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[Anticitizen One]

Ran into 3 others that i'm not understanding:

Rewrite the expression as an equivalent expression that does not contain powers of trigonometric functions greater than:

Cos^4(X)

I understand you have to use the power reducing formula cos^2(theta)=(1+cos2(theta))/2 but I have no idea how to solve.

The other set giving me problems look like this.

Solve the equation on the interval [0,2pi):

cos[2(theta)-(pi/2)]=(square root 2)/2

and finally:

An airplane flying faster than the speed of sound creates sound waves that form a cone. If a(alpha) is the vertex angle of the cone and m is the Mach number for the speed of the plane, then sin((alpha)/2) = 1/(m) (m >1). Find the Mach number to the nearest tenth if a(alpha)= 90 degrees.

 

[cpp_is_king]

Ran into 3 others that i'm not understanding:

Rewrite the expression as an equivalent expression that does not contain powers of trigonometric functions greater than:

Cos^4(X)

I understand you have to use the power reducing formula cos^2(theta)=(1+cos2(theta))/2 but I have no idea how to solve.

cos^4(x) = [cos^2(x)]^2 = [(1 + cos(2x))/2]^2 = (1 + 2cos(2x) + cos^2(2x))/4

Now apply your identity another time, for the cos^2(2x) term and simplify.

The other set giving me problems look like this.

Solve the equation on the interval [0,2pi):

cos[2(theta)-(pi/2)]=(square root 2)/2

Recall that cos(x - pi/2) = sin(x). Therefore, cos(2x - pi/2) = sin(2x). So we can re-write this as:

sin(2x) = sqrt(2)/2
2x = arcsin(sqrt(2)/2)
2x = pi/4
x = pi/8

and finally:

An airplane flying faster than the speed of sound creates sound waves that form a cone. If a(alpha) is the vertex angle of the cone and m is the Mach number for the speed of the plane, then sin((alpha)/2) = 1/(m) (m >1). Find the Mach number to the nearest tenth if a(alpha)= 90 degrees.

So you've got sin(x/2) = 1/m. The problem tells you that x = 90 degrees, or pi/2. So:

sin(pi/4) = 1/m
1/sqrt(2) = 1/m
m = sqrt(2) ~= 1.4

 

[Lion Heart]

4=log3(15)^x. Answer is approx 1.623.

That's log 15 base 3. Tia.

 

[cpp_is_king]

4=log3(15)^x. Answer is approx 1.623.

That's log 15 base 3. Tia.

Is there a question here?

 

[opticalmace]

4=log3(15)^x. Answer is approx 1.623.

That's log 15 base 3. Tia.

Recall log(a^x) = x*log(a) for logarithms, and for any base.

Also recall how to change bases (for instance, from base 3 to base 10): http://en.wikipedia.org/wiki/Logarithmic_identities#Changing_the_base

 

[Lion Heart]

Recall log(a^x) = x*log(a) for logarithms, and for any base.

Also recall how to change bases (for instance, from base 3 to base 10): http://en.wikipedia.org/wiki/Logarithmic_identities#Changing_the_base

Thanks, I figured it out.

 

[0xCA2]

Is it a Calc I course (i.e. differential calculus)? What is your math background?

It is Calc I. I just finished Trigonometry and before that I took College Algebra.

http://www.wolframalpha.com/

I used that website for Calculus I and Physics 1 & 2. I was lost with Calculus because my prep for it was pre-Calc and it was mainly trig and geometry. To some this might of helped, but I was still lost.

It takes some time, but Calculus I is not that hard. Learn the concepts forwards and backwards. You have to memorize a few formulas like the multiplication, division, and subtraction methods. Then learning how to solve derivatives.

They sort of teach Calculus in order. I think the last thing we did were integrals and the fundamental theorem of calculus.

Watch youtube videos and visit Khan Academy. Its very easy because like I said Calculus, to me, was taught in a specific order. Know what it means when X approaches 0 or infinity and etc.
Limits - http://www.youtube.com/watch?v=lCSk4Df2x-A
^ huge help if you understand that concept.

I did very well in Physics 1 (2 I did good, but its nothing like Physics 1 bc Physics 2 is all electricity) because I learned a lot from my Calculus class.

http://www.youtube.com/watch?v=GGQngIp0YGI

See if there is an online syllabus for the course. Grab the textbook and review the chapters.

