HP 35s User Manual

							 
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HP 35s  Using the formula solver – part 1 
 
hp calculators - 5 - HP 35s  Using the formula solver – part 1- Version 1.0 
 Figure 7 
 
 This shows that it can be very important to begin with a good guess. Obviously, only positive numbers 
should be used in the search for a solution this time. To make sure that only positive numbers are used in 
the search, two guesses can be given. One is the number in the variable being solved for. The other is the 
number displayed in the lower line of the screen. These can be the same, in which case the HP 35s 
generates its own second guess. In this example, try using 1 and 10 as the guesses: Press the cancel key 
: key at the bottom left of the keyboard twice, to cancel the error message, and to leave equation mode. 
Then enter the two guesses, and solve again. 
 
::%#$0;&4$ 
 
 Figure 8 
  
Answer: Unlike the results of examples 1 and 2, the answer is not a whole number. To 12 significant figures it is 
3.16227766017. The true answer is !10, which is an irrational number and can not be displayed exactly on 
a calculator. 
  
 The Solver provides additional information about the solution it has found. If the example was tried in RPN 
mode, press the back-arrow key < to remove the text “X=” from the upper line. (Figure 9 shows the 
display in the ALL setting). 
 
 Figure 9 
  
 In RPN mode the lower line is stack register X and shows the best value that the solver could find. The 
upper line, stack register Y, shows the previous value tried. If two are the same then this is an exact 
solution. Register Z shows the value of the formula using the best answer. Press = to see the value in 
register Z. 
 
 Figure 10 
  
 The number in Z is now on the upper line. It is zero, which means that 3.16227766017 is an exact solution 
to the 12 digit precision of the HP 35s. 
  
 The Solver works in the same way in Algebraic mode, but to see the previous value tried and the best 
answer, press the = key to see a menu. 
   
						

							 
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HP 35s  Using the formula solver – part 1 
 
hp calculators - 6 - HP 35s  Using the formula solver – part 1- Version 1.0 
 Figure 11 
  
 The number in x is the best answer, the number in y, shown in Figure 11, is the previous value, and the 
number in z is the value of the formula. Use the left and right arrow keys to see each of these numbers, 
and press 3 to copy the number shown into a calculation. 
  
 If the numbers in stack registers X and Y differ by 1 in the last digit then there is no solution exactly correct 
to 12 digits, and the two values are on either side of the exact answer. This is confirmed if the value in 
stack register Z is very small. Press : to cancel the menu. 
 
 If an error condition has occurred, such as the log of a negative number, and the Solver has not yet 
calculated f(x) at two values of x, then the three values will not have been put on the stack or in the menu. 
  
Practice Example: Where there is no solution 
 
Example 4: Sometimes a formula or equation has no exact or approximate solution. For example a²  = -4 clearly has 
two complex roots a = (0,2) and (0,-2), but no real root that the solver can find. Try solving this to see how 
the formula solver handles such cases. 
 
The formula solver always begins by moving everything from the right of the equals sign to the left side, so 
the above equation would become a² + 4 = 0. Then it looks for a value of the variable to make the left hand 
side equal to zero. If the formula to be solved already has zero to the right of the equals sign, then there is 
no need to include “= 0”, only the formula to be solved needs to be typed. 
 
Solution: Go to equation mode and enter the formula A² + 4 
  
 &>(-*.3 
  
 Figure 12 
  
 To solve the formula, press 4 and >. > is on the ? key. 
  
 The word “SOLVING” is shown as before. There is no root, so the search can take some time. To interrupt 
a long-lasting search, press the cancel key :. If the search is not interrupted, it will finally display: 
  
 Figure 13 
 
Answer: The solver indicated that there was no root. 
   
						

							 
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HP 35s  Using the formula solver – part 1 
 
hp calculators - 7 - HP 35s  Using the formula solver – part 1- Version 1.0 
What the Solver can and can not find 
 
The examples above have shown the basics of what the formula solver can find and what it can not find. 
 
In a formula or equation with one unknown variable the solver can find one or more roots if there are any. 
  
If the solution can not be represented exactly, the solver finds the two nearest numbers on either side of it. 
  
The solver can not find a solution if two or more variables are unknown. 
  
The solver can not find complex roots, as these have two unknown variables, the real and imaginary parts. 
Note that the HP 35s manual has a polynomial root finder program that will find complex roots. 
  
The solver can not find roots of matrix equations. 
Note that the HP 35s manual has a matrix program for solving three simultaneous equations. 
  
The solver can not find a root if there is no root, but in this case it can find a minimum. 
  
 If there is a solution that is not zero but is less than 10-499 the solver returns zero. 
  
 If there is a solution that is greater than 10499 the solver gives an OVERFLOW error. 
  
 If an error condition occurs in a calculation, for example the log of a negative number, the solver stops. 
  
 In addition there are some special cases that are explained in the second part of this aid. 
 
The solver has many more features. The second part of this training aid will describe some of them. 
   
