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3 years ago

What are all the subsets of {5,9,13}

Mathematics

2 answers:

UNO [17]3 years ago

7 0

(5),(9),(13),(5,9),(9,13),(5,13),(5,9,13) and empty set

Naddika [18.5K]3 years ago

6 0

All The Subsets

For theset {a,b,c}:

The empty set {} is a subset of {a,b,c}And these are subsets: {a}, {b} and {c}And these are also subsets: {a,b}, {a,c} and {b,c}And {a,b,c} is a subset of {a,b,c}

And when we list all the subsets of S={a,b,c} we get the Power Set of {a,b,c}:

P(S) = { {}, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} }

Think of it as all the different ways we can select the items (the order of the items doesn't matter), including selecting none, or all.

Example: The shop has banana, chocolate and lemon ice cream.

What do you order?

Nothing at all: {}Or maybe just banana: {banana}. Or just {chocolate} or just {lemon}Or two together: {banana,chocolate} or {banana,lemon} or {chocolate,lemon}Or all three! {banana, chocolate,lemon}

Question: if the shop also has strawberry flavor what are your options? Solution later.

How Many Subsets

Easy! If the original set has n members, then the Power Set will have 2n members

Example: in the {a,b,c} example above, there are three members (a,b and c).

So, the Power Set should have 23 = 8, which it does!

Notation

The number of members of a set is often written as |S|, so when S has n members we can write:

|P(S)| = 2n

Example: for the set S={1,2,3,4,5} how many members will the power set have?

Well, S has 5 members, so:

|P(S)| = 2n = 25 = 32

You will see in a minute why the number of members is a power of 2

It's Binary!

And here is the most amazing thing. To create the Power Set, write down the sequence of binary numbers (using n digits), and then let "1" mean "put the matching member into this subset".

So "101" is replaced by 1 a, 0 b and 1 c to get us {a,c}

Like this:

abcSubset0000{ }1001{c}2010{b}3011{b,c}4100{a}5101{a,c}6110{a,b}7111{a,b,c}

Well, they are not in a pretty order, but they are all there.

Another ExampleLet's eat! We have four flavors of ice cream: banana, chocolate, lemon, and strawberry. How many different ways can we have them?Let's use letters for the flavors: {b, c, l, s}. Example selections include:{} (nothing, you are on a diet){b, c, l, s} (every flavor){b, c} (banana and chocolate are good together)etcLet's make the table using "binary": bclsSubset00000{}10001{s}20010{l}30011{l,s}...... etc ..... etc ...121100{b,c}131101{b,c,s}141110{b,c,l}151111{b,c,l,s}

And the result is (more neatly arranged):

P = { {}, {b}, {c}, {l}, {s}, {b,c}, {b,l}, {b,s}, {c,l}, {c,s}, {l,s}, {b,c,l}, {b,c,s},
{b,l,s}, {c,l,s}, {b,c,l,s} }

SymmetryIn the table above, did you notice that the first subset is empty and the last has every member?But did you also notice that the second subset has "s", and the second last subset has everything except "s"? In fact when we mirror that table about the middle we see there is a kind of symmetry.This is because the binary numbers (that we used to help us get all those combinations) have a beautiful and elegant pattern.A Prime Example

The Power Set can be useful in unexpected areas.

I wanted to find all factors (not just the prime factors, but all factors) of a number.

I could test all possible numbers: I could check 2, 3, 4, 5, 6, 7, etc...

That took a long time for large numbers.

But could I try to combine the prime factors?

Let me see, the prime factors of 510 are 2×3×5×17 (using prime factor tool).

So, all the factors of 510 are:

2, 3, 5 and 17,2×3, 2×5 and 2×17 as well, and2×3×5 and 2×3×17 and ..... aha! Just like ice cream I needed a Power Set!

And this is what I got:

2,3,5,17SubsetFactors of 51000000{ }110001{17}1720010{5}530011{5,17}5 × 17 = 8540100{3}350101{3,17}3 × 17 = 51 ... etc ...... etc ...... etc ...151111{2,3,5,17}2 × 3 × 5 × 17 = 510

And the result? The factors of 510 are 1, 2, 3, 5, 6, 10, 15, 17, 30, 34, 51, 85, 102, 170, 255 and 510 (and −1, −2, −3, etc as well). See the All Factors Tool.

Automated

I couldn't resist making Power Sets available to you in an automated way.

So, when you need a power set, try Power Set Maker.

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the graph of y=e^tanx-2 crosses the x axis at one point in the interval [0,1]. What is the slope of the graph at this point? I g

Svetlanka [38]

Y = e^tanx - 2

To find at which point it crosses x axis we state that y= 0

e^tanx - 2 = 0
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x = 0.6061

to find slope at that point first we need to find first derivative of funtion y.

