Find a basis and dimension of a vector space

gruba
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Homework Statement


Find basis and dimension of V,W,V\cap W,V+W where V=\{p\in\mathbb{R_4}(x):p^{'}(0) \wedge p(1)=p(0)=p(-1)\},W=\{p\in\mathbb{R_4}(x):p(1)=0\}

Homework Equations


-Vector spaces

The Attempt at a Solution


Could someone give a hint how to get general representation of a vector in V and W?
 
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Look in your course notes when it talks about "general representation of a vector".
You have definitions of the vector spaces - so start with what the symbols turn into in English.
 
I'm not sure what the operator ^ means in the definition of ##V## so I can't help.
For ##W##, you can notice that ##W = (X-1) \mathbb{R}_3[X] ## is isomorphic to ##\mathbb{R}_3[X]##, so what can you say about the dimension of 2 isomorphic vector spaces ? Secondly, a vector space isomorphism sends a base to a base.
 
The only thing I have a question about is "p'(0)^p(1)= p(0)= p(-1)". The "^" typically means "and" so one condition is that p(1)= p(0)= p(-1) but the "p'(0)" is incomplete- nothing is said about the derivative at 0. What must be true about it? p in R^4 is of the form ax^4+ bx^3+ cx^2+ dx+ e and "p(1)= p(0)= p(-1)" requires that a+ b+ c+ d+ e= e= a- b+ c- d+ e. But, again, what is required of "p'(0)= d"?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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