I think you should have made it more clear that you aim to find approximate solutions in different ways. Perhaps it would also be interesting to compare your estimates with the true solutions obtained numerically. For the point nearest to zero, since you are fine with setting sin(x)≈x, you should also be fine with e^x≈1+x, which gives x≈-0.5
Yes, you're right!!! Sorry... But thanks a lot for your comment!!! I'm still young and I have a lot of things to learn but I promise that every video will get better and better!!! And so do I!!!🤩🤩🤩 Yes, I should have been more clear... Sorry, I apologise for that!!!🙄😉
Well... It isn't!!!😅🤣 Actually, I wanted to solve these two equations but I failed: I tried to expand the two functions in their Maclaurn series and then rearrange the equation to get a combination of Maclaurn expansions of some known functions but it failed... So I decided to try a different approach to approximate the answers and it also failed. After having plotted the two functions, I realised that this was indeed the SIMPLEST and BEST way to approximate the solutions!!! And as x approaches -infinity the answers get more and more precise!!! Isn't this amazing??? So I decided to record this video aimed to teach my subscribers that sometimes a bit of reasoning could make hard problems a lot easier and even, as in this case, trivial but effective!!!😉🤗
I tried solving the first equation algebraically using the complex definition of sine, but I keep getting stuck in a nest of logarithms. If we only care about approximations though, I used the Newton-Raphson numerical method to produce an answer of about -3.183063...
Do you like watching videos about crazy equations???🤩🤩🤩 LET ME KNOW!!!⬇️⬇️⬇️
Btw these results are extremely precise!!!🤩🤩🤩
I think you should have made it more clear that you aim to find approximate solutions in different ways. Perhaps it would also be interesting to compare your estimates with the true solutions obtained numerically.
For the point nearest to zero, since you are fine with setting sin(x)≈x, you should also be fine with e^x≈1+x, which gives x≈-0.5
Yes, you're right!!! Sorry... But thanks a lot for your comment!!! I'm still young and I have a lot of things to learn but I promise that every video will get better and better!!! And so do I!!!🤩🤩🤩 Yes, I should have been more clear... Sorry, I apologise for that!!!🙄😉
Is this supposed to be some sort of joke?
Well... It isn't!!!😅🤣 Actually, I wanted to solve these two equations but I failed: I tried to expand the two functions in their Maclaurn series and then rearrange the equation to get a combination of Maclaurn expansions of some known functions but it failed... So I decided to try a different approach to approximate the answers and it also failed. After having plotted the two functions, I realised that this was indeed the SIMPLEST and BEST way to approximate the solutions!!! And as x approaches -infinity the answers get more and more precise!!! Isn't this amazing??? So I decided to record this video aimed to teach my subscribers that sometimes a bit of reasoning could make hard problems a lot easier and even, as in this case, trivial but effective!!!😉🤗
I tried solving the first equation algebraically using the complex definition of sine, but I keep getting stuck in a nest of logarithms.
If we only care about approximations though, I used the Newton-Raphson numerical method to produce an answer of about -3.183063...
Of course, there's an infinite number of solutions, this is only the easiest one to get.
Nice!!! Yes, unfortunately, we cannot find the exact solutions but we could always approximate!!!🤩🤩🤩 Nice job!!!😉