mm-456
===
Subject: Re: Random Variable Questions
alt.math.undergrad:
Posting 11 questions without comment is not terribly useful.
If you're offering them for general amusement, say so. If
you want help, pick one or two of them, and tell us where
you're stuck. On the assumption that you want help, I'll
give you some pointers on the first one.
> a) A total of 4 buses carrying 148 students from the same
school
> arrives at a football stadium. The buses carry,
respectively, 40, 33,
> 25, and 50 students. One of the students is randomly
selected. Let X
> denote the number of students that were on the bus carrying
this
> randomly selected student. One of the 4 bus drivers is also
randomly
> selected. Let Y denote the number of students on her bus.
> -which of the E[X] or E[Y] do you think is larger? why?
Call the buses A (40 students), B (33 students), C (25
students), and D (50 students). There are 148 students
altogether. Pr(X = 40) is clearly the same as the
probability that the student chosen is from Bus A, which is
40/148 = 10/37; you should be able to work out Pr(X = 33),
Pr(X = 25), and Pr(X = 50) similarly.
On the other hand, Pr(Y = 40) = 1/4; in fact, the four
possible values of Y are equally likely. (Why?)
You should be able to see without too much trouble that
while the four values of Y are equally likely, the large
values of X are more likely than the small ones; why? What
does this tell you about the relative sizes of E[X] and
E[Y]?
> -compute E[X] and E[Y]
Once you complete the calculations of Pr(X = ...) above,
you'll have all of the information that you need to
calculate both of these directly, and the calculations are
trivial.
> -find Var(X) and Var (Y) for X and Y.
This also is just a matter of plugging into a formula once
you've completed the groundwork.
[...]
===
Subject: Re: Greatest integer function
I would like to show f(x)=[x] is continuous on every
noninteger x in
R
, where [x] denote the greatest integer less than or equal to
x.
By definiton of continuoity,
let c in R-Z,
For any e>0,
If there exist d>0 such that |x-c|<d => |[x] - [c]|<e,
Then this imply f(x)=[x] is continuous on every noninteger x
in R
but,
I don't know how do I put above d=?
> Huh? Don't understand. You want value for d?
> d = min(c - [c], [c]+1 - c)
Sorry, I don't understand.
Your d does not have e.
So, I don't know how |x-c|<min(c - [c], [c]+1 - c) => |[x] -
[c]|<e.
If possible, would you please give me more explanation.
===
Subject: Re: Greatest integer function
> I would like to show f(x)=[x] is continuous on every
noninteger x in
R
> , where [x] denote the greatest integer less than or equal
to x.
> By definiton of continuoity,
> let c in R-Z,
> For any e>0,
> If there exist d>0 such that |x-c|<d => |[x] - [c]|<e,
> Then this imply f(x)=[x] is continuous on every noninteger x
in R
> I don't know how do I put above d=?
Huh? Don't understand. You want value for d?
d = min(c - [c], [c]+1 - c)
> Sorry, I don't understand.
> Your d does not have e.
So what, does the constant function f(x) = 0 have 'x'?
> So, I don't know how |x-c|<min(c - [c], [c]+1 - c) => |[x] -
[c]|<e.
> If possible, would you please give me more explanation.
[c] - c <= -min(..) < x - c < min(..) <= [c}+1 - c
===
Subject: Re: advanced calculus
> I am really in need of help, I have to present this math
problem in
> class, and I have no idea how to present it.
> Consider the set {x + 1/x: x is an element of the reals and
x>0} the
> infimum of this set is
> a. 0, b. 1, c. 2, d. Squareroot of 3, e. e.
> I know the correct answer is c. 2, but I just don't know how
to
> present it to the class, if anyone could help I would
appriciate it!
> :)
> Lisa
Note that x + 1/x is clearly large when x is near 0 and
equally large
when 1/x is equally near 0.
so, by symmetry (consider replacing x by 1/x), should have a
minimum
when x and 1/x are equally far from 0.
