TT's Maths Thread

[TrueTears]

question about multiples of 7: I think it assumes that say, 9 is really 000 000 000 9. I think if you included that smaller numerator then you wouldn't get a lower bound of 15456 but something smaller and hence you will not be able to conclude that it's greater than 10000.

Oh I see, but say if I worked out the actual "number" of 10 digit multiples of 7, it wouldn't really affect anything, just a smaller lower bound heh

yeah but if it is too small, we cannot say at least 10000

Exactly, that's why the answer would be "no".

[kamil9876]

no, the conclusion woudl be "can't tell". (lower bound is too weak for pigeonholing)

[TrueTears]

no, the conclusion woudl be "can't tell". (lower bound is too weak for pigeonholing)

Oh I see then there is no way of doing the question then... unless you assume 0,000,000,009 is a number...?

[TrueTears]

For Q 2 I thought about it this way.

Consider the set

Let denote the subsets of set .



Partitioning into mutually disjoint sets that have elements where . Let denote this.





Notice how the number of subsets with an even number of elements is

Generalizing this result with the set with terms such that

Then

This means that the number of subsets with an even number of elements is

Let and denote the cardinality of the number of subsets with an even and odd number of elements respectively.

Thus



which proves that for any set prove that the number of its subsets with an even number of elements is equal to the number of subsets with an odd number of elements.


I think this is a fairly weak proof as I didn't really show how was derived, I merely used a pattern found from the case

Can anyone actually show that for any set with elements, the number of subsets that contain an even number of elements is ? Thanks!

[kamil9876]

Here is another proof for q2:

Suppose we want all the subsets of {a,b,c,d....}

now we will write a list of all the subsets of the set. On the top row, we will write down all the subsets that do not contain a, and on the bottom we will write down the same subset but with a added in:

null, {b}, {c}.... {b,c}, {b,d}.... {b,c,d...}
{a},{ab},{ac}....{a,b,c},{a,b,d}... {a,b,c,d..}

We can see, this mapping is a bijection. However the bottom row contains one more element (a is introduced) thus two subsets in the same COLUMN have opposite parity, proving the result.

This picture can also be used to prove that the number of subsets of a set is (ie the top row contains the subsets of the set with one less element, and introducing the bottom row doubles the number of subsets thus showing where is the number of subsets of a set with n elements)

[TrueTears]

Here is another proof for q2:

Suppose we want all the subsets of {a,b,c,d....}

now we will write a list of all the subsets of the set. On the top row, we will write down all the subsets that do not contain a, and on the bottom we will write down the same subset but with a added in:

null, {b}, {c}.... {b,c}, {b,d}.... {b,c,d...}
{a},{ab},{ac}....{a,b,c},{a,b,d}... {a,b,c,d..}

We can see, this mapping is a bijection. However the bottom row contains one more element (a is introduced) thus two subsets in the same COLUMN have opposite parity, proving the result.

This picture can also be used to prove that the number of subsets of a set is (ie the top row contains the subsets of the set with one less element, and introducing the bottom row doubles the number of subsets thus showing where is the number of subsets of a set with n elements)

Ahh that's very smart and answers why it is quite well. Thanks

[TrueTears]

3. Use a combinatorial argument to show that with

I think I got the combinatorial proof but if anyone can show the algebraic proof it would be very much appreciated

My working is as follows:

Consider picking fruits from apples and oranges. (Assume that each apple and orange is distinguishable)

Suppose the following:

If we pick apples then we must pick oranges.

If we pick apple then we must pick oranges.

If we pick apples then we must pick oranges.

.
.
.

If we pick apples then we must pick oranges.

Summing up the different cases we get:



But




So yeah if anyone can explain an algebraic method of doing it, that would be good

[addikaye03]

1) When two resistors of resistances z1 and z2 are connected in series, their equivalent resistance z is given by z=z1+z2, and when they are connected in parallel their equivalent resistance z is given by 1/z=1/z1+1/z2.

In an electric circuit that contains two resistors whose resistances are z1=R1+iwL and z2=R2-i/(wC) respectively, find the value of w so that their equivalent resistance is purely real when their resistors are

i) Connected in Series
ii)Connected in Parallel

2)Decompose the integrand into partial fractions with complex linear denominators, hence, find

int. dx/(x^8-1)

* i got the answer with plenty of tedious algebra + noting (x^4-1)(x^4+1) Hence roots of unity of (x^8-1), but want a quicker way*

3) A car of mass M kg, width w metres and centre of mass h metres above the ground travels round a level curve of radius r metres with a constant speed v m/s such that driver's right-hand side is near centre of the curve.

a) By drawing the front view of the car and two normal reactions of the road's surface on the right and the left wheels (neglect length of car), show that the right and left normal reactions respectively are:

M/2rw(rgw-2hv^2) and M/2rw(rgw+2hv^2)

b) hence, show that the car overturns when v^2>= rgw/2h

[Over9000]

1) When two resistors of resistances z1 and z2 are connected in series, their equivalent resistance z is given by z=z1+z2, and when they are connected in parallel their equivalent resistance z is given by 1/z=1/z1+1/z2.

