Post by: #1procrastinator on December 27, 2012, 03:20:36 am
The inequality signs should be 'or equal to' .
Some hints on where to start be good. im not sure i can start with the second inequality cause of the whole P implies Q doesnt necessarily mean Q implies P thing
Post by: FlorianK on December 27, 2012, 03:47:44 am
Show that |x|<2 => |(x^2+2x+7)/(x^2+1)|<15Do you mean
The inequality signs should be 'or equal to' .
Some hints on where to start be good. im not sure i can start with the second inequality cause of the whole P implies Q doesnt necessarily mean Q implies P thing
Show that if the modulus of x is greater or equal to 2, then the modulus of
Post by: b^3 on December 27, 2012, 03:59:51 am
Since
When we do actually divide by 1, that is when
So as
Anyways, probably not the solution, but it may give you some ideas, so hope it helps.
Post by: #1procrastinator on December 27, 2012, 04:52:36 am
@Indian Battman: thanks, i wouldnt have thought of that. im looking for a method that doesnt use calculus cause this is actually in the preliminaries section of the text lol
Post by: FlorianK on December 27, 2012, 05:28:45 am
The whole function is always smaller than 15 for all values of x ~_~
Post by: #1procrastinator on December 27, 2012, 07:00:14 am
but surely there's another way of proving it
Post by: polar on December 27, 2012, 11:05:57 am
Spoiler
the converse doesn't need to be true for that statement to the true either, it just means you can't put an if and only if sign between them.
Post by: dcc on December 27, 2012, 11:40:07 am
edit: using this, we recover where the mysterious "15" comes from:
Post by: BubbleWrapMan on December 27, 2012, 02:39:29 pm
For all real
Also
If
Hence, whatever the question was
Post by: #1procrastinator on December 28, 2012, 04:18:18 am
@polar: why do you set the expression to 7 and solve for x? im a bit unclear on the.motivation and how you chose that number
@dcc: how did you choose 10 and how do you the inequality is true?
@ClimbTooHigh: think i understand your method, similar to ive seen before :p
Post by: #1procrastinator on December 30, 2012, 06:07:33 am
edit: this is what i got:
which disagrees with what the ti89 gave. even though the calc tends to give different forms of what you usually get if you do it by hand, im not all that confident with my derivation
Post by: FlorianK on December 30, 2012, 07:11:22 am
The answer looks a bit too crazy:
especially because the derivative of arcsin(2x-1) is
Post by: #1procrastinator on December 30, 2012, 07:19:05 am
im not sure if you are meant to integrate but i thought it would be interesting to try to do so
Post by: FlorianK on December 30, 2012, 07:29:06 am
Post by: FlorianK on December 30, 2012, 07:42:11 am
where (a,b) is the point of the center and r is the radius.
So we need to show that
Post by: #1procrastinator on December 30, 2012, 07:54:57 am
any ideas on how to do the nasty integral? one of my mistakes was assuming |sin(x)cos(x)|sin(x)cos(x) is the same as sin(x)^2*cos(x)^2...that's the last thing i integrated before changing variables and getting the above erroneous answer
Post by: FlorianK on December 30, 2012, 08:04:52 am
That confuses me
Post by: #1procrastinator on December 30, 2012, 08:20:08 am
edit: which i think is equivalent to what you got
Post by: b^3 on December 30, 2012, 02:53:01 pm
thanks FlorianKThat took a while, hopefully there are not errors in there...
any ideas on how to do the nasty integral? one of my mistakes was assuming |sin(x)cos(x)|sin(x)cos(x) is the same as sin(x)^2*cos(x)^2...that's the last thing i integrated before changing variables and getting the above erroneous answer
Anyways, the train of thought, try to work on the
There will be domain restrictions that arise when you manipulate the triangle aswell, but anyways... after typing all that out... Anyways, hope it helps :)
Also had a feeling that hyperbolic functions may have helped at one stage... but looking back it it maybe not.
Post by: dcc on December 31, 2012, 09:32:32 pm
@dcc: how did you choose 10 and how do you the inequality is true?
triangle inequality gives you
Post by: #1procrastinator on January 16, 2013, 03:08:45 pm
triangle inequality gives you
Thanks dcc...haven't thought about that one in a while >.<
------
Evaluate the line intgral
I got 2/5, book gives 1638.4 which doesn't look right...
Post by: polar on January 16, 2013, 03:43:00 pm
parametrise it:
x=4cos(t), dx/dt = -4sin(t) for -pi/2 ≤ t ≤ pi/2
and, y=4sin(t), dy/dt = 4cos(t) for -pi/2 ≤ t ≤ pi/2
since, ds = sqrt((dx/dt)^2 + (dy/dt)^2)dt = sqrt(4^2(cos^2(t)+sin^2(t))dt = 4dt
int_c xy^4 ds = int_(-pi/2)^(pi/2) 4(4cos(t))(4sin(t))^4 dt = 1638.4
Post by: #1procrastinator on January 16, 2013, 03:56:27 pm
Post by: #1procrastinator on January 25, 2013, 03:35:01 pm
Find the range of k for which the inequality
Not sure how to find k if cos(x) is an element of [-1, 1]
Post by: BubbleWrapMan on January 25, 2013, 06:48:51 pm
cos^2(x)+cos(x) has a maximum value of 2, so you can have k = -0.5 (which would give you -0.5 x 2 + 1 = 0). If k is greater than this you'll get a positive number, e.g if k = -0.25 you'll have -0.25 x 2 + 1 = 0.5. So we need to have -0.5 ≤ k. There is also an upper bound on k, though, which we need to find.
cos^2(x)+cos(x) reaches a minimum of -1/4, so you can have k = 4 (this would give you 4 x -0.25 + 1 = 0). If k is greater than 4 you will get a negative number, but if it is less than 4 you'll get a positive number, so it holds for k < 4 as well.
So, the range of values of k is [-0.5, 4]
Post by: #1procrastinator on January 26, 2013, 06:30:15 pm
Post by: #1procrastinator on January 30, 2013, 03:42:48 am
This comes from trying to evaluate the improper integral
EDIT: fixed latex
Post by: b^3 on January 30, 2013, 04:57:46 pm
We can rewrite what we have as
Then applying the rule
Now as
Hope I haven't stuffed up, may want to get a more seniour member to check it over.
Post by: #1procrastinator on January 31, 2013, 04:50:48 pm
How would you evaluate
Post by: #1procrastinator on February 04, 2013, 10:17:29 am
Am I supposed to try and get into the form of a geometric series or something? The second term starts on e^(1/2) so the first term of the answer would be e cause there's no e (then the rest cancel), but not sure how the 1 comes in.
(solution is e-1)
Post by: QuantumJG on February 04, 2013, 02:36:46 pm
Let
Then
Claim:
Proof:
N=1
Now assume this holds for N, want to show that
Now
Therefore
So
Post by: #1procrastinator on February 14, 2013, 12:30:20 pm
For your claim and proof, are you proving that
is true?
Cause I think you used that in your proof in this part:
(looks like you subbed in
Post by: QuantumJG on February 14, 2013, 01:33:42 pm
Sorry for the late reply, took me a while to (and attempt to) digest that :P
For your claim and proof, are you proving that
is true?
Cause I think you used that in your proof in this part:
(looks like you subbed in)
What I did is called proof by induction. Proof by induction has three steps:
(i) Show that what you're trying to prove holds for the most basic case. This is called the base "case".
(ii) Assume that what you're trying to prove holds for all $n\le N$ where $N$ is some integer.
