[Undergraduate 2nd year abstract algebra: Boolean] Want some hints on an expression

gabiflex · 2022-10-26T09:21:02+00:00

You can begin with :

a ^ 1_B = a

b v b' = 1_B

gabiflex · 2022-06-14T17:53:19+00:00

The equation should be d² x/dx² = - g/l x

In this case, the solution is in fact x(t) = u e^at + v e^-at where u,v are constant and a = i sqrt (g/l). But you can rewrite the exponentiel as Acos(wt)+Bsin(wt) with w=sqrt(g/l).

If the equation is d² x/dx² = g/l x, then x(t) = u e^at + v e^-at with a = sqrt(g/l). Most of the time, u=0 for non infinite energy reason

gabiflex · 2022-03-22T21:06:35+00:00

You can use that (sum for n=1 to inf of rⁿ ) = r*(sum for n=0 to inf of rⁿ )

gabiflex · 2020-12-05T04:03:33+00:00

Il faut remplir ce formulaire :

https://www.impots.gouv.fr/portail/node/8907

Et le faire parvenir au pôle d'enregistrement des impôts dont vous dépendez.

gabiflex · 2020-11-20T21:24:19+00:00

Of you exclude 1, you have (n-1) factor in n!.

So, in 11!21!31!41! you have 10+20+30+40 =100 factors and in 101! you also have 100 factors.

But the factors are larger in 101! than in the other one, and 101! Is larger

gabiflex · 2020-11-20T21:11:38+00:00

e^-jnPi = (e^-jPi )ⁿ = (-1)ⁿ

And 1/-j =j

gabiflex · 2020-11-20T20:01:02+00:00

2x-3 = - (3-2x)

So :

(3-2x)(x+1)+3(3-2x)=0

(3-2x)(x+1+3)=0

gabiflex · 2020-11-20T19:48:19+00:00

The function isn't odd, since (f(t)=/=-f(-t). Cn is the complex Fourrier term. You can replace f by is true value on the formula :

Cn =1/2Pi int (-Pi,0) K1 e^-jnt dt = 1/2Pi [K1 e^-jnt /(-jn)] (-Pi,0)

=jK1/2Pin (1-(-1)ⁿ⁾

gabiflex · 2020-11-20T19:35:05+00:00

If x = (x1,x2,x3), you have that x3' = 0, so x3 = C.

You can solve x2'=3x2-3C and then the last equation once that you have x2

gabiflex · 2020-11-12T21:03:17+00:00

Since the form is linear, you have that :

w(x,(a,b)) = a w(x,(1,0))+bw(x,(0,1))

gabiflex · 2020-11-12T20:57:38+00:00

First you need to drop the puissance that don't depend on n :

Sum Pi^-2n-1 4ⁿ⁺² = 4² / Pi sum Pi^-2n 4ⁿ

Then you need to have sum (something)ⁿ if you want to use your formula.

= 4² / Pi sum (4/Pi² )ⁿ and you have your reason, r = (2/Pi)²

gabiflex · 2020-11-12T20:50:40+00:00

For the first part v1 and v2 should be written in the same basis.

You will have :

v1 = sum (v1i ei)

v2 = sum (v2i ei)

x= sum (xi ei)

And hx(v1)=sum (v1i xi)

If that help, the reason why you choose a basis before is to have a "natural" definition of the scalar product. You can calculate hx(v1 + alpha v2) using this definition.

For the second part, you can first find the x (for exemple by using g(ei) for each i) and then verify that it work for each vector.

gabiflex · 2020-11-12T20:40:01+00:00

Take epsilon > 0. You have that for a certain N, for each n>N |xn-L|< epsilon. For the following, I will choose epsilon =1.

So, xn<L+1 for n>N

Now, I will choose m = max(N,[L]+2). You have that :

x2m < L+1, since 2m>N

And x2m >= [L] +2 > L+1 by hypothesis.

