Mixing
How to solve piecewise equation with function of a function
$begingroup$
I have got the question below for an assignment, I understand how and what the piecewise equation does, but I am wondering if someone can explain what the $f(x)=f(x+2)$ is referring to? Does this mean function of a function?If so, how would I solve this? Question
calculus
asked Jan 22 at 6:38
mathrook1243mathrook1243
203
$endgroup$
add a comment |
$begingroup$
I have got the question below for an assignment, I understand how and what the piecewise equation does, but I am wondering if someone can explain what the $f(x)=f(x+2)$ is referring to? Does this mean function of a function?If so, how would I solve this? Question
calculus
asked Jan 22 at 6:38
mathrook1243mathrook1243
203
$endgroup$
1
$begingroup$
You can writef(x)=f(x+2)with $, like this: $ f(x)=f(x+2)$ $Big(f(x)=f(x+2)Big)$. See the help or here for more information
$endgroup$
– El borito
Jan 22 at 6:42
add a comment |
$begingroup$
I have got the question below for an assignment, I understand how and what the piecewise equation does, but I am wondering if someone can explain what the $f(x)=f(x+2)$ is referring to? Does this mean function of a function?If so, how would I solve this? Question
calculus
asked Jan 22 at 6:38
mathrook1243mathrook1243
203
$endgroup$
I have got the question below for an assignment, I understand how and what the piecewise equation does, but I am wondering if someone can explain what the $f(x)=f(x+2)$ is referring to? Does this mean function of a function?If so, how would I solve this? Question
calculus
calculus
asked Jan 22 at 6:38
mathrook1243mathrook1243
203
asked Jan 22 at 6:38
mathrook1243mathrook1243
203
edited Jan 22 at 6:48
mathrook1243
asked Jan 22 at 6:38
mathrook1243mathrook1243
203
asked Jan 22 at 6:38
mathrook1243mathrook1243
203
asked Jan 22 at 6:38
mathrook1243mathrook1243
203
203
1
$begingroup$
You can writef(x)=f(x+2)with $, like this: $ f(x)=f(x+2)$ $Big(f(x)=f(x+2)Big)$. See the help or here for more information
$endgroup$
– El borito
Jan 22 at 6:42
add a comment |
1
$begingroup$
You can writef(x)=f(x+2)with $, like this: $ f(x)=f(x+2)$ $Big(f(x)=f(x+2)Big)$. See the help or here for more information
$endgroup$
– El borito
Jan 22 at 6:42
1
1
$begingroup$
You can write
f(x)=f(x+2) with $, like this: $ f(x)=f(x+2)$ $Big(f(x)=f(x+2)Big)$. See the help or here for more information$endgroup$
– El borito
Jan 22 at 6:42
$begingroup$
You can write
f(x)=f(x+2) with $, like this: $ f(x)=f(x+2)$ $Big(f(x)=f(x+2)Big)$. See the help or here for more information$endgroup$
– El borito
Jan 22 at 6:42
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It means if you know $f(0)$, we know $f(2)$, and we also know $f(4)$. The values are equal.
Similarly for $f(-2)$ and $f(-4)$.
In general, if we know $f(x)$, then we know the values of $f(x+2)$ and $f(x-2)$ and they are equal.
This is a periodic function with period $2$.
answered Jan 22 at 6:42
Siong Thye GohSiong Thye Goh
102k1468119
$endgroup$
$begingroup$
So does that mean that the whole function is a period function, including the piecewise equation?
$endgroup$
– mathrook1243
Jan 22 at 6:49
$begingroup$
the whole function? hmmm... we only have one function isn't it. First draw the part from $-1$ to $1$ and then repeat it.
$endgroup$
– Siong Thye Goh
Jan 22 at 6:50
$begingroup$
so how would I sketch the function between -5 and 5??
$endgroup$
– mathrook1243
Jan 22 at 10:06
$begingroup$
First sketch the function from $(-1, 1], then repeat the patterm on the interval $(1,3]$?