(iii) Using the assumption you made in (ii), show that it holds for $N+1$. If you can show that, then you have shown it holds for any value of $n$, since I can arbitrarily make $N$ as large as I want. It's a real, useful and fun proof technique. Specialist maths should include some of these proof techniques (including proof by contrapositive).
Post by: #1procrastinator on February 14, 2013, 09:02:45 pm
Claim:
So that's what we want to prove right?
Proof:
N=1
^ That's the base case?
Now assume this holds for N, want to show that
^ So if we can show that the above is true, then we can show what we're trying to prove is true?
Now
^ The
This is where I get confused cause it I'm not sure what we're proving now. It looks like you put in
But isn't that what we're trying to prove?
Post by: Jeggz on February 14, 2013, 09:59:29 pm
What I did is called proof by induction. Proof by induction has three steps:
(i) Show that what you're trying to prove holds for the most basic case. This is called the base "case".
(ii) Assume that what you're trying to prove holds for all $n\le N$ where $N$ is some integer.
(iii) Using the assumption you made in (ii), show that it holds for $N+1$. If you can show that, then you have shown it holds for any value of $n$, since I can arbitrarily make $N$ as large as I want. It's a real, useful and fun proof technique. Specialist maths should include some of these proof techniques (including proof by contrapositive).
Thanks alot for that Quantum!
I needed help with that for uni maths :)
Post by: kamil9876 on February 16, 2013, 12:30:51 pm
In general notice that
Post by: #1procrastinator on February 16, 2013, 02:52:38 pm
In general it is an idea called telescoping (funnily a lecturer once used this to show that maths has applications to astronomy :P )
In general notice thatwhere
is any sequence. To see this just reorder the terms as follows
Is that what QuantumJG was proving? (the claim part)
Post by: kamil9876 on February 16, 2013, 11:26:43 pm
Post by: #1procrastinator on February 16, 2013, 11:59:09 pm
Cause to me, it looks like he used what he was trying to prove
Post by: Jeggz on February 17, 2013, 11:16:30 am
but can someone please help me with these two questions?
(1) Prove that the cube of an odd integer is odd.
(2) Prove that if n^2 is divisible by three then n is also divisible be 3. (For this question..I tried to do it the same way as if you were to do it for divisible by 2..but it didn't work :( )
Thanks in advance!
Post by: polar on February 17, 2013, 11:32:03 am
Spoiler
the first three terms must be even (or 0 when k=0), hence, adding 1 (an odd number) makes it's odd. hence, the cube of an odd integer is odd
Post by: Jeggz on February 17, 2013, 11:37:38 am
1. hint: write an odd integer as 2k+1Spoiler
the first three terms must be even (or 0 when k=0), hence, adding 1 (an odd number) makes it's odd. hence, the cube of an odd integer is odd
Ohh I get it now!!
Thanks for that :)
Post by: kamil9876 on February 17, 2013, 12:36:56 pm
Hmmm...I'm still confused by this part:
Cause to me, it looks like he used what he was trying to prove
He used induction, so yes he showed that IF it is true for
This is the general scheme for induction:
Show the following:
(1) True for N=1
(2) If true for N=k then true for N=k+1
(1) and (2) imply that it is true for all N as follows: Since by (1) it is true for N=1, then using (2) we get true for N=2. Then using (2) again we show it is true for N=3 etc... i.e it's like a domino thing.
Post by: Jeggz on February 17, 2013, 03:38:32 pm
1. hint: write an odd integer as 2k+1Spoiler
the first three terms must be even (or 0 when k=0), hence, adding 1 (an odd number) makes it's odd. hence, the cube of an odd integer is odd
sorry again polar, but how would you say i communicate to my teacher that the first three terms must always be even..i understand it, but i don't know how to express it?
Post by: polar on February 17, 2013, 03:56:21 pm
sorry again polar, but how would you say i communicate to my teacher that the first three terms must always be even..i understand it, but i don't know how to express it?
an even number can be written in the form
Post by: #1procrastinator on February 17, 2013, 07:32:48 pm
Yeah that's right, he proved exactly that by induction, usually the cleanest way of expressing it but not necessarily the most intuitive.
Ok, in the proof, the claim was for
an even number can be written in the form, if we can write each of first three terms in this form, then they are even:
Would you have to show that the sum of the even numbers is indeed even?
Post by: polar on February 17, 2013, 07:39:59 pm
Post by: QuantumJG on February 18, 2013, 10:12:22 am
Hi #1procrastinator, maybe I should have used some better notation. I claimed that for any
So after doing the base case, I assume that my claim holds for some
Now I use my assumption and get
So now I've shown it holds for
Overall, this is just telescoping a series. In general (sometimes - if it's an infinite series, you should first see if it converges by doing convergence tests) for some sequence
Post by: kamil9876 on February 18, 2013, 11:14:39 am
Ok, in the proof, the claim was forbut then we wanted to show that it was true for
, but how can we assume
Read my most recent post for an explanation of what induction is, or just google "mathematical induction" to see what it is.
Post by: #1procrastinator on February 23, 2013, 01:43:42 am
Also, QuantumJG, in your last step:
Did you really have to take the limit into the exponent or could you just have said that
-----------------------------------------
1) What does 'coordinatewise addition' mean (e.g. 'Define addition of elements in V coordinatewise)? Does it just mean adding the corresponding components of two lists or vectors? So (a, b) + (c, d) = (a+c, b+d)
-----
2) To prove
This is the way the author does it:
---
3) To prove
----
4) How would you write the set of all even functions using set notation?
Post by: kamil9876 on February 23, 2013, 11:46:42 am
Quote
Did you really have to take the limit into the exponent or could you just...
It really depends how much detail you want. Are you satisfied? Or maybe more importantly sometimes the question is are your bosses satisfied. If it was a 1st year course in calculus and you just covered continuity of exponential yesterday, then yeah I would. If however you were writing a research paper then you shouldn't, it would be a waste of paper.
Quote
1) What does 'coordinatewise addition' mean
Yeah it means do it on each coordinate. Again this is slang and if you are ever unsure about slang, look ahead and try to figure out the right definition from the context/examples etc.
Quote
2) To provewould this acceptable?
This depends entirely on your definition of
Quote
3) To prove \frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bc}, he starts with the RHS and shows that it's equal to the LHS. Is this an example of a case where P<->Q? Generally, how would you determine if P<->Q or just P->Q?
I don't understand your concern. Which statements are P and Q supposed to represent in this particular example? As an exercise you may want to prove this identity using the definition of fractions I gave above.
Quote
4) How would you write the set of all even functions using set notation?
How would I write the set of all things that satisfy property P?
sometimes you can use : instead of | if it looks better (for example, when there are absolute values involved in my definitions I obviously prefer to use : )
A more economical and professional way of writing it would be:
Post by: #1procrastinator on February 23, 2013, 05:49:59 pm
This depends entirely on your definition of[/tex]
From the book ('Calculus' by Spivak), he says 'the symbol a/b means ab^-1.
I don't understand your concern. Which statements are P and Q supposed to represent in this particular example? As an exercise you may want to prove this identity using the definition of fractions I gave above.
I was thinking P would be on the LHS and Q would be the RHS but there's a damn equality sign there!
Post by: kamil9876 on February 23, 2013, 08:32:28 pm
Quote
From the book ('Calculus' by Spivak), he says 'the symbol a/b means ab^-1.
