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Hi there.
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In the question, we have to find the root of this equation, y is equal to ln of x square minus sinex using newton -rapson method and we have to calculate the person -related error for each step of our iteration.
00:13
And we have to set x1 is equal to 2 .4 and the percentile error should fall between 0 .01.
00:22
So let y is equal to f of x is equal to the given function, that is ln of x square minus.
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Sine x.
00:32
So that is we are having f of x is equal to ln of x square minus sine x.
00:42
Now we have to find the solution for this equation right.
00:48
So so here d over d x of f of x that is ln of x square minus sine x is equal to x square minus sine x is equal to we'll be having the value as 2x minus cosine x divided by x squared minus sinex.
01:10
So we have f dash of x is equal to 2x minus cosine x divided by x square minus sinex.
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Now, now we said x1 is equal to 2 .4.
01:25
So we'll be finding the first iteration for that f of x is equal to f of 2 .4 is equal to f of 2 .4 is equal to l.
01:33
Of we'll be having the value inside this as x square minus sinex and that is 5 .0845 and that is equal to 1 .6262.
01:47
So similarly we have f dash of x1 is equal to f dash of 2 .4 is equal to 2 multiplied by 2 .4 minus cosine of 2 .4 divided by 2 .4 square minus sine of 5 .4 .2 .2 .2.
02:02
Minus sine of 5 .4 squared minus sine of sin of of 2 .4.
02:05
So this value will be 1 .0891.
02:09
So we have the value of x2 is equal to x2 is equal to x1 minus f of x1 divided by f dash of x1.
02:21
So here we'll be having x2 is equal to x2 is equal to 2 .4 minus f of x1 is 1 .6262 divided divided by 1 .0891.
02:37
So this value is equal to 0 .9068.
02:45
And here we have the percent relative error.
02:49
The percent relative error as we'll be finding that.
02:57
So here for the first iteration, the percent relative error is equal to modulus of x2 minus x1 divided by x1 multiplied by 100.
03:06
So that is equal to here we'll be having modules of x2s 0 .968 minus 2 .4 divided by 2 .4 multiplied by 100.
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And this is equal to 62 .267 percentage.
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So that is the first iteration.
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And now we'll move on to the second iteration.
03:29
So here f of x2 is equal to f of 0 .9068 and that is equal to ln of 0 .0347 and that is equal to negative 3 .3599.
03:45
Now, f dash of x2 means if dash of x2 is equal to 2 multiplied by 0 .9068 minus cosine of 0 .968 divided by 0 .9068 the square minus sign of 0 .9068 and the value is 34 .4655 and here we have the value of x3 is equal to x2 minus f of x2 divided by f dash of x2 so that is equal to 0 .9068 minus minus 3 .39599 divided by 34 .4665.
04:39
That is equal to 1 .0043.
04:44
Now we'll find the percentile relative error here.
04:47
That is ea2 and that is equal to modulus of x3 minus x2 divided by x2 multiplied by 100.
04:57
So this is equal to we'll be getting modeless of x3 was found to be 1 .0043 and x2 was 0 .9068 divided by 0 .9068 multiplied by 100 and the value is 10 .7521 percentage.
05:19
Now we'll move on to the third iteration.
05:23
Third iteration...