Find functions f & g such that f(x+y)=g (x.y) for all x & y?How do you define functions for...
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Find functions f & g such that f(x+y)=g (x.y) for all x & y?
How do you define functions for non-mathematicians?Automated generation of a parametric plot?For each $y in mathbb{R}$ either no $x$ with $f(x) = y$ or two such values of $x$. Show that $f$ is discontinuous.Show that if $C(K)$ is separable, then $K$ is metrisable, for $K$ compact and HausdorffFinding Conjugate harmonic of $u = frac{1}{2} ln(x^2 + y^2)$Asymptotic functions with derivatives that are $1/2^x$Nowhere Differentiable function over a non-compact spaceMethod for deconstructing an integral involving two functionsFinding derivative given $f(x)$ and limitfix point solution or approximation available? logistic regression?
$begingroup$
My approach towards this question was that first I putted x=0 and then y=0 which yields f(x)=f(y)and f(x)=g(0), and again for x=1 & y=1 it gives g (x)=f(1+x), and so on. My query is that what should the compact & rigorous form of proof or solution looks like because I found myself stuck between results and not comprehending results very well. Thanks.
real-analysis calculus functional-analysis functions
asked 2 hours ago
BORN TO LEARNBORN TO LEARN
377
$endgroup$
add a comment |
$begingroup$
My approach towards this question was that first I putted x=0 and then y=0 which yields f(x)=f(y)and f(x)=g(0), and again for x=1 & y=1 it gives g (x)=f(1+x), and so on. My query is that what should the compact & rigorous form of proof or solution looks like because I found myself stuck between results and not comprehending results very well. Thanks.
real-analysis calculus functional-analysis functions
asked 2 hours ago
BORN TO LEARNBORN TO LEARN
377
$endgroup$
add a comment |
$begingroup$
My approach towards this question was that first I putted x=0 and then y=0 which yields f(x)=f(y)and f(x)=g(0), and again for x=1 & y=1 it gives g (x)=f(1+x), and so on. My query is that what should the compact & rigorous form of proof or solution looks like because I found myself stuck between results and not comprehending results very well. Thanks.
real-analysis calculus functional-analysis functions
asked 2 hours ago
BORN TO LEARNBORN TO LEARN
377
$endgroup$
My approach towards this question was that first I putted x=0 and then y=0 which yields f(x)=f(y)and f(x)=g(0), and again for x=1 & y=1 it gives g (x)=f(1+x), and so on. My query is that what should the compact & rigorous form of proof or solution looks like because I found myself stuck between results and not comprehending results very well. Thanks.
real-analysis calculus functional-analysis functions
real-analysis calculus functional-analysis functions
asked 2 hours ago
BORN TO LEARNBORN TO LEARN
377
asked 2 hours ago
BORN TO LEARNBORN TO LEARN
377
edited 1 hour ago
BORN TO LEARN
asked 2 hours ago
BORN TO LEARNBORN TO LEARN
377
asked 2 hours ago
BORN TO LEARNBORN TO LEARN
377
asked 2 hours ago
BORN TO LEARNBORN TO LEARN
377
377
add a comment |
add a comment |
1 Answer
1
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$begingroup$
$$f(x) = f(x + 0) = g(x cdot 0) = g(0)$$ for all $x$. Thus $f(x)$ is constant and always equal to $g(0)$. Moreover, $$g(0) = f(1 + y) = g(1 cdot y) = g(y)$$ for all $y$. Thus $g$ is also constant and equal to $g(0)$. Thus $f, g$ are the same constant function. All constant functions $f = g = c$ will satisfy the relation. This is all you need.
answered 2 hours ago
BenBBenB
46039
$endgroup$
1
$begingroup$
Yeah thanks man m looking for this kind of preciseness.
$endgroup$
– BORN TO LEARN
2 hours ago
$begingroup$
Sorry I edited it because I made a mistake in description that g(x)=f(1+y) instead of g(x)=f(1+x).
$endgroup$
– BORN TO LEARN
1 hour ago
$begingroup$
Now I think your proof also needs editing?
$endgroup$
– BORN TO LEARN
1 hour ago
$begingroup$
@BORNTOLEARN where does it need edited? I only used the relation f(x + y) = g(x y).
$endgroup$
– BenB
1 hour ago
1
$begingroup$
@BORNTOLEARN because f is identically equal to g(0) as I show in the previous part.
$endgroup$
– BenB
1 hour ago
|
show 5 more comments
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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$begingroup$
$$f(x) = f(x + 0) = g(x cdot 0) = g(0)$$ for all $x$. Thus $f(x)$ is constant and always equal to $g(0)$. Moreover, $$g(0) = f(1 + y) = g(1 cdot y) = g(y)$$ for all $y$. Thus $g$ is also constant and equal to $g(0)$. Thus $f, g$ are the same constant function. All constant functions $f = g = c$ will satisfy the relation. This is all you need.
answered 2 hours ago
BenBBenB
46039
$endgroup$
1
$begingroup$
Yeah thanks man m looking for this kind of preciseness.
$endgroup$
– BORN TO LEARN
2 hours ago
$begingroup$
Sorry I edited it because I made a mistake in description that g(x)=f(1+y) instead of g(x)=f(1+x).
$endgroup$
– BORN TO LEARN
1 hour ago
$begingroup$
Now I think your proof also needs editing?
$endgroup$
– BORN TO LEARN
1 hour ago
$begingroup$
@BORNTOLEARN where does it need edited? I only used the relation f(x + y) = g(x y).
$endgroup$
– BenB
1 hour ago
1
$begingroup$
@BORNTOLEARN because f is identically equal to g(0) as I show in the previous part.