Hrmm ok, then what are
So in that case proving
Post by: #1procrastinator on February 23, 2013, 09:51:24 pm
why do we have to write
Could you please point what is wrong with this way:
EDIT:
And also to prove that if
Post by: kamil9876 on February 24, 2013, 06:08:36 pm
Quote
why do we have to write a/b as ab^{-1}?
you told me that was the definition! so I used that definition, if you had given me another definition I would have used that definition. For such simple facts, might as well stick to the definitions and everything will work out.
[1] There are many paths one may take to develop the basics of the reals, I don't know which order Spivak goes in.
Post by: TrueTears on February 24, 2013, 06:11:23 pm
Post by: #1procrastinator on February 25, 2013, 03:37:46 am
Look, when it comes to these sort of foundational things what counts as a proof and what doesn't largely depends on your definitions and what you already know1 (i.e what facts/theorems are you allowed to use). So as for your second query, it depends... sure it is a good enough proof if you already know that the real numbers have multiplicative inverses.
Well, I guess if I'm showing the proof to someone, I'm assuming they're familiar with the axioms I'm working with too :p
you told me that was the definition! so I used that definition, if you had given me another definition I would have used that definition. For such simple facts, might as well stick to the definitions and everything will work out.
[1] There are many paths one may take to develop the basics of the reals, I don't know which order Spivak goes in.
Haha, he mentions it almost in passing, saying that a/b always means ab^(-1) so I thought we could use either one.
you should link arun's site about proofs here lol
This?
http://www.ms.unimelb.edu.au/~ram/Notes/grammarContent.xhtml
Post by: TrueTears on February 25, 2013, 05:27:58 pm
Post by: #1procrastinator on February 26, 2013, 12:07:43 pm
Not sure how to show this, some pointers would be nice :p
Post by: kamil9876 on February 26, 2013, 10:45:06 pm
It may be efficient to plug in values that give zeros. For example:
Now just need one more equation, I'll leave you to find this.
Alternatively, a more systematic way would be to plug in their Taylor series (if you already know what they are) to get some equations by comparing coefficients.
Post by: Jeggz on March 03, 2013, 01:01:55 pm
My teacher said something about real part = (real x real) - (imaginary x imaginary)? But I'm a bit confused..
Post by: b^3 on March 03, 2013, 01:27:57 pm
Do you mean something like
In that case lets look at what happens when we expand it out.
Now we know that
Which is what you're teacher was getting at with real times real minus imaginary times imaginary.
Post by: Jeggz on March 03, 2013, 01:42:54 pm
What do you mean by not in standard form?
Do you mean something like?
In that case lets look at what happens when we expand it out.
Now we know that, which brings us to
Which is what you're teacher was getting at with real times real minus imaginary times imaginary.
Sorry I should have been more specific! What I meant was when we're dealing with things like differentiation and antidifferentiation of exponentials. For examples showing how e^ax sin(bx) = (e^ax / a^2+b^2 ) (asin(bx) - bcos(bx)) ? Im really sorry if this question doesn't make sense.. I don't know how to use latex?!
Post by: b^3 on March 03, 2013, 02:04:30 pm
Now we notice that we can turn our expression into something involving a complex exponent.
Now we know that
Now if we wanted to take the imaginary part of our expression, how would we find it? What multiplied by what will give an imaginary part?
Real
Imaginary
Real
Imaginary
So to find the imaginary part we need to look at the latter two.
If we wanted to take the Real part, when integrating a cosine, we would need to do the first two, that is the Real
Hope that makes sense :)
Post by: Jeggz on March 03, 2013, 02:13:30 pm
I'll start with the sine case, (if we are integrating sine we will use Im part and if we are integrating cosine we will use the Re part, you'll see why later).
Now we notice that we can turn our expression into something involving a complex exponent.
Now we know that
Now if we wanted to take the imaginary part of our expression, how would we find it? What multiplied by what will give an imaginary part?
RealReal gives Real
Imaginaryterms comes out to give a negative)
Real
Imaginary
So to find the imaginary part we need to look at the latter two.
If we wanted to take the Real part, when integrating a cosine, we would need to do the first two, that is the Real
Hope that makes sense :)
Thankyou (x million!)
It totally make sense to me now :)
Post by: #1procrastinator on March 12, 2013, 12:43:15 pm
Thanks
Post by: Lasercookie on March 12, 2013, 01:23:02 pm
Let a = rational number, b = irrational number
So aiming for a proof by contradiction, suppose that
Subtract a, a rational number from both sides
Since a and p are rational numbers, we can express them as ratios of integers
Since this is still a ratio of integers (product of integers are integers, sum of integers are integers - not sure how to prove this, but it seems obvious that it'd be true) this is still rational.
So that means
Edit: Figured out the other one, it was easier than I thought/fairly similar to the one above.
a = rational, b = irrational
So we can show that
Hopefully I haven't done anything iffy.
Post by: kamil9876 on March 12, 2013, 08:07:11 pm
How would you prove the product of a rational number and an irrational number is irrational?
Thanks
FALSE
:P That's the only exception btw, gotta make sure that the rational number you are multiplying by is non-zero.
Post by: #1procrastinator on March 13, 2013, 06:00:22 am
wtf son oO
thanks :p
Post by: #1procrastinator on March 16, 2013, 10:16:17 am
The point of that was to then show that the set of polynomials under the operations of addition and scalar multiplication is a vector space.
My question is how does defining
Post by: #1procrastinator on March 18, 2013, 06:22:45 am
Note that this is an assignment question so I'm only after some hints :p
Post by: kamil9876 on March 18, 2013, 11:40:11 am
My question is how does defininglet us write g(x) in terms of n's rather than m's?
It's a bit too pedantic. I guess what the author must have been worried about is that
How would you prove the sequenceconverges to 0? I'm supposing you need to use the epsilons and deltas but not really sure how to manipulate that inequality to get find N.
Note that this is an assignment question so I'm only after some hints :p
Seems quite weak, both terms go to zero so you can use linearity of limits.
Post by: #1procrastinator on March 19, 2013, 12:28:44 pm
---
I think I'm supposed to prove that it goes to zero ('show that the sequence converges to zero by arguing directly from the definition of convergence'). Worth 3 marks :P
Post by: Lasercookie on March 19, 2013, 12:49:12 pm
I think I'm supposed to prove that it goes to zero ('show that the sequence converges to zero by arguing directly from the definition of convergence'). Worth 3 marks :PI'm assuming that you're just trying to make sure your answer was correct, but just to double check but you did submit the assignment yesterday right?
But yeah the way I approached it was to split it into two parts, and then showed that each part separately converges to zero (using the definition of convergence stuff), and since limits are linear then a_n converges to zero.
The assignment solutions (they're on wattle now) start off similar, but then use the triangle inequality to say that
Post by: #1procrastinator on March 20, 2013, 04:47:16 pm
I'm assuming that you're just trying to make sure your answer was correct, but just to double check but you did submit the assignment yesterday right?:P :-[ :P :-[
I should stop trying to do the assignments on the day they're due haha.
Thanks...I think I'm gonna spend a bit more time trying to do it in a way I understand first but I've got so much other stuff to do. MUST STOP PROCRASTINATING
Post by: FlorianK on March 24, 2013, 09:27:03 am
Post by: kamil9876 on March 24, 2013, 12:27:25 pm
Which is equivalent to:
But then since
Post by: FlorianK on March 26, 2013, 06:39:10 am
got another one :D
Let z be the smallest possible number of a*b*c. a, b and c are all integers and a2=2b3=3c5.
By how many integers is z divisible including 1 and z?