$endgroup$
– Siong Thye Goh
Jan 22 at 10:07
add a comment |
$begingroup$
The function is first defined on $[-1,1]$ and then extended to the whole line by adding the condition $f(x+2)=f(x)$ (periodocity of $f$). For example, if $7 leq x leq 9$ then $x-8$ lies between $-1$ and $1$ and $f(x)$ is defined to be $f(x-8x)$. [Note that $f(x+2)=f(x)$ for all $x$ implies that $f(x+2n)=f(x)$ for all $x$ and for any integer $n$].
answered Jan 22 at 6:44
Kavi Rama MurthyKavi Rama Murthy
65.8k42767
$endgroup$
$begingroup$
I believe your $2$ uses of $x - 8x$ should be just $x - 8$. Also, at the end, the statement $fleft(x + 2nright) = fleft(xright)$ is true for any integers $n$, not just positive ones, as Siong Thye Goh's answer indicates.
$endgroup$
– John Omielan
Jan 22 at 7:43
$begingroup$
@JohnOmielan There was a typo. I meant $x-8$. As far as your second comment is concerned the statement '$f(x+2n)=f(x)$ for all $x$ and all positive integers $n$ ' is equivalent to '$f(x+2n)=f(x)$ for all $x$ and all integers $n$ '. Perhaps I could have said 'all integers' for ease of understanding
$endgroup$
– Kavi Rama Murthy
Jan 22 at 7:47
$begingroup$
Thanks for correcting the first instance. However, you still have a second one, as I said earlier, with it being in the statement "... is defined to be $fleft(x - 8xright)$". As for your suggestion of removing the term "positive", although it's not incorrect, I believe it's more general and it will provide ease of understanding.
$endgroup$
– John Omielan
Jan 22 at 7:50
add a comment |
Your Answer
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2 Answers
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2 Answers
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$begingroup$
It means if you know $f(0)$, we know $f(2)$, and we also know $f(4)$. The values are equal.
Similarly for $f(-2)$ and $f(-4)$.
In general, if we know $f(x)$, then we know the values of $f(x+2)$ and $f(x-2)$ and they are equal.
This is a periodic function with period $2$.
answered Jan 22 at 6:42
Siong Thye GohSiong Thye Goh
102k1468119
$endgroup$
$begingroup$
So does that mean that the whole function is a period function, including the piecewise equation?
$endgroup$
– mathrook1243
Jan 22 at 6:49
$begingroup$
the whole function? hmmm... we only have one function isn't it. First draw the part from $-1$ to $1$ and then repeat it.
$endgroup$
– Siong Thye Goh
Jan 22 at 6:50
$begingroup$
so how would I sketch the function between -5 and 5??
$endgroup$
– mathrook1243
Jan 22 at 10:06
$begingroup$
First sketch the function from $(-1, 1], then repeat the patterm on the interval $(1,3]$?
$endgroup$
– Siong Thye Goh
Jan 22 at 10:07
add a comment |
$begingroup$
It means if you know $f(0)$, we know $f(2)$, and we also know $f(4)$. The values are equal.
Similarly for $f(-2)$ and $f(-4)$.
In general, if we know $f(x)$, then we know the values of $f(x+2)$ and $f(x-2)$ and they are equal.
This is a periodic function with period $2$.
answered Jan 22 at 6:42
Siong Thye GohSiong Thye Goh
102k1468119
$endgroup$
$begingroup$
So does that mean that the whole function is a period function, including the piecewise equation?
$endgroup$
– mathrook1243
Jan 22 at 6:49
$begingroup$
the whole function? hmmm... we only have one function isn't it. First draw the part from $-1$ to $1$ and then repeat it.
$endgroup$
– Siong Thye Goh
Jan 22 at 6:50
$begingroup$
so how would I sketch the function between -5 and 5??
$endgroup$
– mathrook1243
Jan 22 at 10:06
$begingroup$
First sketch the function from $(-1, 1], then repeat the patterm on the interval $(1,3]$?
$endgroup$
– Siong Thye Goh
Jan 22 at 10:07
add a comment |
$begingroup$
It means if you know $f(0)$, we know $f(2)$, and we also know $f(4)$. The values are equal.
Similarly for $f(-2)$ and $f(-4)$.
In general, if we know $f(x)$, then we know the values of $f(x+2)$ and $f(x-2)$ and they are equal.
This is a periodic function with period $2$.
answered Jan 22 at 6:42
Siong Thye GohSiong Thye Goh
102k1468119
$endgroup$
It means if you know $f(0)$, we know $f(2)$, and we also know $f(4)$. The values are equal.
