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2+3*5 17 (2 + 3) * 5 25 Each symbol in Mathematica represents something. Perhaps it is the result of a simple numerical calculation or it may be a complicated mathematical expression.
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a = 3; b= x2 + 1 ; 2x + 3
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Here, a is a symbol representing the numerical value 3 and b is a symbol representing an algebraic expression.
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EXAMPLE 14 (continuation of Example 13)
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1+ x2 3+2 x
To delete a symbol so that it can be used for a different purpose, the Clear or the Remove command can be used.
Clear[symbol] clears symbol s definition and values, but does not clear its attributes, messages, or defaults. symbol remains in Mathematica s symbol list. Typing symbol =. will also clear the definition of symbol. Remove[symbol] removes symbol completely. symbol will no longer be recognized unless it is redefined.
You may have noticed that when you begin to type the name of a symbol, it appears with a blue font until it is recognized as a Mathematica command or symbol (possibly user-defined) having some value. Then it turns black. If the symbol is cleared or removed, all instances of the symbol turn blue once again. Parentheses, brackets, and braces remain purple until completed with a matching mate. Errors caused by having two left parentheses, but only one right parenthesis, for example, can be conveniently spotted.
Getting Acquainted
EXAMPLE 15 (continuation of Example 13)
Clear[a] a
Global`a
a recalls information about the symbol a.
Remove[b] b
Information notfound : Symbol b not found.
(Clicking on gives more information about the error.)
The N command allows you to compute a numerical approximation.
N[expression] gives the numerical approximation of expression to six significant digits (Mathematica s default). N[expression, n] attempts to give an approximation accurate to n significant digits.
A convenient shortcut is to use //N to the right of the expression being approximated. Thus, expression//N is equivalent to N[expression]. // can be used for other Mathematica commands as well.
expression //Command is equivalent to Command[expression] .
Another shortcut is to type a decimal point anywhere in the expression. This will cause Mathematica to evaluate the expression numerically.
EXAMPLE 16
1+1-1 2 3 5 19 30 1+1- 1 2 3 5. 0.633333
EXAMPLE 17
Note the decimal point after the 5.
N[o] or o //N 3.14159 N[ , 50] 3.1415926535897932384626433832795028841971693993751
The Mathematica kernel keeps track of the results of previous calculations. The symbol % returns the result of the previous calculation, %% gives the result of the calculation before that, %%% gives the result of the calculation before that and so forth. Using % wisely can save a lot of typing time.
EXAMPLE 18 To construct + + , we could type: Sqrt[Pi+Sqrt[Pi+Sqrt[Pi]]]. A less confusing way of accomplishing this is to type
Sqrt[Pi]; Sqrt[Pi + %]; Sqrt[Pi + %]
The semicolon suppresses the output of the intermediate calculations.
+ +
Getting Acquainted
Using the Basic Math Input palette, we can type ; +% ; +%
+ +
SOLVED PROBLEMS
1.18 Define a = 3, b = 4, and c = 5. Then multiply the sum of a and b by the sum of b and c. Print only the final answer.
SOLUTION
a = 3; b = 4; c = 5; (a + b) * (b + c) 63
1.19 Let a = 1, b = 2, and c = 3 and add a, b, and c. Then clear a, b, and c from the kernel s memory and add again.
SOLUTION
a = 1; b = 2; c = 3; a+b+c 6 Clear[a,b,c] a+b+c a+b+c
1.20 Obtain a 25-decimal approximation of e, the base of the natural logarithm.
SOLUTION
26 significant digits gives 25 decimal places.
N[E, 26]
N[ , 26]
2.7182818284590452353602875
1 2 3 1 1.21 (a) Express + + as a single fraction. 7 13 19 23 (b) Obtain an approximation accurate to 15 decimal places.
SOLUTION
1/7 + 2/13 3/19 + 1/23 7249 39 767 N[%,15] 0.182286820730757
Getting Acquainted
1.22 Compute
SOLUTION
968 (a) exactly and (b) approximately to 25 significant digits.
968 or Sqrt[968] 22 2 N[%, 25] 31.11269837220809107363715
1.23 Multiply 12 by 6. Then multiply 15 by 7. Then use % and %% to add the two products.
SOLUTION
12 * 6 72 15 * 7 105 % + %% 177
1.24 Compute 1 +
1 1+ 1 1+ 1 1+ 1 2
SOLUTION
1 + 1 2 3 2 1 + 1 % 5 3 1 + 1 % 8 5 1 + 1 % 13 8
1.25 Compute the value of 1 + (1 + (1 + (1 + (1 + 1)2 )2 )2 )2 .
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