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Hi Complex,

I plotted the 3 points and decided that a cubic could go through them if there was a horizontal point of inflection at (2,4)

so the graph goes though (1,1). and (3,7) and (2,4)

also     y'(2)=0   and y''(2)=0

so I started with

\(y=ax^3+bx^2+cx+d\\ y'=3ax^2+2bx+c\\ y''=6ax+2b \)

\(0=6a(2) + 2b\\ 0=12a+2b\\ 0=6a+b\\ b=-6a\\~\\ 0=3a(2^2)+2b(2)+c\\ 0=12a+4b+c\\ 0=12a+4(-6a)+c\\ 0=12a-24a+c\\ 0=-12a+c\\ c=12a\\~\\ y=ax^3+bx^2+cx+d\\ y=ax^3-6ax^2+12ax+d\\\)
 

\(y=ax^3-6ax^2+12ax+d\\ though (1,1)\\ 1=a-6a+12a+d\\ 1=7a+d\\ d=1-7a\\ y=ax^3-6ax^2+12ax+1-7a\\~\\ \)
\(through (2,4)\\ 4=a*2^3-6a*2^2+12a*2+1-7a\\ 4=8a-24a+24a+1-7a\\ 4=a+1\\ a=3\)

\(y=ax^3-6ax^2+12ax+1-7a\\ y=3x^3-18x^2+36x+1-21\\ y=3x^3-18x^2+36x-20\\\)

OR using your letters

\(\boxed{n=3s^3-18s^2+36s-20}\)

Chris, I do not think that yours is a new formula, it is just a mor complicated way of writing exactly the same formula. 

Hi, How is the ventriloquism going ??

You have had a fair bit of time to practice now so I expect that you are getting good   

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Find all 4th roots of z=5000+i.  Show all work and draw a graph to reprecipent the answers.

I think I have worked out how to do this quite neatly I think but I do not know the correct terminology. 

I am not familiar with the correct terms so i used this page to help me there.

http://www.sparknotes.com/math/precalc/complexnumbers/terms.html

The distance (modulus) of this point from 0 is   \(\sqrt{5000^2+1^1}=\sqrt{25000001}\)

The distance (modulus) from 0 that the roots will be is  \((\sqrt{25000001})^{1/4}=25000001^{1/8}=\sqrt[8]{25000001}\)

(25000001)^(1/8) = 8.4089641945819655     approx 8.4090

First, 2pi/4 = pi/2  so the roots will be approx 8.409  units from 0 and pi/2 radians apart.

z=5000+i

I have just drawn a right angle triangle on a scrap of paper to work this out.

\(z=\sqrt{25000001}*(\frac{5000}{\sqrt{25000001}}+\frac{1i}{\sqrt{25000001}})\)

\(cos\theta=\frac{5000}{\sqrt{25000001}}\qquad and \qquad sin\theta=\frac{1}{\sqrt{25000001}}\)

argument (angle) of the first root is     acos(5000/sqrt(25000001) = 0.000199999998

asin(1/sqrt(25000001)) = 0.000199999997

They had to be the same, I was just showing you. :)

So the angle (argument) is very close to 0.0002 radians

The first 4th root angle is \(acos(\frac{5000}{\sqrt{25000001}})\div4\approx 0.0002\div4 = 0.00005 \;radians\)

So the angles of the 4 roots will be

0.00005, 0.00005+pi/2, 0.00005+pi, 0.0005+3pi/2

0.00005+pi/2 = 1.5708463267948966                    approx 1.5708 radians     

0.00005+pi/2+pi/2 = 3.1416426535897932            approx 3.1416  radians

0.00005+pi/2+pi/2+pi/2 = 4.7124389803846899    approx  4.7124  radians

So the 4th roots of  z=5000+i    are

8.409e^(0.00005i), 8.409e^(1.5708i), 8.409e^(3.1416i), 8.409e^(4.712i)

Check the first one.     