$endgroup$
– BenB
1 hour ago
|
show 5 more comments
$begingroup$
$$f(x) = f(x + 0) = g(x cdot 0) = g(0)$$ for all $x$. Thus $f(x)$ is constant and always equal to $g(0)$. Moreover, $$g(0) = f(1 + y) = g(1 cdot y) = g(y)$$ for all $y$. Thus $g$ is also constant and equal to $g(0)$. Thus $f, g$ are the same constant function. All constant functions $f = g = c$ will satisfy the relation. This is all you need.
answered 2 hours ago
BenBBenB
46039
$endgroup$
1
$begingroup$
Yeah thanks man m looking for this kind of preciseness.
$endgroup$
– BORN TO LEARN
2 hours ago
$begingroup$
Sorry I edited it because I made a mistake in description that g(x)=f(1+y) instead of g(x)=f(1+x).
$endgroup$
– BORN TO LEARN
1 hour ago
$begingroup$
Now I think your proof also needs editing?
$endgroup$
– BORN TO LEARN
1 hour ago
$begingroup$
@BORNTOLEARN where does it need edited? I only used the relation f(x + y) = g(x y).
$endgroup$
– BenB
1 hour ago
1
$begingroup$
@BORNTOLEARN because f is identically equal to g(0) as I show in the previous part.
$endgroup$
– BenB
1 hour ago
|
show 5 more comments
$begingroup$
$$f(x) = f(x + 0) = g(x cdot 0) = g(0)$$ for all $x$. Thus $f(x)$ is constant and always equal to $g(0)$. Moreover, $$g(0) = f(1 + y) = g(1 cdot y) = g(y)$$ for all $y$. Thus $g$ is also constant and equal to $g(0)$. Thus $f, g$ are the same constant function. All constant functions $f = g = c$ will satisfy the relation. This is all you need.
answered 2 hours ago
BenBBenB
46039
$endgroup$
$$f(x) = f(x + 0) = g(x cdot 0) = g(0)$$ for all $x$. Thus $f(x)$ is constant and always equal to $g(0)$. Moreover, $$g(0) = f(1 + y) = g(1 cdot y) = g(y)$$ for all $y$. Thus $g$ is also constant and equal to $g(0)$. Thus $f, g$ are the same constant function. All constant functions $f = g = c$ will satisfy the relation. This is all you need.
answered 2 hours ago
BenBBenB
46039
answered 2 hours ago
BenBBenB
46039
answered 2 hours ago
BenBBenB
46039
answered 2 hours ago
BenBBenB
46039
46039
1
$begingroup$
Yeah thanks man m looking for this kind of preciseness.
$endgroup$
– BORN TO LEARN
2 hours ago
$begingroup$
Sorry I edited it because I made a mistake in description that g(x)=f(1+y) instead of g(x)=f(1+x).
$endgroup$
– BORN TO LEARN
1 hour ago
$begingroup$
Now I think your proof also needs editing?
$endgroup$
– BORN TO LEARN
1 hour ago
$begingroup$
@BORNTOLEARN where does it need edited? I only used the relation f(x + y) = g(x y).
$endgroup$
– BenB
1 hour ago
1
$begingroup$
@BORNTOLEARN because f is identically equal to g(0) as I show in the previous part.
$endgroup$
– BenB
1 hour ago
|
show 5 more comments
1
$begingroup$
Yeah thanks man m looking for this kind of preciseness.
$endgroup$
– BORN TO LEARN
2 hours ago
$begingroup$
Sorry I edited it because I made a mistake in description that g(x)=f(1+y) instead of g(x)=f(1+x).
$endgroup$
– BORN TO LEARN
1 hour ago
$begingroup$
Now I think your proof also needs editing?
$endgroup$
– BORN TO LEARN
1 hour ago
$begingroup$
@BORNTOLEARN where does it need edited? I only used the relation f(x + y) = g(x y).
$endgroup$
– BenB
1 hour ago
1
$begingroup$
@BORNTOLEARN because f is identically equal to g(0) as I show in the previous part.
$endgroup$
– BenB
1 hour ago
1
1
$begingroup$
Yeah thanks man m looking for this kind of preciseness.
$endgroup$
– BORN TO LEARN
2 hours ago
$begingroup$
Yeah thanks man m looking for this kind of preciseness.
$endgroup$
– BORN TO LEARN
2 hours ago
$begingroup$
Sorry I edited it because I made a mistake in description that g(x)=f(1+y) instead of g(x)=f(1+x).
$endgroup$
– BORN TO LEARN
1 hour ago
$begingroup$
Sorry I edited it because I made a mistake in description that g(x)=f(1+y) instead of g(x)=f(1+x).
$endgroup$
– BORN TO LEARN
1 hour ago
$begingroup$
Now I think your proof also needs editing?
$endgroup$
– BORN TO LEARN
1 hour ago
$begingroup$
Now I think your proof also needs editing?
$endgroup$
– BORN TO LEARN
1 hour ago
$begingroup$
@BORNTOLEARN where does it need edited? I only used the relation f(x + y) = g(x y).
$endgroup$
– BenB
1 hour ago
$begingroup$
@BORNTOLEARN where does it need edited? I only used the relation f(x + y) = g(x y).
$endgroup$
– BenB
1 hour ago
1
1
$begingroup$
@BORNTOLEARN because f is identically equal to g(0) as I show in the previous part.
$endgroup$
– BenB
1 hour ago
$begingroup$
@BORNTOLEARN because f is identically equal to g(0) as I show in the previous part.
$endgroup$
– BenB
1 hour ago
|
show 5 more comments
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