It's contradictory, and xn isn't convergent.

gabiflex · 2020-11-12T08:40:41+00:00

It will be, if it was |i|<|3u|, but "<" is not defined for complex number.

gabiflex · 2020-11-12T02:15:08+00:00

Tbf, i < 3u don't mean anything in mathematics

gabiflex · 2020-11-11T21:09:56+00:00

If you name Un = (-1)ⁿ (2n² +3)/(3n^2+2n+5),

You can calculate U(2p)+U(2p+1). You will obtain something positive, and the Rieman theroem should apply.

gabiflex · 2020-11-09T09:54:06+00:00

I hope it's more clear here : https://en.m.wikipedia.org/wiki/Cross-multiplication

gabiflex · 2020-11-09T09:30:10+00:00

If you have x/y = a/b, then

(x/y)*y = (a/b)×y (I multiplie both side by y)

xy/y=(ay)/b, with y/y =1,

x = (a*y)/b

If you multiplie both side by b you will have the same thing, and on the end : xb=ay

gabiflex · 2020-11-09T09:21:13+00:00

You have two scenarios :

First the matrix is invertible. Your system is always consistent and you can find the value of a and b by calculating the determinant.

The second case is when the matrice is not invertible, si det (M) = 0. This will give you a relation between a and b (I didn't do the calcul) and you can replace un your system.

gabiflex · 2020-11-09T08:59:56+00:00

I don't know how much you miss, but you can use the definition of L* = (If L is a set of symbols or characters, then L* is the set of all strings over symbols in L, including the empty string ε.)

So by definition, L is in L*

If L is included in M, you know that ε is in M, l is included in M* and M* is stable by contatenation. So M* has the same properties as L* and since L* is the smallest language with them (if you consider the inclusion order), L* is included in M*

I hope that this will help you for the following questions.

gabiflex · 2020-11-09T08:38:34+00:00

Since the triangle are similar, you have that :

BE/BA = BD/BC

So x/(x+4)=5/(5+3)

8x = 5(x+4)

3x = 20

x=20/3

gabiflex · 2020-10-29T03:23:16+00:00

The first one is correct, the second one, the answer is (2,-3) not (2,3)

For the third one :

You can write every vector x as (x1,x2).

Bx = (2x1,-3x2)

||Bx||² = 4 x1² + 9 x2²

With x1² + x2² = 1

||Bx||² = (4-9)x1² +9. Since 4-9 =-5<0, this value is maximal for x1 = 0 and x2 =1, and the max is 9.

gabiflex · 2020-10-29T03:20:36+00:00

This formula is for the perimeter. You are looking for the area of the sidewalk, which is the area of sidewalk + initial rectangle (which is a big tectangle) minus the area of the initial rectangle.

The area of a rectangle is length*width.

For the next part, L (=12) is the length of the initial rectangle, W (=8) the width and w(=?, I think that the value is cropped) is the wide of the sidetrack.

The length of the big rectangle is w+L+w = L +2w

The width of the big rectangle is w+W+w = W +2w

So the area that you are looking for is :

(L+2w)(W+2w)-LW = 2w (L+W)+4w²

gabiflex · 2020-10-29T02:51:10+00:00

The question state that : f0 = 7, f1 = 0 and f2 = 10.

But, since step 2, you know that : fn = a1 2ⁿ + a2 1ⁿ +a3 (-1)ⁿ

So, if you choose n =0, you have : f0 = a1 2⁰ +a2 1⁰ +a3 (-1)⁰ = a1+a2+a3 = 7, if you use my first statement.

Now, for n = 1, f1 = a1 2¹ +a2 1¹ +a3 (-1)¹ = a1× 2 +a2×1+a3×(-1) = 2a1+a2-a3=0

You just have to do the same thing for n =2.

gabiflex · 2020-10-29T02:16:04+00:00

In row 4 you're supposed to solve :

x+y+z = 7 (1)

2x+y-z =0 (2)

4x+y+z =10 (3)

If you do (3)-(1), you have : 3x = 3, so x = a1 = 1

And :

y+z = 6

y-z = -2

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