Wikipedia:Reference desk/Archives/Mathematics/2011 September 26

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September 26

A bit on the Heavi Side

I have a problem to do with a differential equation involving a Heaviside Function. It goes  :${\frac {d^{2}y}{dt^{2}}}+4y=($ ){sin t, 0 ≤ t < 2π ; 0, 2π ≤ t } y(0) = 0 , y'(0) = 0

I have got it down to y(s) = (1 - e^(-2πs))/(s^2+ 1)(s^2+ 4 ) = 1/3(s^2+ 1) - 1/3(s^2+ 4) - ( e^(-2πs))/3(s^2+ 1) + ( e^(-2πs))/3(s^2+ 4) , but now I do not know how to convert the exponential terms into funπctions of t, while I know the ones without the exponentials are variants of sin(t). If anyone knows how to work this out, then I would be pleased to hear that. Thank You. Chris the Russian Christopher Lilly 07:18, 26 September 2011 (UTC)

It appears you are trying to do this with Laplace transforms. (There are other ways.) An exponential factor in the s domain corresponds to a shift operation in the t domain; search for "Frequency shifting" in the Laplace transforms article. So basically just find the inverse transform without the exponential factor and then shift the function to the right by 2π. In the solution I got, y is 0 for y>2π so the shifted components should cancel out the non-shifted ones except on the interval [0,2π].--RDBury (talk) 13:08, 26 September 2011 (UTC)

Thank You. I hope I made this clear that that which is in the curly brackets was a step function, but I suppose that is obvious with the title Heaviside. I wondered why sin ( t - 2π ) should be any different from sin (t) itself. If I took the equation to be ${\frac {d^{2}y}{dt^{2}}}+4y=sint$ on its own, I can work that out, but to be honest, I could not understand how a function could change like symbolically. I get the graph, like a switch or something going on, then all of a sudden the value of some function jumps or suddenly appears different, but it still seems hard to fathom.

I do have a different problem involving an integro differential equation. It goes :

${\frac {d}{dt}}y(t)+\int _{0}^{t}f(u)\,du=sin(t),\qquad y(0)=1,\qquad {\frac {d}{dt}}y(0)=0$

Now I was advised that the assumption is that $y=f(t)$, and to isolate the integral by rearranging the equation, such that I got  :

${\frac {d}{dt}}y(t)-sin(t)=\int _{0}^{t}f(u)\,du$

Next I was told to take the derivative of both sides, and was told that this meant that since

$\int _{0}^{t}f(u)\,du={\frac {d}{dt}}y(t)-sin(t)$

that

$y=\cos(t)-{\frac {d^{2}y}{dt^{2}}}$

But when I carry out the integration on $\int _{0}^{t}f(u)\,du$, according to the Fundamental Theorem of Arithmetic, I get : $\int _{0}^{t}f(u)\,du=f(t)-f(0).$, or have I done the wrong thing, since I thought that taking the derivative of an integral is where they cancel each other out. But also, if it is assumed that

$y=f(t)$, what is

$f(u)$ ?

If I assume that, then I think I get $y-1=\cos(t)-{\frac {d^{2}y}{dt^{2}}}$ since if $y=f(t)$, then $f(0)=y(0)=1$, as stated to begin with. If I do say that $y=\cos(t)-{\frac {d^{2}y}{dt^{2}}}$, this seems correct, since it says that $y(0)=1$, and indeed $y(0)=\cos(0)-{\frac {d^{2}y(0)}{dt^{2}}}$, and since if \qquad \frac{d}{dt}y(0) = 0, being the first derivative, so will be the second, since that will just be the derivative of zero, so this seems right.

To solve this, I tried LaPlace once more, and got

$y=\cos(t)-{\frac {d^{2}y}{dt^{2}}}$ implies that $\cos(t)=y+{\frac {d^{2}y}{dt^{2}}}$, so that Laplace of $\cos(t)$ = Laplace of $y+{\frac {d^{2}y}{dt^{2}}}$, which means after carrying out all the steps that Failed to parse (unknown function "\y"): {\displaystyle \y (s) } = s/(s^2 + 1 )^2 + s/(s+1), which I thought was cos(t) + t sin(t), but it does not seem to work when I check it. I certainly would appreciate any help on this, thank You. Chris the Russian Christopher Lilly 07:24, 27 September 2011 (UTC)

Trig identity and complex numbers.