Thanks for the advice

 

[Lion Heart]

Hi guys, sorry for the basic question, but how do you convert

10+-√500 / 2

to

5+-5√5?

They both have the same answer but I can't seem to do it.

Ah, thanks stephen. Fuck I suck.

 

[stephen08]

Hi guys, sorry for the basic question, but how do you convert

10+-√500 / 2

to

5+-5√5?

They both have the same answer but I can't seem to do it.

sqrt(500) = sqrt(100)*sqrt(5) = 10*sqrt(5)

 

[Leona Lewis]

Need help with two problems. They're part of a GRE review guide I'm trying to work on, and are labeled moderate difficulty.

1. Find the area of the polygon

Um, how on earth do I figure out area only using side length? This isn't a square or rectangle, so there's no formula to draw upon that I know of.

2.

I have no idea how to find the length of arc AD or the measure of angle P. I'm trying everything I can knowing that the two tangent lines must have some relation to figuring out the problem, but I'm lost because there are no central angles (where central angle = intercepted arc) or intercepted arcs (where arc measure = 2x interior angle) to fill in the blanks.

Thanks

 

[luoapp]

Brahmagupta's formula finds the area of any cyclic quadrilateral (one that can be inscribed in a circle) given the lengths of the sides.

http://en.wikipedia.org/wiki/Brahmagupta's_formula

 

[Leona Lewis]

Brahmagupta's formula finds the area of any cyclic quadrilateral (one that can be inscribed in a circle) given the lengths of the sides.

http://en.wikipedia.org/wiki/Brahmagupta's_formula

Thanks for that! It seems a bit high level for GRE (supposed to be SAT-level math), so there must be a more logical way that doesn't involve using a little-taught formula. Glad to have any solution, though.

For the second one, I found a theorem that states that two chords equidistant from the center of a circle share the same corresponding arc lengths/measurements, but since we know nothing of the radius, I'm just...lost.

 

[opticalmace]

Thanks for that! It seems a bit high level for GRE (supposed to be SAT-level math), so there must be a more logical way that doesn't involve using a little-taught formula. Glad to have any solution, though.

For the second one, I found a theorem that states that two chords equidistant from the center of a circle share the same corresponding arc lengths/measurements, but since we know nothing of the radius, I'm just...lost.

Is this for the GRE general? Or Math GRE?

I did the general and the math there was a joke... nothing like the questions you posed.

 

[bachikarn]

Need help with two problems. They're part of a GRE review guide I'm trying to work on, and are labeled moderate difficulty.

1. Find the area of the polygon



Um, how on earth do I figure out area only using side length? This isn't a square or rectangle, so there's no formula to draw upon that I know of.

Do they say XZ or YW form the diameter? If so, you could do the problem because there would have to be a right triangle. From the info you gave us, the only way I can think of it is to use stuff that's way more advanced than a normal GRE math problem. I imagine by drawing a circle around it, they want you to use that information some how to find the area, so it'd make sense if XZ or YW were a diameter.

 

[Leona Lewis]

Do they say XZ or YW form the diameter?

Nope - no indication at all.

After I figure out the second one, I might just dump this book. The fact that the questions aren't multiple choice should've hinted to me that it's not the best prep for the actual test.

I wanted to start at a higher level and then work backwards to boost my confidence, but it's having the opposite effect

 

[Bear]

Anyone have any good online resources covering integration by parts, partial fraction integration, and the integration of rational functions by partial fractions?

Paul's Online Notes is a good resource if you are looking for well written notes with good examples. I mainly used it for differential equations, but it covers several those techniques well in the calculus section. It came in handy whenever I needed to refresh my memory of those integration methods.

 

[Troll]

I am starting to study for the GRE Mathematics Subject Test and so far I have gotten the advice to review the official practice exam put up online, review old algebra/calculus textbooks and get the 4th edition Cracking the GRE Math Subject Test. Is there anyone on here who has taken the exam that can offer me some more advice on how to be as prepared as possible for the exam? Thank you-

 

[cpp_is_king]

I am starting to study for the GRE Mathematics Subject Test and so far I have gotten the advice to review the official practice exam put up online, review old algebra/calculus textbooks and get the 4th edition Cracking the GRE Math Subject Test. Is there anyone on here who has taken the exam that can offer me some more advice on how to be as prepared as possible for the exam? Thank you-

It's been a while since I took it. It's probably much different now. The best advice I can give you is to skip anything that looks even remotely tricky. There is a lot of time pressure. They ask a handful of questions that are pretty elementary but are nevertheless "tricky", like something you'd find on a high school mathematics competition. These are the worst offenders at eating up your time.

