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Approximation

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We think you wrote:

m-1/m=5m2-1m2

This solution deals with approximation.

Solution found

m≓-0.556693077
m≓-0.556693077

Step by Step Solution

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Reformatting the input :

Changes made to your input should not affect the solution:

 (1): "m2"   was replaced by   "m^2".  1 more similar replacement(s).

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

                     m-1/m-(5*m^2-1*m^2)=0 

Step by step solution :

Step  1  :

Equation at the end of step  1  :

        1              
  (m -  —) -  (5m2 -  m2)  = 0 
        m              

Step  2  :

            1
 Simplify   —
            m

Equation at the end of step  2  :

        1     
  (m -  —) -  4m2  = 0 
        m     

Step  3  :

Rewriting the whole as an Equivalent Fraction :

 3.1   Subtracting a fraction from a whole

Rewrite the whole as a fraction using  m  as the denominator :

          m     m • m
     m =  —  =  —————
          1       m  

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

 3.2       Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

 m • m - (1)     m2 - 1
 ———————————  =  ——————
      m            m   

Equation at the end of step  3  :

  (m2 - 1)    
  ———————— -  4m2  = 0 
     m        

Step  4  :

Rewriting the whole as an Equivalent Fraction :

 4.1   Subtracting a whole from a fraction

Rewrite the whole as a fraction using  m  as the denominator :

           4m2     4m2 • m
    4m2 =  ———  =  ———————
            1         m   

Trying to factor as a Difference of Squares :

 4.2      Factoring:  m2 - 1 

Theory : A difference of two perfect squares,  A2 - B2  can be factored into  (A+B) • (A-B)

Proof :  (A+B) • (A-B) =
         A2 - AB + BA - B2 =
         A2 - AB + AB - B2 =
         A2 - B2

Note :  AB = BA is the commutative property of multiplication.

Note :  - AB + AB equals zero and is therefore eliminated from the expression.

Check : 1 is the square of 1
Check :  m2  is the square of  m1 

Factorization is :       (m + 1)  •  (m - 1) 

Adding fractions that have a common denominator :

 4.3       Adding up the two equivalent fractions

 (m+1) • (m-1) - (4m2 • m)     -4m3 + m2 - 1
 —————————————————————————  =  —————————————
             m                       m      

Step  5  :

Pulling out like terms :

 5.1     Pull out like factors :

   -4m3 + m2 - 1  =   -1 • (4m3 - m2 + 1) 

Polynomial Roots Calculator :

 5.2    Find roots (zeroes) of :       F(m) = 4m3 - m2 + 1
Polynomial Roots Calculator is a set of methods aimed at finding values of  m  for which   F(m)=0  

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers  m  which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational number  P/Q   then  P  is a factor of the Trailing Constant and  Q  is a factor of the Leading Coefficient

In this case, the Leading Coefficient is  4  and the Trailing Constant is  1.

 
The factor(s) are:

of the Leading Coefficient :  1,2 ,4
 
of the Trailing Constant :  1

 
Let us test ....

  P  Q  P/Q  F(P/Q)   Divisor
     -1     1      -1.00      -4.00   
     -1     2      -0.50      0.25   
     -1     4      -0.25      0.88   
     1     1      1.00      4.00   
     1     2      0.50      1.25   
     1     4      0.25      1.00   


Polynomial Roots Calculator found no rational roots

Equation at the end of step  5  :

  -4m3 + m2 - 1
  —————————————  = 0 
        m      

Step  6  :

When a fraction equals zero :

 6.1    When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

  -4m3+m2-1
  ————————— • m = 0 • m
      m    

Now, on the left hand side, the  m  cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :
   -4m3+m2-1  = 0

Cubic Equations :

 6.2     Solve   -4m3+m2-1 = 0

Future releases of Tiger-Algebra will solve equations of the third degree directly.

Meanwhile we will use the Bisection method to approximate one real solution.

Approximating a root using the Bisection Method :

We now use the Bisection Method to approximate one of the solutions. The Bisection Method is an iterative procedure to approximate a root (Root is another name for a solution of an equation).

The function is   F(m) = -4m3 + m2 - 1

At   m=   0.00   F(m)  is equal to  -1.00 
At   m=   -1.00   F(m)  is equal to  4.00 

Intuitively we feel, and justly so, that since  F(m)  is negative on one side of the interval, and positive on the other side then, somewhere inside this interval,  F(m)  is zero

Procedure :
(1) Find a point "Left" where F(Left) < 0

(2) Find a point 'Right' where F(Right) > 0

(3) Compute 'Middle' the middle point of the interval [Left,Right]

(4) Calculate Value = F(Middle)

(5) If Value is close enough to zero goto Step (7)

Else :
If Value < 0 then : Left <- Middle
If Value > 0 then : Right <- Middle

(6) Loop back to Step (3)

(7) Done!! The approximation found is Middle

Follow Middle movements to understand how it works :

    Left       Value(Left)     Right       Value(Right)

 0.000000000   -1.000000000 -1.000000000    4.000000000
 0.000000000   -1.000000000 -1.000000000    4.000000000
-0.500000000   -0.250000000 -1.000000000    4.000000000
-0.500000000   -0.250000000 -0.750000000    1.250000000
-0.500000000   -0.250000000 -0.625000000    0.367187500
-0.500000000   -0.250000000 -0.562500000    0.028320312
-0.531250000   -0.118041992 -0.562500000    0.028320312
-0.546875000   -0.046707153 -0.562500000    0.028320312
-0.554687500   -0.009660721 -0.562500000    0.028320312
-0.554687500   -0.009660721 -0.558593750    0.009212255
-0.556640625   -0.000253528 -0.558593750    0.009212255
-0.556640625   -0.000253528 -0.557617188    0.004472028
-0.556640625   -0.000253528 -0.557128906    0.002107418
-0.556640625   -0.000253528 -0.556884766    0.000926487
-0.556640625   -0.000253528 -0.556762695    0.000336365
-0.556640625   -0.000253528 -0.556701660    0.000041390
-0.556671143   -0.000106077 -0.556701660    0.000041390
-0.556686401   -0.000032345 -0.556701660    0.000041390
-0.556686401   -0.000032345 -0.556694031    0.000004522
-0.556690216   -0.000013912 -0.556694031    0.000004522


     Next Middle will get us close enough to zero:

     F( -0.556693077 ) is  -0.000000087  

     The desired approximation of the solution is:

       m ≓ -0.556693077

     Note, ≓ is the approximation symbol

One solution was found :

                         m ≓ -0.556693077

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