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Featured 2 months ago

Please see below.

Given that the axis is

Now use the two points given to get:

From

From

Replace

Using

The equation is

If course, there are other ways to write the answer.

graph{(y+8/21(x+3)^2-95/21)(x+3)((x+1)^2+(y-3)^2-0.02)((x-2)^2+(y+5)^2-0.02)=0 [-12, 12, -6, 6]}

Featured 2 months ago

The solution is

We perform the Gauss Jordan elimination with the augmented matrix

Featured 1 month ago

Let

First, we will put the complex number into polar form:

# |omega| = sqrt((-3)^2+3^2) =3sqrt(2)#

# theta =arctan(3/(-3)) = arctan(-1) = -pi/4#

# => arg \ omega = pi/2+pi/4 = (3pi)/4 #

So then in polar form we have:

# omega = 3sqrt(2)(cos((3pi)/4) + isin((3pi)/4)) #

We now want to solve the equation

# z^8 = 3sqrt(2)(cos((3pi)/4) + isin((3pi)/4)) #

Whenever dealing with complex variable equation such as this it is essential to remember that the complex exponential (and therefore the polar representation) has a period of

# z^8 = 3sqrt(2)(cos((3pi)/4+2npi) + isin((3pi)/4+2npi)) \ \ \ n in ZZ #

By De Moivre's Theorem we can write this as:

# z = 3sqrt(2)(cos((3pi)/4+2npi) + isin((3pi)/4+2npi))^(1/8) #

# \ \ = (3sqrt(2))^(1/8)(cos((3pi)/4+2npi) + isin((3pi)/4+2npi))^(1/8) #

# \ \ = 3^(1/8)2^(1/16) (cos(((3pi)/4+2npi)/8) + isin(((3pi)/4+2npi)/8))#

# \ \ = 3^(1/8)2^(1/16) (cos theta + isin theta ) #

Where:

# theta =((3pi)/4+2npi)/8 = ((3+8n)pi)/32#

And we will get

After which the pattern continues (due the above mentioned periodicity).

We can plot these solutions on the Argand Diagram:

Featured 4 weeks ago

In linear algebra, one studies the notion of linear vector spaces over different scalar fields.

So assuming that you are familiar with the notion of a linear vector space, let us consider an LVS

Then for two elements

1) Complex conjugate of

2)

3) For four vectors

Where

If these properties hold, we call

However, there is no general recipe for defining the inner product of two vectors (or functions). We just have to make sure that the properties 1-3 are satisfied.

For an example, an inner product of two wave functions

One can easily check if the properties 1-3 hold.

Featured 4 weeks ago

Set up the long division like this:

#color(magenta)(x^2)+2x+1bar(|" "color(magenta)(x^4)-2x^3-4x^2+2x+3)#

Divide

#color(white)(x^2+2x+1bar(|" "color(red)(x^2)#

#x^2+2x+1bar(|" "x^4-2x^3-4x^2+2x+3)#

Multiply

#color(white)(x^2+2x+1bar(|" "color(black)(x^2)#

#x^2+2x+1bar(|" "x^4-2x^3-4x^2+2x+3)#

#color(white)(x^2+2x+1bar(|" "color(blue)(x^4+2x^3+color(white)(1)x^2))#

Subtract

#color(white)(x^2+2x+1bar(|" "color(black)(x^2)#

#x^2+2x+1bar(|" "x^4-2x^3-4x^2+2x+3)#

#color(white)(x^2+2x+1bar(|" "color(black)(x^4+2x^3+color(white)(1)x^2))#

#color(white)(x^2+2x+1|)bar(" "color(white)(x^4" ")color(orange)(-4x^3-5x^2)" ")#

Copy the

#color(white)(x^2+2x+1bar(|" "color(black)(x^2)#

#x^2+2x+1bar(|" "x^4-2x^3-4x^2+color(green)(2x)+3)#

#color(white)(x^2+2x+1bar(|" "color(black)(x^4+2x^3+color(white)(1)x^2))#

#color(white)(x^2+2x+1|)bar(" "color(white)(x^4)-4x^3-5x^2" "color(green)(+2x))#

Repeat this process twice, dividing the latest leading term below your line by the leading

#color(white)(x^2+2x+1bar(|" "color(black)(x^2-color(red)(4x))#

#color(magenta)(x^2)+2x+1bar(|" "x^4-2x^3-4x^2+2x+3)#

#color(white)(x^2+2x+1bar(|" "color(black)(x^4+2x^3+color(white)(1)x^2))#

#color(white)(x^2+2x+1|)bar(" "color(white)(x^4)-color(magenta)(4x^3)-5x^2+2x)#

...

#color(white)(x^2+2x+1bar(|" "color(black)(x^2-color(red)(4x))#

#color(red)(x^2+2x+1)bar(|" "x^4-2x^3-4x^2+2x+3)#

#color(white)(x^2+2x+1bar(|" "color(black)(x^4+2x^3+color(white)(1)x^2))#

#color(white)(x^2+2x+1|)bar(" "color(white)(x^4)-4x^3-5x^2+2x)#

#color(white)(x^2+2x+1|bar(" "color(blue)(-4x^3-8x^2-4x)))#

...

