Why does Matlab substitute a number in a formula instead of solving it?

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tommytutone 2019-7-18

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评论： Walter Roberson 2019-7-18

Using the Symbolic toolbox, Matlab seems to arbitrarily determine whether or not to return a solved value when a concrete value is substituted in a formula.

>> syms x
>> f(x) = 1/(5+4*cos(x))
>> f(0)
ans = 1/9
>> f(1)
ans = 1/(4*cos(1) + 5) 

Why isn't it solving for f(1) ? I'm guessing its because I'm working symbolically, so I can substitute a second equation as x, but how I can force Matlab to return a concrete value?

EDIT: I tried double(f(1)) and that seemed to work. Would that be considered best practice here?

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Answer 1

Star Strider 2019-7-18

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The Symbolic Math Toolbox outputs its results as symbolic expressions, unless you ask it to do otherwise. (It assumes you want a symbolic result.) The double function is not always appropriate for symbolic results, since if the symbolic result contains a symbolic variable, double will throw an error.

If you want a numeric result, use the vpa() function:

syms x
f(x) = 1/(5+4*cos(x))
Out1 = f(0)
Out2 = f(1)
Out3 = vpa(f(1))

producing:

Out1 =
1/9
 
Out2 =
1/(4*cos(1) + 5)
 
Out3 = 
0.1396412210276252254724967860432

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Walter Roberson 2019-7-18

The Symbolic Toolbox tries to create closed form expressions of indefinitely precise algebraic numbers when practical. The Symbolic Toolbox works with mathematical results instead of approximations whenever it can.

If 1/(4*cos(1) + 5) "is" 0.139641221027625 then it follows that mod(1e17 * 1/(4*cos(1) + 5),1) should be 0 -- that multiplying by 10^17 should get you an exact integer. But that is clearly not true mathematically: cos(1) is mathematically an infinitely long decimal value.

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