傅里叶分析|MATH10051 Fourier Analysis代写

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Understand the statements and proofs of important theorems and be able to explain the key steps in proofs, sometimes with variation.

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傅里叶分析|MATH10051 Fourier Analysis代写

To do this we introduce complex numbers by adding $i$ times the second equation to the first to obtain the single equation
$$
\frac{d}{d t}(U+i V)=B(V-i U) .
$$
If we set $\Phi=U+i V$ this equation takes the form
$$
\dot{\Phi}=-i B \Phi
$$
This is an equation that we can solve to obtain
$$
\Phi(t)=A e^{-i B t}
$$
where $A$ is a fixed complex number.
How does this solution fit in with our original problem. recall that we can write
$$
A=r \exp i \alpha
$$

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MATH10051 COURSE NOTES :

If $x=R \cos \theta, y=R \sin \theta$ where $R$ is very large then to a very good approximation
$$
S(R \cos \theta, R \sin \theta, t)=R^{-2} \exp (i(\omega t-\lambda R) Q(u)
$$
where
$$
Q(u)=\sum_{k=-N}^{N} A_{k} \exp \left(i\left(k u-\phi_{k}\right)\right)
$$
and $u=\lambda^{-1} l \sin \theta$.
In the discussion that follows we use the notation of Lemma $4.1$ and the discussion that preceded it. We set
$$
P(u)=\sum_{k=-N}^{N} A_{k} \exp (i k u)
$$



傅立叶分析法|Fourier Analysis代写 6CCM318A

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这是一份kcl伦敦大学学院  6CCM318A作业代写的成功案

傅立叶分析法|Fourier Analysis代写 6CCM318A
问题 1.

The general solution of $y^{\prime \prime}+\lambda y=0$ is
$y=A \cos \sqrt{\lambda} x+B \sin \sqrt{\lambda} x$
Then from the condition $y(0)=0$ we find $A=0$, so that
$$
y=B \sin \sqrt{\lambda} x
$$


证明 .

The condition $y^{\prime}(L)+\beta y(L)=0$ gives
$$
B \sqrt{\lambda} \cos \sqrt{\lambda} L+\beta B \sin \sqrt{\lambda} L=0{ }^{\prime} \quad \text { or } \quad \tan \sqrt{\lambda} L=-\frac{\sqrt{\lambda}}{\beta}
$$

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6CCM318ACOURSE NOTES :

$$
R=r^{-1 / 2}\left[A r \sqrt{1 / 4-\lambda^{2}}+B r^{\left.-\sqrt{1 / 4-\lambda^{2}}\right]}\right.
$$
or
$$
R=A r^{-1 / 2}+\sqrt{1 / 4-\lambda^{2}}+B r^{-1 / 2-\sqrt{1 / 4-\lambda^{2}}}
$$
This solution can be simplified if we write
$$
-\frac{1}{2}+\sqrt{\frac{1}{4}-\lambda^{2}}=n
$$
so that
$$
-\frac{1}{2}-\sqrt{\frac{1}{4}-\lambda^{2}}=-n-1
$$
In sueh case $(1)$ becomes
$$
R=A r^{n}+\frac{B}{r^{m+1}}
$$
Multiplying equations $(2)$ and $(s)$ together leads to
$$
\lambda^{2}=-n(n+1)
$$