Use function to find value
Suppose the function ,$f:Bbb RtoBbb R$, satisfies the following conditions:
$$f(4xy)=2y[f(x+y)+f(x-y)]$$
$$f(5)=3$$
Find the value of $f(2015)$.
I have tried to find some other hiding condition, like $f(0)=0,$ but which is useless.
functions
asked 5 hours ago
yuanming luo
503
add a comment |
Suppose the function ,$f:Bbb RtoBbb R$, satisfies the following conditions:
$$f(4xy)=2y[f(x+y)+f(x-y)]$$
$$f(5)=3$$
Find the value of $f(2015)$.
I have tried to find some other hiding condition, like $f(0)=0,$ but which is useless.
functions
asked 5 hours ago
yuanming luo
503
add a comment |
Suppose the function ,$f:Bbb RtoBbb R$, satisfies the following conditions:
$$f(4xy)=2y[f(x+y)+f(x-y)]$$
$$f(5)=3$$
Find the value of $f(2015)$.
I have tried to find some other hiding condition, like $f(0)=0,$ but which is useless.
functions
asked 5 hours ago
yuanming luo
503
Suppose the function ,$f:Bbb RtoBbb R$, satisfies the following conditions:
$$f(4xy)=2y[f(x+y)+f(x-y)]$$
$$f(5)=3$$
Find the value of $f(2015)$.
I have tried to find some other hiding condition, like $f(0)=0,$ but which is useless.
functions
functions
asked 5 hours ago
yuanming luo
503
asked 5 hours ago
yuanming luo
503
edited 5 hours ago
asked 5 hours ago
yuanming luo
503
asked 5 hours ago
yuanming luo
503
asked 5 hours ago
yuanming luo
503
503
add a comment |
add a comment |
1 Answer
1
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oldest
votes
With the provided function,
$$fleft(4xyright) = 2yleft[fleft(x + yright) + fleft(x - yright)right] tag{1}label{eq1}$$
First, substitute $y = 0$ to get
$$fleft(0right) = 0 tag{2}label{eq2}$$
Next, substitute $x = 0$ and use eqref{eq2} to get
$$0 = 2yleft[fleft(yright) + fleft(-yright)right] tag{3}label{eq3}$$
Thus, for all values of $y$ other than $0$, dividing by $y$ gives
$$fleft(yright) = -fleft(-yright) tag{4}label{eq4}$$
In other words, $f$ is an odd function. Next, substituting $y = 1$ into eqref{eq1} gives
$$fleft(4xright) = 2left[fleft(x + 1right) + fleft(x - 1right)right] tag{5}label{eq5}$$
Now, using $x = 1$ in eqref{eq5} gives
$$fleft(4right) = 2left[fleft(2right) + fleft(0right)right] tag{6}label{eq6}$$
Thus, using eqref{eq2} gives
$$fleft(4right) = 2fleft(2right) tag{7}label{eq7}$$
Similarly, using $x = 3$ in eqref{eq5} gives
$$fleft(12right) = 2left[fleft(4right) + fleft(2right)right] tag{8}label{eq8}$$
Thus, using eqref{eq7} gives
$$fleft(12right) = 6fleft(2right) tag{9}label{eq9}$$
Note that eqref{eq2}, eqref{eq4}, eqref{eq7} and eqref{eq9} imply that
$$fleft(nxright) = nfleft(xright) forall ; n in N tag{10}label{eq10}$$
This suggests a linear multiple of $x$, which from the given condition of
$$fleft(5right) = 3 tag{11}label{eq11}$$
gives a solution of
$$fleft(xright) = cfrac{3x}{5} tag{12}label{eq12}$$
Substituting eqref{eq12} into eqref{eq5} gives on the left side
$$cfrac{12x}{5} tag{13}label{eq13}$$
with the right side becoming
$$2left[cfrac{3left(x + 1right)}{5} + cfrac{3left(x - 1right)}{5} right] = 2left[cfrac{left(3x + 3 + 3x - 3right)}{5} right] = cfrac{12x}{5} tag{14}label{eq14}$$
Thus, eqref{eq12} gives a solution to eqref{eq1}. I'm not quite sure how to prove it's unique (note it's been almost 30 years since I've done very much higher level math, and which is one reason this solution is likely more wordy than need be), but which I assume it is. As such, we get a final answer of
$$fleft(2015right) = cfrac{3 times 2015}{5} = 1209 tag{15}label{eq15}$$
Note: The solution could involve just "guessing" the function being a constant multiple of the argument, and then showing this works to determine the final answer, but I thought it might be useful for people to see how one can approach solving this problem somewhat systematically as I did.
answered 4 hours ago
John Omielan
4467
I am trying to give a reason for the 10th step.
