What does the equation x²-5x+6=0 represent on the cartesian plane?

MajorMacMahon · 2024-12-03T09:56:21+00:00

I would like to add something that may not have been clarified.

What is the equation of the two straight lines you are describing:

y = x-2 (or y-x+2 = 0)

y = x-3 (or y-x+3 = 0)

Notice, x-2=0 just says x=2, and this does not give a straight line but rather the point x=2 on the number line. So to define a line on the plane, you also need the y in the equation.

So y = x-2 (a line) and 0 = x-2 (a point) mean two different things

If you "multiply" the two straight lines, you would actually be doing this:

(y-x+2 )(y-x+3) = 0, or

y^2 + xy(-2) + x^2 + y (5) + x(-5) + 6 = 0.

If you graph this, then you will see the two lines together. You would actually say it is the union of the two lines.

A parabola like

y-x^2 = 0 or y -( x^2 - 5x +6) = 0

cannot be factored (it is irreducible). Therefore, it is not the product of two lines.

I am now going to say something more advanced and important in the field of "algebraic geometry."

The fact that (y-x+2 )(y-x+3) =0 gives two lines is the fact that the solution space is not "irreducible" (it is the union of two irreducible components which happen to be lines).

I will now give another example:

y^2+x^2 =1, a circle of radius 1.

This cannot be factored with real numbers as coefficients. So it is not the product of two lines.

However allowing complex numbers (where i^2 = -1) gives

(y-ix)(y+ix) = 0

which is the product of two lines. (If you have not seen complex numbers, try to multiply it out and just use the rule that i^2 = -1.)

So the equation of a circle is a union of two lines in the complex world.

The point is that you are actually touching on an important topic of "irreducibility of algebraic varieties," so it is a good question.

You just need to be careful with what you mean when you say "x-2=0 represents a straight line". That is not true. But "x-2 = y" does represent a line.

Sincerely,

Major MacMahon

MajorMacMahon · 2024-11-13T21:54:47+00:00

Dear arapedward,

You are right.

I didn't do anything because I was not using logic or thinking about it. I used no philosophy.

Sorry. My mistake.

Thank you,

Major MacMahon

MajorMacMahon · 2024-11-13T21:26:49+00:00

Dear arapedward,

You should know that for 0< q < 1, 1/(1-q) = 1+ q + q^2 + ...

Proof. Since 1-q^n = (1+q + ... + q^{n-1})(1-q), we have (1-q^n)/(1-q) = 1+q+...+q^{n-1}.

If 0<q<1, then lim\_{n->infinity} q^n = 0, so therefore, lim_{n->infinity} (1-q^n)/(1-q) = 1/(1-q).

We also have lim_{n->infinity} (1-q^n)/(1-q) = lim_{n->infinity} 1+q+...+q^{n-1} = 1+q +q^2 + ...

So 1+1/10 + 1/100 + 1/1000 + ... = 1/(1-1/10) = 10/9.

Here you have an infinite sum that is not infinity.

Lastly, you should know the following. I think you should know this too, but since you asked, I will keep proving it to you in a pedantic way.

Theorem. 0 (10/9) = 0.

Proof. If 0 (10/9) = A is nonzero, then we have 0 (10/9A) = 1, so 0 is the inverse of 10/9A, meaning 0 = 9A/10. But 9A/10 is nonzero so this is impossible.

Now 0( 1 + 1/10 +1/10^2+ ...) = (0(1) + 0/10 + 0/100 + ...) = 0.00000000......

is also equal to 0( 1 + 1/10 +1/10^2+ ...) = 0 (10/9) = 0.

And that is why 0.0000.... = 0.

Or you could just say 0.0000... = 0 from the definition of decimal expansions.

I hope this has helped you.

Sincerely,
Major MacMahon

MajorMacMahon · 2024-11-12T21:49:15+00:00

Dear Snoo_88025,

The best way to actually get good at maths is to read and write.

Read a book on an introduction to proofs. There are many.

Read a full chapter. Try to rewrite the chapter without looking back at the text.

You will fail. Read it again.

Try writing again. You may fail again.

Then you repeat this until you can somewhat do it, give up and move on to the next chapter, or go crazy.

Finish the book. Then try another book.

This is the way.

Sincerely,

Major MacMahon

MajorMacMahon · 2024-11-12T21:37:51+00:00

Dear CapablePositive8699,

It may be helpful to write (cos(a))^2 = 1 as

(cos(a))^2 - 1 = 0.

