Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

By shifting the index of summation as in equation \((9)\) or \((13)\), rewrite the given power series so that the general term involves \(t^{n}\). $$ \sum_{n=2}^{\infty} n(n-1) a_{n} t^{n-2} $$

Short Answer

Expert verified
Question: Rewrite the power series \(\sum_{n=2}^{\infty} n(n-1) a_{n} t^{n-2}\) such that the general term involves \(t^{n}\) instead of \(t^{n-2}\). Answer: The rewritten power series is \(\sum_{m=0}^{\infty} (m+1)(m+2) a_{m+2} t^{m}\).

Step by step solution

01

Shifting the index

We will shift the index by introducing a new index, \(m = n-2\), so that the general term involves \(t^{m}\) instead of \(t^{n-2}\). We need to express \(n\) in terms of \(m\), which is simply \(n = m + 2\).
02

Updating the summation limits

Now, we need to update the summation limits based on this shift. For the lower summation limit, when \(n=2\), now we have \(m = 2 - 2 = 0\). For the upper summation limit, since it is \(\infty\), it remains the same.
03

Rewriting the general term

With the shifted index and new summation limits, we can rewrite the general term of the power series. Replace \(n\) with \(m+2\) and \(t^{n-2}\) with \(t^{m}\) in the general term: $$ (n-1)(n)a_n t^{n-2} = (m+1)(m+2)a_{m+2} t^{m} $$
04

Rewriting the power series

Finally, we can rewrite the entire power series with the updated general term and the new summation limits: $$ \sum_{n=2}^{\infty} n(n-1) a_{n} t^{n-2} = \sum_{m=0}^{\infty} (m+1)(m+2) a_{m+2} t^{m} $$ Now, the given power series is rewritten with the general term involving \(t^{m}\) instead of \(t^{n-2}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Index of Summation
In the world of mathematics, the index of summation is the variable that represents the position of a term within a series. When we deal with power series, this index is critical because it gets tied to the powers of the variable we are summing over.

Think of the index as the 'running counter' in a summation. Usually denoted by letters like n, i, or m, it changes value with each term of the sum. For instance, in the sum \(\sum_{n=1}^{5} a_{n}\), n is the index of summation that takes on the values 1 through 5 to sum the respective terms a1, a2, ..., until a5.

When we shift the index, like we did in the original exercise, we're simply changing the 'name' and the starting point of this counter. By setting m equal to n-2, we've created a new starting point and a new way to reference each term in the series, facilitating the manipulation of the series to fit a desired form, such as having a power of tm instead of tn-2.
Summation Limits
The summation limits determine where the summation starts and ends. The lower limit indicates the initial value of the index of summation, while the upper limit shows where the summation stops, which might be a finite number or even infinity.

In our example, the original series has lower and upper summation limits of 2 and \(\infty\), respectively, indicated by \(\sum_{n=2}^{\infty}\). After shifting the index, these limits need to be updated to \(m = 0\) and \(\infty\), expressed as \(\sum_{m=0}^{\infty}\), which reflects the change in the index of summation. When the upper limit is infinity, it signifies an infinite series, one that has infinitely many terms.

This update in the summation limits after the index shift is crucial for preserving the integrity of the series. It ensures that we sum over the correct terms and that our series still represents the function or sequence we are interested in analyzing or manipulating.
General Term of Power Series
A power series is a series where each term involves a variable (often x or t) raised to a power. The general term of a power series gives us a formula that describes the nth term in the series.

The general term is often represented as \(a_{n}x^{n}\), where \(a_{n}\) is a coefficient that might depend on n, and \(x^{n}\) signifies that the variable x is being raised to the power equal to the current index of summation. In the given exercise, the power series had a more complex general term \(n(n-1) a_{n} t^{n-2}\).

By shifting the index and updating the summation limits, we can rewrite the general term to involve \(t^{m}\) instead. The process requires substituting n with m+2 to match our new index and adjust the powers of t accordingly. This manipulation is not just symbolic; it can simplify calculations and help in studying the convergence properties of the power series, which are pivotal in the realms of calculus and complex analysis.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For the special case \(v=\frac{1}{2}\), consider the modified Bessel equation for \(t>0\), \(t^{2} y^{\prime \prime}+t y^{\prime}-\left(t^{2}+\frac{1}{4}\right) y=0\). (a) Define a new dependent variable \(u(t)\) by the relation \(y(t)=t^{-1 / 2} u(t)\). Show that \(u(t)\) satisfies the differential equation \(u^{\prime \prime}-u=0\). (b) Show that the differential equation has a fundamental set of solutions $$ \frac{\sinh t}{\sqrt{t}}, \quad \frac{\cosh t}{\sqrt{t}}, \quad t>0 . $$

In each exercise, (a) Verify that the given differential equation has a regular singular point at \(t=0\). (b) Determine the indicial equation and its two roots. (These roots are often called the exponents at the singularity.) (c) Determine the recurrence relation for the series coefficients. (d) Consider the interval \(t>0\). If the two exponents obtained in (c) are unequal and do not differ by an integer, determine the first two nonzero terms in the series for each of the two linearly independent solutions. If the exponents are equal or differ by an integer, obtain the first two nonzero terms in the series for the solution having the larger exponent. (e) When the given differential equation is put in the form \(y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0\), note that \(t p(t)\) and \(t^{2} q(t)\) are polynomials. Do the series, whose initial terms were found in part (d), converge for all \(t, 0

In each exercise, use the stated information to determine the unspecified coefficients in the given differential equation. \(t=0\) is a regular singular point. The roots of the indicial equation at \(t=0\) are \(\lambda_{1}=1\) and \(\lambda_{2}=2\). \(t=0\) is a regular singular point. The roots of the indicial equation at \(t=0\) are \(\lambda_{1}=1+2 i\) and \(\lambda_{2}=1-2 i\). \(t=0\) is a regular singular point. One root of the indicial equation at \(t=0\) is \(\lambda=2\). The recurrence relation for the series solution corresponding to this root is \(\left(n^{2}+n\right) a_{n}-4 a_{n-1}=0, n=1,2, \ldots .\) The recurrence relation for a series solution is \(n^{2} a_{n}-(n-1) a_{n-1}+3 a_{n-2}=0, n=2,3, \ldots .\) $$ t y^{\prime \prime}+(1+\alpha t) y^{\prime}+\beta t y=0 $$

In each exercise, the exponent \(n\) in the given differential equation is a nonnegative integer. Determine the possible values of \(n\) (if any) for which (a) \(t=0\) is a regular singular point. (b) \(t=0\) is an irregular singular point. $$ y^{\prime \prime}+\frac{1}{t^{n}} y^{\prime}+\frac{1}{1+t^{2}} y=0 $$

Solve the given initial value problem. What is the interval of existence of the solution? $$ t^{2} y^{\prime \prime}-t y^{\prime}-3 y=8 t+6, \quad y(1)=1, \quad y^{\prime}(1)=3 $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free