– yuanming luo
3 hours ago
@yuanmingluo I don't offhand see any easy way to directly prove from the provided functional equation that the function is linear. However, the various sets of values all match this so it would have to be a somewhat "perverse" function to not be linear, but the provided equation indicates it is a relatively simple function. I know this is not very precise, but it's the best I can do for now. As I mention in my answer, I am quite rusty at solving these types of problems, but I was once fairly good at it many years ago.
– John Omielan
3 hours ago
Yeah, I know. It was my first time to see this type of questions. So, I also want to do my best.
– yuanming luo
2 hours ago
add a comment |
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1 Answer
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1 Answer
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With the provided function,
$$fleft(4xyright) = 2yleft[fleft(x + yright) + fleft(x - yright)right] tag{1}label{eq1}$$
First, substitute $y = 0$ to get
$$fleft(0right) = 0 tag{2}label{eq2}$$
Next, substitute $x = 0$ and use eqref{eq2} to get
$$0 = 2yleft[fleft(yright) + fleft(-yright)right] tag{3}label{eq3}$$
Thus, for all values of $y$ other than $0$, dividing by $y$ gives
$$fleft(yright) = -fleft(-yright) tag{4}label{eq4}$$
In other words, $f$ is an odd function. Next, substituting $y = 1$ into eqref{eq1} gives
$$fleft(4xright) = 2left[fleft(x + 1right) + fleft(x - 1right)right] tag{5}label{eq5}$$
Now, using $x = 1$ in eqref{eq5} gives
$$fleft(4right) = 2left[fleft(2right) + fleft(0right)right] tag{6}label{eq6}$$
Thus, using eqref{eq2} gives
$$fleft(4right) = 2fleft(2right) tag{7}label{eq7}$$
Similarly, using $x = 3$ in eqref{eq5} gives
$$fleft(12right) = 2left[fleft(4right) + fleft(2right)right] tag{8}label{eq8}$$
Thus, using eqref{eq7} gives
$$fleft(12right) = 6fleft(2right) tag{9}label{eq9}$$
Note that eqref{eq2}, eqref{eq4}, eqref{eq7} and eqref{eq9} imply that
$$fleft(nxright) = nfleft(xright) forall ; n in N tag{10}label{eq10}$$
This suggests a linear multiple of $x$, which from the given condition of
$$fleft(5right) = 3 tag{11}label{eq11}$$
gives a solution of
$$fleft(xright) = cfrac{3x}{5} tag{12}label{eq12}$$
Substituting eqref{eq12} into eqref{eq5} gives on the left side
$$cfrac{12x}{5} tag{13}label{eq13}$$
with the right side becoming
$$2left[cfrac{3left(x + 1right)}{5} + cfrac{3left(x - 1right)}{5} right] = 2left[cfrac{left(3x + 3 + 3x - 3right)}{5} right] = cfrac{12x}{5} tag{14}label{eq14}$$
Thus, eqref{eq12} gives a solution to eqref{eq1}. I'm not quite sure how to prove it's unique (note it's been almost 30 years since I've done very much higher level math, and which is one reason this solution is likely more wordy than need be), but which I assume it is. As such, we get a final answer of
$$fleft(2015right) = cfrac{3 times 2015}{5} = 1209 tag{15}label{eq15}$$
Note: The solution could involve just "guessing" the function being a constant multiple of the argument, and then showing this works to determine the final answer, but I thought it might be useful for people to see how one can approach solving this problem somewhat systematically as I did.
answered 4 hours ago
John Omielan
4467
I am trying to give a reason for the 10th step.