Aside from that, take the practice exam, and focus a lot on problems involving the most important theorems and results from each subject, and less on the lesser theorems. So like fundamental theorem of calculus / algebra, intermediate / mean value theorem, etc you should definitely know really well. More advanced stuff will show up, but if my memory serves me correctly, stuff shows up with a frequency relative to how fundamental it is. Also make sure you know any theorem that was named after a person.

I don't even know if they rate on the same scale as when I took it in 2005, but I think I got like a 750 if that means anything to you. My biggest loss of points was due to time pressure caused by spending too much time on a handful of problems.

 

[nicoga3000]

OK math GAF, time to see if you paid attention in diffEQ (because I know I didn't and I NEVER use this stuff). I imagine this is a super easy question to anyone who understands these stupid things.

The short version of my question is this:

You have a homogeneous equation with a general solution of the following form:

w1 = Ro + sum(from m = 1 to infinity) of (Rm * cos(m*theta)) + sum(from m = 1 to infinity) of (Rm' * sin(m*theta))

For m = 0:

Ro = Ao + Bo*r^2 + Co*log(r) + Do*r^2*log(r)

For m = 1:

R1 = A1*r + B1*r^3 + C1*r^-1 + D1*r*log(r)

A, B, C, D, are arbitrary constants.

Hopefully the above equations and information are enough to answer my VERY simple question:

Is log(r) = log10(r) or ln(r) ?

The difference is HUGE in the end results of the application of the equations the solution puts forth.

 

[DEATH™]

OK math GAF, time to see if you paid attention in diffEQ (because I know I didn't and I NEVER use this stuff). I imagine this is a super easy question to anyone who understands these stupid things.

The short version of my question is this:

You have a homogeneous equation with a general solution of the following form:

w1 = Ro + sum(from m = 1 to infinity) of (Rm * cos(m*theta)) + sum(from m = 1 to infinity) of (Rm' * sin(m*theta))

For m = 0:

Ro = Ao + Bo*r^2 + Co*log(r) + Do*r^2*log(r)

For m = 1:

R1 = A1*r + B1*r^3 + C1*r^-1 + D1*r*log(r)

A, B, C, D, are arbitrary constants.

Hopefully the above equations and information are enough to answer my VERY simple question:

Is log(r) = log10(r) or ln(r) ?

The difference is HUGE in the end results of the application of the equations the solution puts forth.

Is that the general solution already? I honestly want to see the original problem and solve the thing myself...

My educated guess would be ln (r) though, since that's awfully look like a euler equation wrapped in a series... It's very unusual to integrate something and log 10 will come out... especially in your everyday differential equation... even wolfram alpha calls ln as logs... that's how unusual it is...

 

[Memento_Mori15]

I thought I was done with the math help thread on gaf but doesn't look like it. I'm taking discrete math and it really seems...so different from what I'm used to. I kinda get what's going on but then I forget it pretty fast. It's so different from the other math courses I've taken. Linear algebra kinda helped because in this class, we focused more on the theories/proving why things work. However, this is just more abstract(is this the right word? lol). Does anyone know some tips to handle this course? Are there any sites like Khan? This is a 6-weeks course so we're going to go fast. I rather focus on linear algebra for now since I have my final Wednesday and would like to still aim for that A...

 

[Ra1den]

I'm supposed to list the differences and similarities between the Ramp Function and the function

x
----------------
1+e^-20x


(that is, x divided by 1 + e to the -20x power)


When I graph them both, they look identical everywhere. The best difference I can come up with is that the ramp function is often written as a piecewise defined function.

Pretty lame I know. Are there any other differences?

Here is the ramp function

http://en.wikipedia.org/wiki/Ramp_function

 

[Feep]

I'm supposed to list the differences and similarities between the Ramp Function and the function

x
----------------
1+e^-20x


(that is, x divided by 1 + e to the -20x power)


When I graph them both, they look identical everywhere. The best difference I can come up with is that the ramp function is often written as a piecewise defined function.

Pretty lame I know. Are there any other differences?