Post by: kamil9876 on March 26, 2013, 01:05:25 pm
Is
Post by: kamil9876 on March 26, 2013, 10:41:55 pm
Post by: FlorianK on March 27, 2013, 06:30:55 am
z=a*b*c
a,b, and c are positive integers
a² = 2b³ = 3c5
By how many numbers is z divisible including 1 and z.
The answer is 60, but I'm not sure why :/
Post by: Deleted User on March 27, 2013, 10:51:26 am
1+0=1 which is not 0
Now before you judge me, I was half-awake when I posted that question. :P
Post by: Deleted User on March 27, 2013, 09:34:53 pm
Post by: kamil9876 on March 28, 2013, 12:08:06 am
nah no translation error :/
z=a*b*c
a,b, and c are positive integers
a² = 2b³ = 3c5
By how many numbers is z divisible including 1 and z.
The answer is 60, but I'm not sure why :/
Well it seems that the answer could vary as there are many possibilities for
Problem: Let
Now why don't we just find
It is clear that
So we we really need to solve is the following system in the non-negative integers:
So what are the smallest possible non-negative integer solutions to these (i.e such that
As for the second one, we need s to be a multiple of
So in fact:
Now the number of divisors is
(i.e the number of divisors of
This seems like a very contrived problem, where is it from?
Post by: FlorianK on March 28, 2013, 06:06:23 am
77 was an answer as well. Maybe it was just an error. Thx :D
So or my understanding:
If we have n=2x * 3y * 5z the number of divisors is (x+1)*(y+1)*(z+1) ?
From which 'part' of math did you learn this? discrete mathematics?
Post by: kamil9876 on March 28, 2013, 11:43:52 am
Quote
So or my understanding:
If we have n=2x * 3y * 5z the number of divisors is (x+1)*(y+1)*(z+1) ?
From which 'part' of math did you learn this? discrete mathematics?
Yeah it could count(pun!) as discrete mathematics I guess. I guess it is mentioned in some elementary number theory text books. That's really how the proof works, by number theory. A divisor of
Of course this needs some justification, (how do you know every divisor is of such a form? how do you know different choices of f_k give different numbers (i.e no repetition) ?) All of this can be justified with unique prime factorization in the integers, which can be found in any elementary course/book/free website on number theory I guess.
Post by: FlorianK on April 01, 2013, 07:39:45 am
Let Z be the amount of 8 digit numbers of which all 8 digits are different from each other and from 0.
How many of those numbers are divisible by 9?
Z/8 | Z/3 | Z/9 | 8Z/9 | 7Z/8 |
and
a0=1, a1=2, an+2=an+ a²n+1 for n>=0
What is a2009 mod 7 ?
and
You have a 3*3 square (like a 9th of a soduku). In the middle, right field is a 47 and in the middle, bottom field is a 63. The sum of the numbers in each row, collum and diagonal is the same.
Which number is in the top, left field?
and
this one if probs like the one in the post where 77 was the answer:
z has 2 digits
t(z) is the number that is created when you put each divisor of z in a row including 1 and z
So t(14)=12714
How many digits has the largest value of t(z)?
Post by: kamil9876 on April 01, 2013, 02:10:17 pm
Quote
Let Z be the amount of 8 digit numbers of which all 8 digits are different from each other and from 0.
How many of those numbers are divisible by 9?
Z/8 | Z/3 | Z/9 | 8Z/9 | 7Z/8 |
There is a criterion: a number is divisible by 9 if and only if the sum of its digits is divisible by 9.
Hence we wish to find the number of 8 element sequences of distinct numbers in
Post by: Deleted User on April 03, 2013, 11:17:36 am
2^n > n^3
Post by: kamil9876 on April 03, 2013, 01:09:17 pm
You can verify it for small positive integers manually (these will be your base case). Then the following inductive step works whenever
So the point is: Find
Post by: Deleted User on April 03, 2013, 02:46:20 pm
If we subbed in n= k+1 into n^3 < 2^n, how would we get to what you wrote?
Post by: brightsky on April 03, 2013, 03:04:28 pm
Post by: Deleted User on April 03, 2013, 05:13:38 pm
Post by: brightsky on April 03, 2013, 05:42:48 pm
check P(10): 2^10 > 10^3 or 1024 > 1000 is true
now assume P(k) is true, where k >= 10. we now need to prove that if this assumption is made, then P(k+1) is also true.
P(k): 2^k > k^3
P(k+1): 2^(k+1) > (k+1)^3 = k^3 + 3k^2 + 3k + 1
LHS = 2^(k+1)
=2*2^k
>2*k^3 (from inductive hypothesis)
= k^3 + k^3
>k^3 + 3k^2 + 3k + 1 (prove later that k^3 > 3k^2 + 3k + 1 for k>= 10)
=(k+1)^3
if all's well, then P(k+1) is also true, which means, according to the principle of maths induction, P(n) is true.
now we need to prove that k^3 > 3k^2 + 3k + 1, for k>= 10
now since k>= 10
k^3 = k*k^2 >= 10k^2 = 3k^2 + 3k^2 + 4k^2 which is obviously greater than 3k^2 + 3k + 1
QED
Post by: kamil9876 on April 03, 2013, 06:44:16 pm
Post by: #1procrastinator on April 06, 2013, 10:54:47 pm
Post by: brightsky on April 06, 2013, 11:14:38 pm
Post by: #1procrastinator on April 07, 2013, 01:58:51 am
Post by: kamil9876 on April 07, 2013, 01:12:02 pm
How would you show that when you rotate a square, the angles the edges make with the horizontal and vertical are equal? (i'm just assuming it's true :p)
Can you be a bit more specific? Is the rotation by any angle? Which horizontal and vertical are we talking about? And which angle specifically?
Post by: #1procrastinator on April 07, 2013, 06:23:12 pm
Can you be a bit more specific? Is the rotation by any angle? Which horizontal and vertical are we talking about? And which angle specifically?
i suppose so. by horizontal and vertical lines i mean the edges of the square you draw around it after rotation
Post by: Deleted User on April 10, 2013, 08:37:33 pm
Post by: brightsky on April 10, 2013, 09:25:22 pm
= d^4/dt^4 (e^(-t) Re(e^(2it))
= Re [d^4/dt^4 (e^(-t)*e^(2it))]
= Re [d^4/dt^4 (e^(t(-1 + 2i)))]
= Re [(-1 + 2i)^4*e^(-t)*e^(2it)]
= e^(-t) Re [(-7+24i) (cos(2t) + i*sin(2t))]
= e^(-t) [-7cos(2t)-24sin(2t)]
Post by: Jeggz on April 10, 2013, 09:41:23 pm
How do I do d^4/dt^4 (e^(-t) cos(2t))? Thank you
Edit: Majorly Beaten :P
Post by: Deleted User on April 11, 2013, 08:46:02 pm
Re(1+sqrt(3)i)^51 can be written as -2^51
and
Im(1+i)^75 can be written as 2^37?
Post by: polar on April 11, 2013, 08:56:33 pm
Thanks guys. Also, can someone explain to me how
Re(1+sqrt(3)i)^51 can be written as -2^51
and
Im(1+i)^75 can be written as 2^37?
hint: use de moivre's theorem
Post by: #1procrastinator on April 27, 2013, 08:19:04 pm
Using Cauchy's Mean Value Theorem, prove that
Don't know where to start with this one...right-hand side looks like the first few terms of the Taylor series but I don't think that's relevant here :P
Post by: kamil9876 on April 29, 2013, 10:27:37 pm
I know that probably doesn't help (or maybe it does?).
