I need to calculate a taylor polynomial for a function

Maria Huey 2022-01-20 Answered
I need to calculate a taylor polynomial for a function f:RR where we know the following
f(x)+f(x)=exx
f(0)=0
f(0)=2
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lalilulelo2k3eq
Answered 2022-01-20 Author has 38 answers
We have the following f(x)+f(x)=ex
and f(0)=0,f(0)=2.
And thus we need to find f(n)(0) to construct the Taylor series.
Note that we already have two values and can find f(0) since
f(0)+f(0)=e0
f(0)+0=1
f(0)=1
So now we differentiate the original equation and get:
f(x)+f(x)=ex
But since we know f(0)=2, then
f(0)+f(0)=e0
f(0)+2=1
f(0)=3
And we have our third value. Differentiating one more time gives
fIV(x)+f(x)=ex
So again we have
fIV(0)+f(0)=1
fIV(0)+1=1
fIV(0)=0
Using this twice more you'll get
fV(0)=2
fVI(0)=1
fVII(0)=3
In general the equation is saying that
f(2n+2)(0)+f(2n)(0)=1
f(2n+1)(0)+f(2n1)(0)=1
which will allow you to get all values.
A little summary of the already known values:
f(0)=0
f(0)=2
f(0)=1
f(0)=3
fIV(0)=0
fV(0)=2
fVI(0)=1
Jillian Edgerton
Answered 2022-01-21 Author has 34 answers
The Frobenius method for solving differential equations is easily done: assume an ansatz
f(x)=c0+k=1ckxk
and you have the derivativesf(x)=k=1kckxk1=c1+k=1(k+1)ck+1xk
f(x)=k=1k(k+1)ck+1xk1
From these, you have c0=0 and c1=2 (why?); a relation for the other ck can be derived by comparing the series coefficients of f(x)+f(x) with the coefficients of exp(x).
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RizerMix
Answered 2022-01-27 Author has 438 answers

This is a 2nd order linear non-homogeneous ODE with solution f(x)=12cosx+52sinx+12ex. With the Taylor series for sin(x),cos(x), and ex in hand, you should be able to compute the series for f straightforwardly.

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