/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q36E Let S(t)聽be the number of dayli... [FREE SOLUTION] | 91影视

91影视

Let S(t)be the number of daylight hours on the tth day of the year 2012 in Rome, Italy. We are given the following data for S(t):

We wish to fit a trigonometric function of the form

f(t)=a+bsin(2蟿蟿366t)+ccos(2蟿蟿366t)

To these data. Find the best approximation of this form, using least squares. How many daylight hours does your model predict for the longest day of the year 2012? (The actual value is 15 hours, 13 minutes, 39 seconds.)

Short Answer

Expert verified

x*r=12.26100.4310-2.8993and the required hours aref*17315.1913 hrs.

Step by step solution

01

The required hours.

From the given data we get the terms below

a+bsin232366+c232366........1a+bsin277366+c277366........2a+bsin2121366+c2121366........3a+bsin2152366+c2152366........4

Therefore, consider the system,

1sin64366cos1543661sin154366cos643661sin242366cos2423661sin304366cos304366abcxA=10121415


If ker(A)=0then, xr*=ATA-1ATbr*.

Since, the kernel of the matrix A is {0}.

So, it is written as,

ATA=4.00002.8732-0.24752.87322.2341-0.17733183ATA-1=3.2823-4.2183-0.0365-4.21835.8725-0.00160.0365-0.00160.5712ATA-1AT=1.1109-0.7970-0.42421.1104-1.15341.47280.9718-1.23720.52290.1757-0.2420-0.4566ATA-1ATbr*=12.26100.4310-2.8993

Thus, it is written as,

role="math" localid="1659702486449" xr*=12.26100.4310-2.8993

And the trigonometry function that best fits the data points is

f*t=12.610+0.4310sin2366t-2.89993cos2366t

Hence,xr*=12.26100.4310-2.8993 and the required hours will bef*(173)15.1913 hrs.

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 91影视!

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

Consider the subspaceim(A) of 2 . Where A=[2436] . Find a basis of ker(AT) , and draw a sketch illustrating the formula(imAT)=ker(AT)in this case.

(a) Consider an matrix A such that AA=Im. It is necessarily true that? Explain.

(b) Consider an nnmatrix A such that ATA=In. Is it necessarily true that AAT=In? Explain.

Consider an n x m matrix A with rank (A) = m. Is it always possible to write A as A = QL where Q is an n x m matrix with orthonormal columns and L is a lower triangular m x m matrix with positive diagonal entries? Explain.

Let Abe annmmatrix. Is the formula(kerA)=im(AT)necessarily true? Explain.

Question: In Exercises 1 and 2, you may assume that\(\left\{ {{{\bf{u}}_{\bf{1}}},...,{{\bf{u}}_{\bf{4}}}} \right\}\)is an orthogonal basis for\({\mathbb{R}^{\bf{4}}}\).

1.\({{\bf{u}}_{\bf{1}}} = \left[ {\begin{aligned}{*{20}{c}}{\bf{0}}\\{\bf{1}}\\{ - {\bf{4}}}\\{ - {\bf{1}}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{2}}} = \left[ {\begin{aligned}{*{20}{c}}{\bf{3}}\\{\bf{5}}\\{\bf{1}}\\{\bf{1}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{3}}} = \left[ {\begin{aligned}{*{20}{c}}{\bf{1}}\\{\bf{0}}\\{\bf{1}}\\{ - {\bf{4}}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{4}}} = \left[ {\begin{aligned}{*{20}{c}}{\bf{5}}\\{ - {\bf{3}}}\\{ - {\bf{1}}}\\{\bf{1}}\end{aligned}} \right]\),\({\bf{x}} = \left[ {\begin{aligned}{*{20}{c}}{{\bf{10}}}\\{ - {\bf{8}}}\\{\bf{2}}\\{\bf{0}}\end{aligned}} \right]\)

Write x as the sum of two vectors, one in\({\bf{Span}}\left\{ {{{\bf{u}}_1},{{\bf{u}}_2},{{\bf{u}}_3}} \right\}\)and the other in\({\bf{Span}}\left\{ {{{\bf{u}}_{\bf{4}}}} \right\}\).

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.