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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.




int main()


double x, funcValue, approx, difference;

cout<<“\n Enter a value of 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);




// calculate the first approximation

approx = 1;

difference = abs(funcValue – approx);




// calculate the second approximation

approx = approx + x;

difference = abs(funcValue – approx);




// calculate the third approximation

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

difference = abs(funcValue – approx);




// calculate the fourth approximation

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

difference = abs(funcValue – approx);




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:

A sample run produced by the above program

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.

Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
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
I'm not sure why it wrote it the other way
I got X =-6
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?
Commplementary angles
Idrissa Reply
im all ears I need to learn
right! what he said ⤴⤴⤴
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
im not good at math so would this help me
Rachael Reply
I'm not good at math so would you help me
what is the problem that i will help you to self with?
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
many many of nanotubes
what is the k.e before it land
what is the function of carbon nanotubes?
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
what is system testing
what is the application of nanotechnology?
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
anybody can imagine what will be happen after 100 years from now in nano tech world
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
silver nanoparticles could handle the job?
not now but maybe in future only AgNP maybe any other nanomaterials
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Programming fundamentals in c++. OpenStax CNX. Jul 29, 2009 Download for free at http://cnx.org/content/col10788/1.1
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