Post by: #1procrastinator on April 30, 2013, 12:46:55 am
We haven't covered Taylor series yet. I think the only way to do for now is to actually use the hint which suggests to consider F(x)=f(x) - f(c) - f'(c)(x-c) and G(x) = (x-c)^2, but I wanted to see if it could be done (and how) without that.
Post by: kamil9876 on April 30, 2013, 11:46:55 am
Post by: Deleted User on May 02, 2013, 09:01:25 pm
Find the orthogonal projection of (x,y,z) onto the subspace of R^3 spanned by the vectors (1,2,2),(-2,2-1).
I got (5x-2y+4z,-2x+8y+2z,4x+2y+5z), but in the answers there's a 1/9 multiplied to my answer. Maybe I used the wrong formula?
Post by: kamil9876 on May 02, 2013, 11:42:22 pm
Seeing that you actually GOT an answer (and presumably not the right one) you should actually show us WHAT YOU DID so that we know whether you "used the wrong formula".
Post by: #1procrastinator on May 03, 2013, 12:24:24 am
Post by: Deleted User on May 04, 2013, 08:55:28 pm
I havn't done the calculation myself, but I notice that the two vectors are of norm. So perhaps you forgot to divide out by the norm.
Seeing that you actually GOT an answer (and presumably not the right one) you should actually show us WHAT YOU DID so that we know whether you "used the wrong formula".
I don't know how to write maths on my computer.
Is this the right formula? projw(u) = <v,u1>u1 + <v,u2>u2 + ... + <v,uk>uk where {u1,u2,...,u3} is an orthonormal basis for W.
Post by: kamil9876 on May 04, 2013, 10:26:18 pm
Post by: #1procrastinator on May 05, 2013, 06:55:07 pm
If A is a 50x50 matrix then shouldn't it be invertible?
------
2) If A is an nxn matrix such that
It was suggested I use the geometric series but I haven't learnt that yet so I'm hoping there's an alternative method.
------
3) Define [itex]e^A=I_n+A+\frac{1}{2!}A^2+\frac{1}{3!}A^3+...+\frac{1}{2012!}A2012[/itex]
where A is an nxn matrix such that [itex]A^{2013}=0[/itex]. Show that [itex]e^A[/itex] is invertible and find an expression for [itex](e^A)^{-1}[/itex] in terms of A.
...'bout to attempt this one again but I just know I'm not gonna get far...so here it is...
------
3) Show that
For the first part, I did used the previous result
I'm not sure how to do the limit part though.
Post by: kamil9876 on May 06, 2013, 08:31:09 pm
2)
Quote
It was suggested I use the geometric series but I haven't learnt that yet so I'm hoping there's an alternative method.
Would proving the geometric series identity without using the phrase "geometric series" count as an "alternative" method? Geometric series is just polynomial factorization really. The polynomial
3)
Quote
What's itex? Btw is this
Post by: Deleted User on May 06, 2013, 09:07:34 pm
And by [y z; -x 0] I mean
[y z]
[-x 0]
Post by: humph on May 07, 2013, 12:05:08 pm
1) Let A be a 50x49 matrix and B be a 49x50 matrix. Show that the matrix AB is not invertible.Bahahaha. Are these MATH1115 questions? I remember teaching these two years ago :P (the only difference was that 2013 was replaced by 2011, for obvious reasons...).
If A is a 50x50 matrix then shouldn't it be invertible?
------
2) If A is an nxn matrix such that, show that
is invertible and find an expression for [itex](A+I_n)^{-1}[/itex]
It was suggested I use the geometric series but I haven't learnt that yet so I'm hoping there's an alternative method.
------
3) Define [itex]e^A=I_n+A+\frac{1}{2!}A^2+\frac{1}{3!}A^3+...+\frac{1}{2012!}A2012[/itex]
where A is an nxn matrix such that [itex]A^{2013}=0[/itex]. Show that [itex]e^A[/itex] is invertible and find an expression for [itex](e^A)^{-1}[/itex] in terms of A.
...'bout to attempt this one again but I just know I'm not gonna get far...so here it is...
------
3) Show thatand use this to show that the limit as
approaches infinity is infinity. n is any positive integer.
For the first part, I did used the previous resultand inserted an 'n' with no justification of where I put it and then multiplied by
. Is this valid?
I'm not sure how to do the limit part though.
For what it's worth, most people got these wrong (though that may have been my fault for talking about exponentials and logarithms - people tried to think about those without realising that they are quite different notions when defined for matrices).
Post by: Deleted User on May 14, 2013, 10:36:29 pm
2 -3 6
0 5 -6
0 1 0
I got 2,2,3,5 but the answer only accepts 2,2,3 and not 5.
Post by: b^3 on May 14, 2013, 10:41:49 pm
(expanded down the first column)
Not sure where you got the
Post by: #1procrastinator on June 20, 2013, 08:26:14 am
Post by: Leip on June 20, 2013, 11:40:49 am
If you are using the argument that
In any case, I would start by taking the x in the denominator outside the limit. From there, if you want to show things more explicitly, maybe you could consider applying l'Hopitals rule and induct (i.e. show that upon repetition of l'Hopitals rule the numerator becomes constant in t).
Post by: #1procrastinator on June 20, 2013, 03:17:52 pm
I guess it is valid, but what argument are you using to show that the denominator goes to infinty?
Bah, that gets me no where then.
If you are using the argument thatis exponential in t and
is polynomial in t (monomial even), couldn't you do that without rewriting it?
In any case, I would start by taking the x in the denominator outside the limit. From there, if you want to show things more explicitly, maybe you could consider applying l'Hopitals rule and induct (i.e. show that upon repetition of l'Hopitals rule the numerator becomes constant in t).
I was a bit hesitant in using l'Hopital's rule cause I thought that came after the chapter I'm on but turns out it was a couple of chapters back :p
So we would apply the rule x number of times until the numerator goes to 1 (taking out the constants of course) and then have 1/e^t, which would go to 0
Post by: kamil9876 on June 20, 2013, 06:38:53 pm
Post by: Leip on June 20, 2013, 09:18:37 pm
I was a bit hesitant in using l'Hopital's rule cause I thought that came after the chapter I'm on but turns out it was a couple of chapters back :p
So we would apply the rule x number of times until the numerator goes to 1 (taking out the constants of course) and then have 1/e^t, which would go to 0
L'Hopital's is so often useful, you just have to be careful to remember what it does so that you don't overuse it. I like to think of it as comparing similar order terms in the Taylor expansion of both the numerator and denominator. Sometimes it is better to use the Taylor expansions directly.
But it's fairly straight forward in this case, and you can even leave the constants in since they work out nicely to x!.
Btw, I sure hope x is a positive integer! :)
Post by: #1procrastinator on June 21, 2013, 02:26:27 pm
@kamil: nah, it's from integrating the gamma function . am not up to power series yet
Post by: #1procrastinator on July 06, 2013, 07:10:09 am
Edit fixed missing power
Post by: kamil9876 on July 06, 2013, 05:51:25 pm
If it was true it would imply that
Post by: #1procrastinator on July 07, 2013, 09:06:26 pm
Post by: kamil9876 on July 07, 2013, 09:24:53 pm
Now use the relation
Post by: #1procrastinator on July 09, 2013, 08:50:45 am
I see, so in other words we want to see that the following is the identity matrix:
Now use the relationwhich follows because the entries of
are
Thanks - I thought we had to somehow derive that equation/identity(?) from scratch.