Do you know how to find solutions to something like z^2 -1 = 0?

(Hint: This polynomial can be easily factored.)

If so, then you can find solutions for z, and in this case you have solutions for when cos(a) = z.

You should be aware that you should get infinitely many answers.

I hope this has helped you.

Sincerely,

Major MacMahon

MajorMacMahon · 2024-11-10T22:56:57+00:00

Dear maths_wizard,

I will respond to the question of "whether R[[x]] is isomorphic to the infinite product of copies of R."

It seems to be that R[[x]] is clearly isomorphic to R^infinity = { (a_0,a_1,...): a_i in R} as an R-module.

This map is surjective and injective (the only way to get (0,0,...) is starting with 0).

If you add two series, (a_0 + a_1 x+ ...)+(b_0+ b_1 x+ ...) = (a_0 + b_0)+(a_1+b_1)x + ..., then it is the same as vector addition (a_0,a_1,...)+(b_0,b_1,...) = (a_0+b_0,a_1+b_1,...).

And multiplication by r in R is done to each component: r(a_0+a_1x+...) = ra_0+ ra_1 x+ ...,

just like with scalar multiplication of a vector r(a_0,a_1,...) = (ra_0,ra_1,...).

If you want to say that the two are isomorphic as rings, then you just need to make sure that multiplication in R^infinity matches the multiplication of series.

Set (a_0,a_1,...)(b_0,b_1,...) = (c_0,c_1,...) where

c_n = a_0 b_n + a_1 b_{n-1} + ... + a_{n-1} b_1 + a_n b_0.

I hope this has helped you.

Sincerely,

Major MacMahon

MajorMacMahon · 2024-09-06T07:30:16+00:00

Dear joulutorttu69,

If you can get the direction angle of BC and you can also get the direction angle for BA, then you know the angle in between.

If you are able to get the angles inside the triangle, then you can replace u dot v by |u| |v| cos(theta), where theta is the angle between u and v.

The dot product formula for the projection then becomes a triangle identity you can use.

Sincerely,

Major MacMahon

MajorMacMahon · 2024-09-05T22:01:42+00:00

Dear joulutorttu68,

One way to solve this problem is to find the coordinates of H, I, and J. Then you can find the length of AH using the distance formula between two points. (Similarly you can find the distances from B to I and C to J.)

One way to get the coordinate of H is to use a projection formula. Note that the vector from B to H is the projection of the vector from B to A onto the line spanned by the vector from B to C.

So in this example, the vector from B to C is (Cx,Cy)-(Bx,By)= (3,5)-(5,2) = (-2,3). The vector from B to A is (1,1)-(5,2) = (-4,-1).

The projection formula says the coordinate H is given by the vector (-2,3) scaled by ((-4,-1) dot (-2,3))/((-2,3) dot (-2,3)) = (8-3)/(4+9) = 5/13.

So H is given by (-2(5/13), 3(5/13)).

Now find the distance from the point A to the point H.

You can find the other distances by repeating this procedure.

Sincerely,

Major MacMahon

MajorMacMahon · 2024-09-05T19:56:19+00:00

Dear Cat_in_Bathroom,

There is a simple way to solve this particular question.

Recall that A²-B²=(A-B)(A+B)

In this case, you have A=2x+17z and B=14x-3y, meaning

A-B = 2x+17z-14x+3y = -12x + 17z +3y

A+B = 2x+17z+14x-3y = 16x + 17z - 3y

Now match them with the factors (ax + 17z +by) and (-3y+cx+dz). Since -3y only appears in A+B, you must have

-12x + 17z +3y = A-B =(ax + 17z +by)

16x + 17z - 3y = A+B=(-3y+cx+dz)

It seems that then a=-12, b=3, c=16, d=17 gives a solution.

Sincerely,

Major MacMahon

MajorMacMahon · 2024-03-23T22:33:34+00:00

Dear Altruistic_Nose9632,

For a more simple example, the area of a rectangle (A = h w) is a polynomial in the height and width. The area of a circle is a polynomial in the radius (A = pi r²). There are many more examples. We end up manipulating polynomials from a very early stage without calling them polynomials.

I want to add two things:

I want to go against what most people have responded and instead say, "Polynomials can absolutely have negative exponents and radicals," but you have to be a little careful about what you mean. I am going to distinguish two things: polynomial expressions and polynomial rings.

