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**zetafunc.****Guest**

Hello,

Originally I posted a thread that was accidentally deleted about the Laplace Transform of tan(t). I've decided not to post my original working because it's wrong (the first integral diverges so it doesn't really make sense) and so I've left it a bit more open now.

The problem is to simply compute the Laplace transform of tan(t). This is what Wolfram|Alpha gives (with credit to Au101 who typed it up):

I've started to approach the problem differently by using the definition of tan(t);

and I'm just trying to integrate that at the moment and I've ended up with some more complicated integrals after using integration by parts. I was wondering what your thoughts on this are and if this is a step in the right direction, or just a push in which direction I should go next.

One question I have though is if this even makes sense... in that, using the definition of the Laplace transform, the integral is divergent in the first place. So how does Wolfram|Alpha get an answer, and will my method of using the definition of tan(t) in terms of e and i work?

Thanks.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,306

Hi zetafunc.;

What are you entering into Alpha that is getting that answer?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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**zetafunc.****Guest**

I'm entering "Laplace transform of tan(t)".

When I just enter the integral itself, which is divergent, it doesn't give me an answer... it just says operation timed out.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,306

May I have the exact way you are entering it. Because I am timing out in Alpha and at home.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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**zetafunc.****Guest**

I'm entering "Laplace transform of tan(t)" and getting that. What's in the quotes is what I'm entering...

Sorry for delays between replies, I am dealing with rioters/people...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,306

Hi;

I do not think I would trust how he is interpreting that input:

Try entering 'Laplace transform of t"

He gags on it. Seems to me since mathematica can not do that

transform, how can Alpha?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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**zetafunc.****Guest**

Laplace transform of t just brings up a bunch of text, but entering "Laplace transform of sin(t)" or "Laplace transform of cos(t)" gives the right answer as a function of s.

**zetafunc.****Guest**

It can do Laplace transform of t if you enter "Laplace transform of (t)" but not "Laplace transform of t". I think you have to encase it in brackets.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,306

Is that not the most squirrely thing I have ever seen?

First their top of the line package, mathematica 8.01 does not

do it on a fast machine. Then Alpha which is supposed to be freeform input needs a parentheses...

Do you have a table of L transforms? None that I have do tan(t).

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**zetafunc.****Guest**

Nope, sorry... I asked on another forum and they don't seem to know either. I'm not sure what to do because it's an interesting problem especially given what Wolfram|Alpha spat out, what with the digamma function being there...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,306

That is the problem, that is not an elementary function. It is not likely that hand methods are going to get that answer.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**zetafunc.****Guest**

So you are saying I should give up?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,306

Hi;

I have checked two tables of L transforms. None of them have L(tan(t)). If it were me I would give up on it and do something that was possible to do. This should be a rule:

**Just because Mathematica ( Alpha ) does it does not mean a human can.**

If you are going to continue, first find a table ( an old one ) that has it. That would mean some very clever person found a way.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Hi,

Using expansion, I found the series:

But I don't know for now whether it matches with the wolfram's output.

I'll continue trying for some more time!

"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**zetafunc.****Guest**

Okay...

I did a Google search for "laplace transform of tan(at)" and there is an exercise right at the top of the first page that asks you to find the transforms of sin(at), cos(at) and tan(at). In that document the solution says that the Laplace transform doesn't exist because the function tan(t) is not piecewise-continuous for all values of t, which makes sense...

Now my only question is where the heck Wolfram|Alpha got that weird output from.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,306

Check what gAr did it might be your only hope.

As I said I did not trust Alpha's answer.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**zetafunc.****Guest**

gAr wrote:

Hi,

Using expansion, I found the series:

But I don't know for now whether it matches with the wolfram's output.

I'll continue trying for some more time!

Thanks for this, can you talk me through how you found that series?

**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Hi zetafunc.,

Put that in the integral and integrate term by term...

Now we need to check what would be the expansion for the wolfram's output. The output was instantaneous, looked like a cached result!

edit: I read #15 again, now I doubt W|A's output, as well as mine!

And cannot find it in any table also. So, may not exist.

*Last edited by gAr (2011-08-12 05:08:44)*

"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,306

Hi all;

Math 8.01 after 20 minutes just spits it out meaning it can not find any transform. Maple spits it out immediately. Alpha is probably making a mistake in translating what you asked it to do.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**zetafunc.****Guest**

Thanks for the replies. I like gAr's method of finding a series then integrating each term.

I guess the only problem now is to figure out why Wolfram|Alpha gives that weird result?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,306

Only thing is it cannot reduce it to an elementary form. If the transform does not exist then it does not exist.

Mathematica ( Alpha ) makes mistakes just like people do. Do not confuse useful with perfect.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**zetafunc.****Guest**

Someone on another forum started talking about Cauchy principle value integrals and converting a Lebesgue divergent integral (this one) into a Cauchy principle value integral to get the Laplace transform, but I don't know how to do that and I'm trying to make sense of the Wikipedia article.

Maybe I should just stop altogether... the rest of my class next year are going to start learning how to find the nth term of a sequence.

**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Hi bobbym,

Yes, probably doesn't exist...

http://docs.google.com/viewer?a=v&q=cac … fN7lKCb0GQ

page 20

I guess zetafunc. refered that.

But again, Digamma's graph looks somewhat similar to that of tan ??

Hi zetafunc.,

This is still confusing me. I don't know whether to believe the document or W|A's output!

"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**zetafunc.****Guest**

I just took a look at the graph of the digamma function and it does look very very similar for negative integer values of x... there probably is a connection. The positive side of the graph looks nothing like that though.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,306

Hi;

That input looked fishy to me. I did not trust the answer Wolfram gave. I still do not.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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