– yuanming luo
3 hours ago
@yuanmingluo I don't offhand see any easy way to directly prove from the provided functional equation that the function is linear. However, the various sets of values all match this so it would have to be a somewhat "perverse" function to not be linear, but the provided equation indicates it is a relatively simple function. I know this is not very precise, but it's the best I can do for now. As I mention in my answer, I am quite rusty at solving these types of problems, but I was once fairly good at it many years ago.
– John Omielan
3 hours ago
Yeah, I know. It was my first time to see this type of questions. So, I also want to do my best.
– yuanming luo
2 hours ago
add a comment |
With the provided function,
$$fleft(4xyright) = 2yleft[fleft(x + yright) + fleft(x - yright)right] tag{1}label{eq1}$$
First, substitute $y = 0$ to get
$$fleft(0right) = 0 tag{2}label{eq2}$$
Next, substitute $x = 0$ and use eqref{eq2} to get
$$0 = 2yleft[fleft(yright) + fleft(-yright)right] tag{3}label{eq3}$$
Thus, for all values of $y$ other than $0$, dividing by $y$ gives
$$fleft(yright) = -fleft(-yright) tag{4}label{eq4}$$
In other words, $f$ is an odd function. Next, substituting $y = 1$ into eqref{eq1} gives
$$fleft(4xright) = 2left[fleft(x + 1right) + fleft(x - 1right)right] tag{5}label{eq5}$$
Now, using $x = 1$ in eqref{eq5} gives
$$fleft(4right) = 2left[fleft(2right) + fleft(0right)right] tag{6}label{eq6}$$
Thus, using eqref{eq2} gives
$$fleft(4right) = 2fleft(2right) tag{7}label{eq7}$$
Similarly, using $x = 3$ in eqref{eq5} gives
$$fleft(12right) = 2left[fleft(4right) + fleft(2right)right] tag{8}label{eq8}$$
Thus, using eqref{eq7} gives
$$fleft(12right) = 6fleft(2right) tag{9}label{eq9}$$
Note that eqref{eq2}, eqref{eq4}, eqref{eq7} and eqref{eq9} imply that
$$fleft(nxright) = nfleft(xright) forall ; n in N tag{10}label{eq10}$$
This suggests a linear multiple of $x$, which from the given condition of
$$fleft(5right) = 3 tag{11}label{eq11}$$
gives a solution of
$$fleft(xright) = cfrac{3x}{5} tag{12}label{eq12}$$
Substituting eqref{eq12} into eqref{eq5} gives on the left side
$$cfrac{12x}{5} tag{13}label{eq13}$$
with the right side becoming
$$2left[cfrac{3left(x + 1right)}{5} + cfrac{3left(x - 1right)}{5} right] = 2left[cfrac{left(3x + 3 + 3x - 3right)}{5} right] = cfrac{12x}{5} tag{14}label{eq14}$$
Thus, eqref{eq12} gives a solution to eqref{eq1}. I'm not quite sure how to prove it's unique (note it's been almost 30 years since I've done very much higher level math, and which is one reason this solution is likely more wordy than need be), but which I assume it is. As such, we get a final answer of
$$fleft(2015right) = cfrac{3 times 2015}{5} = 1209 tag{15}label{eq15}$$
Note: The solution could involve just "guessing" the function being a constant multiple of the argument, and then showing this works to determine the final answer, but I thought it might be useful for people to see how one can approach solving this problem somewhat systematically as I did.
answered 4 hours ago
John Omielan
4467
I am trying to give a reason for the 10th step.
– yuanming luo
3 hours ago
@yuanmingluo I don't offhand see any easy way to directly prove from the provided functional equation that the function is linear. However, the various sets of values all match this so it would have to be a somewhat "perverse" function to not be linear, but the provided equation indicates it is a relatively simple function. I know this is not very precise, but it's the best I can do for now. As I mention in my answer, I am quite rusty at solving these types of problems, but I was once fairly good at it many years ago.
– John Omielan
3 hours ago
Yeah, I know. It was my first time to see this type of questions. So, I also want to do my best.