Here is the ramp function

http://en.wikipedia.org/wiki/Ramp_function

They're completely different. The high coefficient on the exponent in the denominator makes it extremely large as soon as x goes negative, which brings it extremely close to zero, but it's never ACTUALLY zero, as the ramp function is. On the other side, that exponent brings that part EXTREMELY close to zero and thus the y value extremely close to the x value, but once again, y never actually equals x.

It's a very good approximation, which is useful since piecewise functions aren't wholly differentiable, but those two functions never have the same value at all except for x = 0.

 

[elitehebrew]

I'm about to take elementary statistics for the whole next school year and i was wondering if anyone has a recommendation on how to get a headstart. I have no idea what text we will be using but i've heard many conflicting things about elementary stats being hard or really easy. I'm sure it depends on the person, but i'm semi scared. Please help and give stories about personal experiences with elem. stats. Thanks Gaffers!

 

[Ra1den]

They're completely different. The high coefficient on the exponent in the denominator makes it extremely large as soon as x goes negative, which brings it extremely close to zero, but it's never ACTUALLY zero, as the ramp function is. On the other side, that exponent brings that part EXTREMELY close to zero and thus the y value extremely close to the x value, but once again, y never actually equals x.

It's a very good approximation, which is useful since piecewise functions aren't wholly differentiable, but those two functions never have the same value at all except for x = 0.

Thanks, you saved my bacon!

 

[Memento_Mori15]

Linear algebra question:
Can anyone see if my answers are correct? It's a practice for a final
Let A be a 7x7 matrix. Suppose A has eigenvalues  λ= -4,-4, 2,3, 10, 10, 10 with
multiplicities included.
A) What conditions must A satisfy in order to be diagonalisable?

My answer: For every nxn matrix, we need to have n eigenvalues. Thus with a 7x7 matrix, we have 7 values. The dimension of the eigen space must also be equal to the eigenvalue's multiplicity.

B)Suppose A meets your criteria of part A), ie. suppose A is now diagonalisable, A and λ as before. Find det A; explain your solution.

My answer: Wouldn't this just be the eigenspace's dim of each of the eigenvalues? So would it be dim, or rather dim(7)?
C)Now find a general formula for det A^n where n = 1; 2; 3,... for A as above, assuming A is still diagonalisable.

I don't know what this question is asking so I'm skipping this for now. A hint would be nice.

 

[rjc571]

That's kind of a weird question. The determinant of a matrix is always the product of its eigenvalues, regardless of whether or not it's diagonalizable.

Since the determinant is a multiplicative function (i.e. det(A*B) = det(A) * det(B)), it's easy to figure out what det(A^n) should be. (Again, without requiring A to be a diagonal matrix.)

 

[Memento_Mori15]

That's kind of a weird question. The determinant of a matrix is always the product of its eigenvalues, regardless of whether or not it's diagonalizable.

Since the determinant is a multiplicative function (i.e. det(A*B) = det(A) * det(B)), it's easy to figure out what det(A^n) should be. (Again, without requiring A to be a diagonal matrix.)

Ah, I see. I was also wrong about my part B. So redoing part B, A=PDP^-1. It'll just be det(D). det(eigenvalues multiplied with each other). Then part c would simply be det(eigenvalues^n). I was supposed to find in part B the det, not the dim. I was getting the two confused...

Thanks for the reply.

 

[RyanDG]

Before I go any further, I'm kind of curious if someone can glance at these integrals really quick and just verify that I have them set up correctly with the proper axis of rotations and all that jazz.

 

[Troll]

It's been a while since I took it. It's probably much different now. The best advice I can give you is to skip anything that looks even remotely tricky. There is a lot of time pressure. They ask a handful of questions that are pretty elementary but are nevertheless "tricky", like something you'd find on a high school mathematics competition. These are the worst offenders at eating up your time.

Aside from that, take the practice exam, and focus a lot on problems involving the most important theorems and results from each subject, and less on the lesser theorems. So like fundamental theorem of calculus / algebra, intermediate / mean value theorem, etc you should definitely know really well. More advanced stuff will show up, but if my memory serves me correctly, stuff shows up with a frequency relative to how fundamental it is. Also make sure you know any theorem that was named after a person.

I don't even know if they rate on the same scale as when I took it in 2005, but I think I got like a 750 if that means anything to you. My biggest loss of points was due to time pressure caused by spending too much time on a handful of problems.

