Show that $f(x,y) = cos(x+y)+8$ is continuous at $(0,0).$
I tried the following for this problem:
We first compute $f(0,0) = 9.$ Next we look at
$$|f(x,y)- f(0,0)| = |cos(x+y)-1|leq frac{(x+y)^2}{2}$$
when $(x,y)$ is in the neighbourhood of $(0,0).$ Since $frac{(x+y)^2}{2}to 0$ as $(x,y)to (0,0)$ we conclude that $f(x,y)$ is continuous at $(0,0).$
I am not sure whether this Taylor series type of argument can be applied for functions of several variables. Therefore, I would be grateful if someone could give feedback regarding this problem.
real-analysis
asked Dec 12 '18 at 19:19
Hello_WorldHello_World
4,12321630
add a comment |
I tried the following for this problem:
We first compute $f(0,0) = 9.$ Next we look at
$$|f(x,y)- f(0,0)| = |cos(x+y)-1|leq frac{(x+y)^2}{2}$$
when $(x,y)$ is in the neighbourhood of $(0,0).$ Since $frac{(x+y)^2}{2}to 0$ as $(x,y)to (0,0)$ we conclude that $f(x,y)$ is continuous at $(0,0).$
I am not sure whether this Taylor series type of argument can be applied for functions of several variables. Therefore, I would be grateful if someone could give feedback regarding this problem.
real-analysis
asked Dec 12 '18 at 19:19
Hello_WorldHello_World
4,12321630
It's a composition of continuous functions.
– Callus
Dec 12 '18 at 19:22
Even though it is very intuitive, perhaps one should add explanation for why $frac{(x+y)^2}{2}to 0$?
– Anurag A
Dec 12 '18 at 19:23
add a comment |
I tried the following for this problem:
We first compute $f(0,0) = 9.$ Next we look at
$$|f(x,y)- f(0,0)| = |cos(x+y)-1|leq frac{(x+y)^2}{2}$$
when $(x,y)$ is in the neighbourhood of $(0,0).$ Since $frac{(x+y)^2}{2}to 0$ as $(x,y)to (0,0)$ we conclude that $f(x,y)$ is continuous at $(0,0).$
I am not sure whether this Taylor series type of argument can be applied for functions of several variables. Therefore, I would be grateful if someone could give feedback regarding this problem.
real-analysis
asked Dec 12 '18 at 19:19
Hello_WorldHello_World
4,12321630
I tried the following for this problem:
We first compute $f(0,0) = 9.$ Next we look at
$$|f(x,y)- f(0,0)| = |cos(x+y)-1|leq frac{(x+y)^2}{2}$$
when $(x,y)$ is in the neighbourhood of $(0,0).$ Since $frac{(x+y)^2}{2}to 0$ as $(x,y)to (0,0)$ we conclude that $f(x,y)$ is continuous at $(0,0).$
I am not sure whether this Taylor series type of argument can be applied for functions of several variables. Therefore, I would be grateful if someone could give feedback regarding this problem.
real-analysis
real-analysis
asked Dec 12 '18 at 19:19
Hello_WorldHello_World
4,12321630
asked Dec 12 '18 at 19:19
Hello_WorldHello_World
4,12321630
asked Dec 12 '18 at 19:19
Hello_WorldHello_World
4,12321630
asked Dec 12 '18 at 19:19
Hello_WorldHello_World
4,12321630
asked Dec 12 '18 at 19:19
Hello_WorldHello_World
4,12321630
4,12321630
It's a composition of continuous functions.
– Callus
Dec 12 '18 at 19:22
Even though it is very intuitive, perhaps one should add explanation for why $frac{(x+y)^2}{2}to 0$?
– Anurag A
Dec 12 '18 at 19:23
add a comment |
It's a composition of continuous functions.
– Callus
Dec 12 '18 at 19:22
Even though it is very intuitive, perhaps one should add explanation for why $frac{(x+y)^2}{2}to 0$?
– Anurag A
Dec 12 '18 at 19:23
It's a composition of continuous functions.
– Callus
Dec 12 '18 at 19:22
It's a composition of continuous functions.
– Callus
Dec 12 '18 at 19:22
Even though it is very intuitive, perhaps one should add explanation for why $frac{(x+y)^2}{2}to 0$?
– Anurag A
Dec 12 '18 at 19:23
Even though it is very intuitive, perhaps one should add explanation for why $frac{(x+y)^2}{2}to 0$?
– Anurag A
Dec 12 '18 at 19:23
add a comment |
1 Answer
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$$0le 1-cos(x+y)=2sin^2(frac{x+y}{2})$$
$$le frac{(x+y)^2}{2}$$
since
$$(forall Xin Bbb R) ;;; |sin(X)|le |X|$$
answered Dec 12 '18 at 19:26
hamam_Abdallahhamam_Abdallah
38k21634
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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active
oldest
votes
$$0le 1-cos(x+y)=2sin^2(frac{x+y}{2})$$
$$le frac{(x+y)^2}{2}$$
since
$$(forall Xin Bbb R) ;;; |sin(X)|le |X|$$
answered Dec 12 '18 at 19:26
hamam_Abdallahhamam_Abdallah
38k21634
add a comment |
$$0le 1-cos(x+y)=2sin^2(frac{x+y}{2})$$
$$le frac{(x+y)^2}{2}$$
since
$$(forall Xin Bbb R) ;;; |sin(X)|le |X|$$
answered Dec 12 '18 at 19:26
hamam_Abdallahhamam_Abdallah
38k21634
add a comment |
$$0le 1-cos(x+y)=2sin^2(frac{x+y}{2})$$
$$le frac{(x+y)^2}{2}$$
since
$$(forall Xin Bbb R) ;;; |sin(X)|le |X|$$
answered Dec 12 '18 at 19:26
hamam_Abdallahhamam_Abdallah
38k21634
$$0le 1-cos(x+y)=2sin^2(frac{x+y}{2})$$
$$le frac{(x+y)^2}{2}$$
since
$$(forall Xin Bbb R) ;;; |sin(X)|le |X|$$
answered Dec 12 '18 at 19:26
hamam_Abdallahhamam_Abdallah
38k21634
answered Dec 12 '18 at 19:26
hamam_Abdallahhamam_Abdallah
38k21634
answered Dec 12 '18 at 19:26
hamam_Abdallahhamam_Abdallah
38k21634
answered Dec 12 '18 at 19:26
hamam_Abdallahhamam_Abdallah
38k21634
38k21634
add a comment |
add a comment |
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It's a composition of continuous functions.
– Callus
Dec 12 '18 at 19:22
Even though it is very intuitive, perhaps one should add explanation for why $frac{(x+y)^2}{2}to 0$?
– Anurag A
Dec 12 '18 at 19:23