The result follows immediately.
It can then be validated by taking derivatives of (x + 1/x)
with respect
to x.
===
Subject: Re: advanced calculus
>Consider the set {x + 1/x: x is an element of the reals and
x>0} the
>infimum of this set is
>a. 0, b. 1, c. 2, d. Squareroot of 3, e. e.
>I know the correct answer is c. 2, but I just don't know how
to
>present it to the class, if anyone could help I would
appriciate it!
> Just look at the differeence:
> x+(1/x)-2 = [(x-1)^2]/x
> and this is always nonnegative. Done.
You're not done with 2 <= x + 1/x.
All you've shown is 2 <= inf.
You need one more thing to show inf <= 2.
===
Subject: Slightly Complicated RREF (i think)
I an undergrad student taking Linear Algebera, and I had this
question: (no its not a homework problem... even though it is
related
to a computer project for another class)
Lets define 4 integers a,b,c,d. Now these variables need to
satisfy
these equations:
(a + b + c) + (b + c + d) = 9
(a + b) + (b + c) + (c + d) = 9
a + b + c + d = 7
Also a >= 1, b>=1, c>=1, d>=1
The equations simplify to:
a + 2b + 2c + d = 9
a + 2b + 2c + d = 9
a + b + c + d = 7
a >= 1, b>=1, c>=1, d>=1
We are able to see by observation that 1st and 2nd rows are
multiples
of each other...
So we just take one of them out, and compute the rref of the
following:
1 2 2 1 9
1 1 1 1 7
rref :
1 0 0 1 5
0 1 1 0 2
Notice that I did not include the a,b,c,d >=1... because I
didn't know
how...
In any case I'm totally stuck...Even in parametric form, the
answer
doesn't make any sense.
I know for a fact that the only two answers should be:
4 1
1 1
1 1
1 4
Please advise on how I can include the a,b,c,d >= 1 in
matrix...
===
Subject: Re: Slightly Complicated RREF (i think)
> Lets define 4 integers a,b,c,d. Now these variables need to
satisfy
> these equations:
> (a + b + c) + (b + c + d) = 9
> (a + b) + (b + c) + (c + d) = 9
> a + b + c + d = 7
> Also a >= 1, b>=1, c>=1, d>=1
> The equations simplify to:
> a + 2b + 2c + d = 9
> a + b + c + d = 7
> a >= 1, b>=1, c>=1, d>=1
b + c = 2; b = c = 1
a + d = 5; d = 5 - a; 1 <= a <= 4
===
Subject: More mathematical induction help requested
X-No-Archive: ?yes
Hello folks:
I don't quite know why but I have spent literally hours this
week trying
to understand how to effectively use mathematical induction to
prove
statements. Unfortunately, it has not clicked in 100%. Here's a
problem I'm having. Maybe it's just the confusing way this
book tries
to explain. It's a certain line I'm having a problem with.
This is an
example from the book:
Prove that n < 2^n for all positive integers n
Solution:
1. For n = 1 or 2, the statement is true
2. Assuming that
k < 2^k
you need to show that (k+1) < 2^(k+1)
3. For n=k you have
2^(k+1) > 2(2k) > 2(k) = 2k (by assumption)
Because 2k=k + k > k + 1 for all k > 1, it follows that
2^(k+1) > 2k > k + 1
or
k + 1 < 2^(k + 1)
4. So n < 2^n for all integers >= 1
=============================
Where I get lost is in step 3 where they do the ('by
assumption'):
For n = k
2^(k+1) = 2(2^k) > 2(k) = 2k (by assumption)
I understand (by properties of exponents) that 2^(k+1) = 2x2^k.
But why are they doing for n = k again for the by assumption
step.
Are they simply implying that if 2^(k+1) > k + 1, then 2^(k+1)
> k ????
Also (I'm sorry if this seems to painfully rudimentary) why do
they have
the > 2(k) in the
2(2^k) > 2(k) portion?
