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Actually,every value repeats infinite times.If you look at the graph,which is this,you'll notice that every value of the function repeats infinitely.Let's try to prove this mathematically.Let $f(x)={ e }^{ x }\sin { x }$ and lets choose $3=x_1$ for an example, we need to find another value $x_2$ such that $f(3)=f(x_2)$ with $x_2>2$ as specified by you and of course,$3\neq x_2$.So we have${ e }^{ 3 }\sin { 3 } ={ e }^{ { x }_{ 2 } }\sin { { x }_{ 2 } }$

taking the natural logarithm on both sides,$3+\ln { \sin { 3 } } ={ x }_{ 2 }+\ln { \sin { { x }_{ 2 } } }$

approximating $\sin{3}$ and plugging it in,

$1.041={ x }_{ 2 }+\ln { \sin { { x }_{ 2 } } }$

the function on the $\text{RHS}$ repeats values too(we can check this graphically or mathematically,but mathematically proving it will take a while)

so there's an infinite number of solutions for $x_2$.We can do this for any value of $x_1$ and still end up with infinitely many values of $x_2$

Thank you. After posting this question, I got my answer,not completely. Firstly I observed that $e^x$ increases for all $x$ and $sin(x)$ increases and decreases. So there will be such values such that function will repeat its value. But I found some problems with this. Lets say, for $x=3$, $f(x) = e^3*sin(3)$. $3≈π$, and so $f(x)$ will be very less. Now for $x = 2k\pi+l$ function will repeat its value, $l$ depends on power of $e$.If power of $e$ is larger, value of $l$ will be smaller. Now for repetition, $e^{2\pi+l}*sin(l) = e^3*sin(3)$, $l< \pi-3$. But the problem is this only, is this condition be satisfied? As it is a function with exponents, so it is not necessary that it will satisfy the above condition, or I may be wrong.

@Hamza A
–
If $x= 3$ which is $≈\pi$ , then $f(x)$ will be very less. Let this value be $f(3) = A$. Now I want to get $A$ for some higher value than $3$.
For this we have to take the value of $x$ greater than $2k\pi$ or less $(2k+1)\pi$, as $sin$ will be positive for such values. Let for $x=2\pi+l$ ,$f(x)$ repeats its value as that of $f(3)$. Here $l$ must be such that, it should satisfy, $e^{(2\pi+l)}*sin(l) = e^{(3)}*sin(3)$. Here value of $x$ is related with both the $e^x$ and $sin(x)$. So it will be very difficult to have same values as $f(3)$, for different values of $x$. As $e^x$ increases in a different way and $sin(x)$ changes in different way. But it is possible to have $e^{(2\pi+l)}*sin(t) = e^{(3)}*sin(3)$, where t is nearer to $l$ or different than $l$

@Akash Shukla
–
Ok,although i don't think a nice closed form for $l$ exists(if there is one),it's still there.You can approximate it by your method at the end by making another variable $t$ which is very near to $l$,but to claim that it is not possible to have such $l$ just because the equation is very hard to solve exactly isn't a valid conclusion.

@Hamza A
–
'Very hard' I mean to say that there will be hardly few values of $l$ to repeat the value, because it has to satisfy two different functions.

@Hamza A
–
I think either you are not getting me or I'm not writing it properly. But anyway,now I will write it elaborately.

Now I will give you my calculative part done by calculator. I have first found the value of $e^3*sin(3)$ which comes to be $1.051195..$. Now I have taken $l = 0.01$, and found the value of $e^{2\pi+0.01}*sin(0.11355..) = f(3)$. So $l$ cannot be $0.01$. Now here what I see is we have to increase the value of $l$. But here $e^x$ function rapidly increase but sin(x) won't. So we couldn't get such value of $l$ to repeat the function.

@Akash Shukla
–
The function is continuous, so you can use the Intermediate Value Theorem.

Let $f(x) = e^x \sin x$. Then $f(2\pi ) = 0$ and $f(2\pi + \frac{\pi}{2} ) = e^{2 \pi + \frac{\pi}{2}} > e^6 > 1.051195 = f(3)$.

Thus by IVT, we can conclude there exists an $x_0$ between $2\pi$ and $2\pi + \frac{\pi}{2}$ such that $f(x_0) = f(3)$.