Yeah I have heard that the time really kills everyone's score. Thank you for the advice, it is very useful.

 

[Memento_Mori15]

Anyone good at logical equivalences? Need help with proving [(p ->q) ∧ (q ->r)] -> (p->r) using logical equivalence
~=negation.
My work:
(~pVq)∧(~qVr)->(~pVr)
~[(~pVq)∧(~qVr)]V(~pVr)
Apply De Morgan's
p∧~q V q∧~r V~pVr)

And I have no idea where to go from there. Any hint/help would be nice. I was doing this for at least an hour playing around with it. In my last expression, can I just freely move the variables?

 

[Enkidu]

Anyone good at logical equivalences? Need help with proving [(p ->q) ∧ (q ->r)] -> (p->r) using logical equivalence
~=negation.
My work:
(~pVq)∧(~qVr)->(~pVr)
~[(~pVq)∧(~qVr)]V(~pVr)
Apply De Morgan's
p∧~q V q∧~r V~pVr)

And I have no idea where to go from there. Any hint/help would be nice. I was doing this for at least an hour playing around with it. In my last expression, can I just freely move the variables?

I'm not exactly an expert but can't you just use Karnaugh maps?

 

[Memento_Mori15]

I'm not exactly an expert but can't you just use Karnaugh maps?

Can't use anything but logical equivalences. If I could, I would use truth tables but I definitely think when exam time comes(which is next week :c), they will ask me to use logical equivalences to prove something.

 

[Leezard]

you should be able to move around the expressions separated by "or", since A v B == B v A, if it helps you.
p∧~q V q∧~r V~pVr
make it ((p∧~q) v ~p) v ((q∧~r) v r).

(p∧~q) v ~p should be the same as (~q v ~p).
Likewise, (q∧~r) v r becomes (r v q).

Which leads to ~q v ~p v r v q. (q v ~q ) can be removed.
=> (~p v r), which is the same as p -> r, right?

It has been a while since I did logic, so this might not be entirely proper.

 

[Lonely1]

Anyone good at logical equivalences? Need help with proving [(p ->q) ∧ (q ->r)] -> (p->r) using logical equivalence
~=negation.
My work:
(~pVq)∧(~qVr)->(~pVr)
~[(~pVq)∧(~qVr)]V(~pVr)
Apply De Morgan's
p∧~q V q∧~r V~pVr)

And I have no idea where to go from there. Any hint/help would be nice. I was doing this for at least an hour playing around with it. In my last expression, can I just freely move the variables?

I'm not exactly sure what you mean by "using logical equivalences". You mean Natural Deduction? If so, is very easy, but I guess that's not it.

Your approach doesn't seems right. Are you supposing "(~pVq)∧(~qVr)->(~pVr)"? You need a supposition in order to introduce an implication, since you don't have an equivalence, you can't use "logical equivalence". Unless you mean you start with a tautology and transform it via equivalences to your objective (another tautology)? If so, the exercise doesn't make much sense to me.

Anyway, Natural Deduction Proof:

1: -  (p->q) & (q->r)				Assumption
2: - - p                                 	Assumption
3: - - (p->q)                           	&EL (1)
4: - - (q->r)                            	&ER (1)
5: - - q                                    	->EL (2,3)
6: - - r                                    	->EL (4,5)
7: - p -> r                                 	->I(6)
8: [(p ->q) & (q ->r)] -> (p->r)  		->I(7)

Where &E, ->E, ->I are Natural deduction Inference rules.

Edit: Sorry, I had a couple of typos on the proof. Is correct now.

 

[Memento_Mori15]

you should be able to move around the expressions separated by "or", since A v B == B v A, if it helps you.
p∧~q V q∧~r V~pVr
make it ((p∧~q) v ~p) v ((q∧~r) v r).

(p∧~q) v ~p should be the same as (~q v ~p).
Likewise, (q∧~r) v r becomes (r v q).

Which leads to ~q v ~p v r v q. (q v ~q ) can be removed.
=> (~p v r), which is the same as p -> r, right?

It has been a while since I did logic, so this might not be entirely proper.

