Mixing
How do I show that this function is not bijective?
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My question is sample: I want to show that there is no open $V subset mathbb{R}$ around $x=0$ such the function:
$$f(x)= x^2sin(1/x), mbox{ }xin V- {0}$$
$$f(0)=0$$
is bijective.
I just want a hint.
calculus
asked Jan 15 at 12:30
tererecomchimarraotererecomchimarrao
617
$endgroup$
add a comment |
$begingroup$
My question is sample: I want to show that there is no open $V subset mathbb{R}$ around $x=0$ such the function:
$$f(x)= x^2sin(1/x), mbox{ }xin V- {0}$$
$$f(0)=0$$
is bijective.
I just want a hint.
calculus
asked Jan 15 at 12:30
tererecomchimarraotererecomchimarrao
617
$endgroup$
5
$begingroup$
Show that $f$ is $0$ on more than one input.
$endgroup$
– SmileyCraft
Jan 15 at 12:34
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Remark: You need to specify the codomain if you want to prove that a function is bijective. However in this ecample it is sufficient to show that the function is not injective.
$endgroup$
– maxmilgram
Jan 15 at 16:12
add a comment |
$begingroup$
My question is sample: I want to show that there is no open $V subset mathbb{R}$ around $x=0$ such the function:
$$f(x)= x^2sin(1/x), mbox{ }xin V- {0}$$
$$f(0)=0$$
is bijective.
I just want a hint.
calculus
asked Jan 15 at 12:30
tererecomchimarraotererecomchimarrao
617
$endgroup$
My question is sample: I want to show that there is no open $V subset mathbb{R}$ around $x=0$ such the function:
$$f(x)= x^2sin(1/x), mbox{ }xin V- {0}$$
$$f(0)=0$$
is bijective.
I just want a hint.
calculus
calculus
asked Jan 15 at 12:30
tererecomchimarraotererecomchimarrao
617
asked Jan 15 at 12:30
tererecomchimarraotererecomchimarrao
617
asked Jan 15 at 12:30
tererecomchimarraotererecomchimarrao
617
asked Jan 15 at 12:30
tererecomchimarraotererecomchimarrao
617
asked Jan 15 at 12:30
tererecomchimarraotererecomchimarrao
617
617
5
$begingroup$
Show that $f$ is $0$ on more than one input.
$endgroup$
– SmileyCraft
Jan 15 at 12:34
$begingroup$
Remark: You need to specify the codomain if you want to prove that a function is bijective. However in this ecample it is sufficient to show that the function is not injective.
$endgroup$
– maxmilgram
Jan 15 at 16:12
add a comment |
5
$begingroup$
Show that $f$ is $0$ on more than one input.
$endgroup$
– SmileyCraft
Jan 15 at 12:34
$begingroup$
Remark: You need to specify the codomain if you want to prove that a function is bijective. However in this ecample it is sufficient to show that the function is not injective.
$endgroup$
– maxmilgram
Jan 15 at 16:12
5
5
$begingroup$
Show that $f$ is $0$ on more than one input.
$endgroup$
– SmileyCraft
Jan 15 at 12:34
$begingroup$
Show that $f$ is $0$ on more than one input.
$endgroup$
– SmileyCraft
Jan 15 at 12:34
$begingroup$
Remark: You need to specify the codomain if you want to prove that a function is bijective. However in this ecample it is sufficient to show that the function is not injective.
$endgroup$
– maxmilgram
Jan 15 at 16:12
$begingroup$
Remark: You need to specify the codomain if you want to prove that a function is bijective. However in this ecample it is sufficient to show that the function is not injective.
$endgroup$
– maxmilgram
Jan 15 at 16:12
add a comment |
2 Answers
2
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oldest
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Hint: $f(frac{1}{n pi})=0$ for all $n in mathbb N.$ Can you proceed ?
answered Jan 15 at 12:55
FredFred
46.9k1848
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add a comment |
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Hint:
A bijective function is a which is both one-one and onto function.
Now you can prove that the function is not bijective by proving that the function is not one-one or the function is not onto.
It would be much easier to prove that the function is one-one.The various methods to check injectivity of a function are:
1) For a one one function it gives same output for same number or for one input it gives only one output this can be exploited by proving either one of the following:
$$ if x_1 =x_2, then f(x_1) ne f(x_2) $$
$$ if x_1 ne x_2, then f(x_1) = f(x_2) $$
2)you can check the function's slope(Slope can be found out by finding the derivative of the function or simply differentiating it once)-If the slope is independent of x hence the function is either increasing or decreasing so it is one-one else not.
answered Jan 15 at 14:18
Harsh WasnikHarsh Wasnik
13
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add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
Hint: $f(frac{1}{n pi})=0$ for all $n in mathbb N.$ Can you proceed ?
