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
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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 \( e^{2n\pi + \frac{\pi}{2}} > f(3) \)

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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 \(x_2\)

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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.

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

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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.

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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 \( e^{2n\pi + \frac{\pi}{2}} > f(3) \)

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