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I found another easier way of solving y=x^k.

  1. Oct 19, 2014 #1
    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 theorem





    Is 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)
     
  2. jcsd
  3. Oct 19, 2014 #2

    Mark44

    Staff: Mentor

    How is this an improvement over multiplying 57 by 57 or 1234 times 1234 times 1234?

    I realize that you're doing more than just squaring or cubing numbers, but the explanation seems very long winded for doing something that can be done more quickly using plain old arithmetic.

    Don't write stuff like the above. 57 ##\neq## 5 + 7. Instead, say that you're adding the digits.
    Don't write the above in mathematical symbols. Anybody reading this would think you're nuts, since 57 - 7 = 50, not 5.
    Just solve for y by dividing both sides by 5. This results in y = 10.
    No. ##10^4 \neq 10^2##, nor are the next two equations true (or even meaningful).
    No. None of the three equations above is true.
    These should be a tipoff that you're doing something wrong.
     
  4. Oct 20, 2014 #3
    sorry for this horrid proof and I will send you a better one and this formula maybe useless compared to a calculator or the good old grade 3 math way but it's another way of getting to it you see.
     
  5. Oct 20, 2014 #4
    sorry for the bad proof I made it up on the spot since I found this formula by experimenting with numbers but that's no excuse so I'll send you a better one later today and yes it is useless compared to Grade 3 algebra way of doing this 452=2025 or today with a calculator or in your head which I do.Also I have language problems so easier wasn't the write word I meant easy
     
    Last edited: Oct 20, 2014
  6. Oct 21, 2014 #5

    Mentallic

    User Avatar
    Homework Helper

    The binomial theorem is a useful way of solving squared numbers in your head more simply, but it's very well known.
     
  7. Oct 22, 2014 #6
    here's the better proof
    Let's start with the finding of x = 10n+L by writing x in a form that can be expanded to calculate easier.
    Let n be all digits except last (ex:x=1234,n=123)
    Let L be last digit(ex:x=1234,L=4)

    x=Cn+L
    where C stands for constant
    (x-L)=Cn
    therefore
    any (number x)-(last digit)=n digits with a 0 as the last digit
    x-L=n0
    n0=Cn
    n0/n=C
    10=C
    so x=10n+L
    xk=(10n+L)k

    so what about decimal numbers
    (0.25)1=0.25 starts with 2 digits to the right of decimal and ends with 2
    (0.25)2=0.0625 starts with 2 digits to the right of decimal and ends with 4
    (0.25)3=0.015625 starts with 2 digits to the right of decimal and ends with 6
    so therefore
    (decimal places of x)(the exponent k)=decimal places of xk
    (ex:0.252, (2)(2)=4)
    so ((25)(10-2))2=0.0625
    hence (25)2)(10-4=0.0625
    hence 625/104=0.0625
    therefore 0.0625=0.0625
    so if you look at the 2nd last line
    625/104
    you can write this for (0.25)3
    as ((25)(10-2))3=0.015625
    hence ((25)3)(10-6)=0.015625
    hence 15625/106=0.015625
    therefore 0.015625=0.015625
    so xk/10dk
    which is
    (10n+L)k/10dk=A
    where A is answer
    n is all digits except L
    L is last digit
    d is decimal places moved to make x whole (ex: (0.9056)2=((9056)(10-4))2
    k is the exponent
     
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