All Questions
6
questions
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How to prove exponential functional identity knowing that it is a solution to a first order ODE and knowing its Taylor expansion
Establish the identity
$$E(ax)E(bx) = E[(a+b)x]$$
knowing that
$y = E(px)$ satisfies $y' - py = 0$ and
$E(px) = \sum_{n=0}^\infty\frac{(px)^n}{n!}$
An additional hint the textbook gives ; "...
- 1
0
votes
2
answers
115
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multiplicative property of complex exponentials
How can I demonstrate the property $e^u\cdot e^v$ = $e^{u+v}$ for complex $u,v$ using the summation definition of $\exp(z)$. Specifically this is the definition saying that $\exp(z) = \sum_{k=0}^{\...
- 37
1
vote
1
answer
55
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Calculate the functions $e^{kd/dt}f(t)$ and $e^{td/dt}f(t)$
The problem reads as follows:
A slight generalization of the Taylor expansion is
$$f(x+a)=\sum^\infty_{n=0}\frac{a^n}{n!}f^{(n)}(x)=f(x+a)$$
Calculate the functions $e^{kd/dt}f(t)$ and $e^{td/dt}f(...
- 350
2
votes
2
answers
161
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Finding sum of infinite series $1+\frac{x^3}{3!}+\frac{x^6}{6!}+\frac{x^9}{9!}+\ldots $
So the question is 'express the power series $$1+\frac{x^3}{3!}+\frac{x^6}{6!}+\frac{x^9}{9!}+\ldots $$
in closed form'.
Now we are allowed to assume the power series of $e^x$ and also we derived the ...
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1
vote
1
answer
344
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Solution of $d^2u/dx^2 + u/A = 0 \ (\text{or } \ C),$ with conditions
Does the following ODE:
$$d^2u/dx^2 + u/A = 0 \quad (\text{or } \ C),$$
have a solution with the conditions:
$$
\left.\frac{d^2u}{dx^2}\right|_{x=0} = 0,
$$
$$u(x=0) = B$$ and
$$
\left.\frac{du}{...
- 343
1
vote
1
answer
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Should I use Taylor Expansion or the expansion of $e^x$ to express a second order differential (in this case)
I've been given this equation:
$(1+x^2)\dfrac{d^2y}{dx^2} + 4x\dfrac{dy}{dx} + 2y = 0$
I've also been told that:
$y=1, \dfrac{dy}{dx} = 1$, at $x=-1$
I've been asked to find a series solution of ...
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