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Digiron [165]
3 years ago
10

Prove this polynomial identity

Mathematics
1 answer:
77julia77 [94]3 years ago
8 0

Answer:

{(x - y)}^{2}  = (x - y)(x - y) \\  = ( {x}^{2}  - xy - xy +  {y}^{2} ) \\  = ( {x}^{2}  - 2xy +  {y}^{2} )

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Please help me with the below question.
VMariaS [17]

By letting

y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}

we get derivatives

y' = \displaystyle \sum_{n=0}^\infty (n+r) c_n x^{n+r-1}

y'' = \displaystyle \sum_{n=0}^\infty (n+r) (n+r-1) c_n x^{n+r-2}

a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to

5r(r-1) c_0 x^{r-1} + \displaystyle \sum_{n=1}^\infty \bigg( (n+r+1) c_n + (n + r + 1) (5n + 5r + 1) c_{n+1} \bigg) x^{n+r} = 0

Examine the lowest degree term \left(x^{r-1}\right), which gives rise to the indicial equation,

5r (r - 1) + r = 0 \implies 5r^2 - 4r = r (5r - 4) = 0

with roots at r = 0 and r = 4/5.

b) The recurrence for the coefficients c_k is

(k+r+1) c_k + (k + r + 1) (5k + 5r + 1) c_{k+1} = 0 \implies c_{k+1} = -\dfrac{c_k}{5k+5r+1}

so that with r = 4/5, the coefficients are governed by

c_{k+1} = -\dfrac{c_k}{5k+5} \implies \boxed{g(k) = -\dfrac1{5k+5}}

c) Starting with c_0=1, we find

c_1 = -\dfrac{c_0}5 = -\dfrac15

c_2 = -\dfrac{c_1}{10} = \dfrac1{50}

so that the first three terms of the solution are

\displaystyle \sum_{n=0}^2 c_n x^{n + 4/5} = \boxed{x^{4/5} - \dfrac15 x^{9/5} + \frac1{50} x^{13/5}}

4 0
2 years ago
The response to a question has three alternatives: A, B, and C. A sample of 120 responses provides 60 A, 12 B, and 48 C. Show th
zimovet [89]

Answer:

CLASS     FREQUENCIES     RELATIVE FREQUENCIES

A                        60                                 0.5

B                        12                                  0.1

C                        48                                 0.4

TOTAL              120                                  1

Step-by-step explanation:

Given that;

the frequencies of there alternatives are;

Frequency A = 60

Frequency B = 12

Frequency C = 48

Total = 60 + 12 + 48 = 120

Now to determine our relative frequency, we divide each frequency by the total sum of the given frequencies;

Relative Frequency A = Frequency A / total = 60 / 120 = 0.5

Relative Frequency B = Frequency B / total = 12 / 120 = 0.1

Relative Frequency C = Frequency C / total = 48 / 120 = 0.4

therefore;

CLASS     FREQUENCIES     RELATIVE FREQUENCIES

A                        60                                 0.5

B                        12                                  0.1

C                        48                                 0.4

TOTAL              120                                  1

5 0
2 years ago
What is the REMAINDER of 15,908 divided by 13?
stellarik [79]
The remainder is 9........
6 0
2 years ago
Read 2 more answers
-<br>4(5x - 1) – 8x+11 + -17-3(2x + 13) + 24<br>simplify​
navik [9.2K]

Answer:

Step-by-step explanation:

need to multiply outsides into the parenthesis

and then combine like terms and you get

-34x+71

6 0
3 years ago
Write the number thirty five million thirty thousand
Shtirlitz [24]
35,030,000 is the correct answer, hope this helps.

7 0
3 years ago
Read 2 more answers
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