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Old 12.05.2008, 07:13   #1
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Default помогите решить

помогите пожалуйста
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Old 12.05.2008, 07:22   #2
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Sin1/x в аргументе арктангенса или просто умножается на него ?
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Old 12.05.2008, 07:35   #3
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Странно, вроде бы решение в случае, когда arctg(x)*sin(1/x) тривиальное, но Matematica 5.0 не считает его.....
Короче, когда x ->0, аргумент синуса стремитсья к бесконечности, но синус ограниченная функция, а арктангенс в нуле дает ноль. А произведение ограниченной функции на ноль тоже ноль, поэтому в предле то, что у тебя под корнем дает ноль и ты получаешь вответе Ln2.
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Old 12.05.2008, 07:37   #4
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Во втормо случае, когда у тебя arctg(x*sin(1/x)) овтет тот же. Но теперь у тебя x*sin(1/x)) в пределе дает произведеие ограниченной функции на ноль, что есть ноль, а арктангенс нуля опять ноль.
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Old 12.05.2008, 07:46   #5
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С Matematica 5.0, проблема решена, я ошибся в написании. Так что мое решение верно.
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Old 12.05.2008, 08:45   #6
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большое спосибо очень памог
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Old 12.05.2008, 13:04   #7
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Да не за что....обращайся, если что
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Old 05.06.2008, 00:49   #8
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Ребята может кто нибудь знает как решить энто?

1. (i) Suppose f : R^n -> R and g : R^n ->R are two functions such that both are homogenous of degree r: Let h : R^n ->R be a function such that for each x E R^n; h(x) = f(x) + g(x): Show whether the function h is also homogenous of degree r:
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Old 05.06.2008, 11:29   #9
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Ребята может кто нибудь знает как решить энто?

1. (i) Suppose f : R^n -> R and g : R^n ->R are two functions such that both are homogenous of degree r: Let h : R^n ->R be a function such that for each x E R^n; h(x) = f(x) + g(x): Show whether the function h is also homogenous of degree r:
Lets take arbitrary point x from R^n. Then h(x)=f(x)+g(x). Using homogenousy of functions f and g one has f(a*x)=a^r*f(x) and g(a*x)=a^r*g(x), therefore h(a*x)=f(a*x)+g(a*x)=a^r*f(x)+a^r*g(x)=a^r*(f(x)+g(x))=a^r*h(x).
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Old 05.06.2008, 13:21   #10
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Lets take arbitrary point x from R^n. Then h(x)=f(x)+g(x). Using homogenousy of functions f and g one has f(a*x)=a^r*f(x) and g(a*x)=a^r*g(x), therefore h(a*x)=f(a*x)+g(a*x)=a^r*f(x)+a^r*g(x)=a^r*(f(x)+g(x))=a^r*h(x).
What concerned me in this problem is that h(x) is given as a function of a sum of two other functions of one variable. According to homogenity rule the sum of the exponents of the polynomal should be homogenous of some degree.
I 'coined' a function for f(x) and a function of g(x) then differentiated and messed up everything.

So say we had a product of two functions h(x)=f(x)*g(x), would we again choose an arbitrary point p from the real line and follow the same logic multiplying the two functions?
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Old 05.06.2008, 14:38   #11
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Originally Posted by Meme View Post
What concerned me in this problem is that h(x) is given as a function of a sum of two other functions of one variable. According to homogenity rule the sum of the exponents of the polynomal should be homogenous of some degree.
I 'coined' a function for f(x) and a function of g(x) then differentiated and messed up everything.

So say we had a product of two functions h(x)=f(x)*g(x), would we again choose an arbitrary point p from the real line and follow the same logic multiplying the two functions?
Well, we have not a "function of a sum" as you said, we have a function which is given as a sum of two other functions. But, these functionas are both homogeneous functions of the same dergee r. This fact provide the h(x) function to be hmogeneous again with the same degree r.
In case of produc h(x)=f(x)*g(x) we get homogeneous function again, but the dergee will be different.
Suppose f is homogeneous function with degree r, which means f(a*x)=a^r*f(x), and g is homogeneious function with degree q, i.e. g(a*x)=a^q*g(x), then
h(a*x)=f(a*x)g(a*x)=a^rf(x)a^qg(x)=a^(r+q)f(x)g(x)=a^(r+q)h(x).
Therefore, h is homogeneous function with degree r+q.
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Old 05.06.2008, 18:40   #12
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Originally Posted by Monopole View Post
Well, we have not a "function of a sum" as you said, we have a function which is given as a sum of two other functions. But, these functionas are both homogeneous functions of the same dergee r. This fact provide the h(x) function to be hmogeneous again with the same degree r.
In case of produc h(x)=f(x)*g(x) we get homogeneous function again, but the dergee will be different.
Suppose f is homogeneous function with degree r, which means f(a*x)=a^r*f(x), and g is homogeneious function with degree q, i.e. g(a*x)=a^q*g(x), then
h(a*x)=f(a*x)g(a*x)=a^rf(x)a^qg(x)=a^(r+q)f(x)g(x)=a^(r+q)h(x).
Therefore, h is homogeneous function with degree r+q.
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