The matrix is
The inverse matrix is
A linear transformation such as
#f(x,y,z) = (x-y,"Â Â "x+y,"Â Â "z-x)#
can be expressed as matrix multiplication. Just like how the function
#F xx stackrelrarrx=[(x-y),(x+y),(z-x)]# .
Let's remember how matrix multiplication works: when multiplying a
#[(a,b,c),(d,e,f),(g,h,i)]_ color(red)(3 xx 3) xx [(x),(y),(z)]_ color(red)(3xx1) = [(ax+by+cz),(dx+ey+fz),(gx+hy+iz)]_ color(red)(3 xx 1)#
We want to find what
In other words, we want
It may still look like a jumble of letters, but rememberâ€”we're going to choose what
#ax+by+cz=x-y#
What do
#1x+(â€“1)y+0z=x-y#
The process is the same for the other two rows. The matrix is
#F=[(1,â€“1,0),(1,1,0),(â€“1,0,1)].#
Notice:
the first row contains the coefficients of
#x-y,#
the second row contains the coefficients of#x+y,# and
the third row contains the coefficients of#z-x,#
all in
The inverse matrix
#F^(â€“1) xx Fstackrelrarrx = stackrelrarrx#
or
#F^(â€“1) xx [(x-y),(x+y),(z-x)] = [(x),(y),(z)].#
You can find this inverse the same way as before; this one's not too hard to visualize, though others may take some effort. You can also find it by computing the inverse of
#F^(â€“1)=[(1/2,1/2,0),(â€“1/2,1/2,0),(1/2,1/2,1)]#
First note that
Second observe that
So we have
So the graph is:
graph{abs(x^2-1)/(x^2-1) [-4.995, 4.87, -1.794, 3.14]}
It is now clear that there are
two discontinuities (jump),
no vertical asymptotes,
no
depending on your definition of horizontal asymptote, either none or
C
So let's do this two ways: using transformations of a function, and using the logic behind those transformations.
Transformations
We know that the transformation
Logic
The only thing we know about the function
Well, if we want to mess with the given equation for y
so that the argument of
When we plug in the value for
Manifold...
There are many things we can say about the relationship between the roots of a polynomial and its coefficients. Here are just a few...
Rational roots theorem
Given a polynomial:
#a_n x^n + a_(n-1) x_n-1 + ... + a_1 x + a_0" "#
with integer coefficients and
any rational zeros are expressible in the form
The same is true if we specify Gaussian integers instead of integers, or indeed the elements of any integral domain.
Descartes' Rule of Signs
For any polynomial with real coefficients written in standard form, the number of changes in the signs of the coefficients gives the maximum number of positive real zeros. If the number of zeros is less, then it is less by a multiple of
For example:
#x^4+x^3-4x^2+2x-5#
has coefficients with signs
Furthermore, if you invert the signs on the terms of odd degree, then the number of changes in the resulting pattern indicates the number of negative real zeros. In our example, we would get
Discriminant
You are almost certainly familiar with the formula for the discriminant
#Delta = b^2-4ac#
Assuming that
If
If
If
If
Polynomials of higher degree also have discriminants. They are not particularly useful for quartics and above, except to identify when they have repeated zeros, but for cubics they are very useful.
The discriminant of
#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#
If
If
If
A few more...
Elementary symmetric polynomials
The coefficients of a monic polynomial are (modulo alternating signs), the elementary symmetric polynomials in the zeros.
For example:
#(x-alpha)(x-beta) = x^2-(alpha+beta)x+alphabeta#
#(x-alpha)(x-beta)(x-gamma) = x^3-(alpha+beta+gamma)x^2+(alphabeta+betagamma+gammaalpha)x-alphabetagamma#
etc.
In particular, given a polynomial:
#a_n x_n + a_(n-1) x_(n-1) + ... + a_1 x + a_0#
the sum of its zeros is
Since any symmetric polynomial can be constructed from the elementary symmetric polynomials, we can find a polynomial that has zeros that are (say) the squares of the zeros of another polynomial - without finding what the zeros actually are.
For a substantial and important application of this, see https://socratic.org/s/aPGxwybx
Coefficient sum shortcuts
If the sum of the coefficients of a polynomial is
If inverting the signs on terms of odd degree results in coefficients that sum to
Reversing the order and reciprocals
Given a polynomial:
#a_n x_n + a_(n-1) x^(n-1) + ... + a_1 x + a_0#
with
Then the polynomial:
#a_0 x_n + a_1 x_(n-1) + ... + a_(n-1) x + a_n#
has zeros which are reciprocals of the original polynomial.