#color(white)(x^2+2x+1bar(|" "color(black)(x^2-4x)#

#x^2+2x+1bar(|" "x^4-2x^3-4x^2+2x+color(green)(3))#

#color(white)(x^2+2x+1bar(|" "color(black)(x^4+2x^3+color(white)(1)x^2))#

#color(white)(x^2+2x+1|)bar(" "color(white)(x^4)-4x^3-5x^2+2x)#

#color(white)(x^2+2x+1|bar(" "color(black)(-4x^3-8x^2-4x)))#

#color(white)(x^2+2x+1|" ")bar(color(white)(-4x^3+)color(orange)(3x^2+6x)+color(green)(3))#

...

#color(white)(x^2+2x+1bar(|" "color(black)(x^2-4x" "+color(red)3)#

#color(magenta)(x^2)+2x+1bar(|" "x^4-2x^3-4x^2+2x+3)#

#color(white)(x^2+2x+1bar(|" "color(black)(x^4+2x^3+color(white)(1)x^2))#

#color(white)(x^2+2x+1|)bar(" "color(white)(x^4)-4x^3-5x^2+2x)#

#color(white)(x^2+2x+1|bar(" "color(black)(-4x^3-8x^2-4x)))#

#color(white)(x^2+2x+1|" ")bar(color(white)(-4x^3+)color(magenta)(3x^2)+6x+3)#

...

#color(white)(x^2+2x+1bar(|" "color(black)(x^2-4x" "+color(red)(3))#

#color(red)(x^2+2x+1)bar(|" "x^4-2x^3-4x^2+2x+3)#

#color(white)(x^2+2x+1bar(|" "color(black)(x^4+2x^3+color(white)(1)x^2))#

#color(white)(x^2+2x+1|)bar(" "color(white)(x^4)-4x^3-5x^2+2x)#

#color(white)(x^2+2x+1|bar(" "color(black)(-4x^3-8x^2-4x)))#

#color(white)(x^2+2x+1|" ")bar(color(white)(-4x^3+)3x^2+6x+3)#

#color(white)(x^2+2x+1|" "bar(color(blue)(" "3x^2+6x+3)#

...

#color(white)(x^2+2x+1bar(|" "color(black)(x^2-4x" "+3)#

#x^2+2x+1bar(|" "x^4-2x^3-4x^2+2x+3)#

#color(white)(x^2+2x+1bar(|" "color(black)(x^4+2x^3+color(white)(1)x^2))#

#color(white)(x^2+2x+1|)bar(" "color(white)(x^4)-4x^3-5x^2+2x)#

#color(white)(x^2+2x+1|bar(" "color(black)(-4x^3-8x^2-4x)))#

#color(white)(x^2+2x+1|" ")bar(color(white)(-4x^3+)3x^2+6x+3)#

#color(white)(x^2+2x+1|" "bar(color(black)(" "3x^2+6x+3)#

#color(white)(x^2+2x+1|" ")bar(color(white)(3x^2+6x+color(orange)0)#

Featured 2 days ago

Note that

This is a geometric series, with common ratio

The general term of a geometric series can be written:

#a_n = ar^(n-1)#

where

Then we find:

#(1-r)sum_(n=1)^N a_n = (1-r)sum_(n=1)^N a r^(n-1)#

#color(white)((1-r)sum_(n=1)^N a_n) = sum_(n=1)^N a r^(n-1) - rsum_(n=1)^N a r^(n-1)#

#color(white)((1-r)sum_(n=1)^N a_n) = a + color(red)(cancel(color(black)(sum_(n=2)^N a r^(n-1)))) - color(red)(cancel(color(black)(sum_(n=2)^N a r^(n-1)))) - ar^N#

#color(white)((1-r)sum_(n=1)^N a_n) = a(1 - r^N)#

So dividing both ends by

#sum_(n=1)^N a_n = (a(1 - r^N))/(1-r)#

In our example

#sum_(n=1)^N a_n = (color(blue)(1)(1 - color(blue)(2)^color(blue)(11)))/(1-color(blue)(2)) = 2^11-1 = 2047#

Alternatively, in this particular example there is a simpler solution...

Note that:

#1+(1+2+4+8+...+1024) = 2+2+4+8+...+1024#

#color(white)(1+(1+2+4+8+...+1024)) = 4+4+8+16+...+1024#

#color(white)(1+(1+2+4+8+...+1024)) = 8+8+16+32+...+1024#

#color(white)(1+(1+2+4+8+...+1024) =) vdots#

#color(white)(1+(1+2+4+8+...+1024)) = 1024+1024#

#color(white)(1+(1+2+4+8+...+1024)) = 2048#

So:

#1+2+4+8+...+1024 = 2048-1 = 2047#

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