(8.409e^(0.00005i))^4 = 5000.08516065925605241+1.000017045882085632i   near enough

z = (5000+i)^(1/4)

Algebraic form:
z = 8.4089642+0.0004204i

Exponential form:
z = 8.4089642 × ei 0°10″

Trigonometric form:
z = 8.4089642 × (cos 0°10″ + i sin 0°10″)

Polar form:
r = |z| = 8.40896
φ = arg z = 0.00286° = 0°10″ = 2.0E-5π

All 4th roots are:

13^(1/4) 147929^(1/8) e^(1/4 i tan^(-1)(1/5000))=8.4090+0.0004 i  (principal root)

13^(1/4) 147929^(1/8) e^((i pi)/2+1/4 i tan^(-1)(1/5000))=-0.0004+8.4090 i

13^(1/4) 147929^(1/8) e^(i pi+1/4 i tan^(-1)(1/5000))=-8.4090-0.0004 i

13^(1/4) 147929^(1/8) e^((3 i pi)/2+1/4 i tan^(-1)(1/5000))=0.0004-8.4090 i

1  4  7  10  13

How about this pattern ????

1^2  - 0^2 + 0   =  1 + 0   = 1

2^2 - 1^2 + 1   =  3 + 1   = 4

3^2 - 2^2 + 2 =  5 + 2  = 7

4^2 - 3^2 + 3 = 7 + 3  = 10

5^2 - 4^2 + 4 = 9 + 4  = 13

6^2 - 5^2 + 5  =  11 + 5  = 16

=

[ (n)^2 - (n-1)^2] + (n - 1)  =  3n - 2

Hi Amazinggirl, welcome to the forum :)

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Thank you CPhill. Brilliant!.

Pi is 3.14 etc.

My bro is fricking named H I t l e r it ain't not a cuss word...

I guess not I'll go practice ventriloquism.

cot^-1( 2 + sqrt(2) + sqrt(3) + sqrt(6) )  = pi/24

cot [ pi/24]  = 2 + sqrt(3) + sqrt(2) + sqrt(6)

cot [pi/24]  =

cot [(1/2)pi/12]] =

1/ tan [(1/2)pi/12)] =

1/ [sqrt ( 1 - cos(pi/12))/ (1 + cos(pi/12)]  =  

sqrt( 1 + cos(pi/12)) / sqrt( 1 - cos(pi/12) ) =  [multiply top/bottom  by sqrt( 1 + cos(pi/12)) ]

[1 + cos (pi/12)] / sqrt( 1 - cos^2(pi/12)) =

[1 + cos(pi/12)] / sqrt( sin^2(pi/12)) =

[1 + cos(pi/12) ] / sin(pi/12)  =

1/sin(pi/12) + cos(pi/12) / sin(pi/12)  =

csc[pi/12] + cot[pi/12] =

1/ sin [ pi/12] + 1 / [tan [pi/12]  =

1 / [ sin[ pi/4 - pi/6]] + 1 / [ tan[ pi/4 - pi/6] ]

1/ [ sin(pi/4)cos(pi/6) - sin(pi/6)cos(pi/4)]  +  1/ ( [tan(pi/4] - tan[pi/6]] / [1 + tan(pi/4)tan(pi/6)] )  =

1 / [ sqrt(2)sqrt(3)/4] - (sqrt(2)/4) ] + [ 1 + 1/sqrt(3)]/ [1 - 1/sqrt(3)]  =

4 / [ sqrt(6) - sqrt(2)]  +  [ (sqrt(3) + 1)/ sqrt(3))]   / ( [ sqrt(3) - 1]/ sqrt(3) )

4 [ sqrt(6) + sqrt(2)]/ 4  +  [ sqrt(3) + 1] / [sqrt(3) - 1]  =  [ multiply the second term by [sqrt(3) + 1] on top/bottom ]

sqrt(6) + sqrt(2)  + [sqrt(3) + 1]^2 / 2  =

sqrt(6) + sqrt(2) + [3 + 2sqrt(3) + 1] / 2  =

sqrt(6) + sqrt(2) + [ 4 + 2 sqrt(3)] / 2  =

sqrt(6) + sqrt(2) + 2 + sqrt(3)  =

2 + sqrt(3) + sqrt(2) + sqrt(6)    ..........and the left side  =  the right side

HahahhahhahahhHahhHHHhahahahahahahahaahahaha, I just felt like putting that.

I'll take yo silence as a no?

Do yo like to read bro?

He might no that...

IDFGI

你没有笔

WEL

I wast talkin to whatevergoddy.