Ok, I need a few math questions

[Wacky]

A question I was doing the other day:

i) Explaain why for a> 0

[integral from 0 to 1](x^a)*(e^x)dx < 3 / (a+1)

(you may assume e < 3)

ii) Show by induction that for n = 0, 1, 2... there exist integers a_n and b_n such that

[integral from 0 to 1](x^n)*(e^x)dx = a_n + b_ne

(EWP: by <sub>tags I mean subscript)

iii) Suppose that r is a positive rational, so that r = p/q where p and q are positive integers. Show that for all integers a and b either |a + br| = 0 or |a + br| >= 1/q

iv) Prove that e is irrational.


You might want to write out the symbols first before attempting it.</sub>

[mathcam]

Did you end up getting it? I don't think it's too bad. I hadn't seen this particular proof of the irrationality of e. Cool.

Cam

[mathcam]

I couldn't resist.

Solutions

Cam

[d8P]

Don't pretend you tried

That's really elegant math. That's really elegant, math. Elegant math. Whatever. [/sb_email]

[Wacky]

I can't really say I got ALL of it. I did part 1 in class ages ago, then got distracted. I tried to do part 1 again a few days ago and forgot what I did earlier and didn't see it. I made an algebraic error in part 2 and couldn't be bothered but I got the rest.

But its not really bad, which is good because its like the third or second hardest question ever to pop up in my maths final exams. Another contender for about second place is another proof on the irrationality of e, which I'll post later.

[mathcam]

You mean pi? We just did "e" above.

Cam

[Wacky]

It was another year, and another e proof. But it is pretty much the same except with things phrased differently, rearranged, and divided by n!, and with a very slightly different integral. So never mind.

Proof of pi irrationality is slightly beyond the scope of the course. Here, however, is the hardest question ever to appear in that exam in any year. I can't do the last part or understand exactly what the solution means (I keep on wondering, whether I'm reading too fast)

8 B)

The difference between a real number r and the greatest integer less than or equal to r is called the fractional part of r, F(r), thus F(3.45) = 0.45 Note that for all real numbers r, 0=<F(r)<1

i. Let a = 2136log₁₀2

Given that F(a) = 7.0738... x 10^-5
F(2a) = 14.176... x 10^-5
F(3a) = 21.2214 x 10^-5

a) Use your calculator to show log₁₀ 1.989 < F(4223a) < log₁₀ 1.990

b) Hence Calculate an integer m such that the decimal representation of 2^m begins with 1989. Thus 2^m = 1989......

ii. Let r be a real number and m, n be nonzero integers m != n.

a) show that if F(mr) = 0, then r is rational

b) show that if F(mr) = F(nr) then r is rational

iii) suppose r is an irrational number. Let N be a positive integer and consider the fractional parts

hmm... didn't realise I could copy and paste.... Lets try this again:


8(b). The difference between a real number r are the greatest integer less than or
equal to r is called the fractional part of r, F(r). Thus F(3.45) = 0.45. Note that
for all real numbers r, 0 F(r) < 1.
(i) Let a = 2136 log10 2.
Given that F(a) = 7.0738 · · ·×10−5
observe that F(2a) = 14.1476 · · ·×10−5
F(3a) = 21.2214 · · ·×10−5
(α) Use your calculator to show that log10 1.989 < F(4223a) < log10 1.990.
(β) Hence calculate an integer M such that the ordinary decimal representation of
2M begins with 1989. Thus 2M=1989 . . . .
(ii) Let r be a real number and let m and n be non-zero integers with m = n.
(α) Show that if F(mr) = 0, then r is rational.
(β) Show that if F(mr) = F(nr), then r is rational.
(iii) Suppose that b is an irrational number. Let N be a positive integer and consider
the fractional parts F(b), F(2b), . . . , F ((N + 1)b).
(α) Show that these N + 1 numbers F(b), . . . , F ((N + 1)b) are all distinct.
(β) Divide the interval 0 x < 1 into N subintervals each of length 1/N and show
that there must be integers m and n with m = n and 1 m, n N + 1 such that
F((m − n)b) < 1/N.

(In case that didn't show up, m!= n, 1=< m, n =< N+1)

(iv) Given that log10 2 is irrational, choose any integer N such that 1/N < log10 (1990/1989) ;
note that in (i) , F(a) < log10 (1990/1989 .)
Use (iii) to decide whether there exists another integer M such that 2M = 1989 . . . .

[mathcam]

Whew, that looks like a mess. Do you mean log base 2, maybe? Or maybe a 10^m? I haven't thought about it too carefully yet, but it would seem that you'd want either a 2 in both cases, or a 10 in both cases, whereas you now have one of each.

Cam

[Wacky]

I'm pretty sure it starts off with log base 10 of 2

and the last part has decide whether there exists another integer M such that 2^M = 1989...

sorry about the mess, I'll try and find some link or something.