Using the above format, you can find infinite values of $x_0$ between $2n\pi$ and $2n\pi + \frac{\pi}{2}$ as long as $e^{2n\pi + \frac{\pi}{2}} > f(3)$

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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TopNewestActually,every value repeats infinite times.If you look at the graph,which is this,you'll notice that every value of the function repeats infinitely.Let's try to prove this mathematically.Let $f(x)={ e }^{ x }\sin { x }$ and lets choose $3=x_1$ for an example, we need to find another value $x_2$ such that $f(3)=f(x_2)$ with $x_2>2$ as specified by you and of course,$3\neq x_2$.So we have${ e }^{ 3 }\sin { 3 } ={ e }^{ { x }_{ 2 } }\sin { { x }_{ 2 } }$

taking the natural logarithm on both sides,$3+\ln { \sin { 3 } } ={ x }_{ 2 }+\ln { \sin { { x }_{ 2 } } }$

approximating $\sin{3}$ and plugging it in,

$1.041={ x }_{ 2 }+\ln { \sin { { x }_{ 2 } } }$

the function on the $\text{RHS}$ repeats values too(we can check this graphically or mathematically,but mathematically proving it will take a while)

so there's an infinite number of solutions for $x_2$.We can do this for any value of $x_1$ and still end up with infinitely many values of $x_2$

Log in to reply

Thank you. After posting this question, I got my answer,not completely. Firstly I observed that $e^x$ increases for all $x$ and $sin(x)$ increases and decreases. So there will be such values such that function will repeat its value. But I found some problems with this. Lets say, for $x=3$, $f(x) = e^3*sin(3)$. $3≈π$, and so $f(x)$ will be very less. Now for $x = 2k\pi+l$ function will repeat its value, $l$ depends on power of $e$.If power of $e$ is larger, value of $l$ will be smaller. Now for repetition, $e^{2\pi+l}*sin(l) = e^3*sin(3)$, $l< \pi-3$. But the problem is this only, is this condition be satisfied? As it is a function with exponents, so it is not necessary that it will satisfy the above condition, or I may be wrong.

Log in to reply

Sorry,but i'm not understanding what you're saying in the third line,can you please elaborate?

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$x= 3$ which is $≈\pi$ , then $f(x)$ will be very less. Let this value be $f(3) = A$. Now I want to get $A$ for some higher value than $3$. For this we have to take the value of $x$ greater than $2k\pi$ or less $(2k+1)\pi$, as $sin$ will be positive for such values. Let for $x=2\pi+l$ ,$f(x)$ repeats its value as that of $f(3)$. Here $l$ must be such that, it should satisfy, $e^{(2\pi+l)}*sin(l) = e^{(3)}*sin(3)$. Here value of $x$ is related with both the $e^x$ and $sin(x)$. So it will be very difficult to have same values as $f(3)$, for different values of $x$. As $e^x$ increases in a different way and $sin(x)$ changes in different way. But it is possible to have $e^{(2\pi+l)}*sin(t) = e^{(3)}*sin(3)$, where t is nearer to $l$ or different than $l$

IfLog in to reply

$l$ exists(if there is one),it's still there.You can approximate it by your method at the end by making another variable $t$ which is very near to $l$,but to claim that it is not possible to have such $l$ just because the equation is very hard to solve exactly isn't a valid conclusion.

Ok,although i don't think a nice closed form forLog in to reply

$l$ to repeat the value, because it has to satisfy two different functions.

'Very hard' I mean to say that there will be hardly few values ofLog in to reply

$f(3)=f(2\pi+l)$

how will it need to satisfy 2?it only needs to satisfyLog in to reply

$e^x$ and $sin(x)$

two different functions :-Log in to reply

Log in to reply

Now I will give you my calculative part done by calculator. I have first found the value of $e^3*sin(3)$ which comes to be $1.051195..$. Now I have taken $l = 0.01$, and found the value of $e^{2\pi+0.01}*sin(0.11355..) = f(3)$. So $l$ cannot be $0.01$. Now here what I see is we have to increase the value of $l$. But here $e^x$ function rapidly increase but sin(x) won't. So we couldn't get such value of $l$ to repeat the function.

Log in to reply

Let $f(x) = e^x \sin x$. Then $f(2\pi ) = 0$ and $f(2\pi + \frac{\pi}{2} ) = e^{2 \pi + \frac{\pi}{2}} > e^6 > 1.051195 = f(3)$.

Thus by IVT, we can conclude there exists an $x_0$ between $2\pi$ and $2\pi + \frac{\pi}{2}$ such that $f(x_0) = f(3)$.

Using the above format, you can find infinite values of $x_0$ between $2n\pi$ and $2n\pi + \frac{\pi}{2}$ as long as $e^{2n\pi + \frac{\pi}{2}} > f(3)$

Log in to reply