Post by: kamil9876 on July 09, 2013, 10:30:33 am
One possible way is to find the eigenspaces of
It follows that:
So
Now you just need to find what
Post by: #1procrastinator on July 23, 2013, 11:04:39 am
Quote
Theorem:
For each positive integer n, let P(n) be a statement. If
1) P(1) is true and
2) the implication if P(k), then P(k+1) is rue for every positive integer k,
then P(n) is true for every positive integer n.
Proof:
Assume that the theorem is false. Then conditions 1 and 2 are satisfied but there exists some positie integers n for which P(n) is a false statement. Let S={n∈N: P(n) is false}. Since S is a nonempty subset of N, it follows by the well-ordering principle that S contains a least element s. Since P(1) is true, 1∉S. Thus s≧2 and (s-1) is in N. Therefore (s-1)∉S and so P(s-1) is a true statement. By condition (2), P(s) is also true and s∉S. However, this contradicts our assumption that s∈S.
I don't get the bolded bit. Is there any significance to saying (s-1) is in N? Are they not all in N? I don't see how (s-1)∉S follows - I get it if s is 2 because 1 isn't in S but how do you know it holds for s>2?
Post by: mark_alec on July 23, 2013, 11:11:13 am
Post by: #1procrastinator on July 24, 2013, 08:07:57 am
s is the smallest natural number for which the statement P(s) is false. Since P(1) is true (statement #1), s must be greater than or equal to 2. Since s >=2, s-1 is a natural number and we know that P(s-1) is also true; but by statement #2 if P(s-1) is true, then P(s) must also be true. Hence we have a contradiction and S must be an empty set.
How do we know P(s-1) is also true for all s>2? And why is it necessary to state s-1 is a natural number, aren't they all natural numbers?
Post by: kamil9876 on July 24, 2013, 11:53:33 am
Quote
I don't get the bolded bit. Is there any significance to saying (s-1) is in N? Are they not all in N?
If
Quote
For each positive integer n, let P(n) be a statement.
Post by: #1procrastinator on July 25, 2013, 09:13:52 pm
------
IN evaluating
I get
Post by: lzxnl on July 25, 2013, 10:13:35 pm
Now cot y = 1/(tan y)
Therefore arccot (cot y) = y = arcccot(1/tan y) = arccot (cot y)
But tan y = 1/ cot y so y=arctan(1/cot y)
So arctan(1/cot y) = arccot (cot y)
arctan 1/y = arccot y = pi/2 - arctan y
Therefore arctan 1/y and -arctan y differ by a constant, so when integrating these two forms are equivalent.
Post by: kamil9876 on July 26, 2013, 04:40:23 pm
But didn't we already say s>=2?
Yes that was the reason why
Quote
Since s >=2, s-1 is a natural number...
Post by: #1procrastinator on August 06, 2013, 09:45:31 am
Firstly, e^x > 0. I'll denote e^x with y and y>0.
Now cot y = 1/(tan y)
Therefore arccot (cot y) = y = arcccot(1/tan y) = arccot (cot y)
But tan y = 1/ cot y so y=arctan(1/cot y)
So arctan(1/cot y) = arccot (cot y)
arctan 1/y = arccot y = pi/2 - arctan y
Therefore arctan 1/y and -arctan y differ by a constant, so when integrating these two forms are equivalent.
Thanks - so the missing C has been the culprit all along? :P
---
Show that the sequence
Can we solve for
The solution given does it another way, which was demand that
Post by: kamil9876 on August 06, 2013, 03:20:35 pm
And sure, since we only care about how it behaves as
Post by: Deleted User on August 07, 2013, 09:15:16 pm
I know that I need to show that |sn+1 - sn| ≤ k|sn - sn-1|.
So I've managed to get up to this point
sn+1 - sn = (sn^3 - sn-1^3)/7.
How do I proceed from here? And what would the value of k be?
Thanks in advance.
Post by: kamil9876 on August 07, 2013, 10:20:43 pm
There's a factorization:
So we want to study
Alternatively using a bit more theory (the mean value theorem). One knows that
and of course we are thinking here of
Post by: Deleted User on August 08, 2013, 07:58:30 pm
And sorry I don't really understand the contraction theorem either so could you explain why
Post by: #1procrastinator on August 09, 2013, 08:02:38 pm
for
And sure, since we only care about how it behaves asis large we could have restricted ourselves to
and obtained a similair inequality. But that isn't necessary, this was good enough.
Thanks...if we didn't use this 'trick', is it possible to solve for
Post by: #1procrastinator on August 10, 2013, 02:42:59 pm
Post by: BubbleWrapMan on August 10, 2013, 04:47:40 pm
Does it make sense to talk of a union of a set with one of its subsets?
It makes sense, it's just redundant because
Post by: #1procrastinator on August 11, 2013, 01:04:56 am
---
How do you evaluate
Post by: #1procrastinator on August 12, 2013, 03:56:37 am
You can't assume P(n-1) as well...just want to make sure cause only way it's the only way I know how do one of the assignment questions :D
Post by: Lasercookie on August 12, 2013, 05:35:52 am
When proving by induction, when you assume P(n) is true, this only applies for P(n), right?There's the idea of strong induction where you show that P(k), P(k+1), ..., P(n) are true, then P(n+1) is true for some
You can't assume P(n-1) as well...just want to make sure cause only way it's the only way I know how do one of the assignment questions :D
I like this analogy from here http://mathcircle.berkeley.edu/BMC4/Handouts/induct/node6.html
Quote
The difference is in what you need to assume in the induction step. For ordinary induction--in the ladder metaphor--you simply go from the rung you are on up to the next one. For strong induction, you need to know that all the rungs below the rung you are on are solid in order to step up. As a practical matter, both have the same logical strength when you apply them - since as you climb up the ladder from the bottom rung, you sweep through all the intermediate rungs anyway.
Post by: #1procrastinator on August 13, 2013, 08:19:01 am
Post by: Lasercookie on August 13, 2013, 07:39:18 pm
Ah thanks laser. How're you finding the calculus part by the way? the notes look insane lolYeah, I'm finding it pretty interesting. Its mostly the suggested questions from that Trench Real Analysis book where I've been spending the most time on. I've also been reading through Spivak Calculus (the last part of it is basically what we're covering now) and a couple of other books too.
I wish the Linear Algebra part of the course would speed up the pace though, I really hope we're done with chapter 4 of the book now.
Post by: #1procrastinator on August 15, 2013, 08:43:58 pm
Yeah, I'm finding it pretty interesting. Its mostly the suggested questions from that Trench Real Analysis book where I've been spending the most time on. I've also been reading through Spivak Calculus (the last part of it is basically what we're covering now) and a couple of other books too.
I wish the Linear Algebra part of the course would speed up the pace though, I really hope we're done with chapter 4 of the book now.
Haha, most of the time I've spent so far on the calculus part is trying to decipher the notes!
One plus is that the pace of LA gives me time to wrap my head round the calc stuff :P
Post by: #1procrastinator on August 17, 2013, 04:48:11 am
Question 7 from 4.1
Quote
7) Suppose that(finite) and, for each
for large n. Show that
Der Versuch
By hypothesis,
Post by: #1procrastinator on October 06, 2013, 01:32:33 am
(http://img94.imageshack.us/img94/6560/yf6j.jpg)
If I'm not wrong, we're supposed to prove this
Post by: Lasercookie on October 06, 2013, 11:56:17 am
Assignment question:I was stuck on this one too. The way I've gone about it is to have a close look at the hint (you can see what it says in the hint is true by substituting in the equation we have for the unit tangent vector and using the various properties of dot products), the equation we're told to consider and also look at the results you proved in the previous parts of the question (you can see they carry over the multivariable case if you start off with x(t,s) etc. instead of x(s) with your proof of those).
