Power

Author

Eric W. Weisstein
December 3, 2008

This notebook downloaded from http://mathworld.wolfram.com/notebooks/SpecialFunctions/Power.nb.

For more information, see Eric's MathWorld entry http://mathworld.wolfram.com/Power.html.

©2008 Wolfram Research, Inc. except for portions noted otherwise

Plots

V5

Show[Plot[Evaluate[Table[x^n,{n,-3,3}]],Evaluate[{x,Sequence@@#}],DisplayFunction->Identity,PlotStyle->{Red,Yellow,Green,Black,Blue,Indigo,Violet},AxesLabel->TraditionalForm/@{x,x^n},PlotRange->{Automatic,{-10,10}}]&/@{{-3,0},{0,3}},DisplayFunction->$DisplayFunction]

Plot[Evaluate[Table[x^n,{n,0,3,.5}]],{x,0,3},PlotStyle->{Black,Red,Yellow,Green,Blue,Indigo,Violet},AxesLabel->TraditionalForm/@{x,x^n}]

Plot[Table[x^n,{n,-4,4,.5}]//Evaluate,{x,0.01,2},
AspectRatio->Automatic,Ticks->None,PlotRange->{Automatic,{-.05,2}}]

V6

Needs["Graphics`Colors`"]//Quiet

Plot[Table[x^n,{n,0,3,.5}],{x,0,3},PlotStyle->{Black,Red,Yellow,Green,Blue,Indigo,Violet},AxesLabel->{x,x^n},Evaluated->True]

Sums

(∑_(b=2)^∞)(∑_(a=2)^∞)(1/(a^b))==1

Sum[1/a^b,{b,2,∞},{a,2,∞}]

V7

Sum[1/a^b,{b,2,∞},{a,2,∞}]

(∑_(n=0)^(N-1))(p^n) (e^(i n x))==((((((e^(i x)) p))^N)-1)/((e^(i x)) p-1))

Sum[p^n Exp[I n x],{n,0,N-1}]

V7

Sum[p^n Exp[I n x],{n,0,N-1}]

FullSimplify[%]

(∑_(n=0)^∞)(p^n) (e^(i n x))==((1-p (e^(-i x)))/((p^2)-2 cos(x) p+1))

s=Sum[p^n Exp[I n x],{n,0,Infinity}]

Simplify[s-(1-p Exp[-I x])/(1-2 p Cos[x]+p^2)]

V7

Sum[p^n Exp[I n x],{n,0,Infinity}]

(∑_k^n)(∑_j^k)(((-1))^(k-j)) (j^n) (({{k}, {j}})) (({{x}, {k}}))==(x^n)

Table[Sum[(((-1))^(k-j))(j^n)Binomial[k,j]Binomial[x,k],{k,n},{j,k}],{n,10}]//FunctionExpand//Expand

V6

Sum[(((-1))^(k-j))(j^n)Binomial[k,j]Binomial[x,k],{k,n},{j,k}]

V7

Sum[(((-1))^(k-j))(j^n)Binomial[k,j]Binomial[x,k],{k,n},{j,k}]

(((∑_k^(p+1))(n^k) (BernoulliB[-k+p+1]) (Binomial[p+1, k]) (((-1))^(DiscreteDeltaSeq[k,p])))/(p+1))==(HarmonicNumber2[n, -p])

Sum[(k^p),{k,n}]

(1/(p+1))Sum[(((-1))^(DiscreteDelta[k,p]))Binomial[p+1,k]BernoulliB[p+1-k](n^k),{k,p+1}]

(1/(p+1))Sum[Binomial[p+1,k]BernoulliB[p+1-k](n^k),{k,p+1}]

Solutions of (a^b)==(b^a)

FullSimplify[Solve[a^b==b^a,a]]

FullSimplify[Solve[a^b==b^a,b]]

b=2

FullSimplify[Solve[a^2==2^a,a]]

N[%,10]

b=3

FullSimplify[Solve[a^3==3^a,a]]

N[%]

Roots of -1

Show[GraphicsArray[
MapIndexed[Graphics[{Circle[{0,0},1],PointSize[.05],#1,Red,Point[Through[{Re,Im}[(((-1))^(1/#2[[1]]))]]]},AspectRatio->Automatic]&,Map[Point,
Map[{Re[#],Im[#]}&,Table[x/.{ToRules[Roots[x^n==-1,x]]},{n,5}],{2}],{2}]]]
]