Download Development of Algebra:Cardaro, Tartaglia and Ferrari in 1300`s in

Development of Algebra:Cardaro, Tartaglia and Ferrari in 1300’s in Italy found radical formulas for solving 3 rd and 4th order
polynomial eqns. Tried to find one for 5 th order. No one could find a 5th order formula. These formulas and the quadratic formula
need complex #s. since you encounter roots of negative #s. the complex # issue is similar to that of the problem with irrationals
which the Greeks saw. Complex #s. and irrationals where in the 1800’s properly defined and understood. In 1600’s Descartes
found the connection between algebra and geometry. He saw that the best way to do geometry problems is to turn them into
algebra problems. He had the first idea of x-y coordinates in the plane, but initially his axes where not perpendicular. He quickly
changed them to perpendicular since it makes everything simple. Descartes noticed that curves(includes lines) in the xy-planes are
the points(x,y) which satisfies a particular eqn. In the variables x and y.
(X-H)^2+(Y-K)^2=R^2 is the eqn. For the circle with (H,K) center and radius R. Fermat helped out and showed how to get
rectangular eqns. For all conic sections.(the issue was the proper def. Of each type of conic section.)Fermat also saw in doing
this idea of function f(x) (Liebniz’s notation), which is any formula in x, and then by setting ‘y’ equal to f(x) and plotting the
eqn. Y=f(x) we get the geometric realization of the formula f(x). Due to Fermat and Descartes, geometry and algebra are
becoming the same thing and that to solve a geometry problem turn it into an algebra problem. You really want to go back and
forth between the two types to solve a problem-then calculus comes in with Newton and Liebniz in the late 1600’s and early
1700’s it is the ultimate way right now for connecting algebra with geometry. Fermat and tangent lines.
Fermat had the idea of approximating as accurately as you wish the slope of
this tangent line. Slope of the line thru P,Q = slope of tangent line thru P.
Analytic Geometry. (Turning curves in xy-coord. Plane into algebraic eqns. In x an y variables) And coordinate geometry
(turning geometric structures in the plane or space into algebraic eqns. Or inequalities in x, y or x,y,z variables for space) where
discovered in the early to middle 1600’s by both Descartes and Fermat. These ideas weren’t fully accepted until the Bernoulli
brothers and Euler in the 1700’s fully developed calculus past the original work of Newton and Liebniz. These ideas of
Descartes and Fermat lead directly to the development of calculus. By Newton and Liebniz in the middle to late 1600’s calculus
is the ultimate incorporation of geometry into algebra. All calculus follow tangents lines (slopes). Fermat is the grandfather of
calculus with Newton-Liebniz its fathers. The bed rock of calculus is the ancient Greek problem of tangent lines to curves. To
know a line(tangent line) you need a point (p) that it goes thru and its slope “m”. The tangent problem is how to get its slope.
Fermat’s idea. Slope of tangent line thru p. Q->P. Q is on the same curve as P. From Archimides’s day, people knew how to get
the slopes of tangent lines to parabolas. For the parabola y=x^2 they knew m=2x is the slope of tangent line formula.
M=(x+H)^2/x+H-x; Q=(x+H)^2; Q->P = H->O. (x+H)^2-x^2/x+H-x = 2x+H; m(tangent line) = 2x;
Newton developed Fermat’s idea for the tangent line problem because he relize that it gave him a way of computing rates such
as velocity and acceleration. Liebniz developed calculus to solve all kinds of geometry problems including the tangent line
problem. Calculus is the incorporation of algebra and geometry into a single subject. Alg. And geom. (trigonometry got better
developed in this time period (1600’s) because people needed more than just the basic for right triangles to do calculus->double
angle identities inverse trigonometry functions law of sines and cosines. Show up to do any triangle problem. Newton-Raphson
method: This description of the method is due to Raphson in 1690 following a purely algebraic idea that Newton published in
1687 in one of his 1st papers on calculus. Solve approx. f(x)=0; This method gives solutions to whatever accuracy you want, but
never exactly. Graphing you get the solutions. Solve f(x)=0 Geometrically requires your
graph y=f(x) which Is very hard to do by hand. X0 is a
starting guess chosen so that f(x0) = 0. x1=x intercept of the tangent line to y=f(x) at x=x0. Equation of the tangent line: to
y=f(x) at x=x0; slope m=f’(x0) point it: goes thru: (x0, f(x0)); y-f(x0)=f’(x0)(x-x0) eq. Of tangent line find x1, its x- intercept.
f(x0)/f’(x0) = x-x0; x=x0-f(x0)/f’(x0); x2=x1-f(x1)/f’(x1)……xn+1=xn-f(xn)/f’(xn). Iterating function =g(x)=x-f(x)/f’(x);
f’(x)=f(x+H)-f(x)/H=f(x+10^-5)-f(x)/10^-5; H=10^-5 is optimal if you work to 10 decimals for float. EXAMPLE: Use N-R to
find pi to 10 decimal places. Sin x=0 has pi as a solution f(x). Let x0=4; g(x)=x-f(x)/f’(x); f’(x)=f(x+10^-5)-f(x)/10^-5
y1=sin x; y2=(y1(x+10^-5)-y1)/10^-5; y2=(y1(x+10^(-5))-y1)/10^(-5); y3=x-y1/y2;
call y3(4) 2.84=x1; y3(4)->y3(Anws); x2=3.15; ENTER=> 4 times, x5=3.14159..
