Mathematica Help


by vp43
Tags: mathematica
vp43
vp43 is offline
#1
Dec2-04, 08:02 PM
P: 1
Here is my problem: I got A right, but B is wrong... Can point out what I'm doing wrong? Thanks!

Develop a Mathematica procedure to find 5 roots of the following equations.

a.) x^5 + 5x^4 + 4x^3 + 3x^2 + 2x + 1 = 0

b.) e^x sin^2 x - cos x = 0

For part A, I got NSolve[x^5 + 5x^4 + 4x^3 + 3x^2 + 2x + 1 == 0, x]

And got 5 roots of the following:
{{x -> -4.19273}, {x -> -0.564099 -
0.390903 \[ImaginaryI]}, {x -> -0.564099 + 0.390903 \[ImaginaryI]}, {x -> \
0.160462\[InvisibleSpace] - 0.693272 \[ImaginaryI]}, {x -> 0.160462\
\[InvisibleSpace] + 0.693272 \[ImaginaryI]}}

But for part B, I did the procedure:

FindRoot[E^x Sin^2 x - Cos x == 0, x]

And gave me: FindRoot::fdss: Search specification x should be a list with a 2-5 elements. (FindRoot[\[ExponentialE]\^x\ Sin\^2\ x - Cos\ x == 0, x]\)

~TRI~
Phys.Org News Partner Science news on Phys.org
SensaBubble: It's a bubble, but not as we know it (w/ video)
The hemihelix: Scientists discover a new shape using rubber bands (w/ video)
Microbes provide insights into evolution of human language
Justin Lazear
Justin Lazear is offline
#2
Dec2-04, 11:55 PM
P: 291
You might also be interested in the TeXForm command if you're going to be posting output from Mathematica much. That way, you only need put in the [ tex ] and [ /tex ] tags.

[tex]\{ \{ {x\rightarrow {-2.96732}}\} ,
\{ {x\rightarrow {-0.652083 - 0.707484\,\imag }}\} ,
\{ {x\rightarrow {-0.652083 + 0.707484\,\imag }}\} ,
\{ {x\rightarrow {0.135744 - 0.587885\,\imag }}\} ,
\{ {x\rightarrow {0.135744 + 0.587885\,\imag }}\} \}[/tex]

is output from

NSolve[x^5 + 5x^4 + 4x^3 + 3x^2 + 2x + 1 == 0, x] // TeXForm

or

TeXForm[NSolve[x^5 + 5x^4 + 4x^3 + 3x^2 + 2x + 1 == 0, x]]

--J
rayveldkamp
rayveldkamp is offline
#3
Dec3-04, 09:00 AM
P: 60
What you do is replace sin^2(x) by 1-cos^2(x) and you get the DE:
e^x cos^2(x) +cos(x)==e^x,

using the Solve function on Mathematica, BUT dont solve in terms of x solve for Cos[x]
i.e: Solve[e^x cos^2(x) +cos(x)==e^x, Cos[x] ]

You then get solutions in terms of Cos[x], which you can then solve trigonometrically, the reason Mathematica doesnt like the equation is because the there are infintite solutions, and the Solve function cant handle these.

Hope this helps
Ray


Register to reply

Related Discussions
mathematica help. Math & Science Software 9
Mathematica 6.0 Changes Everything Math & Science Software 37
Mathematica Help Math & Science Software 5
Where can I download Mathematica Math & Science Software 3
mathematica bug? Math & Science Software 5