I tried applying the following trig identity $a\cos bx+c\sin bx={\sqrt {a^{2}+c^{2}}}\sin \left(bx+\arctan \left({\dfrac {c}{b}}\right)\right)$ to the right hand side of Euler's Formula $e^{ix}=\cos x+i\sin x$. Interestingly, one gets $e^{ix}={\sqrt {1^{2}+i^{2}}}\sin \left(x+\arctan i\right)$, but simplifying the square root gives $e^{ix}={\sqrt {0}}\sin \left(x+\arcsin i\right)$, which can't be true. Any insight as to why this identity fails here? I also found through wolframalpha.com that $\arctan i=i\infty$. It almost seems as if it's trying to "cancel out" the zero with an infinity, though doesn't really make complete sense since it's composed inside another function. — Trevor K. — 16:55, 26 September 2011 (UTC) — Preceding unsigned comment added by Yakeyglee (talk • contribs)

In your identity $a\cos bx+c\sin bx={\sqrt {a^{2}+c^{2}}}\sin \left(bx+\arctan \left({\dfrac {c}{b}}\right)\right)$ substitute a=b=1, c=0 to get $\cos x=\sin(x+\arctan(0))$ which is not true. Bo Jacoby (talk) 17:20, 26 September 2011 (UTC).

As Bo points out, your formula is a bit off. See List of trigonometric identities#Linear combinations for the right identity. But the question still stands: why does that one still seemingly fail? If you work through a proof of the identity, you'll see that you end up dividing by zero at some point. So, strictly speaking, the identity should come with a disclaimer about what values of a and b are allowed. Your comment about the zero and infinity "trying to cancel each other out" is on the right track: see L'Hopital's_rule. 130.76.64.109 (talk) 18:02, 26 September 2011 (UTC)

The L'Hopital part could have been true, but I don't think that's what happening here. Take $a=b=1,\ c=i/2$, then you have the same mess with the arctan, but without the multiplication by 0. -- Meni Rosenfeld (talk) 18:26, 26 September 2011 (UTC)

[ec] Should be $a\cos bx+c\sin bx={\sqrt {a^{2}+c^{2}}}\sin \left(bx+\arctan \left({\dfrac {a}{c}}\right)\right)$ (note argument of arctan). But that's not the problem. The problem is that some identities developed for real numbers only work because we have sloppily assumed that squares are always nonnegative (in particular, wherever the square root symbol appears it should raise red flags with regards to extension to complex numbers. With complex numbers you also need to worry about branches and stuff). The generalization of squaring in its capacity as something nonnegative to real matrices is $A^{T}A$ and to complex numbers is $z{\bar {z}}=|z|^{2}$. If some version of the identity works for complex numbers, I'll bet it involves the factor ${\sqrt {|a|^{2}+|c|^{2}}}$. -- Meni Rosenfeld (talk) 18:12, 26 September 2011 (UTC)

That equation still doesn't work with a=0, c=-1. Probably the most concise formula that actually works uses the atan2 function.--RDBury (talk) 19:34, 26 September 2011 (UTC)

It's true that complex numbers sometimes break the rules we learned with real numbers, Meni, but this one does work (assuming you mind the quadrants, as RDBury points out). 130.76.64.119 (talk) 21:27, 26 September 2011 (UTC)

It's easier to see what is going on here, if you approach this using complex numbers from the start (trigonometry shouldn't be taught before complex numbers). For real a and b, we have:

a cos(x) + b sin(x) =

a/2 [exp(i x) + exp(-i x)] + b/(2i) [exp(i x) - exp(-i x)] =

(a - b i)/2 exp(i x) + (a + bi)/2 exp(-i x) =

|a + b i| cos(x + phi)

So, the trig identity works because you have an expression involving exp(i x) and exp(-ix). While they are multiplied by complex numbers of equal modulus, you can generalize it. For arbitrary a and b not equal to zero, we can write:

a exp(i x) + b exp(-i x) =

a exp[i (x + p) - i p] + b exp[-i (x+p) + i p] =

a exp(-ip) exp[i(x+p)] + b exp(i p) exp[-i (x+p)]

You then choose p such that

|a exp(-ip)| = |b exp(i p)|

If |a| is not the same as |b|, then we can't choose p real, but that's not a problem. We can always choose p = 1/(2 i) Log(a/b) to make both coefficients equal, but of course, both a and b have to be nonzero.

The fundamental issue here is that if you only have exp(i x), you can't magically get exp(-ix) out of nowhere. Count Iblis (talk) 21:49, 26 September 2011 (UTC)

What do you mean? Just compute $e^{-ix}={\frac {1}{e^{ix}}}$. Bo Jacoby (talk) 07:37, 27 September 2011 (UTC).