Euler greatest mathematician in all times more than Gauss. Euler did more and very friendly loved traveling. Euler’s kids->pile
of math works->publishing. Euler in 1700’s used complex #s. as much as real #s. and he first saw that the N-R method also
finds complex solutions to eqns. He probably realized it will also work to solve square systems of eqns. EX) Use N-R to solve
f(x)=y1=(5-3i)x^3+(7+I)x^2-ix+3i;
y2=y1(x+10^-5)-y1/10^-5=f’(x); iteration funct. y3=x-y1/y2;
y1=(5,-3)*x^3+(7,1)*x^2-(0,1)*x+(0,3); let x0=2->y3(2)=>(1.25155.., -.06475); y3((Answ)); y2=evalF(y1,x,x+10^(-5))y1)/10^(-5); evalF(y3,x,(1,1)) (.558977, .5948)
y3=x1-y1/y2; evalF(y3,x,(2,0)) (1.251559.., -.064); evalF(y3,x,Anws) (.775026, -.125)....
After 10 enters: (.449897.., -.314)
Mathematician French named Julia looking for thinks related to iterating functions. At the time of WWI was interested in what
happen to the behavior that might occur when you iterate functions. Prisioner in German Campus where he did math(reaserch).
This work (graphical information) was done in the prision camp, made him famous. These graphs where first produced in
1970’s by John Hubbard. EX) Let f(x)=x^3-1=0; it has 3 solutions. Pick up any complex # x0 and iterate Newton’s method.
Boundary Set, called by Hubbard the Julia Set. Regions.- basins of attraction for the solutions.Engineers build complex #s “i”
into the machine and problems IT-83.
Y1=(5-3*i)*x^3+(7+I)*x^2-I*x+3*I; y2=3*(5-3*I)*x^2 +2*(7+I)*x-I;y3=x-Y1/y2 (iterator function); starting point 2 sto=>x
alfa: y3; 2=>x:y3; (1.25151566, _.065); 2=>x:y3 (copy); ^ns=>x:y3 delete 2; (.775 -.125); enter (.508 -.21); (.438 -.30); (.45 .31); (.449 -.31); (.4498 -.3146); Ans=> store =>x:y1 Ans=>x:y1; 1.6E-1.3power; Math=>solver Doesn’t woprk for 83 for
complex #5. Equation Solver eqn:0=y1 enter y1+0;x+2, bound={-1E99,1 E99}’ 2=>x:y3; 1.251556-.06…;2nd
enter;2=>x:y3;2nd delete “2”; 2nd INS (insert) 2nd ans; Ans=>x:y3; (.7750216998-.12…); .5082 -.21; .43837 -.30; .4500 -.31493;
.44989 -.31462; .44989 -.3146; .449892 -.31462,etc. Euler’s best mathematicin 1700’s like Mozart w/symphonies, travel a lot.
Gauss less famous.Newton just respect. Enter=>no money PP fascinated by polynominals. Gauss at the end of the 1700’s (Math
student in Germany)became interested in proving tghe3 fundamental theorem of algebra, which states that a polynominal of
degree “N” with coplex coeffs. Has “N” complex solutions. Euber and many others tried to prove it and failed. Gauss thought
he had a proof in 1799(at 19 yrs of age) and got itpublished.(some holes)unfortunately it had many geometry holes (a few
holes) which weren’t immediately found but were taken care of later. We will see the proof due to G.H. Hardy and Charles
Feffermna-published by Fefferman in 1967, this proof is short (2 paragraphs) and almost completely algebraic.; Polynominal
Division Algorithm (Gauss) Let “F” be a field from which our coefficients are chosen. Let P(x) and g(x) be two polys over “F”;
P(x)
P(x)/g(x)=Q(x)+ r(x)/g(x) quotient reminder; degree<degreeg; POLYNOMINal Division Algorithm; Def.: p(x)=sum (k=0 to N)
ak x^k is a poly, over the field “F: of degree “N” if ak element of F for all k.; PDA: Let p(x), g(x) be two polys. Over the field
“F” with degree p(x)> degree g(x).Then p/g=q+r/g pqg+r; There exists unique polys q(x), r(x) over the field “F: where
p(x)+x(x).g(x)+r(x) and degree r(x)<degree g(x).; This algorithm also describes how to get the quotient q (x). Fundamental
Theorem of Algebra. Real version: Let p(x) be a polynominal of degree “N” over|R (real #s). Then p(x) can be factored
completed over real #’s as p(x)=a.(x-ri)…(x-rk) (x^2+bix=ci)…(x^2+b2xic2) For the leading coeff. Of p(x), all r’s, b’s,c’s are
REAL, N==K+2L. Comples version: same initial statement w/field complex, then p(x)=a.(x-r1)…(x-rk); For “a” the leading
coeff. Of p(x) and r’s complex #s. Note: these factorizations of p(x) are unique up to order of the product(due to using monic
factors)Note: if “c” is a complex number, then “c and C’ c=a+bI and C’=a-bi; Property that (x-c)(x-C’) is a real poly, if
multiplied out. Also, if p (x) is a real poly. With complex root c, then C’ is also a root. Preliminaries for proving the fund. Them
of alg. (Complex Version. Max-min principle (complex version) Simple Real Versionl Let fi[a,b]=>real numbers be a
continuous function.