Omg, thank you so much. I worked on this for hours and just moved on. SI didn't know I can move them so I was just stuck trying to do an alternative approach. What I did was from your suggestion(which is basically what you did):
From p^~q V q∧~r V~pVr
(p∧~q) V ~p V (r V(q^~r)
Distribute
(~pVp) ^ ~pV~q V rVq ^ rV~r
T^ ~pV~q V rVq^T
Apply associative law and then you'll just get (~p v r) and a bunch of trues, which is p->r. Thank you!

@Lonely1
Since this problem is before that was introduce, I was assuming I couldn't use it. Thanks for the post though since I do I have to study that as well and it would be good practice using that method on this problem.

Edit: knowing that one little rule is really helpful. Made the next problem so easy even though there were more variables.

 

[Memento_Mori15]

Sorry for the double post but I need help in another problem.
((∀x.P (x) -> ∀y.Q(x, y)) ∧ ∀z.P (z)) then ∀w.Q(w, w)
I need to use natural deduction. We're given a hint that:
To get ∀w Q(w, w) you need to get ∀y P first and you can get that from the premise by ^ elimination.
However, how are you supposed to get ∀y P when you don't even have that from the above? If someone could point me to getting that, I'm sure I could do the rest.

Then this problem:

Write the numbers 1, 2, . . . , 2n on a blackboard, where
n is an odd integer. Pick any two of the numbers, j and
k, write |j − k| on the board and erase j and k. Continue
this process until only one integer is written on the board.
Prove that this integer must be odd.

I thought because of the 1, no matter what you subtract it with, you'll just get an odd no matter what. So I look at the odd-solution manual and it said:

The parity (oddness or evenness) of the sum
of the numbers written on the board never changes, because
j + k and |j − k| have the same parity (and at each step we
reduce the sum by j + k but increase it by |j − k|). Therefore
the integer at the end of the process must have the same
parity as 1 + 2 + · · · + (2n) = n(2n + 1), which is odd
because n is odd.

I don't get the term parity nor do I even get how they got n(2n+1) on the right side. I could kinda see how they got 2n+1 because for an odd, that's how you represent it. But having an n there seems out of place.

 

[Therion]

I don't get the term parity nor do I even get how they got n(2n+1) on the right side. I could kinda see how they got 2n+1 because for an odd, that's how you represent it. But having an n there seems out of place.

Parity is just referring to whether the sum is odd or even, and it will not change by carrying out the given operation. The sum of the numbers 1,...,n is given by n(n+1)/2. So the sum of the numbers up to 2n is 2n(2n+1)/2 = n(2n+1).

Edit: For the first question, ∀z.P(z) means for any choice of z, P(z) is true. You could just as easily say ∀y.P instead, as you are just changing an arbitrary label, and not the content of the statement.

 

[Memento_Mori15]

Parity is just referring to whether the sum is odd or even, and it will not change by carrying out the given operation. The sum of the numbers 1,...,n is given by n(n+1)/2. So the sum of the numbers up to 2n is 2n(2n+1)/2 = n(2n+1).

Edit: For the first question, ∀z.P(z) means for any choice of z, P(z) is true. You could just as easily say ∀y.P instead, as you are just changing an arbitrary label, and not the content of the statement.

Either I forgot all about the summation of numbers or I didn't learn it. I had to google it after you posted this. But after reading on wiki, it made sense and I was able to prove this problem.

However, still having trouble with the first question. :\

 

[cpp_is_king]

Either I forgot all about the summation of numbers or I didn't learn it. I had to google it after you posted this. But after reading on wiki, it made sense and I was able to prove this problem.

However, still having trouble with the first question. :\

The sum of all the numbers is initially odd. You can compute the sum of the first k integers by the well known formula k(k+1)/2. Let k = 2n. (2n)(2n+1)/2 = n(2n+1). We were given that n is odd, so we've got odd*odd=odd, verifying that the sum is odd.

When you choose two numbers from the board, there are only 3 possibilities. odd+odd, odd+even, or even+even.

At each step we will subtract (j+k) and then add |j-k|, so the net effect is that we will compute Sum := Sum - (j+k) + |j-k|

odd+odd: j+k=even, |j-k|=even, -even+even = even
odd+even: j+k=odd, |j-k|=odd, -odd+odd = even
even+even: j+k=even, |j-k|=even, -even+even = even

Thus, every step of the way, the sum will decrease by an even number. Since an odd minus an even is always odd, the sum will always be odd. Since it started odd, it will be odd when there is only 1 number left, hence the last number must be odd.

