# 2.2 Completing the basics  (Page 6/6)

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The second output line illustrated in Table 3.4 displays the same type of items as the first line, except that the approximation to ex requires the two of two terms of the approximating polynomial. Notice also that the first item on the second line, the value obtained by the exp() function, is the same as the first item on the first line. This means that this item does not have to recalculated; the value calculated for the first line can simply be displayed a second line. Once the data for the second line have been calculated, a single cout statement can again be used to display the required values.

Finally, only the second and third items on the last two output lines shown in Figure 1 need to be recalculated because the first item on these lines is the same as previously calculated for the first line. Thus, for this problem, the complete algorithm described in pseudocode is:

`Display a prompt for the input value of x.`

`Read the input value.`

`Display the heading lines.`

`Calculate the exponential value of x using the exp() function.`

`Calculate the first approximation.`

`Calculate the first difference.`

`Print the first output line.`

`Calculate the second approximation.`

`Calculate the second difference.`

`Print the second output line.`

`Calculate the third approximation.`

`Calculate the third difference.`

`Print the third output line.`

`Calculate the fourth approximation.`

`Calculate the fourth difference.`

`Print the fourth output line.`

To ensure that we understand the processing used in the algorithm, we will do a hand calculation. The result of this calculation can then be used to verify the result produced by the program that we write. For test purposes, we use a value of 2 for x, which causes the following approximations:

Using the first term of the polynomial, the approximation is

e^2 = 1

Using the first two terms of the polynomial, the approximation is

e^2 = 1 + 2/1 = 3

Using the first three terms of the polynomial, the approximation is

e^2 = 3 + 2^2/2 = 5

Using the first four terms of the polynomial, the approximation is

e^2 = 5 + 2^3/6 = 6.3333

Notice that the first four terms of the polynomial, it was not necessary to recalculate the value of the first three terms; instead, we used the previously calculated value.

## Step 3: code the algorithm

The following program represents a description of the selected algorithm in C++.

`// This program approximates the function e raised to the x power`

`// using one, two, three, and four terms of an approximating polynomial.`

`#include<iostream.h>`

`#include<iomanip.h>`

`#include<math.h>`

`int main()`

`{`

`double x, funcValue, approx, difference;`

`cout<<“\n Enter a value of x: “;`

`cin>>x;`

`// print two title lines`

`cout<<“ e to the x Approximation Difference\n”`

`cout<<“------------ --------------------- --------------\n”;`

`funcValue = exp(x);`

`// calculate the first approximation`

`approx = 1;`

`difference = abs(funcValue – approx);`

`cout<<setw(10)<<setiosflags(iso::showpoint)<<funcValue`

`<<setw(18)<<approx`

`<<setw(18)<<difference<<endl;`

`// calculate the first approximation`

`approx = 1;`

`difference = abs(funcValue – approx);`

`cout<<setw(10)<<setiosflags(iso::showpoint)<<funcValue`

`<<setw(18)<<approx`

`<<setw(18)<<difference<<endl;`

`// calculate the second approximation`

`approx = approx + x;`

`difference = abs(funcValue – approx);`

`cout<<setw(10)<<setiosflags(iso::showpoint)<<funcValue`

`<<setw(18)<<approx`

`<<setw(18)<<difference<<endl;`

`// calculate the third approximation`

`approx = approx + pow(x,2)/2.0;`

`difference = abs(funcValue – approx);`

`cout<<setw(10)<<setiosflags(iso::showpoint)<<funcValue`

`<<setw(18)<<approx`

`<<setw(18)<<difference<<endl;`

`// calculate the fourth approximation`

`approx = approx + pow(x,3)/6.0;`

`difference = abs(funcValue – approx);`

`cout<<setw(10)<<setiosflags(iso::showpoint)<<funcValue`

`<<setw(18)<<approx`

`<<setw(18)<<difference<<endl;`

`return 0;`

`}`

In reviewing the program, notice that the input value of x is obtained first. The two title lines are then printed prior to any calculations being made. The value of ex is then computed using the exp() library function and assigned to the variable funcValue. This assignment permits this value to be used in the four difference calculations and displayed four times without the need for recalculation.

Since the approximation to the ex is “built up” using more and more terms of the approximating polynomial, only the new term for each approximation is calculated and added to the previous approximation. Finally, to permit the same variables to be reused, the values in them are immediately printed before the next approximation is made.

## Step 4: test and correct the program

The following is the sample run produced by the above program is:

The first two columns of output data produced by the sample run agree with our hand calculation. A hand check of the last column verifies that it also correctly contains the difference in values between the first two columns.

can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
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