I understood that 2(2^k) is the same as 2^(k + 1) but why the
need to
also multiply the other side k by 2 ?? I'm just not seeing the
connection. Please help.
thanks!!
===
Subject: Re: More mathematical induction help requested
alt.math.undergrad:
> Hello folks:
> I don't quite know why but I have spent literally hours this
week trying
> to understand how to effectively use mathematical induction
to prove
> statements. Unfortunately, it has not clicked in 100%.
Here's a
> problem I'm having. Maybe it's just the confusing way this
book tries
> to explain. It's a certain line I'm having a problem with.
This is an
> example from the book:
> Prove that n < 2^n for all positive integers n
> Solution:
> 1. For n = 1 or 2, the statement is true
> 2. Assuming that
> k < 2^k
> you need to show that (k+1) < 2^(k+1)
> 3. For n=k you have
> 2^(k+1) > 2(2k) > 2(k) = 2k (by assumption)
Typo (which I see you corrected below): 2^(k+1) > 2(2^k)
> Because 2k=k + k > k + 1 for all k > 1, it follows that
> 2^(k+1) > 2k > k + 1
> or
> k + 1 < 2^(k + 1)
> 4. So n < 2^n for all integers >= 1
> =============================
> Where I get lost is in step 3 where they do the ('by
assumption'):
> For n = k
> 2^(k+1) = 2(2^k) > 2(k) = 2k (by assumption)
> I understand (by properties of exponents) that 2^(k+1) =
2x2^k.
> But why are they doing for n = k again for the by assumption
step.
> Are they simply implying that if 2^(k+1) > k + 1, then
2^(k+1) > k ????
While I tend to take complaints about textbooks with a grain
of salt, if this is basically a copy of your book's
presentation, then I will agree that it's poorly presented.
The 'For n=k you have' isn't really wanted. To make it
clearer I would write the argument out in a bit more detail,
something like this:
We are assuming that k > 1 and that k < 2^k, and we
want to show that k+1 < 2^(k+1). We know from
ordinary algebra that 2^(k+1) = 2(2^k); by the
induction hypothesis 2^k > k, so 2^(k+1) = 2(2^k) > 2k.
Now you can go on with the rest of the book's argument,
which is much clearer:
> Because 2k=k + k > k + 1 for all k > 1, it follows that
> 2^(k+1) > 2k > k + 1
> or
> k + 1 < 2^(k + 1)
Does that help?
[...]
===
Subject: Re: More mathematical induction help requested
X-No-Archive: yes
> While I tend to take complaints about textbooks with a grain
> of salt, if this is basically a copy of your book's
> presentation, then I will agree that it's poorly presented.
Yes, it's exactly how it was worded :-(
> The 'For n=k you have' isn't really wanted. To make it
> clearer I would write the argument out in a bit more detail,
> something like this:
> We are assuming that k > 1 and that k < 2^k, and we
> want to show that k+1 < 2^(k+1). We know from
> ordinary algebra that 2^(k+1) = 2(2^k); by the
> induction hypothesis 2^k > k, so 2^(k+1) = 2(2^k) > 2k.
> Now you can go on with the rest of the book's argument,
> which is much clearer:
> Does that help?
Hi :
It does help. However, I'm still a little fuzzy about one
thing (I
don't now why it's taking me so long to understand this).
I'm having a difficult time understanding why the induction
hypothesis:
>so 2^(k+1) = 2(2^k) > 2k
would not be:
2(2^k) > (k+1)
rather than:
2(2^k) > 2k
?
I can easily see why the left hand side became 2(2^k).
However, it's
the right hand side that's confusing me here.
I'm still confused about how the right hand side > 2k became >
2k
rather than > (k+1)? Is the right hand side 2k expressed in
terms
of k+1???
thanks again!!!
===
Subject: Re: More mathematical induction help requested
Content-transferncoding: 8bit
> X-No-Archive: yes
While I tend to take complaints about textbooks with a grain
of salt, if this is basically a copy of your book's
presentation, then I will agree that it's poorly presented.
