Re: Re: Solutions for functions containing jump discontinuities
- To: mathgroup at smc.vnet.net
- Subject: [mg39005] Re: [mg38982] Re: Solutions for functions containing jump discontinuities
- From: Dr Bob <drbob at bigfoot.com>
- Date: Fri, 24 Jan 2003 05:05:49 -0500 (EST)
- References: <b0lvo5$5b2$1@smc.vnet.net> <200301231305.IAA11755@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
You missed two roots (0 and 1500). Here's the case Mod[x, 1760] > 0:
f[x_] := 5(x - 1500 Ceiling[x/1760])
xRule = x -> 1760 k + y;
Simplify[f@x /. xRule, {0 < y < 1760, k â?? Integers}]
% /. Ceiling[a_] -> 1
yRule = First@Solve[% == 0, y]
Flatten[k /. Solve[y == # /. yRule, k] & /@ {1, 1759}]
kValues = Range[Ceiling@Min@%, Floor@Max@%]
yValues = y /. (yRule /. List /@ Thread[k -> kValues])
1760kValues + yValues
f /@ %
and here's the case Mod[x, 1760] == 0:
xRule = x -> 1760 k + y;
Simplify[5(x - 1500 Ceiling[x/1760]) //. {xRule, y -> 0}, {k â?? Integers}]
yRule = First@Solve[% == 0, k]
1760k /. %
f@%
Bobby
On Thu, 23 Jan 2003 08:05:15 -0500 (EST), Orestis Vantzos
<atelesforos at hotmail.com> wrote:
> Your function simplifies to:
> f[x_]:=5(x - 1500 Ceiling[x/1760])
>
> Now assume that [First Case] x==1760 k + y , 0<y<=1760 and k Integer
> Then Ceiling[x/1760]== Ceiling[k + y/1760]== k+1
> so that f[x]==5 (-1500 + 260 k + y)
>
> If f[x]==0 then 5 (-1500 + 260 k + y)==0 and we solve for y:
> y== 1500-260 k
>
> 0<y<=1760 =>
> 0< 1500-260k <=1760
> -1500<-260 k <= 260
> 5.77 > k >= 1
>
> So k ranges from 1 to 5 and since x==1500(k+1) the roots are:
> Table[1500(k + 1),{k,1,5}]
> {3000, 4500, 6000, 7500, 9000}
>
> Orestis Vantzos
>
> newspostings at burkert.de (Burkert, Philipp) wrote in message
> news:<b0lvo5$5b2$1 at smc.vnet.net>...
>> Hi folks,
>>
>> we are searching all solutions where the function f results null.
>>
>> f[x_] := -7500 * Ceiling[(0.5 * x) / 880] + (5 * x)
>> Solve[{f[x] == 0}, x]
>>
>> As f contains jump discontinuities, we recieved the following error:
>>
>> InverseFunction::"ifun": "Inverse functions are being used. Values may
>> be \
>> lost for multivalued inverses."
>>
>> Solve::"tdep": "The equations appear to involve the variables to be
>> solved \
>> for in an essentially non-algebraic way."
>>
>> We would be pleased if anybody could help us.
>>
>> Regards,
>> Philipp Burkert
>> Carsten Siegmund
>
>
--
majort at cox-internet.com
Bobby R. Treat
- References:
- Re: Solutions for functions containing jump discontinuities
- From: atelesforos@hotmail.com (Orestis Vantzos)
- Re: Solutions for functions containing jump discontinuities