(Due to Weierstrass in 1850’s) Then, there is at least one pt. C is element of [a,b]at which f(x_ achieves its maximum value and
at least one pt. D is element of [a,b]at which f(x)achieves its minimum value.; More complex version: (Due to poincare(French)
in late 1800’s to early 1900’s, he is the founder of topology.). Let f:=.complex version be a congtinuous func. Then the maxmin statement follows: interms of modules dist. To origen Domain=>interior at the circle and everything inside. D=Disk of
complex version #’s this is a circle and everything inside it.Fundamental Thm. Of Algebra” Any complex coefficient nonconstant polynomial p(x)has at least one root z0 in the complex #s. p(z0)=0; z-z0 is a facgtor of p(z)=(z-z0)q(z); Proof of the
complex factoring version of the F.T.A., (From the Root Version) You are given P(z) w/complex coeffs. And degree
p(z)=N>=1. By the root version, we have z0 is elements of complex with p(z0)=0. By the division alg., p(z)=(z-z0) q (z)+R(z),
R(z)=0.; p(z)=(z-z0)q(z), degreeq(z)=N-1 if degree Q(z)=N-1>=1, then by the root version of F.T.A., Q(z1)=0,Z1 is element of
complex. So q(z)=(z-z1)q*(z).; p(z)=(z-z0) (z-z1)Q*(z). This process can continue until the quotient has degr.0, then: p(z)=(zz0) (z-z1)… (z-zN-1)a last quotient, ‘a’ is element of complex. Proof of the root version of F.T.A. Let p(z) be a complex
coeff. Poly w/degree at least 1.z is a complex variable. “defintion” Lim p(z) “def” lim|Z|->inf p(z)= 1|Z|=>infinite.
LimZ^N(aN+AN-1/Z+aN-2/Z^2+-+a0/Z^N) when |Z|->inf.; In the real case, we have two infinites plus/minus inf. With inf.
Which indicates something is getting larger and larger in pos. value. And smaller, smaller in negative value.
Lim –8/x^5=0 when x->+-inf; Lim alpha/xK=0 when x->+-inf; alpha is a real #, k is pos. int); Lim-7x^3+2x-6=+-inf when x>+-inf; Lim –7x^3+2x-6 when x->inf = Limx^3(-7+2/x^2-6/x^3) when x->inf = “inf (-7)” = -inf; PROOF OF FTA: There is
some disk “D” where for all Z is not an element of D, |p (z)|>=|p(0)|. (It follows from lim p) z(Z)=inf, when z->inf; Let’s use
the max-min principles on p(z) as p:D=>complex version. Then there is some values z0 and z1 in “D: where |p(z)| for all values
z is element of D has its minimum value at z0 and its maximum value at z1. |p(z)|>=|p(z0)|,|p(z)|<=|p(z1)|for all Z elements of
D. We want to use the minimum part involving z0. |p(z)|>=|p(0)|for all Z no elements of D. |p(z)|>=|p(z0)| for all Z no elements
of D. |p(z)|>=|[(z0)| for all Z no elements of complex version. Note: if p(z) has any roots, then z0 is one of these roots. Assume
z0 is not a root of p(z), p(z0) not equal to 0. If we can find a contradiction, then this assumption must be false. Let q(z)
=p(z+z0); q(z) is just another polynomial g(o)=p(z0)., This says that 0 is not a root of g(z).;
g(z)=co=C1z+C2Z^2+…+CN.Z^N=C0+CjZj+r(Z); r(Z) is another polynomial.
Now: |q(alpha w)|<|C0|; Let Z=alpha w +Z0; |q(alpha w)| < |C0| = |q(0)= |p(Z0)| <= |p(Z)| = p(alpha w+ Z0); for all Z Elems of
Complex #s; p(alpha w + Z0)=|q(alpha w)|; q(alpha w)=p(alpha w+Z0) Contradiction. So we assume is false, p(Z0) no equals to
zero.