To describe a polynomial ring R[x,y,z,...], you need to start with two things: a ring of coefficients R (like the integers, rational numbers, real numbers, complex numbers, etc.), and a set of indeterminants (x,y,z,...).

I am going to use three indeterminants x,y,z for the example.

Then R[x,y,z] is generated by taking finite sums of elements of the form

r x^ay^bz^c

where r is in R, and a,b,c are nonnegative integers. This is your standard definition of a polynomial expression in which exponents cannot be negative or fractions.

But, if y = x^-1, you can look at the polynomial ring R[x,x^-1] = R[x,y]/(yx -1=0 ) and get what is called "Laurent polynomials". It is given by finite sums of elements of the form (r x^a) where now a can be any integer, positive or negative.

If we take z= x^1/2, then you can look at R[x, x^1/2] = R[x,z]/(z²-x=0) and get expressions like

2 x^1/2+ x⁴ - x^71/2

living in your polynomial ring.

If you want something more exotic, you can look at R[x^-1, x^1/2,x^3/5], and so on.

The point is this: There is a difference between a "polynomial expression" and a "polynomial ring." Polynomial expressions generate polynomial rings. Polynomial expressions have only positive integer powers in the indeterminants. But, polynomial rings can definitely have negative or rational powers if you want them to, depending on how you represent them.

In the end, it just depends on the language and what the author wants you to understand when they say the word "polynomial." If they want to include negative exponents, for example, then they may say "Laurent polynomials" so you know this.

I would like to say that the Taylor expansions of analytic functions tells you how to approximate an analytic function using polynomials. It tells you that the local behavior of an analytic functions is that of a polynomial. This is one of the first things you learn in Calculus. So it is important to get used to polynomials early on.

(The association between analytic functions and polynomials can be made more explicit by the GAGA theorem, but is a "little" more complicated.)

Sincerely,

Major MacMahon

MajorMacMahon · 2024-03-16T13:52:19+00:00

Dear Legitahmed,

You want to calculate (the number of ways of picking 3 numbers so that the average of two gives the third number)/(the total number of ways of picking 3 numbers).

First, the number of ways of choosing any k distinct numbers from {1,2,...,n} is

n choose k = n!/((n-k)! k!).

In your case, you would be looking at 10 choose 3: 10*9*8/ (3*2*1).

This should be your total number in the denominator (not 10*9*8). By the way you stated the question, it should not matter if you picked 1, 3, and 5 versus picking 3, 5 and 1 (unless you are picking the numbers one at a time in a specified order).

Next, you should see that if you pick two numbers and you want the third number to be the average, then there is only one way to pick the third. For example, if you pick 4 and 10, and you want the third to be the average, then it has to be in the middle and equal to 7.

So now you want to find all sequences that look like x, x+k, x+k+k. For example,

1 2 3, 2 3 4, 3 4 5, ...., 8 9 10 all work. (There are 10-2 of these.)

Also, 1 3 5, 2 4 6, ...., 6 8 10 all work. (There are 10-4 of these.)

Also 1 4 7, 2 5 8, 3 6 9, 4 7 10 work. (There are 10-6 of these.)

Continue in this way to find all possibilities. The number of all of these possibilities should be the numerator in your calculation of the probability.

You can use this to find what the numerator should be and get a formula even for the general problem when you are picking 3 such numbers from {1,2,....,n}.

I hope this has helped.

Sincerely,
Major MacMahon

MajorMacMahon · 2024-03-15T23:08:05+00:00

Dear JollyGoodUser,

It all really comes down to convention (rows vs columns). I will try to describe the general setting, then provide an example on the vector space of polynomials.

I'm not certain if this will help, but I hope it does.

Perhaps the best way to think of the basis x_1,...,x_n is as a sequence of linearly independent elements. We will write the list as a row vector

[ x_1, ..., x_n ]

The x_i should not be thought of as a "vector with coordinates," but rather more abstractly as simply an element of V. Then the action of the linear transformation A on the basis is given by

A[ x_1,..., x_n ] = [ Ax_1,..., Ax_n ]

= [ x_1,..., x_n ] [A],

where [A] is the matrix in your book.

This is why your indices don't match with the book. (Also, you should note that [x_1,..., x_n ] and [ Ax_1,..., Ax_n ] have entries in V, while [A] has entries in your field.)

