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Ambrose 2014
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The proof of the
Exponential Equation
For starts this proof and example will give you tools and tricks to find any number squared or cubed etc like 57^2=A and even 1234^3
So x^2=A
Where A is the Answer
and x is the #
now think of n being every digit except L where L is last digit
Ex: 57 n=5,L=7 1234 n=123,L=4
now (n+L)=x is not true So what do you do since L is separated as a one digit so times n by 10 would give you (10n+L)=x since taking
x-L=ny where y is a constant coefficient stating that x's digits are involved as the maxed # of digits so if n's digits -1digit gives 1 less digit then x were when x-L would give n digits with 0 on the end as the x # of digits which want coefficient of y would give a value times n give a number with n digits with 0 10 would.
Ex: 57=(5+7)=12
so to make this true move 7 over
57-7=5+7-7
50=5y
so to make this a true equation 5 has to be multiplied by 10 as y
50=5•10
50=50
Ex2: 1234=10n+L
1234=(10•123)+(4)
1234=1230+4
1234=1234
So if x=10n+L
therefore x^2=(10n+L)^2=A
Which is the same as say A=(10n+L)(10n+L)=
100n^2+(10nL)+(10nL)+L^2=100n^2+20nL+L^2
now let's get the answer first use a calculator and square 57
giving 3249 and done by my equation gives the same answer proven by the Example below
Ex: A=100n^2+20nL+L^2
where n=5 and L=7
A=100(5)^2+20(5)(7)+(7)^2
A=100(25)+20(35)+49
A=2500+700+49
A=3249
Eureka! I know this is one example so I'll do one more
345,679^2=A
Which 34,567=n and L=9
A=100n^2+20nL+L^2
A=100(34,567)^2+20(34,567)(9)+81
now your thinking that's to large of a number to square so you take 34,567^2
3456=n,L=7
100(3456)^2+20(3456)(7)+49
and to make easier repeat
3456^2
n=345,L=6
100(345)^2+20(345)(6)+36
so you can keep going down
345^2
n=34,L=5
100(34)^2+20(34)(5)+25
And again a final time
34^2
n=3,L=4
100(3)^2+20(3)(4)+16
100(9)+20(12)+16
900+240+16
1156
Put back into the previous part as 34^2
100(1156)+20(170)+25
115,600+3,400+25
119,025
which we do again
For 345^2
100(119,025)+20(2,070)+36
11,902,500+41,400+36
11,943,936
and for 3,456^2
100(11,943,936)+20(24,192)+49
1,194,393,600+483,840+49
1,194,877,489
And for 34,567^2
100(1,194,877,489)+20(311,103)+81
119,487,748,900+6,222,060+81
119,493,971,041
which is proven by Putting 345,679^2= giving 119,493,971,041
what about decimals like 0.25^2 or 0.39^2 how can you calculate the number you add a new thing to the theorem which will give us a decimal number so for first 0.25^2=.0625 which reminds me of 25^2=625 and for .39^2=.1521 which reminds me of 39^2=1521. Therefore for both these answers I see that if I take the # of decimals place and move it to the left until there's a whole I can use the exponential formula then after evaluating I have to move it 4 spaces right so it would be for squared #'s with two decimal places would be taking exponential formula answer and divid by 10^4 but in this proof we have to create general so now we'll look at 3 and 4 decimal places like .025^2 or .0025^2 which evaluate to .000625 and .00000625 which we see would be for putting two decimal placed # which would be evaluate as a whole then divided by 10,000 or 10^4 and for three decimal placed # would evaluate as a whole # then divid by 1,000,000 or 10^6 and four decimal placed # which would be evaluate as a whole # divided by 100,000,000 or as 10^8 which if we write in groups to find unknown variable to make this equation true as the # to divid and = 10^d were d is amount of decimal places moved.
10^4=10^d1
10^6=10^d2
10^8=10^d3
we're d1=2,d2=3,d3=4
10^4=10^2
10^6=10^3
10^8=10^4
Where we take out 10^ by doing the log function
Log(10^4)=Log(10^2)
Log(10^6)=Log(10^3)
Log(10^8=Log(10^4)
Which gives
4=2
6=3
8=4
by divid the right would give a constant of 2
2=2
2=2
2=2
Which proves that by evaluating the decimal # by moving d decimal places to the left until it's a whole # and putting the whole # though the exponential formula to get a answer divid by 10^2d will get the value of the decimal # squared
Ex: a).45^2 and b)0.0003^2
a) A=100n^2+20nL+L^2/10^2d
n=4,L=5,d=2
A=100(4)^2+20(4)(5)+(5)^2/10^2(2)
A=100(16)+20(20)+25/10^4
A=1600+400+25/10,000
A=2.025(10^3)/10^4
A=2.025(10^-1) or .2025
b) A=100n^2+20nL+L^2/10^2d
n=0,L=3,d=4
A=100(0)^2+20(0)(3)+(3)^2/10^2(4)
A=0+0+9/10^8
A=9/10^8
A=9(10^-8) or 0.00000009
This part of the exponential formula
proven by using binomial theorem
First starting with what we know as (10n+L)^2/(10^d)^2 which can be generalized by
(10n+L)^k/(10^d)^k or (10^kd)
(The true Exponential Theorem)
like I said the the Exponential can proven by a similar Binomial theoremIs the Summation of the theorem which is different from the normal binomial theorem that you do 10n^k
Which is the reason for this to be different. For starts theirs three ways of using Binomial Theorem
1. Pascals Triangle shows what the coefficients are and where to put them
Ex: (10n+L)^3/10^3d
1 1
1 2 1
1 3 3 1
so coefficients are 1, 3,3,1
So from summation we see that 10 and n have same exponent
10^3n^3+10^2n^2L+10nL^2+L^3
then put in coefficients
1000n^3+300n^2L+30nL^2+L^3
2. Factorials which were written where the summation is as
(n!)/(n-k)!(k!) which find each of the coefficients.
