I need to create a loop to evaluate a function at decreasing points.

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DJ Mckenna
DJ Mckenna 2023-5-18
评论： DJ Mckenna 2023-5-18
I need to evaluate the function ( (e^x)-e^(-x))/x at values starting from 1e-6, 1e-7, 1e-8... to 1e-17 and have each output be stored in a table. How can I do this?

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Dyuman Joshi
Dyuman Joshi 2023-5-18
编辑：Dyuman Joshi 2023-5-18
Ran in:
You can use a for loop to do that, evaluate the function at each point and store the result.
Another way is vectorize the operation, read more - Vectorization.
If you want to store the output in a specific way, please mention the output format.
format long
%define x
x = 10.^-(6:17)
x = 1×12
1.0e-06 * 1.000000000000000 0.100000000000000 0.010000000000000 0.001000000000000 0.000100000000000 0.000010000000000 0.000001000000000 0.000000100000000 0.000000010000000 0.000000001000000 0.000000000100000 0.000000000010000
f = (exp(x)-exp(-x))./x
f = 1×12
1.999999999946489 1.999999998947288 1.999999987845058 2.000000054458440 2.000000165480742 2.000000165480742 2.000066778862220 1.999511667349907 1.998401444325282 2.109423746787797 1.110223024625157 0
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Torsten
Torsten 2023-5-18
编辑：Torsten 2023-5-18
@DJ Mckenna
Is it the purpose of the exercise to see that usual floating point arithmetic as above gives "wrong" results ?
DJ Mckenna
DJ Mckenna 2023-5-18
Error is part of it, but I am needing to compare it to some other functions as well

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Torsten
Torsten 2023-5-18
移动：Torsten 2023-5-18
Ran in:
digits(100)
vpa((exp(sym('1/10'))-exp(sym('-1/10')))/sym('1/10'))
ans = 
2.003335000396880516474587670438100470298418337571177667366059723851880014034059419766398729346581964
vpa((exp(sym('1/1000000'))-exp(sym('-1/1000000')))/sym('1/1000000'))
ans = 
2.000000000000333333333333350000000000000396825396825402336860670194053631553631553952734508290065375
vpa((exp(sym('1/10000000'))-exp(sym('-1/10000000')))/sym('1/10000000'))
ans = 
2.000000000000003333333333333335000000000000000396825396825396880511463844797183140933140933141254322
vpa((exp(sym('1/100000000000000000'))-exp(sym('-1/100000000000000000')))/sym('1/100000000000000000'))
ans = 
2.000000000000000000000000000000000033333333333333333333333333333333333499999999999999999999781236209
I think you should be able to read the documentation on how to create a table for the results.

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