## 105 Reputation

10 years, 288 days

## Sorry for confusion....

@Carl Love Sir,

I was not ignoring your solution even that one is really worked out for me. I just wanted to look for other solutions.

Thanks a lot sir for your suggestion to my problem. It really worked out.

Thanks and regards

Sunit

## It worked....

@Carl Love Thanks a lot sir for the suggestion. It really worked. But I am having a doubt that if i put T[0],T[1] and T[2] instead of a, b and c, respectively then it is giving an error.

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Just being curious that why it is not working.

Regards

Sunit

## Not working!...

@John Fredsted Thanks a lot sir. But this is not working for the following equation.

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I tried to substitute for the x[1](T[0],T[1],T[2]), but it is not evaluating.

Please see and help me out for the same,

Regards

Sunit

## Thank a lot....

@John Fredsted Sir,

Thanks a lot for helping me out. But just one doubt,will this also work for the following equation.question3.mw

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I need to substitute x[1](T[0]-tau_1,T[1],T[2])=R(T[1],T[2],tau_1)*sin(omega*T[0]+phi(T[1],T[2])). I am asking it, because i tried it but again not performing the subsititution for the 'D'.

Thanks a lot sir.

Sunit

## Fixing the type of data...

@dharr Thanks for clarifying it. But cann't we fix the type of data or structure of a defined function?

## Works fine....

@dharr Hi,

Thanks for the help. But during running i encountered an issue and thought probably you might help on it. If in the expression for sin if i substitute phi instead of Phi, then map(expand,q) just expanding everything.

I did not get how phi or Phi is making a difference?

## Sorry for confusion....

@taro Hi, I think i could not pose my question properly. I have to expand sin in terms of omega*T0+phi and T,

something like sin(omega*T0+phi)*cos(T)-sin(T)*cos(omega*T0+phi).

## Solved it....

@John Fredsted Hi Carl,

I am really thankful for your help. However, i solved it using a little trick. At first i substitute D(lamda)(t-tau_1)=diff(lambda1(t),t) in the expression. After that I put the expression as mentioned above in place of lamda1(t). And it worked.

Thanks

Regards

Sunit

## Still not clear...

@John Fredsted Hi John,

I am really thankful for putting your effort in my problem. But I think i could not pose my problem properly. Here it is.

I am doing Multiple scale method, and in doing so in one of the steps i am getting D(lamda)(t-tau_1).

Now in further steps, i have to substitute the above mentioned expression  instead of lamda(t-tau_1), and thats why i was asking for the substitution.

I hope this time I made it clear the problem statement.

regards

Sunit

## Attached maple sheet....

@John Fredsted Yes, tau_1 is same as t_1. Please find the attached sheet.

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## For another case....

@John Fredsted Thanks for the help. but my question was for in general.  What if lambda(t-t1) will be

I am really sorry for the confusion, i should have asked  it at the first place.

Regards

Sunit

## Not working...

@Doug Meade Thanx for help. But i run the code on maple that you provided and it is giving same result, i.e., neither it is taking out epsilon nor substituting eta(t)=epsilon*z(t).

## Thanx!...

@Kitonum Thanx and i really appreciate ur help, but please help me for one more thing. When i am trying to multiply two terms under square root like sqrt(a)*sqrt(b) then it is not giving me sqrt(a*b), how to simplify this expression.

Regards

Sunit

## Thanx!...

@Alejandro Jakubi Thanx a lot for helping, but here comes another question. If by default the integration method is cook, then to get correct answer doi have to always use method=nocook whenever i have to integrate a function like  that?

Regards

Sunit

## Ok.....

@J4James Ok, here it is..

> restart;
> int(BesselJ(1, x), x = 0 .. infinity);
-1
> evalf(Int(BesselJ(1, x), x = 0 .. 2500));
0.9987629907

Regards

Sunit

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