Ex: (10n+L)^3/10^3d
Where n! is the exponent of 3
(3!)/0!(3-0)!=1
(3!)/1!(3-1)!=(3•2•1)/1•2•1=3
(3!)/2!(3-2)!=3
(3!)/3!(3-3)!=1
so the Coefficients are 1,3,3,1
now down what we know from the summation
10^3n^3+10^2n^2L+10nL^2+L^3
Put in coefficients
1000n^3+300n^2L+30nL^2+L^3
3. This way has no name it's
a technique which is hard to explain except by variables so let's do an example:(10n+L)^3/10^3d
So we start with putting down what we know from the Summation
1 2 3 4
10^3n^3+10^2n^2L+10nL^2+L^3
so the reason for #ing them since the phase stats that n's exponent of k times the value of last variables coefficient divid by # of section
(k)(c)/s(#)=s(#+1)
where for the first assume 1•1/1
1 1 2 3 4
1 (10n)^3+(10n)^2L+10nL^2+L^3
1 1 3 3 1
The ones do not have to be involved it's there to show why the coefficient is one
1 2 3 4
(10n)^3+(10n)^2L+10nL^2+L^3
1 3 3 1
Which is written as
1000n^3+300n^2L+30nL^2+L^3.
So this is my Exponential Theorem
(10n+L)^k/(10^d)^k
or
(10n+L)^k/(10^kd)
Exponential Equation
For starts this proof and example will give you tools and tricks to find any number squared or cubed etc like 57^2=A and even 1234^3
So x^2=A
Where A is the Answer
and x is the #
now think of n being every digit except L where L is last digit
Ex: 57 n=5,L=7 1234 n=123,L=4
now (n+L)=x is not true So what do you do since L is separated as a one digit so times n by 10 would give you (10n+L)=x since taking
x-L=ny where y is a constant coefficient stating that x's digits are involved as the maxed # of digits so if n's digits -1digit gives 1 less digit then x were when x-L would give n digits with 0 on the end as the x # of digits which want coefficient of y would give a value times n give a number with n digits with 0 10 would.
Ex: 57=(5+7)=12
so to make this true move 7 over
57-7=5+7-7
50=5y
so to make this a true equation 5 has to be multiplied by 10 as y
50=5•10
50=50
Ex2: 1234=10n+L
1234=(10•123)+(4)
1234=1230+4
1234=1234
So if x=10n+L
therefore x^2=(10n+L)^2=A
Which is the same as say A=(10n+L)(10n+L)=
100n^2+(10nL)+(10nL)+L^2=100n^2+20nL+L^2
now let's get the answer first use a calculator and square 57
giving 3249 and done by my equation gives the same answer proven by the Example below
Ex: A=100n^2+20nL+L^2
where n=5 and L=7
A=100(5)^2+20(5)(7)+(7)^2
A=100(25)+20(35)+49
A=2500+700+49
A=3249
Eureka! I know this is one example so I'll do one more
345,679^2=A
Which 34,567=n and L=9
A=100n^2+20nL+L^2
A=100(34,567)^2+20(34,567)(9)+81
now your thinking that's to large of a number to square so you take 34,567^2
3456=n,L=7
100(3456)^2+20(3456)(7)+49
and to make easier repeat
3456^2
n=345,L=6
100(345)^2+20(345)(6)+36
so you can keep going down
345^2
n=34,L=5
100(34)^2+20(34)(5)+25
And again a final time
34^2
n=3,L=4
100(3)^2+20(3)(4)+16
100(9)+20(12)+16
900+240+16
1156
Put back into the previous part as 34^2
100(1156)+20(170)+25
115,600+3,400+25
119,025
which we do again
For 345^2
100(119,025)+20(2,070)+36
11,902,500+41,400+36
11,943,936
and for 3,456^2
100(11,943,936)+20(24,192)+49
1,194,393,600+483,840+49
1,194,877,489
And for 34,567^2
100(1,194,877,489)+20(311,103)+81
119,487,748,900+6,222,060+81
119,493,971,041
which is proven by Putting 345,679^2= giving 119,493,971,041
what about decimals like 0.25^2 or 0.39^2 how can you calculate the number you add a new thing to the theorem which will give us a decimal number so for first 0.25^2=.0625 which reminds me of 25^2=625 and for .39^2=.1521 which reminds me of 39^2=1521. Therefore for both these answers I see that if I take the # of decimals place and move it to the left until there's a whole I can use the exponential formula then after evaluating I have to move it 4 spaces right so it would be for squared #'s with two decimal places would be taking exponential formula answer and divid by 10^4 but in this proof we have to create general so now we'll look at 3 and 4 decimal places like .025^2 or .0025^2 which evaluate to .000625 and .00000625 which we see would be for putting two decimal placed # which would be evaluate as a whole then divided by 10,000 or 10^4 and for three decimal placed # would evaluate as a whole # then divid by 1,000,000 or 10^6 and four decimal placed # which would be evaluate as a whole # divided by 100,000,000 or as 10^8 which if we write in groups to find unknown variable to make this equation true as the # to divid and = 10^d were d is amount of decimal places moved.