In an electric circuit that contains two resistors whose resistances are z1=R1+iwL and z2=R2-i/(wC) respectively, find the value of w so that their equivalent resistance is purely real when their resistors are

i) Connected in Series
ii)Connected in Parallel

2)Decompose the integrand into partial fractions with complex linear denominators, hence, find

int. dx/(x^8-1)

* i got the answer with plenty of tedious algebra + noting (x^4-1)(x^4+1) Hence roots of unity of (x^8-1), but want a quicker way*

3) A car of mass M kg, width w metres and centre of mass h metres above the ground travels round a level curve of radius r metres with a constant speed v m/s such that driver's right-hand side is near centre of the curve.

a) By drawing the front view of the car and two normal reactions of the road's surface on the right and the left wheels (neglect length of car), show that the right and left normal reactions respectively are:

M/2rw(rgw-2hv^2) and M/2rw(rgw+2hv^2)

b) hence, show that the car overturns when v^2>= rgw/2h

Are u joking?
What part of TT'S Maths thread dont you understand

[TrueTears]

1) When two resistors of resistances z1 and z2 are connected in series, their equivalent resistance z is given by z=z1+z2, and when they are connected in parallel their equivalent resistance z is given by 1/z=1/z1+1/z2.

In an electric circuit that contains two resistors whose resistances are z1=R1+iwL and z2=R2-i/(wC) respectively, find the value of w so that their equivalent resistance is purely real when their resistors are

i) Connected in Series
ii)Connected in Parallel

2)Decompose the integrand into partial fractions with complex linear denominators, hence, find

int. dx/(x^8-1)

* i got the answer with plenty of tedious algebra + noting (x^4-1)(x^4+1) Hence roots of unity of (x^8-1), but want a quicker way*

3) A car of mass M kg, width w metres and centre of mass h metres above the ground travels round a level curve of radius r metres with a constant speed v m/s such that driver's right-hand side is near centre of the curve.

a) By drawing the front view of the car and two normal reactions of the road's surface on the right and the left wheels (neglect length of car), show that the right and left normal reactions respectively are:

M/2rw(rgw-2hv^2) and M/2rw(rgw+2hv^2)

b) hence, show that the car overturns when v^2>= rgw/2h

Hey addikaye03 didn't you already advertise your questions in this thread just on the page before?

http://vcenotes.com/forum/index.php/topic,20009.0.html

I asked some more Q here, if people would like to attempt them, they're interesting imo

So I would appreciate it if you would stop spamming. Thanks

[TrueTears]

4. We are given points arranged around a circle and the chords connecting each pair of points are drawn. If no three chords meet in a point, how many points of intersection are there? For example, when , there are intersections.

Looking at when there are no intersections.

When , there is only one intersection. At the intersection point 1, it is formed by the 4 points a,b,c,d.

When , there are five intersections.

1 is formed by e,a,b,c.

2 is formed by d,a,b,c.

3 is formed by e,b,c,d.

4 is formed by e,a,c,d.

5 is formed by e,a,b,d.

We can see that every intersection is formed by 4 different points. For , there are a total of combinations of a,b,c,d,e.

for points, there are intersections.

[TrueTears]

A few more interesting Q's.

1. In an office, at various times during the day, the boss gives the secretary a letter to type, each time putting the letter on top of the pile in the secretary's in-box. When there is time, the secretary takes the top letter off the pile and types it. There are nine letters to be typed during the day and the boss delivers them in the order 1,2,3,4,5,6,7,8,9. While leaving for lunch, the secretary tells a colleague that letter 8 has already been typed, but says nothing else about the morning's typing. The colleague wonders which of the nine letters remain to be typed after lunch and in what order they will be typed. Based upon the information given, how many such after-lunch typing orders are possible? (There are no letters left to typed is one of the possibilities.)

2. Let be the set of subsets of . Let be the number of functions such that . Prove that .

[zzdfa]

Hey, with the 2nd question, I think you made a mistake somewhere, you have c as a function of n and m but I don't see an m anywhere.

[TrueTears]

Hmm, I rechecked it just then, I typed the exact thing from art n craft

TBH I don't really understand the notation used in this question, what does mean?

[zzdfa]

I found the error, it should be f: Pn -> {1,2,....,m}

it means that the functions we are counting have to satisfy the property that:

if you have two sets A and B,

and you apply f to the intersection of them,

then the result is the same as if you

applied f to both of them seperately,
then picked the smallest output.