From $cos(x-y)$ to $f(x+y)$?
$begingroup$
Is it possible to transform $cos(x-y)$ into a function $f=f(x+y)$ to have:
$$cos(x-y)=f(x+y)$$?
real-analysis complex-analysis trigonometry change-of-variable
asked Dec 5 '18 at 16:10
Alessio BocciAlessio Bocci
567
$endgroup$
add a comment |
$begingroup$
Is it possible to transform $cos(x-y)$ into a function $f=f(x+y)$ to have:
$$cos(x-y)=f(x+y)$$?
real-analysis complex-analysis trigonometry change-of-variable
asked Dec 5 '18 at 16:10
Alessio BocciAlessio Bocci
567
$endgroup$
add a comment |
$begingroup$
Is it possible to transform $cos(x-y)$ into a function $f=f(x+y)$ to have:
$$cos(x-y)=f(x+y)$$?
real-analysis complex-analysis trigonometry change-of-variable
asked Dec 5 '18 at 16:10
Alessio BocciAlessio Bocci
567
$endgroup$
Is it possible to transform $cos(x-y)$ into a function $f=f(x+y)$ to have:
$$cos(x-y)=f(x+y)$$?
real-analysis complex-analysis trigonometry change-of-variable
real-analysis complex-analysis trigonometry change-of-variable
asked Dec 5 '18 at 16:10
Alessio BocciAlessio Bocci
567
asked Dec 5 '18 at 16:10
Alessio BocciAlessio Bocci
567
asked Dec 5 '18 at 16:10
Alessio BocciAlessio Bocci
567
asked Dec 5 '18 at 16:10
Alessio BocciAlessio Bocci
567
asked Dec 5 '18 at 16:10
Alessio BocciAlessio Bocci
567
567
add a comment |
add a comment |
2 Answers
2
active
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votes
$begingroup$
If it is true for $x=y$ you have that
$1=f(2x)$ so
$f(t)=f(2(frac{t}{2}))=1$
But it is not possibile because $f=1$ not verify the condition $cos(x-y)=f(x-y)$.
So there is not a function that verify your condition
answered Dec 5 '18 at 16:13
Federico FalluccaFederico Fallucca
1,90219
$endgroup$
add a comment |
$begingroup$
No, because $cos(1-1)$ is different from $cos(2-0)$, so $f(2)$ can't be both of them at once.
answered Dec 5 '18 at 16:13
ArthurArthur
113k7115197
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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$begingroup$
If it is true for $x=y$ you have that
$1=f(2x)$ so
$f(t)=f(2(frac{t}{2}))=1$
But it is not possibile because $f=1$ not verify the condition $cos(x-y)=f(x-y)$.
So there is not a function that verify your condition
answered Dec 5 '18 at 16:13
Federico FalluccaFederico Fallucca
1,90219
$endgroup$
add a comment |
$begingroup$
If it is true for $x=y$ you have that
$1=f(2x)$ so
$f(t)=f(2(frac{t}{2}))=1$
But it is not possibile because $f=1$ not verify the condition $cos(x-y)=f(x-y)$.
So there is not a function that verify your condition
answered Dec 5 '18 at 16:13
Federico FalluccaFederico Fallucca
1,90219
$endgroup$
add a comment |
$begingroup$
If it is true for $x=y$ you have that
$1=f(2x)$ so
$f(t)=f(2(frac{t}{2}))=1$
But it is not possibile because $f=1$ not verify the condition $cos(x-y)=f(x-y)$.
So there is not a function that verify your condition
answered Dec 5 '18 at 16:13
Federico FalluccaFederico Fallucca
1,90219
$endgroup$
If it is true for $x=y$ you have that
$1=f(2x)$ so
$f(t)=f(2(frac{t}{2}))=1$
But it is not possibile because $f=1$ not verify the condition $cos(x-y)=f(x-y)$.
So there is not a function that verify your condition
answered Dec 5 '18 at 16:13
Federico FalluccaFederico Fallucca
1,90219
answered Dec 5 '18 at 16:13
Federico FalluccaFederico Fallucca
1,90219
answered Dec 5 '18 at 16:13
Federico FalluccaFederico Fallucca
1,90219
answered Dec 5 '18 at 16:13
Federico FalluccaFederico Fallucca
1,90219
1,90219
add a comment |
add a comment |
$begingroup$
No, because $cos(1-1)$ is different from $cos(2-0)$, so $f(2)$ can't be both of them at once.
answered Dec 5 '18 at 16:13
ArthurArthur
113k7115197
$endgroup$
add a comment |
$begingroup$
No, because $cos(1-1)$ is different from $cos(2-0)$, so $f(2)$ can't be both of them at once.
answered Dec 5 '18 at 16:13
ArthurArthur
113k7115197
$endgroup$
add a comment |
$begingroup$
No, because $cos(1-1)$ is different from $cos(2-0)$, so $f(2)$ can't be both of them at once.
answered Dec 5 '18 at 16:13
ArthurArthur
113k7115197
$endgroup$
No, because $cos(1-1)$ is different from $cos(2-0)$, so $f(2)$ can't be both of them at once.
answered Dec 5 '18 at 16:13
ArthurArthur
113k7115197
answered Dec 5 '18 at 16:13
ArthurArthur
113k7115197
answered Dec 5 '18 at 16:13
ArthurArthur
113k7115197
answered Dec 5 '18 at 16:13
ArthurArthur
113k7115197
113k7115197
add a comment |
add a comment |
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