 

[Memento_Mori15]

The sum of all the numbers is initially odd. You can compute the sum of the first k integers by the well known formula k(k+1)/2. Let k = 2n. (2n)(2n+1)/2 = n(2n+1). We were given that n is odd, so we've got odd*odd=odd, verifying that the sum is odd.

When you choose two numbers from the board, there are only 3 possibilities. odd+odd, odd+even, or even+even.

At each step we will subtract (j+k) and then add |j-k|, so the net effect is that we will compute Sum := Sum - (j+k) + |j-k|

odd+odd: j+k=even, |j-k|=even, -even+even = even
odd+even: j+k=odd, |j-k|=odd, -odd+odd = even
even+even: j+k=even, |j-k|=even, -even+even = even

Thus, every step of the way, the sum will decrease by an even number. Since an odd minus an even is always odd, the sum will always be odd. Since it started odd, it will be odd when there is only 1 number left, hence the last number must be odd.

Thanks for the detail reply, but I already solved with Therion's post. Thanks though. I had trouble with my initial problem but I solved it earlier today.

Thanks guys.

 

[Deadly]

Could anyone help me with a finance related problem?

Your Aunt Tillie is going to retire on January 1, 2012 (her 65th birthday). She faces the following choices for funding her retirement. Her company pension plan will pay her $2,500 per month starting January 31, 2012. Every year, these annual payments will increase by 2% (all payments made in 2012 are $2,500, all payments made in 2013 are $2,500 x (1.02), and so on.) Her alternative is to take a lump of cash from the company when she leaves. Aunt Tillie expects to live until December 31, 2022 (and she gets the December cheque). Assume an APR of 6% with monthly compounding.

How big does the cash payment need to be to make her accept it instead of the promised monthly payments?

I know its a growing annuity but I can't quite figure how to compute it =/

 

[cpp_is_king]

Could anyone help me with a finance related problem?

I know its a growing annuity but I can't quite figure how to compute it =/

I never remember the formulas for this crap so let's just work it out from the basics.

Option 1:

2012: (2500)(12)
2013: (2500)(12)(1.02)
2014: (2500)(12)(1.02)(1.02)
...
2022: (2500)(12)(1.02)(1.02)...(1.02)

We want to compute the sum of these. If we choose 2012 to be t=0 then we can use a nice elegant formula to express the entire sum.

Sum[t=0, 10] (2500)(12)(1.02)^t
= (2500)(12) Sum[t=0, 10] (1.02)^t

The sum is just the sum of a geometric series, and I do remember that formula since it's so common. So the sum is equal to:

(2500)(12) (1 - 1.02^11) / (1 - 1.02) = (2500)(12)(12.1687) = $365,061.46

Can you take it from here to figure out how the cash value option grows with the monthly compound?

 

[Moobabe]

Quick question (I'm sure it's been gone over a hundred times) but I want to start studying maths properly, how?

I've a University degree in another discipline and a GCSE in Maths, but I've started to watch numb3rs (I know, not a great way to start) but some of the higher level stuff in that is really interesting.

I've downloaded a brief history of mathematics, a series focusing on some of the figureheads of maths over the past few centuries, and then Mathematics for primates which I'm yet to properly get into - but it's all fascinating to me.

Can any of you guys recommend me a textbook that goes into depth (in an approachable way)? Or is it necessary that I go back, rethink about GCSE level and then maybe head into A level maths before approaching the higher level stuff?

 

[cpp_is_king]

Quick question (I'm sure it's been gone over a hundred times) but I want to start studying maths properly, how?

I've a University degree in another discipline and a GCSE in Maths, but I've started to watch numb3rs (I know, not a great way to start) but some of the higher level stuff in that is really interesting.

I've downloaded a brief history of mathematics, a series focusing on some of the figureheads of maths over the past few centuries, and then Mathematics for primates which I'm yet to properly get into - but it's all fascinating to me.

Can any of you guys recommend me a textbook that goes into depth (in an approachable way)? Or is it necessary that I go back, rethink about GCSE level and then maybe head into A level maths before approaching the higher level stuff?

I don't know what a GCSE is or how it relates to the US system, but the standard "Everything you could ever want to know about mathematics" is The Princeton Companion to Mathematics.

