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#1 2017-12-30 11:44:30
- Hannibal lecter
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what is the solution for f(x) = exp(-x) ?
Hi, is there a root for the f(x) = e^-x ???
is it close to 0.571143115080177? or that wrong there is no any root?
please help me
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#2 2017-12-30 14:51:23
- Jai Ganesh
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Re: what is the solution for f(x) = exp(-x) ?
Hi,
Hope this graph helps:
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#3 2017-12-31 05:10:30
Re: what is the solution for f(x) = exp(-x) ?
Hi, is there a root for the f(x) = e^-x ???
is it close to 0.571143115080177? or that wrong there is no any root?
please help me
Where did you get 0.571143115080177 from? Are you stating the problem correctly? Or is something else meant by the word 'root' here?
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#4 2018-01-04 09:05:55
- Hannibal lecter
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Re: what is the solution for f(x) = exp(-x) ?
I tried a matlab code to solve it, but thanks for helping I know it's nor have a root now..
but what about exp(-x)=x?
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#5 2018-01-04 23:00:18
- iamaditya
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Re: what is the solution for f(x) = exp(-x) ?
Where did you get 0.571143115080177 from?
See the graph which Ganesh posted. You will see that it converges down to X-axis and almost touches it at that value.
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#6 2018-01-05 01:42:19
- Bob
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Re: what is the solution for f(x) = exp(-x) ?
See here:
http://www.mathsisfun.com/data/function … nc1=e^(-x)
Just add y = x as the second function and you'll see it does have that value as the solution.
Bob
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#7 2018-01-06 01:24:06
Re: what is the solution for f(x) = exp(-x) ?
See the graph which Ganesh posted. You will see that it converges down to X-axis and almost touches it at that value.
No, it doesn't ever touch the x-axis. The exponential function is strictly positive: it can't have any roots.
However, if we take , then there is indeed a root. In fact, if we allow complex solutions, there are infinitely many of them, and they are precisely the nth values of the Lambert-W function, . This sequence generates all the complex roots.There is one real root, the so-called Omega constant, . There are several exact forms for the Omega constant, such as the 'power tower':and a nice integral relation is:
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