> Yes, it's exactly how it was worded :-(
The 'For n=k you have' isn't really wanted. To make it
clearer I would write the argument out in a bit more detail,
something like this:
We are assuming that k > 1 and that k < 2^k, and we
want to show that k+1 < 2^(k+1). We know from
ordinary algebra that 2^(k+1) = 2(2^k); by the
induction hypothesis 2^k > k, so 2^(k+1) = 2(2^k) > 2k.
Now you can go on with the rest of the book's argument,
which is much clearer:
Does that help?
> Hi :
> It does help. However, I'm still a little fuzzy about one
thing (I
> don't now why it's taking me so long to understand this).
> I'm having a difficult time understanding why the induction
hypothesis:
>so 2^(k+1) = 2(2^k) > 2k
> would not be:
> 2(2^k) > (k+1)
> rather than:
> 2(2^k) > 2k
Maybe this will help.
One form of induction says:
Given staements P(n) where n is an integer >= m,
If P(m) is true and P(k + 1) is true whenever P(k) is true,
then P(n)
is true for all n >= m.
For your problem, P(n) := n < 2^n for n > 0
So, you must show P(1) := 1 < 2^1 is true and
if you assume P(k) := k < 2^k is true then show
P(k + 1) := k + 1 < 2^(k + 1) is also true.
--
Paul Sperry
Columbia, SC (USA)
===
Subject: Re: More mathematical induction help requested
>Hello folks:
>I don't quite know why but I have spent literally hours this
week trying
>to understand how to effectively use mathematical induction
to prove
>statements. Unfortunately, it has not clicked in 100%. Here's
a
>problem I'm having. Maybe it's just the confusing way this
book tries
>to explain. It's a certain line I'm having a problem with.
This is an
>example from the book:
>Prove that n < 2^n for all positive integers n
>Solution:
>1. For n = 1 or 2, the statement is true
>2. Assuming that
> k < 2^k
> you need to show that (k+1) < 2^(k+1)
Complete solution sent to the e-mail address you had the
decency to provide.
Hint: 2^(k+1) = 2^k + 2^k
>3. For n=k you have
> 2^(k+1) > 2(2k) > 2(k) = 2k (by assumption)
> Because 2k=k + k > k + 1 for all k > 1, it follows that
> 2^(k+1) > 2k > k + 1
> or
> k + 1 < 2^(k + 1)
>4. So n < 2^n for all integers >= 1
>=============================
>Where I get lost is in step 3 where they do the ('by
assumption'):
>For n = k
>2^(k+1) = 2(2^k) > 2(k) = 2k (by assumption)
>I understand (by properties of exponents) that 2^(k+1) =
2x2^k.
>But why are they doing for n = k again for the by assumption
step.
>Are they simply implying that if 2^(k+1) > k + 1, then
2^(k+1) > k ????
>Also (I'm sorry if this seems to painfully rudimentary) why
do they have
>the > 2(k) in the
>2(2^k) > 2(k) portion?
>I understood that 2(2^k) is the same as 2^(k + 1) but why the
need to
>also multiply the other side k by 2 ?? I'm just not seeing the
>connection. Please help.
>thanks!!
===
Subject: Re: More mathematical induction help requested (small
typo error)
X-No-Archive: yes
> 3. For n=k you have
> 2^(k+1) > 2(2k) > 2(k) = 2k (by assumption)
That meant to say:
2^(k+1) = 2(2k) > 2(k) = 2k (by assumption)
Please help.
thanks...
===
Subject: Re: Beware undergrads, copiers of your posts
Innocent question: Is it unethical to use someone else's work,
and
expand upon it, or revise it, if due credit is given?
I'd think if he'd attempt to publish it, or otherwise profit
from it,
that'd be unethical, but to merely point at flaws and correct
them has
been happening in the sciences for ages. No one's perfect.
===
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