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Featured 3 months ago

The answer is

We must factorise the denominator

Let

Therefore,

To find the other factors, we have to do a long division

Therefore,

Now, we can perform our decomposition into partial fractions

Therefore,

Let

Let

Coefficients of

So,

Featured 1 month ago

The general term of a geometric series can be represented by the formula:

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

where

We find:

#(1-r) sum_(n=1)^N ar^(n-1) = sum_(n=1)^N ar^(n-1) - r sum_(n=1)^N ar^(n-1)#

#color(white)((1-r) sum_(n=1)^N ar^(n-1)) = sum_(n=1)^N ar^(n-1) - sum_(n=2)^(N+1) ar^(n-1)#

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

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

Dividing both ends by

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

If

#sum_(n=1)^oo ar^(n-1) = lim_(N->oo) (a(1 - r^N))/(1-r) = a/(1-r)#

In the given example,

#sum_(n=1)^oo 6*(1/2)^(n-1) = 6/(1-1/2) = 12#

Featured 3 weeks ago

Given:

#{ (-5 = -64a+16b-4c+d), (-4 = -27a+9b-3c+d), (-3 = -8a+4b-2c+d), (4 = -a+b-c+d) :}#

Consider the function:

#f(x) = ax^3+bx^2+cx+d#

Note that the given system of equations is equivalent to:

#{ (f(-4) = -5), (f(-3) = -4), (f(-2) = -3), (f(-1) = 4) :}#

Since the sampling points are at equal intervals of size

Start with the original values:

#color(blue)(-5), -4, -3, 4#

Write down the sequence of differences between successive terms:

#color(blue)(1), 1, 7#

Write down the sequence of differences between successive terms:

#color(blue)(0), 6#

Write down the sequence of differences between successive terms:

#color(blue)(6)#

Then we can write down a formula for

#f(x) = color(blue)(-5)/(0!)+color(blue)(1)/(1!)(x+4)+color(blue)(0)/(2!)(x+4)(x+3)+color(blue)(6)/(3!)(x+4)(x+3)(x+2)#

#color(white)(f(x)) = -5+x+4+x^3+9x^2+26x+24#

#color(white)(f(x)) = x^3+9x^2+27x+23#

So:

#{ (a=1), (b=9), (c=27), (d=23) :}#

**Footnote**

Note that:

#color(purple)(1/(0!))_(color(white)(1/1))# is a constant function taking the value#1# when#x = -4#

#color(purple)(1/(1!)(x+4))_(color(white)(1/1))# is a linear function taking the value#0# when#x = -4# and#1# when#x = -3#

#color(purple)(1/(2!)(x+4)(x+3))_(color(white)(1/1))# is a quadratic function taking the value#0# when#x = -4# or#x = -3# and the value#1# when#x = -2#

#color(purple)(1/(3!)(x+4)(x+3)(x+2))_(color(white)(1/1))# is a cubic function taking the value#0# when#x = -4# ,#x = -3# or#x = -2# and the value#1# when#x = -1#

So as we add suitable multiples of these functions in turn, we get a sequence of polynomials of increasing degree that match each of the sample points in turn. The suitable multiples are the differences we found.

Featured 3 weeks ago

Let us recall that, for vectors **Vector**

**(Cross) Product** , i.e., **Orthogonal** to both

The Desired **Unit Vectors,** then, can be obtained by,

Now, with

Hence, the desired vectors are,

**Enjoy Maths.!**

Featured 2 weeks ago

Minor:

The **minor** of a determinant is created by deleting all elements in the same row or column as the selected element:

The **co-factor** of a determinant is created by multiplying the **minor** by

In this case

(The top left is always row

So for this example

and the **co-factor** is

Featured 1 week ago

See below

To diagonalise, we need the matrix's eigenvectors. **And** .... the matrix needs to offer up 2 distinct eigenvectors in order to be diagonalisable.

Handy trick: **For a triangular matrix, the eigenvalues are its diagonal entries.**

But we can show that here by going the long way round: they are also the solutions to the characteristic equation, which for a

The eigenvectors (

# lambda = 1#

From the first equation,

# lambda = 4#

From the first equation,

The diagonalisation matrix P is then simply:

The matrix D is a **diagonal** matrix that has the eigenvalues as it only non-zero entries:

We are asserting that:

Now, we have everything apart from

So our assertion here is that:

Try it and see.

You can then do some seriously cool stuff with matrix. There is a point to diagonalisation :)

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