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The general term of a geometric series with initial term
#a_k = a*r^(k-1)#
The sum to
#sum_(k=1)^n = (a(1-r^n))/(1-r)#
Note that:
#2/3 = 1 - 1/3#
#8/9 = 1 - 1/9#
#26/27 = 1 - 1/27#
So it seems that the general term of the given series is:
#a_k = 1 - 1/3^k#
We have:
#sum_(k=1)^n 1/3^k = (1/3(1-(1/3)^n))/(1-1/3)#
#color(white)(sum_(k=1)^n 1/3^k) = (1-(1/3)^n)/2#
#color(white)(sum_(k=1)^n 1/3^k) = 1/2-1/2(1/3)^n#
So:
#sum_(k=1)^n a_k = sum_(k=1)^n (1-1/3^k)#
#color(white)(sum_(k=1)^n a_k) = sum_(k=1)^n 1 - sum_(k=1)^n 1/3^k#
#color(white)(sum_(k=1)^n a_k) = n - (1/2-1/2(1/3)^n)#
#color(white)(sum_(k=1)^n a_k) = n - 1/2 + 1/2(1/3)^n#
Footnote
Given:
#a_k = ar^(k-1)#
We find:
#(1-r)sum_(k=1)^n a_k = (1-r)sum_(k=1)^n ar^(k-1)#
#color(white)((1-r)sum_(k=1)^n a_k) = sum_(k=1)^n ar^(k-1) - rsum_(k=1)^n ar^(k-1)#
#color(white)((1-r)sum_(k=1)^n a_k) = sum_(k=1)^n ar^(k-1) - sum_(k=2)^(n+1) ar^(k-1)#
#color(white)((1-r)sum_(k=1)^n a_k) = a+color(red)(cancel(color(black)(sum_(k=2)^n ar^(k-1)))) - color(red)(cancel(color(black)(sum_(k=2)^n ar^(k-1)))) - ar^n#
#color(white)((1-r)sum_(k=1)^n a_k) = a(1-r^n)#
So dividing both ends by
#sum_(k=1)^n a_k = (a(1-r^n))/(1-r)#
See below.
5)
Let
5.1)
5.2)
6)
Let
then
but
so
and
so the prime numbers are
or
A few thoughts...
I am not exactly sure what you are asking, but here's something of what logarithms are about...
Let
Consider the function:
#f(x) = b^x#
This is a continuous, one to one, strictly monotonically increasing function from
graph{2^x [-9.58, 10.42, -2.4, 7.6]}
It has the interesting property that:
#f(x+y) = b^(x+y) = b^x b^y = f(x)f(y)#
for any
The inverse of
graph{2^y-x = 0 [-4.08, 15.92, -4.12, 5.88]}
which is a continuous, one to one, strictly monotonically increasing function from
Given any
#x = f^(-1)(u)#
#y = f^(-1)(v)#
Then we find:
#f^(-1)(uv) = f^(-1)(f(x)f(y)) = f^(-1)(f(x+y)) = x+y = f^(-1)(u)+f^(-1)(v)#
This inverse function
and this property we have found is:
#log_b(uv) = log_b(u)+log_b(v)#
a fundamentally useful property of logarithms.
This property of logarithms was discovered by John Napier in the 17th century and used in the form of slide rules and tables to help perform calculations.
A few thoughts...
The Real numbers are usually thought of as constituting a line which we call the Real line.
This is the
Addition of complex numbers is two dimensional vector addition, that is:
#(a+bi)+(c+di) = (a+c)+(b+d)i#
Multiplication of complex numbers is defined as:
#(a+bi)(c+di) = (ac-bd)+(ad+bc)i#
This has some interesting properties:
Real numbers are complex numbers of the form
Multiplication by Real numbers is scalar multiplication:
#(a+0i)(c+di) = ac+adi#
The square of
#i^2 = (0+1i)(0+1i) = ((0)(0)-(1)(1))+((0)(1)+(1)(0))i = -1#
The modulus of a complex number is its length as a vector, which can be deduce from Pythagoras:
#abs(a+bi) = sqrt(a^2+b^2)#
Hence we find that any Complex number of modulus
#cos theta + i sin theta#
Multiplication by such a number is pure rotation by
#(cos theta + i sin theta)(a+bi)#
#= (a cos theta - b sin theta)+(b cos theta + a sin theta)i#
By the time we get to this point, some readers might be thinking "matrices". Indeed, complex numbers have a natural representation in terms of
#((a, b),(-b, a))+((c,d),(-d,c)) = ((a+c,b+d),(-(b+d),a+c))#
#((a, b),(-b,a))((c,d),(-d,c)) = ((ac-bd,ad+bc),(-(ad+bc),ac-bd))#
Notice that complex numbers of modulus
#((cos theta, sin theta),(-sin theta, cos theta))#
which is recognisable as the matrix representing a rotation through angle
In general, multiplication by a complex number is a combination of rotation about the origin and scaling.
Take the graph of
Do you know what the graph of
graph{y=e^x [-10, 10, -5, 5]}
Next note that
So the graph of
graph{y=ln x [-10, 10, -5, 5]}
Note that:
Finally, to get the graph of
graph{y=-ln x [-10, 10, -5, 5]}
Note that:
(Sorry, but I had to do this as a graphic image to get the arrows in.)
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