#Mathematical Functions

print "Enter - 1(1/x), 2(sqrtx), 3(sin), 4(cos), 5(tan),"

Print "6(expx), 7(lnx), 8(10**x), 9(log10x), 10(x**p)."

input k

print "Enter value to be evaluated ";: input x

if (k-1)=0 then goto k01

if (k-2)=0 then goto k02

if (k-3)=0 then goto k03

if (k-4)=0 then goto k04

if (k-5)=0 then goto k05

if (k-6)=0 then goto k06

if (k-7)=0 then goto k07

if (k-8)=0 then goto k08

if (k-9)=0 then goto k09

if (k-10)=0 then goto k10

k01:

y=1/x

print "1/x = ";:print y

goto Done

k02:

gosub sqrt

print "sqart = ";:print sqrtx

goto Done

k03:

#Convert degree to radian

x=0.0174533*x

y=sin(x)

print "sinx = ";:print y

goto Done

k04:

#Convert degree to radian

x=0.0174533*x

y=cos(x)

print "cosx = ";:print y

goto Done

k05:

#Convert degree to radian

x=0.0174533*x

y=tan(x)

print "tanx = ";:print y

goto Done

k06:

gosub expe

print "expx = ";:print expx;:print " x 10**";:print k

goto Done

k07:

gosub log

print "logex = ";:print logx

goto Done

k08:

#Reduce input number by 1/10**n, then multiply 10**n at the end

for j=0 to 1000

if x<1 then goto Endrun

x=x-1

next j

Endrun:

#Convert to expanding variable in the series: x loge10=2.3025851*x

x=2.3025851*x

gosub expe

print "10**x = ";:print expx;:print " x10**";:print (k+j)

goto Done

k09:

gosub log

#log10x=0.43429*logex

y=0.43429*logx

print "log10x = ";:print y

goto Done

k10:

print "Enter Value of the Exponental = ";:input p

#y=x**p=expe(p*logex)

gosub xp

print "x**p = ";:print expx;:print "x10**";:print k

goto Done

xp:

gosub log

x=p*logx

gosub expe

return

expe:

#Limit for input is x=abs(78)

#expex=1 + x/1! + x**2/2! + … + x**n/n!

#Initialization

expx=1

xp=1

n=0

nf=1

delta=0.00000001

m=0

#Change sign if x<0

x0=x

if x<0 then x=abs(x)

#Sum over series

SubLoop2:

m=m+1

exp0=expx

xp=xp*x

n=n+1

nf=nf*n

expx=expx+xp/nf

if abs(expx-exp0)<delta then goto End2

if m>1000 then goto error

refresh

goto SubLoop2

End2:

#Move decimal point to produce exponential representation: 1234=1.234*10**3

for k=0 to 1000

if expx<10 then goto Quit

expx=expx/10

next k

Quit:

#Reverse exp(x) back to exp(-x) in case of x<0

if x0>0 then goto Finish

expx=1/expx

k=-k

Finish:

Return

log:

#For x>0, log10x=0.43429*logx

#Check number of digits in the input

xint=int(x)

x$=string(xint)

L=length(x$)

logint=0

if L<1 then goto Cont1

if L=1 then goto Cont1

#If the number of digits > 1, then move the decimal point,

#for example: loge1234=loge1.234+3*loge10=loge1.234+3*2.3025851

#This step is necessary since the expanding series converges very slow for large x

logint=(L-1)*2.3025851

x0=x

xfrc=x

#Move the decimal point

for k=1 to (L-1)

xfrc=xfrc/10

next k

x=xfrc

Cont1:

#Initialization

m=0

logx=0

n=1

x1=x-1

x1n=x1

x2=x+1

x2n=x2

x3=x1n/x2n

delta=0.00000001

#Sum over the series: logex=2*[(x-1)/(x+1) + (x-1)**3/3*(x+1)**3 + …]

SubLoop:

logx=logx+x3

if abs(x3)<delta then goto Endlog

if m>1000 then goto error

n=n+2

x1n=x1n*x1*x1

x2n=x2n*x2*x2

x3=x1n/(n*x2n)

refresh

goto SubLoop

Endlog:

logx=2*logx

logx=logx+logint

return

sqrt:

#This method was developed by Heron of Alexandria in AD 60

#It starts with an area A= 1*x, then varying the sides while keeping A constant,

#finally the rectangle is reduced to a square: A=x=y*y, thus, sqrtx=y

#Initialization

A=x

xx=x

y=1

delta=0.00000001

#Algorithm to produce a square

MainLoop:

xx=(xx+y)/2

#The following statement is for keeping the area constant while varying y

y=A/xx

if abs(xx-y)<delta then goto Ende

goto MainLoop

Ende:

sqrtx=y

return

error:

print "Error – out of computational range"

Done: