Re: A very difficult equation ...

• To: mathgroup at smc.vnet.net
• Subject: [mg14072] Re: [mg14063] A very difficult equation ...
• From: BobHanlon at aol.com
• Date: Tue, 22 Sep 1998 03:25:05 -0400
• Sender: owner-wri-mathgroup at wolfram.com

f[x_] := -2 + 2 * Cos[x] + x * Sin[x]

Clearly there will be a root whenever both Sin[x] = 0 and Cos[x] = 1,
that is, for (x = 2 n Pi).

Table[f[x], {x, -6 Pi, 24 Pi, 2 Pi}]

{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}

Plot[f[x], {x, -2Pi, 5Pi},
Ticks -> {Table[n Pi/2, {n, -4, 10}], Automatic}];

The function, f[x], has even symmetry, that is,

f[-x] == f[x]

True

To numerically find the roots in between the roots at (2 n Pi):

Table[{(2n+1) Pi, FindRoot[f[x] == 0, {x, (2n+1) Pi}]}, {n, 5}]

{{3*Pi, {x -> 8.986818916171502}}, {5*Pi, {x -> 15.45050367387554}},
{7*Pi, {x -> 21.80824331885779}}, {9*Pi, {x -> 28.13238784166647}},
{11*Pi, {x -> 34.44151054776893}}}

Roots[f[x] == 0, x]

x == 2*Tan[x/2]

g[x_] := x - 2 Tan[x/2]

g[x] has odd symmetry, that is,

g[-x] == -g[x]

True

eps = 10^-3;
plots = Table[Plot[g[x], {x, (2n+1) Pi + eps,
(2n+3) Pi - eps}, DisplayFunction -> Identity,
Ticks -> {Table[n Pi/2, {n, -6, 10}], Automatic}],
{n, -2, 1}];
Show[plots, DisplayFunction -> \$DisplayFunction];

g[x] has roots at zero and at the "in between"  roots of f[x]

Table[{(2n+1) Pi, FindRoot[g[x] == 0, {x, (2n+1) Pi - eps}]}, {n, 5}]
{{3*Pi, {x -> 8.986818921478805}}, {5*Pi, {x -> 15.45050367387541}},
{7*Pi, {x -> 21.80824331889424}}, {9*Pi, {x -> 28.13238782566299}},
{11*Pi, {x -> 34.44151054626311}}}

As to whether f[x] == Sin[x](Sin[x](x^2 + 4) - 4x).  Since Sin[x] has a
zero at all integer multiples of Pi, these cannot be equal.  Also,
plotting the difference shows that they are not equal:

Plot[Sin[x](Sin[x](x^2 + 4) - 4x) - f[x], {x, -2Pi, 5Pi}];

Bob Hanlon

In a message dated 9/19/98 5:23:50 AM, tobi.kamke at t-online.de wrote:

>f[x_] := -2 + 2 * Cos[x] + x * Sin[x]
>
>How can I solve the equation f[x] == 0 in a symbolic way? Solve[f[x] ==
>0, x] says, that it isn't possible to solve this equation! Furthermore
>I can't simplify this equation. But I think, that x = z * 2 * Pi (z =
>.. -2, -1, 0, 1, 2 ...) are some of the possible solutions. Is there a
>trick or a possibility to find out this solutions and the rest of all
>solutions with Mathematica??
>
>
>
>Thanks,
>
>  Tobi
>
>
>P. S. Is f[x] == Sin[x](Sin[x](x^2 + 4) - 4x)??? How can I find this out
>with Mathematica??

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