I need to calculate a taylor polynomial for a function f:R→R where we know the followingf″(x)+f(x)=e−x∀xf(0)=0f′(0)=2

Name: I need to calculate a taylor polynomial for a function f:R→R where we know the followingf″(x)+f(x)=e−x∀xf(0)=0f′(0)=2
Uploaded: 2022-02-23T17:22:57+00:00
Description: I need to calculate a taylor polynomial for a function f:R→R where we know the followingf″(x)+f(x)=e−x∀xf(0)=0f′(0)=2

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lalilulelo2k3eq · Accepted Answer

We have the following f″(x)+f(x)=e−xand 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) sincef″(0)+f(0)=e−0f″(0)+0=1f″(0)=1So now we differentiate the original equation and get:f‴(x)+f′(x)=−e−xBut since we know f′(0)=2, thenf‴(0)+f′(0)=−e−0f‴(0)+2=−1f‴(0)=−3And we have our third value. Differentiating one more time givesfIV(x)+f″(x)=e−xSo again we have fIV(0)+f″(0)=1fIV(0)+1=1fIV(0)=0Using this twice more you&#039;ll getfV(0)=2fVI(0)=1fVII(0)=−3In general the equation is saying thatf(2n+2)(0)+f(2n)(0)=1f(2n+1)(0)+f(2n−1)(0)=−1which will allow you to get all values.A little summary of the already known values:f(0)=0f′(0)=2f″(0)=1f‴(0)=−3fIV(0)=0fV(0)=2fVI(0)=1

Jillian Edgerton · Accepted Answer

The Frobenius method for solving differential equations is easily done: assume an ansatzf(x)=c0+∑k=1∞ckxkand you have the derivativesf′(x)=∑k=1∞kckxk−1=c1+∑k=1∞(k+1)ck+1xkf″(x)=∑k=1∞k(k+1)ck+1xk−1From 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).

RizerMix · Accepted Answer

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

I need to calculate a taylor polynomial for a function

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