{"id":47877,"date":"2025-07-02T16:44:38","date_gmt":"2025-07-02T16:44:38","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=47877"},"modified":"2025-07-02T16:44:39","modified_gmt":"2025-07-02T16:44:39","slug":"find-a-power-series-representation-for-fxln%e2%81%a11x","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/find-a-power-series-representation-for-fxln%e2%81%a11x\/","title":{"rendered":"\u00a0Find a power series representation for\u00a0f(x)=ln\u2061(1+x)."},"content":{"rendered":"\n
\"\"<\/figure>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Here are the correct expressions for the power series representations.<\/p>\n\n\n\n

(a) Find a power series representation for f(x) = ln(1 + x).<\/strong>
The general term for the series is:
(-1)^(n-1) * x^n \/ n<\/p>\n\n\n\n

(b) Use part (a) to find a power series for f(x) = x ln(1 + x).<\/strong>
The general term for the series is:
(-1)^(n-2) * x^n \/ (n-1)<\/p>\n\n\n\n

(c) Use part (a) to find a power series for f(x) = ln(x\u00b2 + 1).<\/strong>
The general term for the series is:
(-1)^(n-1) * x^(2n) \/ n<\/p>\n\n\n\n

Explanation<\/h3>\n\n\n\n

The solutions for all parts are derived from the basic power series for ln(1+x), which itself comes from the geometric series.<\/p>\n\n\n\n

(a) Power series for f(x) = ln(1 + x)<\/strong>
We begin with the geometric series formula: 1\/(1\u2212u) = \u03a3 u\u207f for |u|<1. By substituting u = -t, we get the series for 1\/(1+t) = \u03a3 (-1)\u207ft\u207f.
Since ln(1+x) is the integral of 1\/(1+t) from 0 to x, we can integrate the series term by term:
ln(1+x) = \u222b \u03a3 (-1)\u207ft\u207f dt = \u03a3 \u222b (-1)\u207ft\u207f dt = \u03a3 (-1)\u207f t\u207f\u207a\u00b9\/(n+1).
The summation starts from n=0. To match the problem’s format starting from n=1, we re-index the series. This gives the well-known Maclaurin series for ln(1+x):
f(x) = \u03a3 (from n=1 to \u221e) [(-1)\u207f\u207b\u00b9x\u207f\/n].
The radius of convergence, R, remains 1, the same as the original geometric series.<\/p>\n\n\n\n

(b) Power series for f(x) = x ln(1 + x)<\/strong>
To find this series, we multiply the series from part (a) by x:
f(x) = x * [\u03a3 (from n=1 to \u221e) (-1)\u207f\u207b\u00b9x\u207f\/n] = \u03a3 (from n=1 to \u221e) (-1)\u207f\u207b\u00b9x\u207f\u207a\u00b9\/n.
The problem requires the series to start from n=2. We re-index by setting a new index k=n+1, which means n=k-1. The sum becomes:
\u03a3 (from k=2 to \u221e) (-1)\u207d\u1d4f\u207b\u00b9\u207e\u207b\u00b9x\u1d4f\/(k-1) = \u03a3 (from k=2 to \u221e) (-1)\u1d4f\u207b\u00b2x\u1d4f\/(k-1).
Replacing k with n, the term is (-1)\u207f\u207b\u00b2x\u207f\/(n-1). Multiplying by a polynomial does not change the radius of convergence, so R remains 1.<\/p>\n\n\n\n

(c) Power series for f(x) = ln(x\u00b2 + 1)<\/strong>
We use the series from part (a) and substitute x with x\u00b2:
ln(1+u) = \u03a3 (from n=1 to \u221e) (-1)\u207f\u207b\u00b9u\u207f\/n.
Let u = x\u00b2.
f(x) = ln(1+x\u00b2) = \u03a3 (from n=1 to \u221e) (-1)\u207f\u207b\u00b9(x\u00b2)\u207f\/n = \u03a3 (from n=1 to \u221e) (-1)\u207f\u207b\u00b9x\u00b2\u207f\/n.
The original series converges for |u|<1. For this new series, convergence requires |x\u00b2|<1, which simplifies to |x|<1. Thus, the radius of convergence R is still 1.<\/p>\n\n\n\n

\"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

The Correct Answer and Explanation is: Here are the correct expressions for the power series representations. (a) Find a power series representation for f(x) = ln(1 + x).The general term for the series is:(-1)^(n-1) * x^n \/ n (b) Use part (a) to find a power series for f(x) = x ln(1 + x).The general […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-47877","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/47877","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=47877"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/47877\/revisions"}],"predecessor-version":[{"id":47880,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/47877\/revisions\/47880"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=47877"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=47877"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=47877"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}