To see why that is so, note that:
#1/x_n (a_n x_n + a_(n-1) x^(n-1) + ... + a_1 x + a_0)#
#=a_n + a_(n-1) 1/x + ... + a_1 1/x^(n-1) + a_0 1/x^n#
Symmetry and reciprocals
From the preceding property, we can deduce that:
If the coefficients of a polynomial are symmetrical then you can infer that the reciprocal of any zero is also a zero.
For example:
#6x^4+5x^3-38x^2+5x+6#
has zeros
Let us locate the outer corner of the corridor at
Consider lines of negative slope passing through
The minimum distance then models the maximum feasible length of beam.
graph{((x-14)^100+(y-13)^100-10^100)((x-20)^100+(y-20)^100-20^100)(y+x/2-5)((x-4)^2+(y-3)^2-0.02)(x^2+(y-5)^2-0.02)((x-10)^2+y^2-0.02) = 0 [-5.625, 14.375, -3, 7]}
The equation of such a line can be written in point slope form as:
#y-b = m(x-a)#
The
#x = a-b/m#
and the
#y = b-am#
The square of the distance between these intercepts is:
#(a-b/m)^2+(b-am)^2 = a^2-2ab 1/m + b^2 1/m^2 + b^2-2ab m + a^2 m^2#
The minimum will occur when the derivative of this with respect to
#0 = 2ab 1/m^2 - 2b^2 1/m^3-2ab+2a^2m#
Multiplying by
#0 = abm-b^2-abm^3+a^2m^4#
#color(white)(0) = a^2m^4-abm^3+abm-b^2#
#color(white)(0) = am^3(am-b)+b(am-b)#
#color(white)(0) = (am^3+b)(am-b)#
#color(white)(0) = (a^(1/3)m+b^(1/3))(a^(2/3)m^2-a^(1/3)b^(1/3)m+b^(2/3))(am-b)#
Note here that the last linear factor gives
Also the quadratic factor has only non-real solutions.
So we require
With this value of
#(a-b/m)^2+(b-am)^2 = (a+a^(1/3)b^(2/3))^2+(b+a^(2/3)b^(1/3))^2#
#color(white)((a-b/m)^2+(b-am)^2) = a^(2/3)(a^(2/3)+b^(2/3))^2+b^(2/3)(b^(2/3)+a^(2/3))^2#
#color(white)((a-b/m)^2+(b-am)^2) = (a^(2/3)+b^(2/3))^3#
So the distance is:
#sqrt((a^(2/3)+b^(2/3))^3) = (a^(2/3)+b^(2/3))^(3/2)#
Cartesian Coordinates to Polar Form:
Given the Cartesian Form:
Find the Polar Form:
Let us examine some of the relevant formula in context:
Plot the coordinate point
Indicate the known values, as appropriate:
Let
Let
Use the formula:
Consider the following triangle with known values:
Hence,
To find the value of
Since the angle
Hence, the required Polar Form:
Hope it helps.
The graphic of the function
Let the graphic of our function be
There are two possible situations; the graphic is either symmetric with respect to the axis
These two have different definitions, as follow:
A graph
We can see a visual representation of this for parabolas:
graph{(y-x^2)((x-2)^2+(y-4)^2-0.03)((x+2)^2+(y-4)^2-0.03)=0 [-10.005, 9.995, -0.8, 9.2]}
The axis of symmetry for this graph (
Basically, in other words, if
Before trying it out, let's define what the center of symmetry is.
A point
We define the mathematical relation so show this the following:
While we could try this out for both relations and do lots of calculations, one simpler way is to obverse that the function
This slightly resembles our second relation, doesn't it? If we put them side to side, we can see the similarity:
In order for these to be the same, we must have
Thus, for an odd function
Since the function we have,
Bonus: If
Create a data table to graph the exponential function :
Create a data table:
Closely examine the values for
No value of
So, the graph gets closer and closer to the x-axis but never touches the x-axis.
Hence,
Using the data table, graph the function
For easy comprehension, graph the exponential function
Compare both the graphs to understand the behavior of the given exponential function
The graph of
See below.
Since we have two different situations, let's break down this into two different statements,
if
if
Let's prove
Now, let
We must prove if the next case holds. However, the next case is not
Notice how the part highlighted in red is the former sum,
We know that
So by substituting this into the statement, we get
This is fairly easy to find. Multiply both sides by
Which is true. Thus, by mathematical induction,
The second statement,
If you do all that, you will find that
Hence, by Mathematical induction, the statement
is true for all