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3 years ago

Solve the following differential equation using using characteristic equation using Laplace Transform i. ii y" +y sin 2t, y(0) 2

kifflom [539]

Answer:

The solution of the differential equation is $y(t)= - \frac{1}{3} Sin(2t)+2 Cos(t)+\frac{5}{3} Sin(t)$

Step-by-step explanation:

The differential equation is given by: y" + y = Sin(2t)

i) Using characteristic equation:

The characteristic equation method assumes that y(t)= $e^{rt}$ , where "r" is a constant.

We find the solution of the homogeneus differential equation:

y" + y = 0

$y'=re^{rt}$

$y"=r^{2}e^{rt}$

$r^{2}e^{rt}+e^{rt}=0$

$(r^{2}+1)e^{rt}=0$

As $e^{rt}$ could never be zero, the term (r²+1) must be zero:

(r²+1)=0

r=±i

The solution of the homogeneus differential equation is:

$y(t)_{h}=c_{1}e^{it}+c_{2}e^{-it}$

Using Euler's formula:

$y(t)_{h}=c_{1}[Sin(t)+iCos(t)]+c_{2}[Sin(t)-iCos(t)]$

$y(t)_{h}=(c_{1}+c_{2})Sin(t)+(c_{1}-c_{2})iCos(t)$

$y(t)_{h}=C_{1}Sin(t)+C_{2}Cos(t)$

The particular solution of the differential equation is given by:

$y(t)_{p}=ASin(2t)+BCos(2t)$

$y'(t)_{p}=2ACos(2t)-2BSin(2t)$

$y''(t)_{p}=-4ASin(2t)-4BCos(2t)$

So we use these derivatives in the differential equation:

$-4ASin(2t)-4BCos(2t)+ASin(2t)+BCos(2t)=Sin(2t)$

$-3ASin(2t)-3BCos(2t)=Sin(2t)$

As there is not a term for Cos(2t), B is equal to 0.

So the value A=-1/3

The solution is the sum of the particular function and the homogeneous function:

$y(t)= - \frac{1}{3} Sin(2t) + C_{1} Sin(t) + C_{2} Cos(t)$

Using the initial conditions we can check that C1=5/3 and C2=2

ii) Using Laplace Transform:

To solve the differential equation we use the Laplace transformation in both members:

ℒ[y" + y]=ℒ[Sin(2t)]

ℒ[y"]+ℒ[y]=ℒ[Sin(2t)]

By using the Table of Laplace Transform we get:

ℒ[y"]=s²·ℒ[y]-s·y(0)-y'(0)=s²·Y(s) -2s-1

ℒ[y]=Y(s)

ℒ[Sin(2t)]= $\frac{2}{(s^{2}+4)}$

We replace the previous data in the equation:

s²·Y(s) -2s-1+Y(s) = $\frac{2}{(s^{2}+4)}$

(s²+1)·Y(s)-2s-1= $\frac{2}{(s^{2}+4)}$

(s²+1)·Y(s)= $\frac{2}{(s^{2}+4)}+2s+1=\frac{2+2s(s^{2}+4)+s^{2}+4}{(s^{2}+4)}$

Y(s)= $\frac{2+2s(s^{2}+4)+s^{2}+4}{(s^{2}+4)(s^{2}+1)}$

Y(s)= $\frac{2s^{3}+s^{2}+8s+6}{(s^{2}+4)(s^{2}+1)}$

Using partial franction method:

$\frac{2s^{3}+s^{2}+8s+6}{(s^{2}+4)(s^{2}+1)}=\frac{As+B}{s^{2}+4} +\frac{Cs+D}{s^{2}+1}$

2s^{3}+s^{2}+8s+6=(As+B)(s²+1)+(Cs+D)(s²+4)

2s^{3}+s^{2}+8s+6=s³(A+C)+s²(B+D)+s(A+4C)+(B+4D)

We solve the equation system:

A+C=2

B+D=1

A+4C=8

B+4D=6

The solutions are:

A=0 ; B= -2/3 ; C=2 ; D=5/3

So,

Y(s)= $\frac{-\frac{2}{3} }{s^{2}+4} +\frac{2s+\frac{5}{3} }{s^{2}+1}$

Y(s)= $-\frac{1}{3} \frac{2}{s^{2}+4} +2\frac{s }{s^{2}+1}+\frac{5}{3}\frac{1}{s^{2}+1}$

By using the inverse of the Laplace transform:

ℒ⁻¹[Y(s)]=ℒ⁻¹[ $-\frac{1}{3} \frac{2}{s^{2}+4}$ ]-ℒ⁻¹[ $2\frac{s }{s^{2}+1}$ ]+ℒ⁻¹[ $\frac{5}{3}\frac{1}{s^{2}+1}$ ]

$y(t)= - \frac{1}{3} Sin(2t)+2 Cos(t)+\frac{5}{3} Sin(t)$

3 0

3 years ago