Similarly for $f(-2)$ and $f(-4)$.
In general, if we know $f(x)$, then we know the values of $f(x+2)$ and $f(x-2)$ and they are equal.
This is a periodic function with period $2$.
answered Jan 22 at 6:42
Siong Thye GohSiong Thye Goh
102k1468119
answered Jan 22 at 6:42
Siong Thye GohSiong Thye Goh
102k1468119
answered Jan 22 at 6:42
Siong Thye GohSiong Thye Goh
102k1468119
answered Jan 22 at 6:42
Siong Thye GohSiong Thye Goh
102k1468119
102k1468119
$begingroup$
So does that mean that the whole function is a period function, including the piecewise equation?
$endgroup$
– mathrook1243
Jan 22 at 6:49
$begingroup$
the whole function? hmmm... we only have one function isn't it. First draw the part from $-1$ to $1$ and then repeat it.
$endgroup$
– Siong Thye Goh
Jan 22 at 6:50
$begingroup$
so how would I sketch the function between -5 and 5??
$endgroup$
– mathrook1243
Jan 22 at 10:06
$begingroup$
First sketch the function from $(-1, 1], then repeat the patterm on the interval $(1,3]$?
$endgroup$
– Siong Thye Goh
Jan 22 at 10:07
add a comment |
$begingroup$
So does that mean that the whole function is a period function, including the piecewise equation?
$endgroup$
– mathrook1243
Jan 22 at 6:49
$begingroup$
the whole function? hmmm... we only have one function isn't it. First draw the part from $-1$ to $1$ and then repeat it.
$endgroup$
– Siong Thye Goh
Jan 22 at 6:50
$begingroup$
so how would I sketch the function between -5 and 5??
$endgroup$
– mathrook1243
Jan 22 at 10:06
$begingroup$
First sketch the function from $(-1, 1], then repeat the patterm on the interval $(1,3]$?
$endgroup$
– Siong Thye Goh
Jan 22 at 10:07
$begingroup$
So does that mean that the whole function is a period function, including the piecewise equation?
$endgroup$
– mathrook1243
Jan 22 at 6:49
$begingroup$
So does that mean that the whole function is a period function, including the piecewise equation?
$endgroup$
– mathrook1243
Jan 22 at 6:49
$begingroup$
the whole function? hmmm... we only have one function isn't it. First draw the part from $-1$ to $1$ and then repeat it.
$endgroup$
– Siong Thye Goh
Jan 22 at 6:50
$begingroup$
the whole function? hmmm... we only have one function isn't it. First draw the part from $-1$ to $1$ and then repeat it.
$endgroup$
– Siong Thye Goh
Jan 22 at 6:50
$begingroup$
so how would I sketch the function between -5 and 5??
$endgroup$
– mathrook1243
Jan 22 at 10:06
$begingroup$
so how would I sketch the function between -5 and 5??
$endgroup$
– mathrook1243
Jan 22 at 10:06
$begingroup$
First sketch the function from $(-1, 1], then repeat the patterm on the interval $(1,3]$?
$endgroup$
– Siong Thye Goh
Jan 22 at 10:07
$begingroup$
First sketch the function from $(-1, 1], then repeat the patterm on the interval $(1,3]$?
$endgroup$
– Siong Thye Goh
Jan 22 at 10:07
add a comment |
$begingroup$
The function is first defined on $[-1,1]$ and then extended to the whole line by adding the condition $f(x+2)=f(x)$ (periodocity of $f$). For example, if $7 leq x leq 9$ then $x-8$ lies between $-1$ and $1$ and $f(x)$ is defined to be $f(x-8x)$. [Note that $f(x+2)=f(x)$ for all $x$ implies that $f(x+2n)=f(x)$ for all $x$ and for any integer $n$].
answered Jan 22 at 6:44
Kavi Rama MurthyKavi Rama Murthy
65.8k42767
$endgroup$
$begingroup$
I believe your $2$ uses of $x - 8x$ should be just $x - 8$. Also, at the end, the statement $fleft(x + 2nright) = fleft(xright)$ is true for any integers $n$, not just positive ones, as Siong Thye Goh's answer indicates.
