What is the sum? 1/x+2+1/x+3+1/x^2+5x+6

1 answer:

4 0

Kids these days, dont know how to add fractions..
(1/x) + (2/1)`+ (1/x) + (3/1) + (1/x^2) + (5x/1) + (6/1)
Notice every number like..2, 3 and 6 I put over 1, becuase it does not change its value..3/1 = 3 for example[m any number divided by 1 = the same number].
but I did this, so I could express all the terms in the question as fractions.

Now, to add fraction we have to find a common denominator. but all our denominators (the number at the bottom of the fraction.i.e. below the line) are different.

However, x^2 being the highest power seems to solve this. so , here are the steps to solve this:
1. look at the first term (1/x), the denominator [the number below the line] is x. Now if we divide that into our common denominator, the result is x. Let me show you:

x ^2/x = x [ because x multiplied by x or x squared is equal to x^2]

Now since we know, that x into x^2 is x we multiply that by the numerator [the number on top of the line] which in this case is 1, so so far we have:

x
---
x^2

next : we have the term (2/1). Well, the denominator 1 divides into the common denominator x^2 times, so the result is now:

x + 2(x^2) +
--------------
x^2

and on and on...till we get to ..

x + 2x^2 + x + 3x ^2 + 1 + 5x ^3 + 6x ^2
---------------------------------------------------------
x^2

b

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Newtons method is a way to approximate the zero of a function.
It is a recursive method such that the output becomes the new input, the goal is to approach the zero by having the inputs change by a smaller amount each iteration until that change is nearly 0.

First, let me explain where the method comes from.
The formula is really just a manipulation of the slope formula using 2 points; the point tangent to the curve, and the point where the tangent line crosses the x-axis.
If (x,y) is point on curve, then (x*,0) is point on tangent line at x-axis.
$m = \frac{dy}{dx} = \frac{y - 0}{x - x^*}$

Rearranging for x*:
$x^* = x - \frac{y}{dy/dx}$

This is the formula for newtons method. y = f(x), dy/dx = f'(x)
Now what happens is (x*,y*) becomes new point on curve with new tangent line with different slope.
This new line will cross x-axis at different point x** and so on until eventually f(x) gets really really close to 0.

Now for the example:
you need to take derivative f'(x) using product rule:
$f(x) = x tan(x) - 1 \\ \\ f'(x) = tan(x) + x sec^2 (x)$

Then its just a matter of plugging in values for x, and repeating until we get close to a zero.

First plug in x = 1
$x_1 = 1 - \frac{(1)(tan 1) - 1}{tan 1 + (1)sec^2 (1)} = 0.888136 \\ \\ x_2 = .888136 - \frac{f(.888136)}{f'(.888136)} = 0.861465 \\ \\ x_3 = .861465 - \frac{f(.861465)}{f'(.861465)} = 0.860335 \\ \\ x_4 = .860335 - \frac{f(.860335)}{f'(.860335)} = 0.860334 \\ \\ x_5 = .860334 - \frac{f(.860334)}{f'(.860334)} = 0.860334$

Now we can stop because x5 = x4 to 6 decimals, this means f(x) is very close to 0 and will serve as a good approximation for a solution.

Final Answer:
x = 0.860334

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muminat

Answer-

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Solution-

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The effective/equivalent capacitance of the circuit is,

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As, when capacitors are connected together in parallel, the equivalent capacitance in the circuit is equal to the sum of capacitance of all the individual capacitor.

We know that,

$X_c=\dfrac{1}{2\pi fC}$