						

							 
 
hp calculators 
 
 
 
 
HP 35s  Using the formula solver – part 2 
 
 
 
 
Overview of the formula solver 
 
Practice Example: A formula with several variables 
 
Practice Example: A direct solution 
 
Practice Example: Where two functions intersect 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
   
						

							 
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HP 35s  Using the formula solver – part 2 
 
hp calculators - 2 - HP 35s  Using the formula solver – part 2 - Version 1.0 
Overview of the Formula Solver 
 
Given an expression of the form: 
 
f(x) = y 
 
The HP Solve Application searches for a value of x that gives: 
 
f(x) = y = 0 
 
A value of x for which this is true is called a root, and it provides a solution of the equation f(x) = 0. The graph in 
Figure 1 shows this graphically – there is a root at the value of x where f(x) is zero. 
 
 Figure 1 
 
On the HP 35s, f(x) can be typed as a formula, in equation mode, or it can be typed as a program. When the Solver is 
used to find a root of a formula or equation typed in equation mode, it is referred to as the Formula Solver. 
 
Part 1 of this training aid provided an introduction to the Formula Solver, using a few simple examples. This second part 
explains how the Solver works, and shows some more examples. 
 
The Formula Solver works with f(x) as a formula containing x, for example 
 
3x² - 3x - 15   or   5sin(x) – 7log(x) 
 
If the Formula Solver is given an equation with terms on both sides of the equals sign, such as: 
 
3x² + 4x = 7x +15 
 
then it begins by moving everything to one side of the equals sign, so the above equation would become the formula: 
 
f(x) = 3x² - 3x - 15 = 0 
 
The Formula Solver ignores the = 0 part, as it is trying to find a value for x to make the formula zero. So there is no need 
to type an equation with = 0 in it; it is enough to type the formula. 
 
The variable x in the above is called the unknown variable. It can be represented by any of the HP 35s variables A 
through Z. A formula can contain more than one variable, the Solver will ask which is the unknown variable and will then 
ask for the known values of all the other variables. 
 
The Formula Solver then tries to rearrange the equation f(x) = 0 to give a direct solution for x. An example will be 
shown later.   
						

							 
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HP 35s  Using the formula solver – part 2 
 
hp calculators - 3 - HP 35s  Using the formula solver – part 2 - Version 1.0 
If a direct solution is not found, the Solver begins by first trying two guesses for the unknown variable. The user can give 
one or two guesses for the Solver to start from. The first part showed that this can be very useful, either to direct the 
Solver towards one root of several, or to direct the Solver away from values that would cause an error. Good guesses 
can also speed up the search for a root. In some cases the function varies very slowly over some values, and a good 
guess is needed to direct the Solver away from them and towards the range of values where a solution is expected. 
 
Beginning from the values obtained for the first two guesses, the Solver searches for values of the unknown variable that 
make f(x) smaller. If two guesses have the opposite sign, the Solver tries to narrow down the region between them until 
it finds where the sign changes and f(x) is zero. If two guesses have the same sign, the Solver uses the difference 
between them to pick the direction in which to change x to look for a third value closer to zero. 
 
The process is repeated until one of the following happens. 
 
A value of x is found for which f(x) is exactly zero. 
 
Two neighboring values of x are found, differing by 1 or 2 in the twelfth significant digit, such that f(x) changes 
sign between them. An example of this was given in part 1.  
 
No value can be found, but the Solver finds a minimum. An example was given in part 1. 
 
No value can be found because the Solver is looking at values of x for which f(x) is constant. 
 
No value can be found because f(x) is decreasing asymptotically towards a non-zero value. 
 
Two more cases arise because the HP 35s works with a finite range of numbers, negative numbers between 
-10500 and -10-499, 0 and positive numbers between +10-499 and +10500. This range is sufficient to cover all 
physical measurements, and even all numbers in government finances, but problems in number theory and in 
combinatorial operations may require numbers outside this range. (To be exact, the largest absolute value that 
the HP 35s can work with is 9.99999999999E499.) 
 
No value can be found because the root is at a value of x that is not zero but lies between -10-499 and +10-499. In 
such a case the Solver gives a result of 0. 
 
No value can be found because the root is at x that is more negative than, or equal to, -10500 or greater than or 
equal to +10500 . In such a case the Solver stops with an OVERFLOW error. 
 
To help the user distinguish between the above, the Solver returns the last value it tried for x, the last but one value tried, 
and the value of f(x) at the last value. Part 1 showed how these values can be seen in RPN mode and in Algebraic 
mode. 
 
The following examples show some of the features that were not included in part 1. 
   
						

							 
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HP 35s  Using the formula solver – part 2 
 
hp calculators - 4 - HP 35s  Using the formula solver – part 2 - Version 1.0 
Practice Example: A formula with several variables 
 
Example 1:  A factory is to produce tin cans with a volume of 100 cubic centimeters. The designer estimates that the 
height should be 10 cm and the radius about 2 cm. Calculate the exact volume of this can, and if it is not 
close to 100 cubic centimeters then recalculate the radius to give the required volume. 
 