$endgroup$
– John Omielan
Jan 22 at 7:43
$begingroup$
@JohnOmielan There was a typo. I meant $x-8$. As far as your second comment is concerned the statement '$f(x+2n)=f(x)$ for all $x$ and all positive integers $n$ ' is equivalent to '$f(x+2n)=f(x)$ for all $x$ and all integers $n$ '. Perhaps I could have said 'all integers' for ease of understanding
$endgroup$
– Kavi Rama Murthy
Jan 22 at 7:47
$begingroup$
Thanks for correcting the first instance. However, you still have a second one, as I said earlier, with it being in the statement "... is defined to be $fleft(x - 8xright)$". As for your suggestion of removing the term "positive", although it's not incorrect, I believe it's more general and it will provide ease of understanding.
$endgroup$
– John Omielan
Jan 22 at 7:50
add a comment |
$begingroup$
The function is first defined on $[-1,1]$ and then extended to the whole line by adding the condition $f(x+2)=f(x)$ (periodocity of $f$). For example, if $7 leq x leq 9$ then $x-8$ lies between $-1$ and $1$ and $f(x)$ is defined to be $f(x-8x)$. [Note that $f(x+2)=f(x)$ for all $x$ implies that $f(x+2n)=f(x)$ for all $x$ and for any integer $n$].
answered Jan 22 at 6:44
Kavi Rama MurthyKavi Rama Murthy
65.8k42767
$endgroup$
$begingroup$
I believe your $2$ uses of $x - 8x$ should be just $x - 8$. Also, at the end, the statement $fleft(x + 2nright) = fleft(xright)$ is true for any integers $n$, not just positive ones, as Siong Thye Goh's answer indicates.
$endgroup$
– John Omielan
Jan 22 at 7:43
$begingroup$
@JohnOmielan There was a typo. I meant $x-8$. As far as your second comment is concerned the statement '$f(x+2n)=f(x)$ for all $x$ and all positive integers $n$ ' is equivalent to '$f(x+2n)=f(x)$ for all $x$ and all integers $n$ '. Perhaps I could have said 'all integers' for ease of understanding
$endgroup$
– Kavi Rama Murthy
Jan 22 at 7:47
$begingroup$
Thanks for correcting the first instance. However, you still have a second one, as I said earlier, with it being in the statement "... is defined to be $fleft(x - 8xright)$". As for your suggestion of removing the term "positive", although it's not incorrect, I believe it's more general and it will provide ease of understanding.
$endgroup$
– John Omielan
Jan 22 at 7:50
add a comment |
$begingroup$
The function is first defined on $[-1,1]$ and then extended to the whole line by adding the condition $f(x+2)=f(x)$ (periodocity of $f$). For example, if $7 leq x leq 9$ then $x-8$ lies between $-1$ and $1$ and $f(x)$ is defined to be $f(x-8x)$. [Note that $f(x+2)=f(x)$ for all $x$ implies that $f(x+2n)=f(x)$ for all $x$ and for any integer $n$].
answered Jan 22 at 6:44
Kavi Rama MurthyKavi Rama Murthy
65.8k42767
$endgroup$
The function is first defined on $[-1,1]$ and then extended to the whole line by adding the condition $f(x+2)=f(x)$ (periodocity of $f$). For example, if $7 leq x leq 9$ then $x-8$ lies between $-1$ and $1$ and $f(x)$ is defined to be $f(x-8x)$. [Note that $f(x+2)=f(x)$ for all $x$ implies that $f(x+2n)=f(x)$ for all $x$ and for any integer $n$].
answered Jan 22 at 6:44
Kavi Rama MurthyKavi Rama Murthy
65.8k42767
edited Jan 22 at 7:50
answered Jan 22 at 6:44
Kavi Rama MurthyKavi Rama Murthy
65.8k42767
answered Jan 22 at 6:44
Kavi Rama MurthyKavi Rama Murthy
65.8k42767
answered Jan 22 at 6:44
Kavi Rama MurthyKavi Rama Murthy
65.8k42767
65.8k42767
$begingroup$
I believe your $2$ uses of $x - 8x$ should be just $x - 8$. Also, at the end, the statement $fleft(x + 2nright) = fleft(xright)$ is true for any integers $n$, not just positive ones, as Siong Thye Goh's answer indicates.