– yuanming luo
2 hours ago
add a comment |
With the provided function,
$$fleft(4xyright) = 2yleft[fleft(x + yright) + fleft(x - yright)right] tag{1}label{eq1}$$
First, substitute $y = 0$ to get
$$fleft(0right) = 0 tag{2}label{eq2}$$
Next, substitute $x = 0$ and use eqref{eq2} to get
$$0 = 2yleft[fleft(yright) + fleft(-yright)right] tag{3}label{eq3}$$
Thus, for all values of $y$ other than $0$, dividing by $y$ gives
$$fleft(yright) = -fleft(-yright) tag{4}label{eq4}$$
In other words, $f$ is an odd function. Next, substituting $y = 1$ into eqref{eq1} gives
$$fleft(4xright) = 2left[fleft(x + 1right) + fleft(x - 1right)right] tag{5}label{eq5}$$
Now, using $x = 1$ in eqref{eq5} gives
$$fleft(4right) = 2left[fleft(2right) + fleft(0right)right] tag{6}label{eq6}$$
Thus, using eqref{eq2} gives
$$fleft(4right) = 2fleft(2right) tag{7}label{eq7}$$
Similarly, using $x = 3$ in eqref{eq5} gives
$$fleft(12right) = 2left[fleft(4right) + fleft(2right)right] tag{8}label{eq8}$$
Thus, using eqref{eq7} gives
$$fleft(12right) = 6fleft(2right) tag{9}label{eq9}$$
Note that eqref{eq2}, eqref{eq4}, eqref{eq7} and eqref{eq9} imply that
$$fleft(nxright) = nfleft(xright) forall ; n in N tag{10}label{eq10}$$
This suggests a linear multiple of $x$, which from the given condition of
$$fleft(5right) = 3 tag{11}label{eq11}$$
gives a solution of
$$fleft(xright) = cfrac{3x}{5} tag{12}label{eq12}$$
Substituting eqref{eq12} into eqref{eq5} gives on the left side
$$cfrac{12x}{5} tag{13}label{eq13}$$
with the right side becoming
$$2left[cfrac{3left(x + 1right)}{5} + cfrac{3left(x - 1right)}{5} right] = 2left[cfrac{left(3x + 3 + 3x - 3right)}{5} right] = cfrac{12x}{5} tag{14}label{eq14}$$
Thus, eqref{eq12} gives a solution to eqref{eq1}. I'm not quite sure how to prove it's unique (note it's been almost 30 years since I've done very much higher level math, and which is one reason this solution is likely more wordy than need be), but which I assume it is. As such, we get a final answer of
$$fleft(2015right) = cfrac{3 times 2015}{5} = 1209 tag{15}label{eq15}$$
Note: The solution could involve just "guessing" the function being a constant multiple of the argument, and then showing this works to determine the final answer, but I thought it might be useful for people to see how one can approach solving this problem somewhat systematically as I did.
answered 4 hours ago
John Omielan
4467
With the provided function,
$$fleft(4xyright) = 2yleft[fleft(x + yright) + fleft(x - yright)right] tag{1}label{eq1}$$
First, substitute $y = 0$ to get
$$fleft(0right) = 0 tag{2}label{eq2}$$
Next, substitute $x = 0$ and use eqref{eq2} to get
$$0 = 2yleft[fleft(yright) + fleft(-yright)right] tag{3}label{eq3}$$
Thus, for all values of $y$ other than $0$, dividing by $y$ gives
$$fleft(yright) = -fleft(-yright) tag{4}label{eq4}$$
In other words, $f$ is an odd function. Next, substituting $y = 1$ into eqref{eq1} gives
$$fleft(4xright) = 2left[fleft(x + 1right) + fleft(x - 1right)right] tag{5}label{eq5}$$
Now, using $x = 1$ in eqref{eq5} gives
$$fleft(4right) = 2left[fleft(2right) + fleft(0right)right] tag{6}label{eq6}$$
Thus, using eqref{eq2} gives
$$fleft(4right) = 2fleft(2right) tag{7}label{eq7}$$
Similarly, using $x = 3$ in eqref{eq5} gives
$$fleft(12right) = 2left[fleft(4right) + fleft(2right)right] tag{8}label{eq8}$$
Thus, using eqref{eq7} gives
$$fleft(12right) = 6fleft(2right) tag{9}label{eq9}$$
Note that eqref{eq2}, eqref{eq4}, eqref{eq7} and eqref{eq9} imply that
$$fleft(nxright) = nfleft(xright) forall ; n in N tag{10}label{eq10}$$
This suggests a linear multiple of $x$, which from the given condition of
$$fleft(5right) = 3 tag{11}label{eq11}$$
gives a solution of
$$fleft(xright) = cfrac{3x}{5} tag{12}label{eq12}$$
Substituting eqref{eq12} into eqref{eq5} gives on the left side
$$cfrac{12x}{5} tag{13}label{eq13}$$
with the right side becoming
$$2left[cfrac{3left(x + 1right)}{5} + cfrac{3left(x - 1right)}{5} right] = 2left[cfrac{left(3x + 3 + 3x - 3right)}{5} right] = cfrac{12x}{5} tag{14}label{eq14}$$
