- #1
Benny
- 584
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Hi I have a few questions regarding the following two part problem.
a) Let a > 0. Use polar coordinates to evaluate [tex]\int\limits_{ - \infty }^\infty {e^{ - ax^2 + bx} } dx[/tex]
[tex]Ans:e^{\frac{{b^2 }}{{4a}}} \sqrt {\frac{\pi }{a}} [/tex]
b) Let [tex]\mathop x\limits^ \to = \left( {x_1 ,...,x_n } \right) \in R^n [/tex], [tex]\mathop b\limits^ \to \in R^n [/tex] and [tex]A = U^T U[/tex] where U is an invertible n * n matrix. Compute
[tex]\int\limits_{R^n }^{} {\exp \left( { - x^T Ax + b^T x} \right)} dx_1 ...dx_n [/tex]
(the x and b in the argument of the exponential are vectors)
Check that your answer for n = 1 reduces to the answer in (a).
Ans: [tex]Ans:\frac{{\pi ^{\frac{n}{2}} }}{{\left| {\det U} \right|}}\exp \left( {\frac{1}{4}b^T A^{ - 1} b} \right)[/tex]
My questions/work done etc. are as follows.
In part (a) I think the idea is to make a substitution which leaves me with an integrand which is a scalar multiple of exp(-y^2). This is so that I can then use the standard 'squaring the integral' method to get the answer.
I had a few problems doing this. Basically after trying a few different methods I just subtituted [tex]x = \frac{1}{{\sqrt a }}\left( {y - c} \right)[/tex] (c is constant) into -ax^2 + bx. Then I selected c so that the coefficient of the resulting y term is zero. I'm wondering if there is a simpler way to go about finding a suitable substitution for the integral.
I don't follow the solution to part b. They start out by saying that a change of coordinates similar to part a needs to be used and come up with [tex]x = U^{ - 1} \left( {y + \frac{1}{2}U^{ - 1} b} \right)[/tex]. I don't understand where they got it from. Could someone suggest how they would start this question? Any help would be good thanks.
a) Let a > 0. Use polar coordinates to evaluate [tex]\int\limits_{ - \infty }^\infty {e^{ - ax^2 + bx} } dx[/tex]
[tex]Ans:e^{\frac{{b^2 }}{{4a}}} \sqrt {\frac{\pi }{a}} [/tex]
b) Let [tex]\mathop x\limits^ \to = \left( {x_1 ,...,x_n } \right) \in R^n [/tex], [tex]\mathop b\limits^ \to \in R^n [/tex] and [tex]A = U^T U[/tex] where U is an invertible n * n matrix. Compute
[tex]\int\limits_{R^n }^{} {\exp \left( { - x^T Ax + b^T x} \right)} dx_1 ...dx_n [/tex]
(the x and b in the argument of the exponential are vectors)
Check that your answer for n = 1 reduces to the answer in (a).
Ans: [tex]Ans:\frac{{\pi ^{\frac{n}{2}} }}{{\left| {\det U} \right|}}\exp \left( {\frac{1}{4}b^T A^{ - 1} b} \right)[/tex]
My questions/work done etc. are as follows.
In part (a) I think the idea is to make a substitution which leaves me with an integrand which is a scalar multiple of exp(-y^2). This is so that I can then use the standard 'squaring the integral' method to get the answer.
I had a few problems doing this. Basically after trying a few different methods I just subtituted [tex]x = \frac{1}{{\sqrt a }}\left( {y - c} \right)[/tex] (c is constant) into -ax^2 + bx. Then I selected c so that the coefficient of the resulting y term is zero. I'm wondering if there is a simpler way to go about finding a suitable substitution for the integral.
I don't follow the solution to part b. They start out by saying that a change of coordinates similar to part a needs to be used and come up with [tex]x = U^{ - 1} \left( {y + \frac{1}{2}U^{ - 1} b} \right)[/tex]. I don't understand where they got it from. Could someone suggest how they would start this question? Any help would be good thanks.