Solution:  The equation for a cylinder’s volume V, given its radius r, and height h, is V =  ! r² h. Enter this as the 
formula ! r² h – V in equation mode and then use the Solver. 
  
Go to equation mode by typing !. If necessary, put the new equation in a particular place in the list of 
equations by moving up or down through the list with the up and down cursor keys below the HP 35s 
screen. 
  
 Enter the formula by typing: 
  
 #$%&($%)*%+, 
  
 As was explained in part 1, to enter a variable into an equation, press the % key and then one of the 
letter keys. As with -., the symbol A..Z at the top of the screen is shown as a reminder that one of 
the keys marked A through Z must be pressed. For example press the / key to enter the variable V. 
  
 Figure 2 
 
 To solve the equation, press the -0 key. The Solver asks which variable to solve for: 
  
 Figure 3 
  
 The symbol A..Z is at the top of the screen again. The variable in this formula is V so press /  again. The 
Solver now knows that V is the unknown variable and it asks for the values of the known variables. 
  
 Figure 4 
  
 The value that is already stored in R is shown too. If this is the required value then it is enough to press 
1. If the variable R has not been used before, then its value is zero. In this example, type the radius 2 
and press 1. 
  
 Figure 5   
						

							 
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HP 35s  Using the formula solver – part 2 
 
hp calculators - 5 - HP 35s  Using the formula solver – part 2 - Version 1.0 
 The Solver asks for the other known variable. Type the height, 10, and press 1 again. The HP 35s 
displays SOLVING for a moment, then the result. 
  
 Figure 6 
  
 The volume is over 125 cubic centimeters, considerably more than the intended 100. Repeat the 
calculation, but this time use the known volume of 100, and solve for the radius. Solve the equation again 
by pressing !-0. The Solver asks for the unknown variable, press &. The Solver then asks 
for the known variables, first H. 
  
 Figure 7 
  
 The present value of H is the value previously given. As this is to remain the same, just press 1 again. 
The Solver now asks for the other variable, V. 
  
 Figure 8 
  
 The present value of V is shown; this is the volume just calculated. As the volume should be 100, type 100 
and press 1. The Solver calculates and displays the radius needed to give the required volume. 
  
 Figure 9 
  
Answer: The cans should have a radius of 1.78 cm. 
 
Practice Example: A direct solution 
 
Example 2:  To show that the HP 35s looks for a direct solution before starting to search for a root, try to solve ln(z) = 0 
beginning from a negative number for the guess. 
 
Solution:  Store –5 in Z. Then store LN(z) as the formula to solve. This means that a solution is wanted for the 
equation LN(z) = 0. 
  
 /2-.3!-4%3, 
    
						

							 
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HP 35s  Using the formula solver – part 2 
 
hp calculators - 6 - HP 35s  Using the formula solver – part 2 - Version 1.0 
 Figure 10 
  
 To solve the equation, press -03. The Solver immediately displays the answer: 
  
 Figure 11 
  
Answer: Z = 1 is the solution to LN(z) = 0. This is obvious, the point of this example is that the answer was found 
immediately, and the negative guess was not tried. If the negative guess had been tried, it would have 
caused a LOG(NEG) error, as in Example 3 of part 1. The Formula Solver recognized that Z appears only 
once in the formula, and that LN(Z) = 0 can therefore be rewritten as Z = exp(0) to solve for Z directly. Such 
direct solutions can speed up the use of Solver, specially when a complicated formula with several 
variables is being solved several times for different variables. 
 
Note: Where more than one solution is possible, for example ASIN(Y)=0, the direct solution is the “principal” value. For 
example, for ASIN(Y)=0, this is 0 degrees, not 180 degrees, or –180 degrees, or any other possible value. In the same 
way, an equation such as X²=4 is solved directly and returns the positive root 2. To find other roots, it is necessary to 
write the expression in such a way that the Solver does not find a direct solution. An easy way to achieve this is to add 
0! the unknown variable into an expression, for example ASIN(Y) + 0!Y = 0 or X² + 0!X = 0. This is because the Solver 
stops looking for a direct solution as soon as it sees the unknown variable more than once in an expression. 
 
Practice Example: Where two functions intersect 
 
The Formula Solver can also be used to solve problems of the form: 
 
g(x) = h(x) 
 
This requires a value of x at which one function g(x) is equal to another function h(x). In other words, the problem is to 
find x at which these functions intersect. 
 
The equation can be rewritten as: 
 
f(x) = g(x) - h(x) = 0 
 
Solving the formula g(x) – h(x) will give the value of x at which the two functions cross over. 
 
Example 3:  The factory from Example 1 is interested in designing spherical containers with the same volume and the 
same radius as their tin cans. This means that they want to find a radius r such that: 
 
V = ! r² h = 4/3 ! r³ 
 
Solution:  Modify the formula from Example 1 to find r such that ! r² h – 4/3 ! r³ is zero.