answered Jan 15 at 12:55
FredFred
46.9k1848
$endgroup$
add a comment |
$begingroup$
Hint: $f(frac{1}{n pi})=0$ for all $n in mathbb N.$ Can you proceed ?
answered Jan 15 at 12:55
FredFred
46.9k1848
$endgroup$
add a comment |
$begingroup$
Hint: $f(frac{1}{n pi})=0$ for all $n in mathbb N.$ Can you proceed ?
answered Jan 15 at 12:55
FredFred
46.9k1848
$endgroup$
Hint: $f(frac{1}{n pi})=0$ for all $n in mathbb N.$ Can you proceed ?
answered Jan 15 at 12:55
FredFred
46.9k1848
answered Jan 15 at 12:55
FredFred
46.9k1848
answered Jan 15 at 12:55
FredFred
46.9k1848
answered Jan 15 at 12:55
FredFred
46.9k1848
46.9k1848
add a comment |
add a comment |
$begingroup$
Hint:
A bijective function is a which is both one-one and onto function.
Now you can prove that the function is not bijective by proving that the function is not one-one or the function is not onto.
It would be much easier to prove that the function is one-one.The various methods to check injectivity of a function are:
1) For a one one function it gives same output for same number or for one input it gives only one output this can be exploited by proving either one of the following:
$$ if x_1 =x_2, then f(x_1) ne f(x_2) $$
$$ if x_1 ne x_2, then f(x_1) = f(x_2) $$
2)you can check the function's slope(Slope can be found out by finding the derivative of the function or simply differentiating it once)-If the slope is independent of x hence the function is either increasing or decreasing so it is one-one else not.
answered Jan 15 at 14:18
Harsh WasnikHarsh Wasnik
13
$endgroup$
add a comment |
$begingroup$
Hint:
A bijective function is a which is both one-one and onto function.
Now you can prove that the function is not bijective by proving that the function is not one-one or the function is not onto.
It would be much easier to prove that the function is one-one.The various methods to check injectivity of a function are:
1) For a one one function it gives same output for same number or for one input it gives only one output this can be exploited by proving either one of the following:
$$ if x_1 =x_2, then f(x_1) ne f(x_2) $$
$$ if x_1 ne x_2, then f(x_1) = f(x_2) $$
2)you can check the function's slope(Slope can be found out by finding the derivative of the function or simply differentiating it once)-If the slope is independent of x hence the function is either increasing or decreasing so it is one-one else not.
answered Jan 15 at 14:18
Harsh WasnikHarsh Wasnik
13
$endgroup$
add a comment |
$begingroup$
Hint:
A bijective function is a which is both one-one and onto function.
Now you can prove that the function is not bijective by proving that the function is not one-one or the function is not onto.
It would be much easier to prove that the function is one-one.The various methods to check injectivity of a function are:
1) For a one one function it gives same output for same number or for one input it gives only one output this can be exploited by proving either one of the following:
$$ if x_1 =x_2, then f(x_1) ne f(x_2) $$
$$ if x_1 ne x_2, then f(x_1) = f(x_2) $$
2)you can check the function's slope(Slope can be found out by finding the derivative of the function or simply differentiating it once)-If the slope is independent of x hence the function is either increasing or decreasing so it is one-one else not.
answered Jan 15 at 14:18
Harsh WasnikHarsh Wasnik
13
$endgroup$
Hint:
A bijective function is a which is both one-one and onto function.
Now you can prove that the function is not bijective by proving that the function is not one-one or the function is not onto.
It would be much easier to prove that the function is one-one.The various methods to check injectivity of a function are:
1) For a one one function it gives same output for same number or for one input it gives only one output this can be exploited by proving either one of the following:
$$ if x_1 =x_2, then f(x_1) ne f(x_2) $$
$$ if x_1 ne x_2, then f(x_1) = f(x_2) $$
2)you can check the function's slope(Slope can be found out by finding the derivative of the function or simply differentiating it once)-If the slope is independent of x hence the function is either increasing or decreasing so it is one-one else not.
answered Jan 15 at 14:18
Harsh WasnikHarsh Wasnik
13
answered Jan 15 at 14:18
Harsh WasnikHarsh Wasnik
13
answered Jan 15 at 14:18
Harsh WasnikHarsh Wasnik
13
answered Jan 15 at 14:18
Harsh WasnikHarsh Wasnik
13
13
add a comment |
add a comment |
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5
$begingroup$
Show that $f$ is $0$ on more than one input.
$endgroup$
– SmileyCraft
Jan 15 at 12:34
$begingroup$
Remark: You need to specify the codomain if you want to prove that a function is bijective. However in this ecample it is sufficient to show that the function is not injective.
$endgroup$
– maxmilgram
Jan 15 at 16:12