Thus, eqref{eq12} gives a solution to eqref{eq1}. I'm not quite sure how to prove it's unique (note it's been almost 30 years since I've done very much higher level math, and which is one reason this solution is likely more wordy than need be), but which I assume it is. As such, we get a final answer of
$$fleft(2015right) = cfrac{3 times 2015}{5} = 1209 tag{15}label{eq15}$$
Note: The solution could involve just "guessing" the function being a constant multiple of the argument, and then showing this works to determine the final answer, but I thought it might be useful for people to see how one can approach solving this problem somewhat systematically as I did.
answered 4 hours ago
John Omielan
4467
edited 3 hours ago
answered 4 hours ago
John Omielan
4467
answered 4 hours ago
John Omielan
4467
answered 4 hours ago
John Omielan
4467
4467
I am trying to give a reason for the 10th step.
– yuanming luo
3 hours ago
@yuanmingluo I don't offhand see any easy way to directly prove from the provided functional equation that the function is linear. However, the various sets of values all match this so it would have to be a somewhat "perverse" function to not be linear, but the provided equation indicates it is a relatively simple function. I know this is not very precise, but it's the best I can do for now. As I mention in my answer, I am quite rusty at solving these types of problems, but I was once fairly good at it many years ago.
– John Omielan
3 hours ago
Yeah, I know. It was my first time to see this type of questions. So, I also want to do my best.
– yuanming luo
2 hours ago
add a comment |
I am trying to give a reason for the 10th step.
– yuanming luo
3 hours ago
@yuanmingluo I don't offhand see any easy way to directly prove from the provided functional equation that the function is linear. However, the various sets of values all match this so it would have to be a somewhat "perverse" function to not be linear, but the provided equation indicates it is a relatively simple function. I know this is not very precise, but it's the best I can do for now. As I mention in my answer, I am quite rusty at solving these types of problems, but I was once fairly good at it many years ago.
– John Omielan
3 hours ago
Yeah, I know. It was my first time to see this type of questions. So, I also want to do my best.
– yuanming luo
2 hours ago
I am trying to give a reason for the 10th step.
– yuanming luo
3 hours ago
I am trying to give a reason for the 10th step.
– yuanming luo
3 hours ago
@yuanmingluo I don't offhand see any easy way to directly prove from the provided functional equation that the function is linear. However, the various sets of values all match this so it would have to be a somewhat "perverse" function to not be linear, but the provided equation indicates it is a relatively simple function. I know this is not very precise, but it's the best I can do for now. As I mention in my answer, I am quite rusty at solving these types of problems, but I was once fairly good at it many years ago.
– John Omielan
3 hours ago
@yuanmingluo I don't offhand see any easy way to directly prove from the provided functional equation that the function is linear. However, the various sets of values all match this so it would have to be a somewhat "perverse" function to not be linear, but the provided equation indicates it is a relatively simple function. I know this is not very precise, but it's the best I can do for now. As I mention in my answer, I am quite rusty at solving these types of problems, but I was once fairly good at it many years ago.
– John Omielan
3 hours ago
Yeah, I know. It was my first time to see this type of questions. So, I also want to do my best.
– yuanming luo
2 hours ago
Yeah, I know. It was my first time to see this type of questions. So, I also want to do my best.
– yuanming luo
2 hours ago
add a comment |
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