<p>find the first 4 nonzero terms in the taylor series expansion about x=0 for the function h such that h’(x) = square root of (1 +(X^3)) AND h(0)=4</p>
<p>anyone please help</p>
<p>1) Find the first 4 nonzero terms in the Taylor series expansion of h’(x).
2) Integrate h’(x) and use h(0) = 4 to find h(x).</p>
<p>the first 4 nonzero terms of h(x) = square root of (1+x^3)
ARE:
1 + (x^3)/2 - (x^6)/8 + (x^9)/16</p>
<p>but how do i find the first 4 nonzero terms of h’(x) = square root of (1 +(X^3))?</p>
<p>Actually, the first 3 terms will suffice since you have the constant 4 also.
T{h’(x)} = T{sqrt(1+x^3)} = 1 + 1/2 x^3 - 1/8 x^6
T{h(x)} = 4 + x + 1/8 x^4 - 1/56 x^7</p>
<p>did you integrate h’(x) to get h(x)</p>
<p>Yes, that’s what I did.</p>