find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that, for all $x\in\mathbb{R}$
\[x^2f(x)+f(1-x)=2x-x^4\]
Find the functions
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nafistiham
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\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Re: Find the functions
Set $x=1-t$ then we get,
$(1-t)^2f(1-t)+f(t)=2(1-t)-(1-t)^4$.................(1)
now set $x=t$ and we found,
$t^2f(t)+f(1-t)=2t-t^4$
$\Rightarrow {(1-t)t}^{2}f(t)+(1-t)^2f(1-t)=(2t-t^4)(1-t)^{2}$................(2)
Now (2)-(1) implies that,
$f(x)=\frac{(1-x)[(2x-x^4)(1-x)+(1-x)^3-2]}{{(1-x)x}^{2}-1}$
I think this is the solution may be
Now do some irritating calculation (if you want.I'm not willing to do this
).
$(1-t)^2f(1-t)+f(t)=2(1-t)-(1-t)^4$.................(1)
now set $x=t$ and we found,
$t^2f(t)+f(1-t)=2t-t^4$
$\Rightarrow {(1-t)t}^{2}f(t)+(1-t)^2f(1-t)=(2t-t^4)(1-t)^{2}$................(2)
Now (2)-(1) implies that,
$f(x)=\frac{(1-x)[(2x-x^4)(1-x)+(1-x)^3-2]}{{(1-x)x}^{2}-1}$
I think this is the solution may be
Now do some irritating calculation (if you want.I'm not willing to do this
).হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
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nafistiham
- Posts:829
- Joined:Mon Oct 17, 2011 3:56 pm
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Re: Find the functions
thanks.
[I am not going to calculate it either. ]
[I am not going to calculate it either.
]\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.