It's very in depth, as you say, so don't expect to understand everything. Because it really is everything. Mathematics is like medicine, in the sense that there's so many fields, and each of those fields has so many subfields, and some of those subfields have deep prerequisite chains on entire other subfields, which themselves have prerequisites (it goes on like this). So not only can a single person not possibly understand everything, a single person can't possibly even understand the prerequisites for understanding the prerequisites of everything

A more gentle and approachable introduction would be Kolmogorov's Mathematics: Its Contents, Methods, and Meanings.

 

[Deadly]

I never remember the formulas for this crap so let's just work it out from the basics.

Option 1:

2012: (2500)(12)
2013: (2500)(12)(1.02)
2014: (2500)(12)(1.02)(1.02)
...
2022: (2500)(12)(1.02)(1.02)...(1.02)

We want to compute the sum of these. If we choose 2012 to be t=0 then we can use a nice elegant formula to express the entire sum.

Sum[t=0, 10] (2500)(12)(1.02)^t
= (2500)(12) Sum[t=0, 10] (1.02)^t

The sum is just the sum of a geometric series, and I do remember that formula since it's so common. So the sum is equal to:

(2500)(12) (1 - 1.02^11) / (1 - 1.02) = (2500)(12)(12.1687) = $365,061.46

Can you take it from here to figure out how the cash value option grows with the monthly compound?

All the available answers are lower than that =/
A. $241,149.00
B. $244,179.39
C. $263,641.08
D. $265,569.78
E. $275,455.75

 

[Moobabe]

I don't know what a GCSE is or how it relates to the US system, but the standard "Everything you could ever want to know about mathematics" is The Princeton Companion to Mathematics.

It's very in depth, as you say, so don't expect to understand everything. Because it really is everything. Mathematics is like medicine, in the sense that there's so many fields, and each of those fields has so many subfields, and some of those subfields have deep prerequisite chains on entire other subfields, which themselves have prerequisites (it goes on like this). So not only can a single person not possibly understand everything, a single person can't possibly even understand the prerequisites for understanding the prerequisites of everything

A more gentle and approachable introduction would be Kolmogorov's Mathematics: Its Contents, Methods, and Meanings.

Thanks for the detailed response!

GCSE is a series of exams you take when you're 16/17 - I'm not sure what the equivalent is for you guys. It's not a very advanced level by any means but I really didn't care about maths at that age because I saw it all as algebra and graphs...

 

[cpp_is_king]

All the available answers are lower than that =/
A. $241,149.00
B. $244,179.39
C. $263,641.08
D. $265,569.78
E. $275,455.75

That's just because you haven't yet done the other half of the problem. The half I've done shows you that at the end of the 10 years, if she goes with the increasing payments, she'll end up with about $365,000.

The question is asking you, if she were to take the lump sum, how big must the lump sum be to end up being greater than $365,000 using the different compounding structure? So the answer to the question is obviously going to be much smaller, because it's asking you for the number before it grows.

This is just a simple compound interest problem where you've got your initial principal, rate of compounding, interest rate, and future value. We want to find the initial value such that the future value is equal to $365,000. You're given the interest rate and the other terms. So solve for the initial principal.


See if that gives you the right answer. If it doesn't then one thing that was a little ambiguous is whether or not she's allowed to invest her monthly annuitized payments at this same 6% APR. If she is, the problem becomes more difficult, but it doesn't say. And it becomes difficult enough that I think they don't aren't expecting you to compute it that way. Just use $365k as the future value in a simple compound interest problem and solve for the initial value.

 

[Leezard]

All the available answers are lower than that =/
A. $241,149.00
B. $244,179.39
C. $263,641.08
D. $265,569.78
E. $275,455.75

I think the 6% APR is what gives a different answer. Do you know what Apr is?

 

[systemfehler]

Sorry not a native speaker so I might be dead wrong but if I understood the question right you want to know how much money you need to offer at the first day to get the same amount in the end with monthyl payments at 2% increase for the monthly payments and 6% interest yearly for the whole sum.

To calculate back you can use:

money_after_x_years_months = starting_money * (1 + interest)^ [n * m + k]

Where starting_money is what you want to know - the lump sum. n are the years you let the money grow, m is the compound period = 12 and k would be for months eg. 10 years 2 months would be:

10 * 12 + 2

interest should be: 0.06/m

After that you should be able to calculate back. However I don't seem to get any of those solutions you mentioned when I take the 365.000 someone posted earlier.