaystyle\sum_{r=0}^ { \infty} P^{-1}diag(\lambda_1^r, . . .,\lambda_N^r)P

.
But

|\lambda_i|<1 \Rightarrow \displaystyle\sum_{r=0}^{\infty} \lambda_i^r = \dfrac{1}{1-\lambda_i}

.
(Assuming

\lambda_i \not= 0

, if so then

\displaystyle\sum_{r=0}^{\infty} \lambda_i^r=0

so still converges.)
Hence we have that:

S=P^{-1} diag \left(\dfrac{1}{1-\lambda_1}, . . . ,\dfrac{1}{1-\lambda_N} \right)P

.
Thus the sum converges.

(edited 8 years ago)

The Proof is Trivial!

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you