10^4=10^d1
10^6=10^d2
10^8=10^d3
we're d1=2,d2=3,d3=4
10^4=10^2
10^6=10^3
10^8=10^4
Where we take out 10^ by doing the log function
Log(10^4)=Log(10^2)
Log(10^6)=Log(10^3)
Log(10^8=Log(10^4)
Which gives
4=2
6=3
8=4
by divid the right would give a constant of 2
2=2
2=2
2=2
Which proves that by evaluating the decimal # by moving d decimal places to the left until it's a whole # and putting the whole # though the exponential formula to get a answer divid by 10^2d will get the value of the decimal # squared
Ex: a).45^2 and b)0.0003^2
a) A=100n^2+20nL+L^2/10^2d
n=4,L=5,d=2
A=100(4)^2+20(4)(5)+(5)^2/10^2(2)
A=100(16)+20(20)+25/10^4
A=1600+400+25/10,000
A=2.025(10^3)/10^4
A=2.025(10^-1) or .2025
b) A=100n^2+20nL+L^2/10^2d
n=0,L=3,d=4
A=100(0)^2+20(0)(3)+(3)^2/10^2(4)
A=0+0+9/10^8
A=9/10^8
A=9(10^-8) or 0.00000009
This part of the exponential formula
proven by using binomial theorem
First starting with what we know as (10n+L)^2/(10^d)^2 which can be generalized by
(10n+L)^k/(10^d)^k or (10^kd)
(The true Exponential Theorem)
like I said the the Exponential can proven by a similar Binomial theoremIs the Summation of the theorem which is different from the normal binomial theorem that you do 10n^k
Which is the reason for this to be different. For starts theirs three ways of using Binomial Theorem
1. Pascals Triangle shows what the coefficients are and where to put them
Ex: (10n+L)^3/10^3d
1 1
1 2 1
1 3 3 1
so coefficients are 1, 3,3,1
So from summation we see that 10 and n have same exponent
10^3n^3+10^2n^2L+10nL^2+L^3
then put in coefficients
1000n^3+300n^2L+30nL^2+L^3
2. Factorials which were written where the summation is as
(n!)/(n-k)!(k!) which find each of the coefficients.
Ex: (10n+L)^3/10^3d
Where n! is the exponent of 3
(3!)/0!(3-0)!=1
(3!)/1!(3-1)!=(3•2•1)/1•2•1=3
(3!)/2!(3-2)!=3
(3!)/3!(3-3)!=1
so the Coefficients are 1,3,3,1
now down what we know from the summation
10^3n^3+10^2n^2L+10nL^2+L^3
Put in coefficients
1000n^3+300n^2L+30nL^2+L^3
3. This way has no name it's
a technique which is hard to explain except by variables so let's do an example:(10n+L)^3/10^3d
So we start with putting down what we know from the Summation
1 2 3 4
10^3n^3+10^2n^2L+10nL^2+L^3
so the reason for #ing them since the phase stats that n's exponent of k times the value of last variables coefficient divid by # of section
(k)(c)/s(#)=s(#+1)
where for the first assume 1•1/1
1 1 2 3 4
1 (10n)^3+(10n)^2L+10nL^2+L^3
1 1 3 3 1
The ones do not have to be involved it's there to show why the coefficient is one
1 2 3 4
(10n)^3+(10n)^2L+10nL^2+L^3
1 3 3 1
Which is written as
1000n^3+300n^2L+30nL^2+L^3.
So this is my Exponential Theorem
(10n+L)^k/(10^d)^k
or
(10n+L)^k/(10^kd)