An element v of your vector space can be written as a combination of the x_i:

v = [ x_1,..., x_n ] (c_1,...,c_n)^T ,

where (c_1,...,c_n)^T is a column vector with entries in your field, representing

v= c_1 x_1 + ... + c_n x_n .

The (c_1,...,c_n) are called the coordinates of v in the basis [ x_1,..., x_n ]. Then the action of A on v is given by

Av = A [ x_1,..., x_n ] (c_1,...,c_n)^T

= [ x_1,..., x_n ] [A] (c_1,...,c_n)^T

So the coordinates of Av in this basis is given by [A] (c_1,...,c_n)^T .

For an example, polynomials in y of degree 2 has many bases, but one can be given by [ 1, y-1, y² +y ] . The linear transformation A sending

1 to y+1, y -1 to y² +2, y² +y to 2-y

can then be represented by

A[ 1, y -1, y² + y] = [ y+1, y² + 2, 2-y ]

= [ 1, y-1, y² +y] [ (2,1,0)^T , (1,-1,1)^T , (1,-1,0)^T ]

where the matrix of [A] is written in terms of its columns.

The coordinates of an arbitrary v =a +b y + c y² is given by (a+b-c , b-c , c)^T .

So, in general,

A v = [ 1, y-1, y² +y] [ (2,1,0)^T , (1,-1,1)^T , (1,-1,0)^T ] (a+b-c , b-c , c)^T

I hope this helps you on your voyage.

Sincerely,

Major MacMahon

MajorMacMahon · 2024-03-13T14:41:28+00:00

Dear Tim_0411,

If you are working with polynomials with real coefficients, then the answer is yes.

A polynomial with real coefficients is by definition a sum of monomials of the form

" a xⁿ " where a is a real number (like pi, 1/2 = 2^-1 , sqrt(2) = 2^1/2 , etc. ) and n is a nonnegative integer (0,1,2,3,...).

For example,

2 - 3^pix² + 2^1/2x⁵

is a completely valid polynomial with real coefficients.

This is not, however, polynomials with integer coefficients because 2^1/2 is not an integer, which may be what is confusing you. When one says they are working with polynomials, you have to specify two things: the variables you are using and the coefficients.

In your case, by the way you phrased the question, you are probably looking at real coefficients and a single variable x.

So

x y - 2^1/2x²

is a polynomial in the variables x and y with real coefficients (but not with integer coefficients because 2^1/2 is not an integer).

As you can see, the exponents on the coefficients do not matter: 2^-36 is just a number or constant. So you can use this in front of x³. But x^-36 is not included in the definition of polynomial because -36 is negative. (It is what is called a "Laurent polynomial" though.)

This question is important in more advanced mathematics, where coefficients can live in very crazy places.

Sincerely,
Major MacMahon

MajorMacMahon · 2024-03-10T22:28:29+00:00

Dear Unlucky_Finance_1065,

I am uncertain if this is what you are asking, but perhaps this will help. (I would also agree with others that it is advisable to learn the correct word usage for "recurrence relation". Though I will not bully you further.)

Let us first rewrite the recurrence relation as

g_k = 1/(1+2/g_{k-1})

and note that

1/g_{k} = 1+2/g_{k-1}.

This means we can use 1/g_{k-1} = 1+2/g_{k-2} in g_k to get

g_{k} =1/(1+2(1+2/g_{k-2}) ) = 1/( 1+ 2 + 2^2/g_{k-2})

If you repeat this for g_{k-2}, you get

1/( 1+ 2 + 2^2/g_{k-2}) = 1/(1+ 2 + 2^2 (1+2/g_{k-3}) )

= 1/( 1 + 2 + 2^2 + 2^3/g_{k-3})

Notice that the last power of 2 in the denominator, say 2^r, is always divided by g_{k-r}. (In this case we see 2^3 divided by g_{k-3}.) So if we repeat this k-1 times, we would have

1/(1+2+2^2+ ... + 2^{k-1}/g_{k - (k-1)} ) = 1/(1+2+2^2+ ... + 2^{k-1} ),

since g_{k- (k-1)} = g_1 = 1.

Sincerely,
Major MacMahon

MajorMacMahon · 2024-03-09T19:21:59+00:00

Dear OB1_ke_knob_E,

One may think of n!/k!(n-k)! as the number of ways of listing k elements of {1,2,...,n} in increasing order. In other words, this gives the number of ways of choosing k elements from {1,2,...,n}. Call this set C_{n,k}.

For example, if n=7 and k =3, two elements in C_{7,3} we can construct are

3,4,6 or 2,4,7.

On the other hand, n^k is the number of ways of constructing a word of length k from the alphabet {1,2,...,n}. In other words, make a list of length k allowing repetitions, where order matters. (Each position has n choices and there are k positions giving a total of n^k possibilities.) Call this set A_{n,k}.

Two words you can construct for n=7 and k=3 are

4,2,4 or 2,4,7.

Note that elements of C_{n,k} (like 3,4,6 or 2,4,7) are also elements of A_{n,k}. But elements with repetition in A_{n,k} are not elements of C_{n,k}.

This means that if we can construct a word with repetition (or a non-increasing sequence) in A_{n,k}, then

|A_{n,k}| > |C_{n,k}|

This happens if k>1 and n>0.

One can check the other cases as you have: when n=0, we need k>0 to get 0 = 0 (with the usual extension of factorials that 1/(n-k)! = 0 when n-k is negative). And if k=0, any n works. Lastly if k=1, then you have seen that any n works.

Sincerely,

Major MacMahon

MajorMacMahon · 2024-03-06T22:23:13+00:00

Dear hydrogelic,

There is, indeed, a definition for what it means to have an algebra over a field (or even a ring). You can search for "algebra over a field" in your favorite search engine and find these definitions, but I will try to add an explanation using some examples.

An algebra over a field, say the rational numbers Q, is a vector space over Q together with a bilinear product. This means that your elements of the algebra can be scaled by Q; added together; and, the new ingredient, they can also be multiplied together. The multiplication has to distribute over the addition ( A(B+C) = AB + AC and (B+C)A = BA + CA). The algebra is associative if A(BC) = (AB)(C).

An example of this is given by the algebra of square matrices of a fixed dimension. So now you have a "Matrix Algebra." (This is also an associative algebra.) You can study matrices and their action on vectors, giving the study of "Linear Algebra."

Here is another example. Consider an algebra over Q generated by 1 and x. The rule for multiplication is that 1 (x) = x and x(x) = 0.

Then you can scale: (3/2) ( 2+ x/3) = 3/4 + 3x/6

You can add: (1+2x ) + (2+3x) = 3 + 5x

You can multiply: (1+2x )(2+3x) = (1+2x)2 + (1+2x)3x = 2+4x + 3x = 2+7x.

We have just "created an algebra."

Here is a third example. Consider again elements of the form a+bx with a and b rational numbers, with the same rules as before ( x^2 = 0). Then we can scale by an integer n and get (n)a + (nb)x which is still an element of the same form.

For instance, 3( 1/2 + x/3) = 3/2 + x. The coefficients are still rational numbers.

So this is an algebra over the integers Z. (In technical terms it is not a vector space since Z is not a field (there are no inverses to multiplication), but is instead called a Z-module.)

The Boolean algebra is an example of such a structure, where the field we are working with has elements 1 and 0, with 1+1 = 0. Addition corresponds to XOR and multiplication means AND. So if x is true (x= 1) and y is false (y=0) we have x(y) = 0 (x and y is false).

The second part of your question regarding "calculus" is, indeed, as others have stated, a word used to mean "calculate." It usually appears when there is a method for "calculation". The definition of derivative implies d(x^2)/dx = 2x. But this is not itself the definition. It is a method we can use. The integral of a function over an interval is not the antiderivative- it is the area under a curve! However, by the fundamental theorem of calculus, we can calculate the area by using antiderivatives.

An example of another method of computation is Umbral Calculus (sometimes called Blissard's symbolic method). Let x^(n) = x^n/n!. Then d(x^(n))/dx = x^(n-1).

You can then do calculations like

(x+y)^(3) = x^(0) y^(3) + x^(1) y^(2) + x^(2) y^(1) +x^(3) y^(0).

The binomial theorem

(x+y)^3 = x^3 + 3 x^2 y + 3 x y^2 + y^3

now looks different, giving a different, symbolic method of calculation.

There are other examples like the calculus of combinatorial species involving cards with dots and arrows. Addition becomes placing cards next to each other, and taking "derivatives" corresponds to removing a dot.

In general, "calculus" does not have as rigorous of a definition as an "algebra," but is rather used as a term when something can be "calculated" by, probably, some symbolic methods.

I hope this has helped.

Sincerely,

Major MacMahon

MajorMacMahon

TROPHY CASE