Arriving at odd possible solutions for functions
$begingroup$
Say $f(x)=dfrac{3x+1}{2}$
I want to find out for which values of $x$ is the value of $f(x)$ an odd number.
So I reframe $f(x)$ to
$dfrac{3x+1}{2}=2k_1+1$
on simplifying further...
$dfrac{3x-1}{4}=k_1$ or $dfrac{4k_1+1}{3}=x$
For $x=3,7,11,15,19,...$ we find that $k_1$ is a natural number and hence we get the solution.
Now I am attempting to find out for which values of $x$ is the value of $f^2(x) $or $f(f(x))$ is an odd number.
The answer is $x=7,15,23,31...$
How do I frame the equation?
I tried putting
$dfrac{3(2k_1+1)+1}{2}=2k_2+1$ and
$f(f(x))=f^2(x)= dfrac{3(frac{3x+1}{2})+1}{2}=2k_2+1$
and both of them lead to incorrect answers.
number-theory functions modular-arithmetic arithmetic
$endgroup$
add a comment |
$begingroup$
Say $f(x)=dfrac{3x+1}{2}$
I want to find out for which values of $x$ is the value of $f(x)$ an odd number.
So I reframe $f(x)$ to
$dfrac{3x+1}{2}=2k_1+1$
on simplifying further...
$dfrac{3x-1}{4}=k_1$ or $dfrac{4k_1+1}{3}=x$
For $x=3,7,11,15,19,...$ we find that $k_1$ is a natural number and hence we get the solution.
Now I am attempting to find out for which values of $x$ is the value of $f^2(x) $or $f(f(x))$ is an odd number.
The answer is $x=7,15,23,31...$
How do I frame the equation?
I tried putting
$dfrac{3(2k_1+1)+1}{2}=2k_2+1$ and
$f(f(x))=f^2(x)= dfrac{3(frac{3x+1}{2})+1}{2}=2k_2+1$
and both of them lead to incorrect answers.
number-theory functions modular-arithmetic arithmetic
$endgroup$
1
$begingroup$
$fcirc f(x)=frac {9x+5}4$. Using that, the method you used first works.
$endgroup$
– lulu
Dec 8 '18 at 10:52
add a comment |
$begingroup$
Say $f(x)=dfrac{3x+1}{2}$
I want to find out for which values of $x$ is the value of $f(x)$ an odd number.
So I reframe $f(x)$ to
$dfrac{3x+1}{2}=2k_1+1$
on simplifying further...
$dfrac{3x-1}{4}=k_1$ or $dfrac{4k_1+1}{3}=x$
For $x=3,7,11,15,19,...$ we find that $k_1$ is a natural number and hence we get the solution.
Now I am attempting to find out for which values of $x$ is the value of $f^2(x) $or $f(f(x))$ is an odd number.
The answer is $x=7,15,23,31...$
How do I frame the equation?
I tried putting
$dfrac{3(2k_1+1)+1}{2}=2k_2+1$ and
$f(f(x))=f^2(x)= dfrac{3(frac{3x+1}{2})+1}{2}=2k_2+1$
and both of them lead to incorrect answers.
number-theory functions modular-arithmetic arithmetic
$endgroup$
Say $f(x)=dfrac{3x+1}{2}$
I want to find out for which values of $x$ is the value of $f(x)$ an odd number.
So I reframe $f(x)$ to
$dfrac{3x+1}{2}=2k_1+1$
on simplifying further...
$dfrac{3x-1}{4}=k_1$ or $dfrac{4k_1+1}{3}=x$
For $x=3,7,11,15,19,...$ we find that $k_1$ is a natural number and hence we get the solution.
Now I am attempting to find out for which values of $x$ is the value of $f^2(x) $or $f(f(x))$ is an odd number.
The answer is $x=7,15,23,31...$
How do I frame the equation?
I tried putting
$dfrac{3(2k_1+1)+1}{2}=2k_2+1$ and
$f(f(x))=f^2(x)= dfrac{3(frac{3x+1}{2})+1}{2}=2k_2+1$
and both of them lead to incorrect answers.
number-theory functions modular-arithmetic arithmetic
number-theory functions modular-arithmetic arithmetic
asked Dec 8 '18 at 10:44
Rahul ShahRahul Shah
2202310
2202310
1
$begingroup$
$fcirc f(x)=frac {9x+5}4$. Using that, the method you used first works.
$endgroup$
– lulu
Dec 8 '18 at 10:52
add a comment |
1
$begingroup$
$fcirc f(x)=frac {9x+5}4$. Using that, the method you used first works.
$endgroup$
– lulu
Dec 8 '18 at 10:52
1
1
$begingroup$
$fcirc f(x)=frac {9x+5}4$. Using that, the method you used first works.
$endgroup$
– lulu
Dec 8 '18 at 10:52
$begingroup$
$fcirc f(x)=frac {9x+5}4$. Using that, the method you used first works.
$endgroup$
– lulu
Dec 8 '18 at 10:52
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Continuing what you have written,
$f^2(x)=frac{9x+5}{4}=2k+1$
$k=frac{9x+1}{8}$, where k is any positive integer.
Note that $x=7$ satisfies this.
Further, let $x=7+y$ also gives integral $k$.
Then, $$k=frac{9(7+y)+1}{8}$$ $$k=8+frac{9y}{8}$$ $$implies 8|y $$.
It gives $x=7,15,23,31...$
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Continuing what you have written,
$f^2(x)=frac{9x+5}{4}=2k+1$
$k=frac{9x+1}{8}$, where k is any positive integer.
Note that $x=7$ satisfies this.
Further, let $x=7+y$ also gives integral $k$.
Then, $$k=frac{9(7+y)+1}{8}$$ $$k=8+frac{9y}{8}$$ $$implies 8|y $$.
It gives $x=7,15,23,31...$
$endgroup$
add a comment |
$begingroup$
Continuing what you have written,
$f^2(x)=frac{9x+5}{4}=2k+1$
$k=frac{9x+1}{8}$, where k is any positive integer.
Note that $x=7$ satisfies this.
Further, let $x=7+y$ also gives integral $k$.
Then, $$k=frac{9(7+y)+1}{8}$$ $$k=8+frac{9y}{8}$$ $$implies 8|y $$.
It gives $x=7,15,23,31...$
$endgroup$
add a comment |
$begingroup$
Continuing what you have written,
$f^2(x)=frac{9x+5}{4}=2k+1$
$k=frac{9x+1}{8}$, where k is any positive integer.
Note that $x=7$ satisfies this.
Further, let $x=7+y$ also gives integral $k$.
Then, $$k=frac{9(7+y)+1}{8}$$ $$k=8+frac{9y}{8}$$ $$implies 8|y $$.
It gives $x=7,15,23,31...$
$endgroup$
Continuing what you have written,
$f^2(x)=frac{9x+5}{4}=2k+1$
$k=frac{9x+1}{8}$, where k is any positive integer.
Note that $x=7$ satisfies this.
Further, let $x=7+y$ also gives integral $k$.
Then, $$k=frac{9(7+y)+1}{8}$$ $$k=8+frac{9y}{8}$$ $$implies 8|y $$.
It gives $x=7,15,23,31...$
answered Dec 8 '18 at 10:56
Ankit KumarAnkit Kumar
1,494221
1,494221
add a comment |
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$begingroup$
$fcirc f(x)=frac {9x+5}4$. Using that, the method you used first works.
$endgroup$
– lulu
Dec 8 '18 at 10:52