(http://img94.imageshack.us/img94/6560/yf6j.jpg)
If I'm not wrong, we're supposed to prove this, right? If so, anyone feel like pointing me in the right direction? I'm thinking perhaps mean value theorem only because I've seen them used to prove inequalities but still not quite sure how to apply that here.
I had this theorem pointed out to me, but you can interchange the order that you take the partial derivatives http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives#Formal_expressions_of_symmetry I haven't been able to prove this yet, I'm assuming it's in the textbook somewhere.
Post by: #1procrastinator on October 06, 2013, 02:38:25 pm
Wasn't that proven in the lectures? I'd thought of that too but it didn't really help, plus I wasn't 100% sure that the functions satisfied the necessary conditions. But x is infinitely differentiable, so that implies continuity therefore the symmetry between the second partials exist...yeah? But perhaps I should do all the previous questions before attempting this one hahaha
---
By the way, in the section on arc length parametrisation in the notes
(http://img28.imageshack.us/img28/2630/ckyl.jpg)
Is the variable 's' meant to be time or length? Because there's
Post by: nubs on October 06, 2013, 09:43:30 pm
Or 1+e^x*sin(x)=0?
Something similar to sin(x)=0, so the general solution would be x=n*pi where n is an element of Z
Thanks!
Post by: #1procrastinator on October 08, 2013, 05:22:35 pm
I was stuck on this one too. The way I've gone about it is to have a close look at the hint (you can see what it says in the hint is true by substituting in the equation we have for the unit tangent vector and using the various properties of dot products), the equation we're told to consider and also look at the results you proved in the previous parts of the question (you can see they carry over the multivariable case if you start off with x(t,s) etc. instead of x(s) with your proof of those).
I had this theorem pointed out to me, but you can interchange the order that you take the partial derivatives http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives#Formal_expressions_of_symmetry I haven't been able to prove this yet, I'm assuming it's in the textbook somewhere.
I got up to a point where I had to show that the
Post by: #1procrastinator on February 13, 2014, 09:29:23 pm
I suppose first thing you to do would be to calculate det(A-eI)=0 to find the eigenvalues e?
I don't know my determinant stuffs well but from playing around with an eigenvalue calculator, would the eigenvalues be 0,...,n?
Post by: kamil9876 on February 14, 2014, 12:31:31 pm
Post by: #1procrastinator on February 17, 2014, 07:09:06 am
Post by: kamil9876 on February 17, 2014, 11:15:27 am
Post by: #1procrastinator on February 24, 2014, 03:36:17 am
Post by: lzxnl on February 24, 2014, 08:06:00 am
This one feels like it should be really easy - expressin terms of hyperbolic functions
Isn't artanh x [tex] ln\frac{1-x }{1+x} [/tex} or something?
Post by: brightsky on February 24, 2014, 09:17:06 am
To find arctanhx, swap x and y.
x = (e^y - e^(-y))/(e^y + e^(-y))
x (e^y + e^(-y)) = (e^y - e^(-y))
x e^(2y) + x = e^(2y) - 1
(1 - x) e^(2y) = 1 + x
y = 1/2 ln((1+x)/(1-x))
So arctanhx = 1/2 ln((1+x)/(1-x))
So it seems that ln((2+2x)/(1-x)) = 2(ln2 + arctanhx)
EDIT: Error corrected.
Post by: lzxnl on February 24, 2014, 01:37:11 pm
Let y = f(x) = tanhx = (e^x - e^(-x))/(e^x + e^(-x))
To find arctanhx, swap x and y.
x = (e^y - e^(-y))/(e^y + e^(-y))
x (e^y + e^(-y)) = (e^y - e^(-y))
x e^(2y) + x = e^(2y) - 1
(1 - x) e^(2y) = 1 + x
y = 1/2 ln((1+x)/(1-x))
So arctanhx = 1/2 ln((1+x)/(1-x))
So it seems that ln((2+2x)/(1-x)) = 2ln2 arctanhx
Crap. I knew I forgot something there. However your final step is flawed. You mean 2 (ln 2 + artanh x) right?
Post by: #1procrastinator on February 25, 2014, 09:01:33 am
Post by: #1procrastinator on February 25, 2014, 03:07:23 pm
Don't know why the tex part isn't working but I'm asking if the union of A and the complement A^c equal to X
Post by: Einstein on February 25, 2014, 03:57:39 pm
Post by: brightsky on February 25, 2014, 06:31:13 pm
Crap. I knew I forgot something there. However your final step is flawed. You mean 2 (ln 2 + artanh x) right?
Woops! Of course. Cheers lzxnl!
Post by: kinslayer on February 27, 2014, 03:37:23 pm
If, and
is the complement of
in
, does
?
Don't know why the tex part isn't working but I'm asking if the union of A and the complement A^c equal to X
Yes. x is in X \ A = A^c if and only if x is in X and x is not in A.
x is in the union of two sets if and only if x is in either one of the sets.
So x is in A union A^c, if and only if 1) x is in X (since A and A^c are both subsets of X) and 2) x is in A, or x is not in A.
2) is a tautology (it is satisfied by any x), so it follows A union A^c is the same set as X.
Post by: idontknow2298 on March 14, 2014, 10:49:35 pm
a positive integer has two digits. The sum of its squares of the digits is 34. The integer formed by reversing the digits is 18 less than the original number. What is the integer?
Post by: RKTR on March 14, 2014, 11:15:26 pm
Post by: kinslayer on March 15, 2014, 12:04:00 am
Can someone please help me with this question?
a positive integer has two digits. The sum of its squares of the digits is 34. The integer formed by reversing the digits is 18 less than the original number. What is the integer?
Let a,b be nonnegative integers and express the number of interest as x = 10a + b
Then a^2 + b^2 = 34 and also 10a + b - 18 = 10b + a, or equivalently, a - b = 2
Solve simultaneously to find a=5, b=3, so that x = 53.
Post by: idontknow2298 on March 15, 2014, 04:46:04 pm
Post by: idontknow2298 on March 15, 2014, 04:47:01 pm
F ind the mean of x,y and z if 2x+2y+18=4z, in terms of z
Post by: RKTR on March 15, 2014, 04:55:31 pm
x+y=2z-9
Mean=(x+y+z)/3
=(3z-9)/3
=z-3
Post by: idontknow2298 on March 15, 2014, 04:59:25 pm
Post by: #1procrastinator on March 17, 2014, 01:23:50 pm
Post by: lzxnl on March 17, 2014, 02:12:11 pm
How do you evaluate? Tried L'hopital's rule but ended with something like
and differentiating top and bottom just gives pretty much the same thing
If you take out a factor of 3^x from the bracket, you get (1+(2/3)^x)^(1/x)×3. The bracket approaches 1 and the power approaches 0 so your final limit is 3.
Alternatively, your function is bounded between (3^x+0)^(1/x) and (3^x+3^x)^(1/x). Either of those limits should be easier to do.
L hopital's rule doesn't solve all limits unfortunately.
Post by: #1procrastinator on March 17, 2014, 08:20:03 pm
Post by: idontknow2298 on March 29, 2014, 09:21:08 pm
How would I go about doing this question?
