I'm not sure, but I'd say \(x \in R^+ - \{1\}\)
–
Harshit Kapur
·
4 years, 1 month ago

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@Harshit Kapur
–
Wait, by taking euler's identity into account, I think we could extend the domain further to imaginary numbers as well ..
–
Harshit Kapur
·
4 years, 1 month ago

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@Harshit Kapur
–
Please avoid taking logarithms of anything other than the positive real numbers. There is more understanding involved when we want to talk about \( \log -1\), in part because it is a multi-valued function, and doesn't behave exactly like what you would expect from your experience working with \(\log\) normally.
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Calvin Lin
Staff
·
4 years, 1 month ago

As an experience for comparing ability of mathematical software, I recommend you that try to plot this function by some popular software such as Mathlab, Maple, Derive, GeoGebra and so on. According to my experience Derive was the best for this especial function. I am waiting for the final answer.
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Adel Shakiba
·
4 years, 1 month ago

I'm not saying it can't be wrong, but its very rare.
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Harshit Kapur
·
4 years, 1 month ago

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Actually, since sin (and /) are defined for (almost) all reals and ln is defined for positive reals (it might be possible to define extent ln to the imaginary numbers, but that's not exactly going to be ln any more, just like how the zeta function is not the factorial function), the domain is actually x>0, since it's defined for all x>0. Oh, excepting 1, since ln (1)=0, and division by 0 is undefined.

But we're not going to be satisfied with the domain, so let's proceed to the codomain.

Basically it has exactly the same codomain as f(x)=sin x/ x for all real values of x (since ln x has the codomain of all reals).
We know that sin(x)/x-->0 as x-->0, and though it might be overkill, we can prove that sin(x)/x<1 using the Taylor series. We can then find the lower limit (as x<0) via differentiation as well*, but since that is quite tedious, and probably inexact, I have used Wolfram Alpha to estimate the lower bound at -0.217233628..., where ln (x)= 4.49340945790906...

*Set the gradient to 0 and argue that it has to lie between x= pi and x=2pi, then solve
–
Ejia Efah
·
4 years, 1 month ago

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@Ejia Efah
–
Thank you for your reply but I think you should look at this problem more easy.
–
Adel Shakiba
·
4 years, 1 month ago

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yes x>0 and x is not equal to 1
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Vaibhav Reddy
·
4 years, 1 month ago

## Comments

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TopNewestI'm not sure, but I'd say \(x \in R^+ - \{1\}\) – Harshit Kapur · 4 years, 1 month ago

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– Harshit Kapur · 4 years, 1 month ago

Wait, by taking euler's identity into account, I think we could extend the domain further to imaginary numbers as well ..Log in to reply

– Calvin Lin Staff · 4 years, 1 month ago

Please avoid taking logarithms of anything other than the positive real numbers. There is more understanding involved when we want to talk about \( \log -1\), in part because it is a multi-valued function, and doesn't behave exactly like what you would expect from your experience working with \(\log\) normally.Log in to reply

– Adel Shakiba · 4 years, 1 month ago

Let's wait for next answers. ;)Log in to reply

As an experience for comparing ability of mathematical software, I recommend you that try to plot this function by some popular software such as Mathlab, Maple, Derive, GeoGebra and so on. According to my experience Derive was the best for this especial function. I am waiting for the final answer. – Adel Shakiba · 4 years, 1 month ago

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With WolphramAlpha

I plug in straight what you ask.

I'm not saying it can't be wrong, but its very rare. – Harshit Kapur · 4 years, 1 month ago

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Actually, since sin (and /) are defined for (almost) all reals and ln is defined for positive reals (it might be possible to define extent ln to the imaginary numbers, but that's not exactly going to be ln any more, just like how the zeta function is not the factorial function), the domain is actually x>0, since it's defined for all x>0. Oh, excepting 1, since ln (1)=0, and division by 0 is undefined.

But we're not going to be satisfied with the domain, so let's proceed to the codomain.

Basically it has exactly the same codomain as f(x)=sin x/ x for all real values of x (since ln x has the codomain of all reals). We know that sin(x)/x-->0 as x-->0, and though it might be overkill, we can prove that sin(x)/x<1 using the Taylor series. We can then find the lower limit (as x<0) via differentiation as well*, but since that is quite tedious, and probably inexact, I have used Wolfram Alpha to estimate the lower bound at -0.217233628..., where ln (x)= 4.49340945790906...

*Set the gradient to 0 and argue that it has to lie between x= pi and x=2pi, then solve – Ejia Efah · 4 years, 1 month ago

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– Adel Shakiba · 4 years, 1 month ago

Thank you for your reply but I think you should look at this problem more easy.Log in to reply

yes x>0 and x is not equal to 1 – Vaibhav Reddy · 4 years, 1 month ago

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– Adel Shakiba · 4 years, 1 month ago

Thank you but not complete.Log in to reply

hello,the domain is :x belong to real such that x is greater than 1 or 0<x<1 – Sayan Chaudhuri · 4 years, 1 month ago

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– Adel Shakiba · 4 years, 1 month ago

let's wait till next days for more answers.Log in to reply