$endgroup$
– John Omielan
Jan 22 at 7:43
$begingroup$
@JohnOmielan There was a typo. I meant $x-8$. As far as your second comment is concerned the statement '$f(x+2n)=f(x)$ for all $x$ and all positive integers $n$ ' is equivalent to '$f(x+2n)=f(x)$ for all $x$ and all integers $n$ '. Perhaps I could have said 'all integers' for ease of understanding
$endgroup$
– Kavi Rama Murthy
Jan 22 at 7:47
$begingroup$
Thanks for correcting the first instance. However, you still have a second one, as I said earlier, with it being in the statement "... is defined to be $fleft(x - 8xright)$". As for your suggestion of removing the term "positive", although it's not incorrect, I believe it's more general and it will provide ease of understanding.
$endgroup$
– John Omielan
Jan 22 at 7:50
add a comment |
$begingroup$
I believe your $2$ uses of $x - 8x$ should be just $x - 8$. Also, at the end, the statement $fleft(x + 2nright) = fleft(xright)$ is true for any integers $n$, not just positive ones, as Siong Thye Goh's answer indicates.
$endgroup$
– John Omielan
Jan 22 at 7:43
$begingroup$
@JohnOmielan There was a typo. I meant $x-8$. As far as your second comment is concerned the statement '$f(x+2n)=f(x)$ for all $x$ and all positive integers $n$ ' is equivalent to '$f(x+2n)=f(x)$ for all $x$ and all integers $n$ '. Perhaps I could have said 'all integers' for ease of understanding
$endgroup$
– Kavi Rama Murthy
Jan 22 at 7:47
$begingroup$
Thanks for correcting the first instance. However, you still have a second one, as I said earlier, with it being in the statement "... is defined to be $fleft(x - 8xright)$". As for your suggestion of removing the term "positive", although it's not incorrect, I believe it's more general and it will provide ease of understanding.
$endgroup$
– John Omielan
Jan 22 at 7:50
$begingroup$
I believe your $2$ uses of $x - 8x$ should be just $x - 8$. Also, at the end, the statement $fleft(x + 2nright) = fleft(xright)$ is true for any integers $n$, not just positive ones, as Siong Thye Goh's answer indicates.
$endgroup$
– John Omielan
Jan 22 at 7:43
$begingroup$
I believe your $2$ uses of $x - 8x$ should be just $x - 8$. Also, at the end, the statement $fleft(x + 2nright) = fleft(xright)$ is true for any integers $n$, not just positive ones, as Siong Thye Goh's answer indicates.
$endgroup$
– John Omielan
Jan 22 at 7:43
$begingroup$
@JohnOmielan There was a typo. I meant $x-8$. As far as your second comment is concerned the statement '$f(x+2n)=f(x)$ for all $x$ and all positive integers $n$ ' is equivalent to '$f(x+2n)=f(x)$ for all $x$ and all integers $n$ '. Perhaps I could have said 'all integers' for ease of understanding
$endgroup$
– Kavi Rama Murthy
Jan 22 at 7:47
$begingroup$
@JohnOmielan There was a typo. I meant $x-8$. As far as your second comment is concerned the statement '$f(x+2n)=f(x)$ for all $x$ and all positive integers $n$ ' is equivalent to '$f(x+2n)=f(x)$ for all $x$ and all integers $n$ '. Perhaps I could have said 'all integers' for ease of understanding
$endgroup$
– Kavi Rama Murthy
Jan 22 at 7:47
$begingroup$
Thanks for correcting the first instance. However, you still have a second one, as I said earlier, with it being in the statement "... is defined to be $fleft(x - 8xright)$". As for your suggestion of removing the term "positive", although it's not incorrect, I believe it's more general and it will provide ease of understanding.
$endgroup$
– John Omielan
Jan 22 at 7:50
$begingroup$
Thanks for correcting the first instance. However, you still have a second one, as I said earlier, with it being in the statement "... is defined to be $fleft(x - 8xright)$". As for your suggestion of removing the term "positive", although it's not incorrect, I believe it's more general and it will provide ease of understanding.
$endgroup$
– John Omielan
Jan 22 at 7:50
add a comment |
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$begingroup$
You can write
f(x)=f(x+2)with $, like this: $ f(x)=f(x+2)$ $Big(f(x)=f(x+2)Big)$. See the help or here for more information$endgroup$
– El borito
Jan 22 at 6:42