0
$\begingroup$

Task from an exam:

Given is the real vector space where $a \in \mathbb{R}$ is fixed:

$V = \left\{ f: \mathbb{R} \rightarrow \mathbb{R}: f(x) = \lambda_{1}e^{x}\sin(ax)+ \lambda_{2}e^{x}\cos(ax) \text{ for } x \in \mathbb{R}, \lambda_{1}, \lambda_{2} \in \mathbb{R} \right\}$

with the basis $A=(f_{1},f_{2})$, whereby $\text{ }$ $\text{ }$$f_{1}(x)=e^{x}\sin(ax), \text{ } \text{ }f_{2}(x)= e^{x}\cos(ax)$.

And because every element of this vector space is characterized by an ordered pair $(\lambda_{1}, \lambda_{2}) \in \mathbb{R}^{2}$ of coefficients, with allocation $f=\lambda_{1}f_{1}+\lambda_{2}f_{2} \mapsto (\lambda_{1}, \lambda_{2})$, every linear mapping of $V$ in $V$ can be seen as a $2\times 2$ matrix.

Let $a=1$. Determine $f \in V$ with $f'(x) = e^{x}\sin(x)$.

So $a = 1$ then we have:

$f(x)= \lambda_{1}e^{x}\sin(x)+ \lambda_{2}e^{x}\cos(x)$

Then $f'(x) = \lambda_{1}(e^{x}\sin{x}+e^{x}+\cos{x}) + \lambda_{2}(e^{x}\cos{x}-e^{x}\sin{x})$

But that is not $f'(x)= e^{x}\sin{x}...$

Or did I understood it wrong completely? :s

  • 0
    In the expression of $f'(x)$ it should be $e^x\cos x$ instead of $e^x+\cos x$. – 2017-01-14

2 Answers 2

1

Your calculation yields \begin{align}f'(x) & = \lambda_1(e^x \sin x+e^x\cos x)+\lambda_2(e^x\cos x-e^x\sin x)\\ & = (\lambda_1-\lambda_2)e^x\sin x + (\lambda_1+\lambda_2)e^x\cos x.\end{align} Now you want this to be equal to $e^x\sin x$; by comparing the decompositions in the basis $f_1,f_2$, you see that this is equivalent to $$\left\{\begin{array}{}\lambda_1-\lambda_2=1\\ \lambda_1+\lambda_2=0\end{array}\right.$$ So all you have to do is solve this system.

  • 0
    Oh alright thank you! So if I solved this system, then is the task really complete? Because they say "determine $f \in V$" but what we did is only the first derivation. – 2017-01-14
  • 1
    You have defined $f$ as $\lambda_1f_1+\lambda_2 f_2$; so once you know $\lambda_1,\lambda_2$, you know $f$... – 2017-01-14
1

Hint:

correct the derivative as suggested in the comment, than reorder as: $$ f'(x)=e^x\sin x(\lambda_1-\lambda_2)+e^x\cos x(\lambda_1+\lambda_2) $$

Now search if there are values of $\lambda_2, \lambda_2$ such that $f'(x)=e^x\sin x$