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?? > >Please help me!!! > > >Thanks, > > Tobi > > >P. S. Is f[x] == Sin[x](Sin[x](x^2 + 4) - 4x)??? How can I find this out >with Mathematica??