Post by: RKTR on March 30, 2014, 12:29:07 am
If the edges of a rectangular box were increased by 1cm, 2cm and 3cm respectively, the box would become a cube and its capacity would be increased by 101 cubic cm. Find the dimensions of the rectangular box.
How would I go about doing this question?
let sides of cube=x ,volume of cube = x^3
dimensions of rectangular box = x-1,x-2,x-3 , volume of rectangular box = (x-1)(x-2)(x-3) = x^3-6x^2+11x-6
V of cube - V of rectangular box =101
6x^2-11x+6=101
6x^2-11x-95=0
(6x+19)(x-5)=0
x=-19/6 or x=5
x>0 therefore x=5
dimensions of rectangular box =4 cm , 3 cm , 2 cm
Post by: idontknow2298 on April 05, 2014, 10:11:04 pm
Post by: idontknow2298 on April 05, 2014, 10:12:17 pm
The volume of a solid sphere varies directly as the cube of its radius. The building material of three spheres with radii 3 cm, 4 cm and 5 cm is melted and a new sphere is then made. Find the radius of this new sphere.
Post by: idontknow2298 on April 05, 2014, 10:23:48 pm
Give that y varies directly with (ax-2) find the value of a if x=3 and y=2; and if x=4 and y=4
Post by: #1procrastinator on April 09, 2014, 08:31:25 am
how do you get
Post by: #1procrastinator on April 09, 2014, 07:41:26 pm
In general, if something is stated for an indexed collection of sets, does it mean it applies to both finitely many and infinitely many sets?
Post by: lzxnl on April 09, 2014, 09:19:20 pm
^ just plug in the given values of x and y and solve for a
how do you getfrom antidifferentiating
i'm getting
from substitution + partial fractions but the plot doesn't fully agree with the solution wolframalpha gave
Erm...I don't think I quite understand your answer. I don't think you're meant to get a spare (1+cos x) term anywhere. Is that under another log or something?
dx/sin x => sin x dx/(1-cos^2 x)
Let u = cos x, du = -sin x dx
Integral becomes du/(u^2-1) => du/2 (1/(u-1) - 1/(u+1)) => 1/2 ln(1-cos x) - 1/2 ln(1+cos x) after integrating and back subbing
That's what you meant, right?
OR you can do this
1/(sin x)
Sub t = tan x/2 (dirtiest substitution trick in the book)
You can show quite easily that tan x = 2t/(1-t^2), sin x = 2t/(1+t^2) and cos x = (1-t^2)/(1+t^2) , dx = 2 dt/(1+t^2)
so our integral becomes (1+t^2)/2t * 2dt/(1+t^2)
The 1+t^2 terms drop off
so our integral is just dt/t
We're left with just ln (tan x/2) which really should appear as ln(sin (x/2)) - ln(cos(x/2))
Post by: #1procrastinator on April 09, 2014, 09:58:19 pm
Erm...I don't think I quite understand your answer. I don't think you're meant to get a spare (1+cos x) term anywhere. Is that under another log or something?
dx/sin x => sin x dx/(1-cos^2 x)
Let u = cos x, du = -sin x dx
Integral becomes du/(u^2-1) => du/2 (1/(u-1) - 1/(u+1)) => 1/2 ln(1-cos x) - 1/2 ln(1+cos x) after integrating and back subbing
That's what you meant, right?
Yeah, typo - was in a hurry when I was writing that
OR you can do this
1/(sin x)
Sub t = tan x/2 (dirtiest substitution trick in the book)
You can show quite easily that tan x = 2t/(1-t^2), sin x = 2t/(1+t^2) and cos x = (1-t^2)/(1+t^2) , dx = 2 dt/(1+t^2)
so our integral becomes (1+t^2)/2t * 2dt/(1+t^2)
The 1+t^2 terms drop off
so our integral is just dt/t
We're left with just ln (tan x/2) which really should appear as ln(sin (x/2)) - ln(cos(x/2))
'World's sneakiest substitution' haha. How come the two aren't 'exactly' the same though?
The bold plot is of the function obtained from the dirty sub.
Post by: lzxnl on April 10, 2014, 05:05:33 pm
Post by: #1procrastinator on April 12, 2014, 02:28:08 am
---
Is the intersection of finitely many closed sets closed? I'm aware it is for a 'collections of sets' but I'm not sure if this means that it has to be an infinite collection or it can be finite.
In general, if something is stated for an indexed collection of sets, does it mean it applies to both finitely many and infinitely many sets?
Post by: idontknow2298 on April 26, 2014, 10:33:40 pm
If z varies directly as x when y is constant and inversely as y when x is constant and if z=8 when x=4 and y=2, find z when x=3 and y= 1/3
Post by: kinslayer on April 26, 2014, 11:09:30 pm
Is the intersection of finitely many closed sets closed? I'm aware it is for a 'collections of sets' but I'm not sure if this means that it has to be an infinite collection or it can be finite.
In general, if something is stated for an indexed collection of sets, does it mean it applies to both finitely many and infinitely many sets?
An intersection of arbitrarily many closed sets is closed. A union of arbitrarily many open sets is open.
An intersection of finitely many open sets is open, while a union of finitely many closed sets is closed.
An interesection of infinitely many open sets need not be open:
Post by: TrueTears on May 01, 2014, 06:15:38 am
Is the intersection of finitely many closed sets closed? I'm aware it is for a 'collections of sets' but I'm not sure if this means that it has to be an infinite collection or it can be finite.Yes. In fact, it holds even stronger than that, the intersection of any arbitrary collection (could be infinite) of closed sets is closed. The proof is very straightforward, just a simple application of De Morgan's laws and the relation between open/closed sets.
Post by: #1procrastinator on July 09, 2014, 05:29:45 pm
Post by: kinslayer on July 09, 2014, 05:42:11 pm
How would you evaluate thiswithout using trigonometric or hyperbolic substitutions? Also for
(I tried u=x^2+a^2 and ended up with an integral of the second form)
Then
Post by: Zlatan on July 09, 2014, 07:42:21 pm
Thanks
Post by: kinslayer on July 10, 2014, 02:23:27 am
A florist has to make a floral arrangement. She has 6 Banksias, 5 wattles and 4 Waratahs. All the flowers of each kind are different.In how many ways can the florist make a bunch of 10 flowers if she has to use at least 3 of each kind ?
Thanks
Combinatorics isn't my strong point, but let's go anyway.
By the multiplication principle, there are:
There are 6 flowers left to choose from, and there is a different floral arrangement for each flower chosen. We've already counted the number of ways these flowers can be included in the initial 9, so the total number of arrangements is simply
e: fixed mistake
Post by: Zlatan on July 10, 2014, 06:49:51 pm
Post by: kinslayer on July 11, 2014, 12:53:52 pm
Ummm .... I'm not too familiar with Combinatorics either. So could you explain it in better detail ? Thanks anyways for this answer anyway !!!!! :)
No problem, I am basically just using combinations:
http://en.wikipedia.org/wiki/Combination
and the multiplication principle:
http://en.wikipedia.org/wiki/Rule_of_product
I looked at the first nine spots because it's simply the product of combinations.
There are
Since each way of rearranging the individuals results in an entirely new arrangement, you must multiply all of them together to get the total number of arrangements with 9 spots. But now you have one more spot with six remaining flowers. We don't need to distinguish between the type now because there aren't any further restrictions and each one will result in a new arrangement. So we just multiply the previous result by 6 to get the final answer.
Post by: #1procrastinator on July 27, 2014, 01:14:15 am
How do you work out these substitutions?