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Science Forum Index » Math - Symbolic Forum » a limit problem in mathematica
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| spmathsp |
 Posted: Sun Mar 30, 2008 3:01 am |
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Guest
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Hi everyone,
I couldn't take the following limit. Can anyone help me to solve indeterminate limit?
Limit[(e^((Sqrt[1 + (a^2 \[Pi]^2)/L^2] (h - 2 y))/(
2 a)) (1 + e^((2 Sqrt[1 + (a^2 \[Pi]^2)/L^2] y)/a))) /((1 + e^((
h Sqrt[1 + (a^2 \[Pi]^2)/L^2])/a)) \[Pi]), a -> 0]
Thank you very much |
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| spmathsp |
| Posted: Sun Mar 30, 2008 4:43 am |
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Thank you Nasser,
I meant by e^ to be Exp[]. But is it possible to get the output 1?
Thanks |
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| SzH |
| Posted: Sun Mar 30, 2008 5:19 am |
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On Mar 30, 4:43 pm, spmathsp <spmat...@gmail.com> wrote:
Quote: Thank you Nasser,
I meant by e^ to be Exp[]. But is it possible to get the output 1?
Thanks
Exp[1] is E in Mathematica. If we substitute explicit values (that
satisfy Nasser's assumptions) for L, y, and h, NLimit gives 0.
expr = (E^((Sqrt[1 + (a^2*Pi^2)/L^2]*(h - 2*y))/(2*a))*
(1 + E^((2*Sqrt[1 + (a^2*Pi^2)/L^2]*y)/a)))/
((1 + E^((h*Sqrt[1 + (a^2*Pi^2)/L^2])/a))*Pi)
<< "NumericalCalculus`"
expr /. {L -> 1, h -> 5, y -> 1}
NLimit[%, a -> 0]
The question is: how are h and y related to each other? |
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| spmathsp |
| Posted: Sun Mar 30, 2008 5:40 am |
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Hi Nesser,
This was a pde question. I simplified it. And the simplified solution is what I am dealing with.
I checked the expr. And it is unfortunately correct that you write.
But I am sorry that I forgot to write the interval. The interval of y is -h/2<y<h/2. But I think this is not goint to change the solution.
Dou you agree?
I would like to thank you very much again to deal with my problem. |
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| SzH |
| Posted: Sun Mar 30, 2008 5:45 am |
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Actually you can rewrite expr as
(x^(h/2 - y) + x^(h/2 + y))/(Pi*(1 + x^h))
where x is Exp[Sqrt[1 + (a^2*Pi^2)/L^2]/a]
If y == h/2 then the expression is just 1/Pi |
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| spmathsp |
| Posted: Sun Mar 30, 2008 5:50 am |
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Yes, you are exactly right. But except y=h/2, I am looking for the solution which must be equal to '1'.
Thanks szhorvat |
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| SzH |
| Posted: Sun Mar 30, 2008 6:38 am |
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Guest
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On Mar 30, 5:50 pm, spmathsp <spmat...@gmail.com> wrote:
Quote: Yes, you are exactly right. But except y=h/2, I am looking for the solution which must be equal to '1'.
Thanks szhorvat
'x' goes to infinity as 'a' goes to 0, so this particular limit is 0
if Abs[y] < h/2, Infinity if Abs[y] > h/2, and 1/Pi if Abs[y] == h/2.
Szabolcs |
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| Nasser Abbasi |
| Posted: Sun Mar 30, 2008 8:38 am |
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"spmathsp" <spmathsp@gmail.com> wrote in message
news:21856719.1206882092360.JavaMail.jakarta@nitrogen.mathforum.org...
Quote: Hi everyone,
I couldn't take the following limit. Can anyone help me to solve
indeterminate limit?
Limit[(e^((Sqrt[1 + (a^2 \[Pi]^2)/L^2] (h - 2 y))/(
2 a)) (1 + e^((2 Sqrt[1 + (a^2 \[Pi]^2)/L^2] y)/a))) /((1 + e^((
h Sqrt[1 + (a^2 \[Pi]^2)/L^2])/a)) \[Pi]), a -> 0]
Thank you very much
I assume you meant by e^ to be Exp[], so I rewrote the above, and used
assumptions to tell Mathematica that (h-2 y) is not zero to help it with the
limit. Here is what I get:
z1 = Exp[Sqrt[1 + (a^2*Pi^2)/L^2]*((h - 2*y)/(2*a))];
z2 = 1 + Exp[2*Sqrt[1 + (a^2*Pi^2)/L^2]*y]/a;
z3 = (1 + Exp[(h*Sqrt[1 + (a^2*Pi^2)/L^2])/a])*Pi;
expr = z1*(z2/z3);
Assuming[h > 2*y && y > 0, Limit[expr, a -> 0]]
Out[96]= 0
Nasser |
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| Nasser Abbasi |
| Posted: Sun Mar 30, 2008 10:05 am |
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"spmathsp" <spmathsp@gmail.com> wrote in message
news:13819569.1206888219767.JavaMail.jakarta@nitrogen.mathforum.org...
Quote: Thank you Nasser,
I meant by e^ to be Exp[]. But is it possible to get the output 1?
Thanks
First, sorry that I seem to have changed your expression slightly when I
wrote in last post.
Here is what I think your original expression rewritten to make it work,
this should match what you had.
z1 = Exp[Sqrt[1 + (a^2*Pi^2)/L^2]*((h - 2*y)/(2*a))];
z2 = 1 + Exp[2*Sqrt[1 + (a^2*Pi^2)/L^2]*(y/a)];
z3 = (1 + Exp[(h*Sqrt[1 + (a^2*Pi^2)/L^2])/a])*Pi;
expr = z1*(z2/z3);
Assuming[h > 2*y && y > 0, Limit[expr, a -> 0]]
Out[36]= 0
And I do not know how to make it '1'. Mathematica says the limit is zero.
Without the above assumptions, it remained unevaluated.
Are you saying the limit should be one? Can you make sure the above 'expr'
is correct as it is meant to be in your textbook or from where you obtained
it?
Nasser |
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